A New Discriminant Algebra Construction

Owen Biesel, Alberto Gioia

Received: December 10, 2015 Revised: April 6, 2016

Communicated by Alexander Merkurjev

Abstract. A discriminant algebra operation sends a commutative ring R and an R-algebra A of rank n to an R-algebra Δ_{A/R} of rank 2 with the same discriminant bilinear form. Constructions of discriminant algebra operations have been put forward by Rost, Deligne, and Loos. We present a simpler and more explicit construction that does not break down into cases based on the parity of n. We then prove properties of this construction, and compute some examples explicitly.

2010 Mathematics Subject Classification: Primary 13B02; Secondary 14B25, 11R11, 13B40, 13C10 Keywords and Phrases: discriminant algebra, discriminant form, algebra of finite rank, étale algebra, polynomial law

Contents

1. Introduction  1052
2. Defining the Ferrand homomorphism and discriminant algebra 1055
3. Understanding the Ferrand homomorphism  1058
4. Proof of Theorem 1  1064
5. Examples of discriminant algebras  1068
6. Proofs of Theorems 2 to 4  1074
7. Functoriality  1078
8. The discriminant algebra of a product  1079
References  1088

1. Introduction

An n-fold covering map of topological spaces X → S has an associated 2-fold “orientation” covering, which on fibers is described by the set of orderings of the n-element fiber of X, up to reorderings by even permutations. The algebraic analogue producing a rank-2 étale algebra from a rank-n one is called the discriminant algebra, and we briefly review its classical constructions below. Let f(x) in ℚ[x] be an irreducible polynomial of degree n. Suppose the Galois group of f is the symmetric group S_{n}, and let x_{1}, …, x_{n} be the roots of f in its splitting field N. The subfield of N consisting of elements invariant under even permutations of those roots is called the discriminant field of L := ℚ[x]/(f(x)). It is formed by adjoining a square root of the discriminant of L over K, hence the name. Following [10, §18], we may extend this operation to general rank-n separable algebras A over a field K of arbitrary characteristic: such an algebra corresponds to an n-element π_{K}-set X, where π_{K} is the absolute Galois group of K, and the discriminant algebra of A is the rank-2 separable algebra Δ_{A/K} corresponding to the 2-element set of orientations of X, orderings of X up to re-orderings by even permutations. (If the characteristic of K is not 2, then the discriminant algebra of A may again be presented as K[x]/(x^{2} − d), where d ∈ K is the discriminant of A with respect to some K-basis; see [10, Proposition 18.24].) The same π-set construction applies more generally whenever R is a connected ring and A is a rank-n projective separable R-algebra, since there is still a contravariant equivalence between projective separable R-algebras and finite sets equipped with an action by a suitable profinite group π_{R}. In [20], William Waterhouse drops the connectedness hypothesis by interpreting étale rank-n algebras as S_{n}-torsors, and the discriminant algebra mapping from S_{n}- torsors to S_{2}-torsors is merely the one coming from the sign homomorphism S_{n} → S_{2}. Our goal in this paper is to extend the discriminant algebra construction even further to the case of R a ring and A a rank-n projective R-algebra, with no separability hypothesis. (In this paper, all rings and algebras are commutative, associative, and unital, so an R-algebra structure on A is merely a homomorphism of commutative rings R → A.) The main feature we wish to keep is that the discriminant algebra of A should have the same discriminant as that of A; this is easy to check in the original motivating case of a finite separable algebra A over K in characteristic other than 2: If A has discriminant d with respect to some K-basis, then the discriminant algebra for A over K is Δ ≅ K[x]/(x^{2}−d). Since 2 is a unit, we may as well write Δ ≅ K[x]/(x^{2}−d/4), which, with respect to the K-basis {1, x}, also has discriminant

det (\begin{matrix}Tr_{Δ/K}(1) & Tr_{Δ/K}(x) \\ Tr_{Δ/K}(x) & Tr_{Δ/K}(x^{2}) \end{matrix}) = (\begin{matrix}2 & 0 \\ 0 & d/2 \end{matrix}) = d.

In this paper, we put forth such a general discriminant algebra construction. In the setting of a ring R and algebra A that is locally free of rank n as an R-module, it still makes sense to talk about the trace and discriminant maps:

A has a discriminant bilinear form δ_{A/R} on ⋀^{n}A defined by

δ_{A/R}(a_{1} ∧ ⋯ ∧ a_{n}, b_{1} ∧ ⋯ ∧ b_{n}) = det(Tr_{A/R}(a_{i}b_{j}))_{ij}.

Here is the first main result of this paper:

Theorem 1 (proven as Theorem 4.1). There is an assignment (R, A) ↦ Δ_{A/R} sending any rank-n algebra A over any ring R to a rank-2 R-algebra Δ_{A/R} that is endowed with a canonical isomorphism ⋀^{2}Δ_{A/R} ≅ ⋀^{n}A identifying the discriminant bilinear form of Δ_{A/R} with that of A.

We call Δ_{A/R} the discriminant algebra of A over R. In a 2005 letter to Manjul Bhargava and Markus Rost, Pierre Deligne suggested a list of four other properties a discriminant algebra operation (R, A) ↦ Δ_{A/R} should have; see [5] for the original formulation. We prove that these properties hold for our construction with the following theorems; the first property is that forming the discriminant algebra should commute with base change:

Theorem 2 (proven as Theorem 6.1). If R is a ring and A is an R-algebra of rank n, and R′ is any R-algebra, let A′ denote R′ ⊗_{R} A, an R′-algebra of rank n. Then there is a canonical isomorphism R′ ⊗_{R} Δ_{A/R} ≅ Δ_{A′/R′}.

Deligne’s second requirement is an explicit description of the discriminant algebra in the case that 2 is invertible, and is a generalization of the idea of adjoining to the base ring a square root of the discriminant of the algebra with respect to some basis.

Theorem 3. If 2 is a unit in R and A is an R-algebra of rank n, then there is an R-algebra isomorphism Δ_{A/R} ⟶^{∼} R ⊕ ⋀^{n}A, where the latter is given the R-algebra structure whose identity element is (1, 0), and where the product of two elements a, b ∈ ⋀^{n}A is via the discriminant form: a · b := δ_{A/R}(a, b) ∈ R.

We will find that it actually makes more sense to include a factor of 1/4 in the multiplication, so that the product of two elements a, b ∈ ⋀^{n}A is \frac{1}{4}δ_{A/R}(a, b). (This is analogous to presenting the discriminant algebra of a separable Kalgebra with discriminant d as K[x]/(x^{2} −d/4) instead of K[x]/(x^{2} −d).) Then Theorem 3 is actually a consequence of Theorem 1; see Corollary 6.3. Deligne’s third requirement is that in the case of R connected and A étale over R, the discriminant algebra for A over R should be the one described at the beginning of the introduction:

Theorem 4 (proven as Theorem 6.8). Let R be a connected ring and A an étale R-algebra of rank n corresponding to a π_{R}-set X. Then Δ_{A/R} is also étale, and corresponds to the 2-element set Or(X) of orientations of X.

Deligne’s fourth requirement was that the two descriptions of Δ_{A/R} from Theorems 3 and 4 should be compatible when both apply, and we omit the statement here. In his letter, Deligne sketched a construction of an operation (R, A) ↦ Δ_{A/R}^{Del} satisfying his four properties. He first reduced to a similar list of properties that a discriminant algebra operation for quadratic forms should have, and extended the corresponding étale and 2-invertible descriptions across their codimension-2 complement in a universal case. This construction only works for odd-rank algebras, so he defined the discriminant algebra of an even-rank A to be that of R×A. However, he does not specifically equip Δ_{A/R}^{Del} with the structure of a discriminant-identifying isomorphism ⋀^{2}Δ_{A/R}^{Del} ≅ ⋀^{n}A. Earlier, Markus Rost had exhibited in [16] a way of producing a natural rank-2 algebra K_{A/R} from a rank-3 R-algebra A, but it admits no natural discriminantidentifying isomorphism ⋀^{2}K_{A/R} ≅ ⋀^{3}A. Rost fixed this defect by the process of Verschiebung or “shifting”: modifying the multiplication of K_{A/R} to obtain a new quadratic algebra Δ_{A/R}^{Rost} with the desired discriminant form. Ottmar Loos later employed this shifting technique to give a discriminant algebra construction in [13] for even-rank algebras, extending to the odd-rank case again by defining Δ_{A/R}^{Loos} := Δ_{(R×A)/R}^{Loos} when A has odd rank. The main advantages of our discriminant algebra construction (R, A) ↦ Δ_{A/R} are that it does not appeal to a separate construction for quadratic forms, it does not split into cases based on whether the rank of A is odd or even, and it does not require first handling special cases (such as 2 being invertible or the algebra being étale). Namely, given a ring R and an R-algebra A of rank n, there is a canonical Ralgebra homomorphism from the S_{n}-invariant tensors of A^{⊗n} to R; we call this the Ferrand homomorphism Φ_{A/R}: (A^{⊗n})^{S_{n}} → R. Then we use the Ferrand homomorphism to define the discriminant algebra of A over R as the tensor product

Δ_{A/R} := (A^{⊗n})^{A_{n}} ⨂_{(A^{⊗n})^{Sn}} R.

In Section 2, we define the Ferrand homomorphism and discriminant algebra associated to a rank-n algebra. We prove Theorem 1 in Section 4, and prove Theorems 2 to 4 in Section 6. In between, in Section 3 we discuss the action of the Ferrand homomorphism, and Section 5 exhibits some examples of discriminant algebras. Section 7 shows that the construction of the discriminant algebra is functorial with respect to algebra homomorphisms that preserve the characteristic polynomial of every element. In Section 8, we analyze the discriminant algebra of a product algebra; our main result is that Δ_{(A×B)/R} ≅ Δ_{A/R} ∗ Δ_{B/R}, where ∗ is the commutative monoid structure on the set of isomorphism classes of quadratic algebras, characterized by John Voight in [19]. We also show that Δ_{(R×A)/R} ≅ Δ_{A/R}, so that if we set Δ_{A/R} := R^{2} for A of rank 0 or 1, we obtain a monoid homomorphism from the set of isomorphism classes of (locally) constant-rank algebras to the set of isomorphism classes of quadratic algebras, carrying × to ∗. In forthcoming work, the authors exhibit isomorphisms between this discriminant algebra construction and those given by Rost and Loos. The isomorphism with Loos’s discriminant algebra is quite subtle, and suggests that there may be only one discriminant algebra operation satisfying Theorems 1 to 4 up to isomorphism. Due to the existence of suitably nice universal cases, the authors have been able to verify this uniqueness in ranks up to 3, but the general case is still unknown. Both authors would like to thank Lenny Taelman for several helpful discussions throughout this project, as well as Darij Grinberg and the anonymous referee for their comments and corrections. The second author would also like to thank the Austrian Science Fund FWF for its support through project P25652.

2. Defining the Ferrand homomorphism and discriminant algebra

Given a ring R with algebra A of rank n, Daniel Ferrand in [6] uses a certain homomorphism Φ_{A/R}: (A^{⊗n})^{S_{n}} → R to construct a functor from A-modules to R-modules. We call this homomorphism the Ferrand homomorphism. In case K is a field and L is a degree-n separable field extension of K, then the Ferrand homomorphism Φ_{L/K} has a simple description, given the Galois closure N of L over K: Compile the n homomorphisms L → N over K into a single homomorphism L^{⊗n} → N, and restrict it to the subalgebra (L^{⊗n})^{S_{n}}. The image in N of any S_{n}-invariant tensor will be invariant under the Galois action on N, and thus belongs to K, giving us the Ferrand homomorphism Φ_{L/K}: (L^{⊗n})^{S_{n}} → K. In this section, we will briefly review Ferrand’s abstract definition of the Ferrand homomorphism in [6], and then we will describe how to compute its action on symmetric tensors. First, we recall the setting in which elements of algebras have norms. Thus an algebra over a ring R is just another ring A with a ring homomorphism R → A.

Definition 2.1. Let R be a ring, and let M be an R-module. We say that M is locally free (of rank n) if there are elements r_{1}, …, r_{k} ∈ R, together generating the unit ideal, such that M_{r_{i}} is free (of rank n) as an R_{r_{i}}-module for each i ∈ {1, …, k}. An R-algebra A is said to be of rank n if A is locally free of rank n as an R-module.

Remark 2.2. Equivalently, an R-module M is locally free of rank n if and only if it is projective and finitely generated (i.e. flat and finitely presented) and the rank of each (necessarily free) R_{𝔭}-module M_{𝔭} is n for each prime ideal 𝔭 ∈ M; see [3, Thm. 2 on p. II.141].

The following observation about locally free algebras and modules will be useful throughout this paper:

Lemma 2.3. Let R be a ring and A an R-algebra of rank n ≥ 2. Then the quotient A/R is a locally-free R-module of rank n − 1, and the natural map

⋀^{n−1}(A/R) → ⋀^{n}A

sending [a_{1}] ∧ ⋯ ∧ [a_{n−1}] to 1 ∧ a_{1} ∧ ⋯ ∧ a_{n−1} is an isomorphism.

In particular, if A is a quadratic (i.e. rank-2) R-algebra then there is a canonical isomorphism A/R → ⋀^{2}A sending the class of a to 1 ∧ a.

Proof. For the first claim, that A/R is locally free, it is enough to show that A/R is flat and finitely presented. First, the structure morphism R → A is injective since the rank of A is everywhere positive, and since the same applies after base change to any R-algebra, we find that R → A is the inclusion of a pure submodule. The quotient of a flat module by a pure submodule is also flat, so the quotient A/R is a flat R-module. It is also finitely presented, since A is, so A/R is locally free. The claims that A/R has rank n−1 and that the given map on exterior powers is an isomorphism can be checked locally, so assume R is a local ring. Then A/R is a free R-module, so let {[θ_{1}], …, [θ_{k}]} be an R-basis for it. Then A is has R-basis {1, θ_{1}, …, θ_{k}}, so since A has rank n we must have k = n − 1 as desired. Finally, note that the indicated homomorphism ⋀^{n−1}(A/R) → ⋀^{n}A maps the singleton basis {[θ_{1}] ∧ ⋯ ∧ [θ_{n−1}]} to the basis {1 ∧ θ_{1} ∧ ⋯ ∧ θ_{n−1}}, so is an isomorphism. □

Definition 2.4. Let R be a ring, and let A be an R-algebra of rank n. For each element a ∈ A, multiplication by a is an R-module homomorphism A → A, and the nth exterior power of this homomorphism is an R-module homomorphism ⋀^{n}A → ⋀^{n}A. Since ⋀^{n}A is locally free of rank 1, this endomorphism is equal to multiplication by a unique element of R, called the norm Nm_{A/R}(a) of a.

What sort of map is Nm_{A/R}: A → R? It is multiplicative, but almost never additive, and thus it is not a ring homomorphism. It is, however, the base component of a multiplicative homogeneous degree-n polynomial law:

Definition 2.5. Let R be a ring and M and N two R-modules. A polynomial law p: M → N is a collection of functions p_{S}: S ⊗_{R} M → S ⊗_{R} N for each R-algebra S, such that for every R-algebra homomorphism f: S → S′ the following square of functions commutes:

\begin{matrix}S ⊗_{R} M & S ⊗_{R} N \\ S′ ⊗_{R} M & S′ ⊗_{R} N \end{matrix} p_{S} f⊗id_{M} f⊗id_{N} p_{S′}

We say that p is homogeneous of degree n if for every R-algebra S, element s ∈ S and element m ∈ S ⊗ M, we have p_{S}(sm) = s^{n}p_{S}(m). If p: A → B is a polynomial law between two R-algebras A and B, we say p is multiplicative if each p_{S}: S ⊗_{R} A → S ⊗_{R} B is a multiplicative function.

For example, if A is an R-algebra, then the diagonal function A → A^{⊗n} sending a to a ⊗ ⋯ ⊗ a extends naturally to a multiplicative homogeneous degree-n polynomial law A → A^{⊗n}. A fundamental result of the theory of polynomial laws, developed by Norbert Roby in [14], is that for each R-module M there is a universal homogeneous degree-n polynomial law γ^{n}: M → Γ_{R}^{n}(M); every homogeneous degree-n polynomial law M → N factors uniquely as γ^{n} followed by an ordinary R-module homomorphism Γ_{R}^{n}(M) → N. Furthermore, Roby shows in [15] that if A is an R-algebra, then Γ_{R}^{n}(A) is also an R-algebra and multiplicative degree-n polynomial laws A → B correspond to R-algebra homomorphisms Γ_{R}^{n}(A) → B. Thus the norm map Nm_{A/R}: A → R corresponds to an R-algebra homomorphism Γ_{R}^{n}(A) → R. A general presentation for Γ_{R}(M) may be found in [14, III.1], but for our purposes it is sufficient to note that the diagonal polynomial law A → A^{⊗n} gives us an R-algebra homomorphism Γ_{R}^{n}(A) → A^{⊗n}, and that if A is a flat R-algebra then this homomorphism restricts to an isomorphism with the subalgebra of S_{n}-invariant tensors (see [4, 5.5.2.5 on p. 123]):

Γ_{R}^{n}(A) ⟶^{∼} (A^{⊗n})^{S_{n}} : γ^{n}(a) ↦ a ⊗ ⋯ ⊗ a.

Thus if A and B are R-algebras with A flat, multiplicative homogeneous degree-n polynomial laws A → B correspond to R-algebra homomorphisms (A^{⊗n})^{S_{n}} → B.

Definition 2.6 (c.f. [6, 3.1.2]). Let R be a ring and A an R-algebra of rank n. Then the norm polynomial law Nm_{A/R}: A → R corresponds to an R-algebra homomorphism

Φ_{A/R}: (A^{⊗n})^{S_{n}} ≅ Γ_{R}^{n}(A) → R,

called the Ferrand homomorphism (of A over R). It is the unique R-algebra homomorphism such that for every R-algebra R′ and every element a ∈ A′ := R′ ⊗_{R} A, the composite

(1)  \begin{matrix}(A′^{⊗_{R′} n})^{S_{n}} & R′ ⊗_{R} (A^{⊗n})^{S_{n}} & R′ ⊗_{R} R & R′ \\ ∼ \\ id_{R′} ⊗Φ_{A/R} \\ ∼ \end{matrix}

sends a ⊗ ⋯ ⊗ a to Nm_{A′/R′}(a) ∈ R′.

As an immediate corollary to this definition, we find for each R-algebra R′ and A′ := R′ ⊗_{R} A that the composite (1) is the Ferrand homomorphism for A′ over R′. We say that the Ferrand homomorphism commutes with base change.

Definition 2.7. Let R be a ring and let A be an R-algebra of rank n with n ≥ 2. Then the discriminant algebra Δ_{A/R} of A over R is the tensor product of (A^{⊗n})^{S_{n}}-algebras

Δ_{A/R} = (A^{⊗n})^{A_{n}} ⨂_{(A^{⊗n})^{Sn}} R

defined by the inclusion (A^{⊗n})^{S_{n}} ↪ (A^{⊗n})^{A_{n}} and the Ferrand homomorphism Φ_{A/R}: (A^{⊗n})^{S_{n}} → R.

If the base ring R is understood, it may be omitted and the discriminant algebra of A denoted Δ_{A}. We will adopt similar conventions for the trace and norm maps, the Ferrand homomorphism, and the discriminant bilinear form.

Remark 2.8. Note that since the Ferrand homomorphism Φ_{A/R} is surjective, so is the homomorphism from (A^{⊗n})^{A_{n}} → Δ_{A/R}. Thus Δ_{A/R} can be understood as the quotient of (A^{⊗n})^{A_{n}} by the ideal generated by elements of the form x − Φ_{A/R}(x) for x ∈ (A^{⊗n})^{S_{n}}. If x is a general element of (A^{⊗n})^{A_{n}}, we will often denote its image in Δ_{A/R} by ẋ.

Remark 2.9. Note that if we replace (A^{⊗n})^{A_{n}} by A^{⊗n} in the definition of Δ_{A/R}, we obtain what Ferrand denotes by P^{(1,...,1)}(A) in [6, 5.2] and what Bhargava calls the S_{n}-closure of A over R in [1]. Denoting the S_{n}-closure by B, we thus obtain an R-algebra homomorphism Δ_{A/R} → B by tensoring the inclusion (A^{⊗n})^{A_{n}} ↪ A^{⊗n} along Φ_{A/R}. In case R → A is a degree-n separable field extension K ↪ L in characteristic other than 2, whose normal closure N has Galois group S_{n}, then B ≅ N and this homomorphism Δ_{A/R} → B is just the inclusion of the discriminant field of L into N.

3. Understanding the Ferrand homomorphism

The two defining facts of the Ferrand homomorphisms are that they send elements of the form a⊗⋯⊗a to the norm of a, and that they commute with base change. Our primary tool for computing the image of an arbitrary symmetric tensor, then, is to identify it as a term in some tensor power of a single element, and then correspondingly break up the norm of that single element. For example, if A is any R-algebra, and a ∈ A, then we have the following elements of (A^{⊗n})^{S_{n}}:

e_{1}(a) := (a ⊗ 1 ⊗ ⋯ ⊗ 1) + (1 ⊗ a ⊗ ⋯ ⊗ 1) + … + (1 ⊗ ⋯ ⊗ 1 ⊗ a)
e_{2}(a) := (a ⊗ a ⊗ 1 ⊗ ⋯ ⊗ 1) + (a ⊗ 1 ⊗ a ⊗ ⋯ ⊗ 1) + …
… + (1 ⊗ ⋯ ⊗ 1 ⊗ a ⊗ a)
…
e_{n}(a) := (a ⊗ a ⊗ ⋯ ⊗ a).

These are the elementary symmetric polynomials in the n elements a^{(1)}, …, a^{(n)} ∈ A^{⊗n}, where a^{(i)} is the element

(2)  a^{(i)} = 1 ⊗ ⋯ ⊗ 1 ⊗ a ⊗ 1 ⊗ ⋯ ⊗ 1

with the a in the ith tensor factor. In case A is a rank-n R-algebra, we can compute the image of e_{k}(a) under Φ_{A/R}: (A^{⊗n})^{S_{n}} → R as follows: First, note that e_{k}(a) is the coefficient of λ^{n−k}μ^{k} in (λ+μa)⊗⋯⊗(λ+μa), as an element of (A[λ, μ]^{⊗_{R[λ,μ]}n})^{S_{n}}. (Thus we sometimes also write e_{0}(a) = 1.) The image of this element under Φ_{A[λ,μ]/R[λ,μ]} is its norm f(λ, μ) ∈ R[λ, μ]. Since the norm map is homogeneous of degree n, we find that f(λ, μ) is a homogeneous polynomial of degree n; denoting by s_{k}(a) the coefficient of λ^{n−k}μ^{k} in f(λ, μ), we obtain

Φ_{A[λ,μ]/R[λ,μ]} : ∑_{k=0}^{n} e_{k}(a)λ^{n−k}μ^{k} ↦ ∑_{k=0}^{n} s_{k}(a)λ^{n−k}μ^{k}.

Since Φ_{A[λ,μ]/R[λ,μ]} is just Φ_{A/R} ⊗ id_{R[λ,μ]}, we thus have

Φ_{A/R}: e_{k}(a) ↦ s_{k}(a)

for each k ∈ {0, …, n}. What are these quantities s_{k}(a) ∈ R? The answer comes by setting μ = −1, so that

∑_{k=0}^{n} s_{k}(a)λ^{n−k}(−1)^{k} = Nm_{A[λ]/R[λ]}(λ − a),

the characteristic polynomial of a. We can thus read off s_{k}(a) as (−1)^{k} times the coefficient of λ^{n−k} in the characteristic polynomial of a. In particular, s_{n}(a) is the norm Nm_{A/R}(a) of a, and s_{1}(a) is its trace Tr_{A/R}(a). We summarize the above discussion in the following lemma:

Lemma 3.1. Let R be a ring and A an R-algebra with an element a ∈ A, and let k ∈ {0, …, n}. Denote by e_{k}(a) the kth elementary symmetric polynomial in the n elements a^{(1)}, …, a^{(n)} ∈ A^{⊗n} defined in (2). Now suppose A is a rank-n R-algebra, and let s_{k}(a) be (−1)^{k} times the coefficient of λ^{n−k} in the characteristic polynomial of a. Then Φ_{A/R}(e_{k}(a)) = s_{k}(a).

We can similarly compute the image under Φ_{A/R} of other elements of (A^{⊗n})^{S_{n}}.

Definition 3.2 (c.f. [6, 2.2.3.1]). Let a = (a_{i})_{i∈I} ∈ A^{I} be a tuple of elements of A, and let α ⊆ I^{n}. We define an element γ^{α}(a) ∈ A^{⊗n} by

γ^{α}(a) := ∑_{(i_{1},...,i_{n})∈α} a_{i_{1}} ⊗ ⋯ ⊗ a_{i_{n}}.

If G is a subgroup of S_{n} and α is invariant under the action of G on I^{n}, then γ^{α}(a) ∈ (A^{⊗n})^{G}. For example, if α is the S_{n}-orbit of (i_{1}, …, i_{n}) ∈ I^{n}, then γ^{α}(a) is the coefficient of λ_{i_{1}} … λ_{i_{n}} in (∑_{i∈I} λ_{i}a_{i}) ⊗ ⋯ ⊗ (∑_{i∈I} λ_{i}a_{i}). In this case Φ_{A/R}(γ^{α}(a)) is the coefficient of λ_{1} … λ_{n} in the norm of ∑_{i∈I} λ_{i}a_{i}, as an element of the rank-n R[λ_{i} : i ∈ I]-algebra A[λ_{i} : i ∈ I]. The case of Φ_{A/R}(e_{k}(a)) = s_{k}(a) is recovered by taking I = {1, 2} and (a_{1}, a_{2}) = (1, a). In general, we will write the set of G-orbits of I^{n} as I^{n}/G, and write α ∈ I^{n}/G to mean that α is one such G-orbit.

We will often consider the case that the tuple I → A is an ordinary ntuple a = (a_{1}, …, a_{n}) ∈ A^{n}, and that the subset α ⊆ {1, …, n}^{n} = Map({1, …, n}, {1, …, n}) is a subgroup G ⊆ S_{n}, or a coset σG of G. (The most common cases are G = S_{n} or A_{n}, or the coset of odd permutations \overline{A}_{n}.) The quantity γ^{σG}(a_{1}, …, a_{n}) is linear in each a_{i}, since each a_{i} appears exactly once per term as one of the tensor factors.

Example 3.3. Consider the case R = ℤ and A = R[x]/(x^{2} +x+2), a quadratic R-algebra. If we set a_{1} = −2x + 1 and a_{2} = 3x + 2, then the element a_{1} ⊗ a_{2} + a_{2} ⊗ a_{1} is an S_{2}-invariant element of A^{⊗2}—what is its image under the Ferrand homomorphism Φ_{A/R}: (A^{⊗2})^{S_{2}} → R? We will calculate the answer three different ways to demonstrate the various methods we will use throughout the paper. First, and simplest, note that since x ∈ A satisfies both the defining equation x^{2}+x+2 = 0 as well as its characteristic polynomial x^{2}−Tr_{A}(x)+Nm_{A}(x) = 0, we find that these two equations are equal because 1 and x are R-linearly independent elements of A. Therefore Tr_{A}(x) = −1 and Nm_{A}(x) = 2. We can then compute the image of a_{1} ⊗ a_{2} + a_{2} ⊗ a_{1} by expanding it in terms of these quantities:

a_{1} ⊗ a_{2} + a_{2} ⊗ a_{1} = (−2x + 1) ⊗ (3x + 2) + (3x + 2) ⊗ (−2x + 1)
= −6(x ⊗ x) − 4(x ⊗ 1) + 3(1 ⊗ x) + 2(1 ⊗ 1)
− 6(x ⊗ x) + 3(x ⊗ 1) − 4(1 ⊗ x) + 2(1 ⊗ 1)
= −12(x ⊗ x) − 1(x ⊗ 1 + 1 ⊗ x) + 4
= −12 · e_{2}(x) − 1 · e_{1}(x) + 4
↦ −12 · s_{2}(x) − 1 · s_{1}(x) + 4
= −12 · Nm_{A}(x) − 1 · Tr_{A}(x) + 4
= −12 · 2 − 1 · −1 + 4 = −19.

We can also determine the image of a_{1} ⊗ a_{2} + a_{2} ⊗ a_{1} knowing only how multiplying by a_{1} and a_{2} act with respect to an R-basis of A. Choosing the R-basis {1, x}, we find that they act by the matrices

a_{1} : (\begin{matrix}1 & 4 \\ −2 & 3 \end{matrix}) a_{2} : (\begin{matrix}2 & −6 \\ 3 & −1 \end{matrix}) .

(For example, a_{2} · x = (3x + 2)(x) = 3x^{2} + 2x = 3(−x − 2) + 2x = −6 − x, whose coefficients are found in the second column of the a_{2} matrix.) Now a_{1} ⊗ a_{2} + a_{2} ⊗ a_{1} is the quantity denoted γ^{S_{2}}(a_{1}, a_{2}), which is the coefficient of λ_{1}λ_{2} in (λ_{1}a_{1} + λ_{2}a_{2}) ⊗ (λ_{1}a_{1} + λ_{2}a_{2}). Thus its image under the Ferrand homomorphism is the coefficient of λ_{1}λ_{2} in the norm of λ_{1}a_{1} + λ_{2}a_{2} (as an element of the quadratic R[λ_{1}, λ_{2}]-algebra A[λ_{1}, λ_{2}]). This element acts by the matrix

λ_{1}a_{1} + λ_{2}a_{2} : (\begin{matrix}λ_{1} + 2λ_{2} & 4λ_{1} − 6λ_{2} \\ −2λ_{1} + 3λ_{2} & 3λ_{1} − λ_{2} \end{matrix}) .

Ignoring all but the λ_{1}λ_{2} terms, this determinant is

(λ_{1})(−λ_{2}) + (2λ_{2})(3λ_{1}) − (4λ_{1})(3λ_{2}) − (−6λ_{2})(−2λ_{1}) = −19λ_{1}λ_{2},

so Φ_{A/R}(a_{1} ⊗ a_{2} + a_{2} ⊗ a_{1}) = −19, as before. Finally, note that we can write a_{1} ⊗ a_{2} + a_{2} ⊗ a_{1} as a polynomial in symmetric tensors of the form e_{k}(f):

a_{1} ⊗ a_{2} + a_{2} ⊗ a_{1} = (a_{1} ⊗ 1 + 1 ⊗ a_{1})(1 ⊗ a_{2} + a_{2} ⊗ 1)
− (a_{1}a_{2} ⊗ 1 + 1 ⊗ a_{1}a_{2})
= e_{1}(a_{1})e_{1}(a_{2}) − e_{1}(a_{1}a_{2})
↦ s_{1}(a_{1})s_{1}(a_{2}) − s_{1}(a_{1}a_{2})
= Tr_{A}(a_{1})Tr_{A}(a_{2}) − Tr_{A}(a_{1}a_{2}).

We can read off Tr_{A}(a_{1}) and Tr_{B}(a_{2}) as the traces of the above matrices, namely 1 + 3 = 4 and 2 − 1 = 1. The trace of a_{1}a_{2} can by computed similarly as the trace of the product matrix:

a_{1}a_{2} : (\begin{matrix}1 & 4 \\ −2 & 3 \end{matrix}) (\begin{matrix}2 & −6 \\ 3 & −1 \end{matrix}) = (\begin{matrix}14 & −10 \\ 5 & 9 \end{matrix}) ,

so Tr_{A}(a_{1}a_{2}) = 14 + 9 = 23. Therefore Φ_{A}(a_{1} ⊗ a_{2} + a_{2} ⊗ a_{1}) = (4)(1) − (23) = −19.

The latter two methods for computing Φ_{A/R} of an S_{n}-invariant tensor— expanding the tensor as a linear combination of the γ^{α}(a) or a polynomial in the e_{k}(a)—will be used throughout this paper. The remainder of this section consists of technically useful results on sets of module and algebra generators for the G-invariants of tensor powers, which, in effect, say that the methods of Example 3.3 will always be sufficient to calculate the image of an S_{n}-invariant tensor under the Ferrand homomorphism.

Lemma 3.4. Let R be a ring and M a projective R-module. Let θ = (θ_{i})_{i∈I} ∈ M^{I} be a generating family (resp. R-basis) for M. Then for each natural number d and subgroup G ⊆ S_{d}, the family {γ^{α}(θ) : α ∈ I^{d}/G} is a generating family (resp. R-basis) for (M^{⊗d})^{G}.

Recall that by α ∈ I^{d}/G we mean that α is an orbit of I^{d} under the action of G, so that the notation γ^{α}(θ) makes sense.

Proof. First consider the case in which θ is an R-basis for M. Then every tensor in M^{⊗d} can be uniquely represented as an R-linear combination of pure tensors of the form θ_{i_{1}} ⊗ ⋯ ⊗ θ_{i_{d}}. Such a linear combination is G-invariant precisely if the coefficients are constant across G-orbits of the pure tensors, i.e. if the tensor is an R-linear combination of the γ^{α}(θ) as α ranges over I^{d}/G. Hence the γ^{α}(θ) generate (M^{⊗d})^{G}. Furthermore, since no pure tensor is a term in more than one of the γ^{α}(θ), they are also R-linearly independent, and hence form an R-basis for (M^{⊗d})^{G}. Second, suppose that M is merely projective and that {θ_{i} : i ∈ I} merely generates M as an R-module. Then let R^{(I)} be the free R-module with basis e = (e_{i})_{i∈I}, and R^{(I)} ↠ M be the surjection sending e_{i} to θ_{i}. Since M is projective, this surjection has a right inverse, which is a property preserved by any functor. Hence after applying the functor ((·)^{⊗d})^{G}, we obtain another surjection ((R^{(I)})^{⊗d})^{G} ↠ (M^{⊗d})^{G}. Since {γ^{α}(e) : α ∈ I^{d}/G} generates ((R^{(I)})^{⊗d})^{G} by the above, its image {γ^{α}(θ) : α ∈ I^{d}/G} generates (M^{⊗d})^{G} as well. □

Thus given a set of R-module generators θ = (θ_{i})_{i∈I} of a rank-n R-algebra A, we find that every element of (A^{⊗n})^{S_{n}} (resp. (A^{⊗n})^{A_{n}}) can be written as an R-linear combination of the γ^{α}(θ), as α ranges over all S_{n}-orbits (resp. A_{n}- orbits) of I^{n}. This fact has a number of consequences that we explore in the remainder of this section.

Proposition 3.5. Let R be a ring, let M be a locally free R-module, let d be a natural number, and let G be a subgroup of S_{d}. Let R′ be any R-algebra, and denote by M′ the locally free R′-module R′ ⊗_{R} M. Then the natural R′-module homomorphism R′ ⊗_{R} (M^{⊗d})^{G} → (M′^{⊗_{R′} d})^{G} is an isomorphism.

Proof. Since (M^{⊗d})^{G} can be written as an equalizer of finitely many maps M^{⊗d} → M^{⊗d}, the functor ((·)^{⊗d})^{G} commutes with localization (see, for example, [9, Prop. A7.1.3]). We therefore only need to check that the given homomorphism is an isomorphism in the case that M is free. Say M has R-basis θ = (θ_{i})_{i∈I}, so that M′ has a free R′-basis θ′ = (1 ⊗ θ_{i})_{i∈I}. Then the canonical map R′ ⊗_{R} (M^{⊗d})^{G} → (M′^{⊗_{R′} d})^{G} carries the R′-basis {1 ⊗ γ^{α}(θ) : α ∈ I^{d}/G} to the R′-basis {γ^{α}(θ′) : α ∈ I^{d}/G}, so is an isomorphism. □

Next we show that elements of the form e_{k}(a) generate (A^{⊗n})^{S_{n}} as an algebra. The proof is algorithmic in nature, and shows us how to write any element of the form γ^{α}(a) as a polynomial in the e_{k}(a).

Lemma 3.6. Let R be a ring and A a rank-n R-algebra. Let Ω ⊆ A be a set of elements of A whose powers together generate A as an R-module. Then the ring (A^{⊗n})^{S_{n}} is generated as an R-algebra by {e_{k}(a) : k ∈ {1, …, n} and a ∈ Ω}.

In particular, the Ferrand homomorphism Φ_{A/R}: (A^{⊗n})^{S_{n}} → R is the unique R-algebra homomorphism sending e_{k}(a) ↦ s_{k}(a) for all k ∈ {1, …, n} and a ∈ Ω. (Note: in case Ω is the set of primitive monomials of a set of algebra generators for A, Lemma 3.6 is a corollary of [17, Theorem 4.10] by an argument similar to the proof of Lemma 3.4.)

Proof. First note that for each m ∈ ℕ and a ∈ Ω, we can express each e_{k}(a^{m}) as a polynomial in {e_{j}(a) : j ∈ {1, …, n}} by the fundamental theorem of symmetric polynomials. Therefore by adding to Ω all powers of its elements, we can assume without loss of generality that Ω is a set of R-module generators for A that contains 1. Therefore the set {γ^{α}(Ω) : α ∈ Ω^{n}/S_{n}} generates (A^{⊗n})^{S_{n}} as an R-module, where we regard Ω as a family of elements of A indexed by itself. Next we show by induction on k that if α ∈ Ω^{n}/S_{n} is the S_{n}-orbit of any tuple with at most k entries not equal to 1, then γ^{α}(Ω) is in the subalgebra of (A^{⊗n})^{S_{n}} generated by {e_{j}(a) : a ∈ Ω and j ∈ {1, …, k}}. The case k = n then implies that the subalgebra generated by all the e_{k}(a) contains a set of R-module generators for (A^{⊗n})^{S_{n}}, and thus is all of (A^{⊗n})^{S_{n}}. The base case k = 0 is clear: if α = {(1, …, 1)}, then γ^{α}(Ω) = 1 is in every subalgebra of (A^{⊗n})^{S_{n}}. Now assume that the hypothesis holds for all j < k, and consider an element of the form γ^{α}(Ω) where α is the S_{n}-orbit of a tuple in Ω^{n} with at most k components unequal to 1. Given a ∈ Ω, let α(a) be the number of times a appears in any single tuple in α; thus α(1) ≥ n − k and ∑_{a∈Ω\{1}} α(a) ≤ k. Construct the element

p = ∏_{a∈Ω\{1}} e_{α(a)}(a) ∈ (A^{⊗n})^{S_{n}};

since α(a) = 0 (and thus e_{α(a)}(a) = 1) for all but finitely many a ∈ Ω, this product is well-defined. Rewrite the product p as follows: first simplify the product and collect terms related by permutations to write p as a sum of terms of the form γ^{β}(A), one of which is the original γ^{α}(Ω). For the remainder of the terms, choose an expression of each tensor factor as an R-linear combination of elements of Ω, taking care to choose the same expression each time the same element of A appears and to leave each tensor factor 1 alone. After collecting like terms again, we have written p as γ^{α}(Ω) plus an R-linear combination of terms of the form γ^{β}(Ω). These other β all have the property that β(1) > n − k, so by the induction hypothesis γ^{α}(a) − p is in the algebra generated by {e_{j}(a) : a ∈ Ω and j < k}. Therefore γ^{α}(a) is in the subalgebra generated by {e_{j}(a) : a ∈ Ω and j ≤ k}, as desired. □

For example, given a tuple a = (a_{1}, …, a_{n}) of elements of an algebra A, we have the following elements of A^{⊗n}:

γ^{A_{n}}(a) = ∑_{σ∈S_{n} even} a_{σ(1)} ⊗ ⋯ ⊗ a_{σ(n)} and
γ^{\overline{A}^{n}}(a) = ∑_{σ∈S_{n} odd} a_{σ(1)} ⊗ ⋯ ⊗ a_{σ(n)},

where \overline{A}_{n} is the nontrivial coset of A_{n}, and their sum

γ^{S_{n}}(a) = ∑_{σ∈S_{n}} a_{σ(1)} ⊗ ⋯ ⊗ a_{σ(n)}.

Finally, we write the product of two elements of the form γ^{G}(a_{1}, …, a_{n}) in terms of more elements of this form.

Lemma 3.7. Let R be a ring and A an R-algebra. Let n be a natural number and G a subgroup of S_{n}. Then for each pair of tuples (a_{1}, …, a_{n}) and (b_{1}, …, b_{n}) in A^{n}, the equation

γ^{G}(a)γ^{G}(b) = ∑_{σ∈G} γ^{G}(a_{1}b_{σ(1)}, …, a_{n}b_{σ(n)}).

holds in (A^{⊗n})^{G}.

If we denote the tuple (b_{σ(1)}, …, b_{σ(n)}) by b_{σ} ∈ A^{n}, and give A^{n} the product algebra structure, then the product ab_{σ} is (a_{1}b_{σ(1)}, …, a_{n}b_{σ(n)}), and we may write the above identity as

γ^{G}(a)γ^{G}(b) = ∑_{σ∈G} γ^{G}(ab_{σ}).

Proof. By definition γ^{G}(a) equals ∑_{τ∈G} ∏_{i=1}^{n} a_{τ(i)}^{(i)}—recalling the ^{(i)} notation for the ith embedding A → A^{⊗n}—so we have

γ^{G}(a)γ^{G}(b) = (∑_{τ∈G} ∏_{i=1}^{n} a_{τ(i)}^{(i)}) (∑_{σ∈G} ∏_{i=1}^{d} b_{σ(i)}^{(i)}) .

Expanding the product we get

∑_{τ∈G} ∑_{σ∈G} ∏_{i=1}^{n} a_{τ(i)}^{(i)}b_{σ(i)}^{(i)} = ∑_{τ∈G} ∑_{σ∈G} ∏_{i=1}^{n} a_{τ(i)}^{(i)}b_{στ(i)}^{(i)}

since for fixed τ the composite στ runs over G as σ does. Now a_{τ(i)}^{(i)}b_{στ(i)}^{(i)} =

(ab_{σ})_{τ(i)}^{(i)}, so by reversing the order of summation we find

γ^{G}(a)γ^{G}(b) = ∑_{σ∈G} ∑_{τ∈G} ∏_{i=1}^{n}(ab_{σ})_{τ(i)}^{(i)} = ∑_{σ∈G} γ^{G}(ab_{σ})

as desired. □

4. Proof of Theorem 1

Recall that given a ring R with a rank-n ≥ 2 algebra A, the discriminant algebra of A (over R) is defined to be the tensor product

Δ_{A} = (A^{⊗n})^{A_{n}} ⨂_{(A^{⊗n})^{Sn}}R,

using the two ring homomorphisms (A^{⊗n})^{S_{n}} ↪ (A^{⊗n})^{A_{n}} and the Ferrand homomorphism Φ_{A}: (A^{⊗n})^{S_{n}} → R. Equivalently, since the Ferrand homomorphism is surjective, we could present Δ_{A} as the quotient of (A^{⊗n})^{A_{n}} by the ideal generated by elements of the form {x − Φ_{A}(x) : x ∈ (A^{⊗n})^{S_{n}}}. In this section, we prove Theorem 1 in the following form:

Theorem 4.1. Let R be a ring, and let A be an R-algebra of rank n, with n ≥ 2.

(a) Its discriminant algebra Δ_{A} fits naturally into a short exact sequence of R-modules

0  R  Δ_{A}  ⋀^{n}A  0.

In particular, Δ_{A} is an R-algebra of rank 2.

(b) The resulting composition of isomorphisms ⋀^{n}A ≅ Δ_{A}/(R·1) ≅ ⋀^{2}Δ_{A} identifies the discriminant bilinear forms of A and Δ_{A}.

In fact, if we denote for each (a_{1}, …, a_{n}) ∈ A^{n} the image of the A_{n}-invariant tensor

γ^{A_{n}}(a_{1}, …, a_{n}) := ∑_{σ∈A_{n}} a_{σ(1)} ⊗ ⋯ ⊗ a_{σ(n)},

under the surjection (A^{⊗n})^{A_{n}} ↠ Δ_{A} by γ̇^{A_{n}}(a_{1}, …, a_{n}), then we can describe the R-module homomorphism Δ_{A} → ⋀^{n}A in the exact sequence as the unique one sending 1 to 0 and each γ̇^{A_{n}}(a_{1}, …, a_{n}) to a_{1} ∧ ⋯ ∧ a_{n}. The isomorphism ⋀^{n}A ≅ ⋀^{2}Δ_{A} then sends a_{1} ∧ ⋯ ∧ a_{n} to 1 ∧ γ̇^{A_{n}}(a_{1}, …, a_{n}). The outline of the proof of Theorem 4.1(a) is as follows. We show first that the (A^{⊗n})^{S_{n}}-linear map A^{⊗n} → (A^{⊗n})^{A_{n}} sending a_{1} ⊗ ⋯ ⊗ a_{n} to γ^{A_{n}}(a_{1}, …, a_{n}) descends to an isomorphism of their quotient modules ⋀^{n}A ≅ (A^{⊗n})^{A_{n}}/(A^{⊗n})^{S_{n}}. This gives us a short exact sequence:

0  (A^{⊗n})^{S_{n}}  (A^{⊗n})^{A_{n}}  ⋀^{n}A  0.

Then we show that base changing this exact sequence along the Ferrand homomorphism Φ_{A/R}: (A^{⊗n})^{S_{n}} → R gives us the desired exact sequence of Rmodules in Theorem 4.1(a). The following lemma gives the technical results needed to make this argument go through.

Lemma 4.2. Let R be a ring and A an R-algebra of rank n ≥ 2.

(1) The R-linear map A^{⊗n} → (A^{⊗n})^{A_{n}} sending a_{1} ⊗ ⋯ ⊗ a_{n} to γ^{A_{n}}(a_{1}, …, a_{n}) descends to an isomorphism ⋀^{n}A ≅ (A^{⊗n})^{A_{n}}/(A^{⊗n})^{S_{n}}. In particular, ⋀^{n}A inherits an (A^{⊗n})^{S_{n}}-module structure from A^{⊗n}.

(2) The resulting homomorphism from (A^{⊗n})^{S_{n}} to End_{R}(⋀^{n}A) ≅ R is the Ferrand homomorphism Φ_{A/R}.

Proof. (1) The R-module homomorphism A^{⊗n} → (A^{⊗n})^{A_{n}} sending a_{1}⊗⋯⊗a_{n} to γ^{A_{n}}(a_{1}, …, a_{n}) is well-defined since γ^{A_{n}}(a_{1}, …, a_{n}) is R-linear in each a_{i}. It is also easily seen to be (A^{⊗n})^{S_{n}}-linear. Therefore we obtain an (A^{⊗n})^{S_{n}}- module homomorphism A^{⊗n} → (A^{⊗n})^{A_{n}}/(A^{⊗n})^{S_{n}} sending a_{1} ⊗ ⋯ ⊗ a_{n} to the class of γ^{A_{n}}(a_{1}, …, a_{n}). But if any a_{i} = a_{j} with i ≠ j, we find that γ^{A_{n}}(a_{1}, …, a_{n}) is S_{n}-invariant and thus equal to zero in the quotient. Thus we obtain a well-defined R-module homomorphism

⋀^{n}A → (A^{⊗n})^{A_{n}}/(A^{⊗n})^{S_{n}}

sending a_{1} ∧ ⋯ ∧ a_{n} to the class of γ^{A_{n}}(a_{1}, …, a_{n}). Since both sides commute with localization (see Proposition 3.5 for the righthand side), it suffices to check that this homomorphism is an isomorphism in the case that A is free as an R-module, say with basis (θ_{1}, …, θ_{n}). Then ⋀^{n}A is free with singleton basis θ_{1} ∧⋯∧θ_{n}. Applying Lemma 3.4, we also find that

(A^{⊗n})^{A_{n}} = ⨁_{α∈{1,...,n}^{n}/A_{n}} R · γ^{α}(θ)  and  (A^{⊗n})^{S_{n}} = ⨁_{α∈{1,...,n}^{n}/S_{n}} R · γ^{α}(θ).

Now any A_{n}-orbit of a tuple with a repeated index is actually its S_{n}-orbit, so these bases are identical except that one uses γ^{A_{n}}(θ) and γ^{\overline{A}^{n}}(θ), and the other their sum γ^{S_{n}}(θ). Thus the quotient is

(A^{⊗n})^{A_{n}}/(A^{⊗n})^{S_{n}} = \frac{R · γ^{A_{n}}(θ) ⊕ R · γ^{\overline{A}^{n}}(θ)}{R · (γ^{A_{n}}(θ) + γ^{\overline{A}^{n}}(θ))},

for which we may take either the class of γ^{A_{n}}(θ) or γ^{\overline{A}^{n}}(θ) as a singleton Rmodule basis. Then under the homomorphism ⋀^{n}A → (A^{⊗n})^{A_{n}}/(A^{⊗n})^{S_{n}}, the free R-module generator θ_{1} ∧ ⋯ ∧ θ_{n} for ⋀^{n}A is transformed into the class of γ^{A_{n}}(θ_{1}, …, θ_{n}), which we now know to also be a free R-module generator for (A^{⊗n})^{A_{n}}/(A^{⊗n})^{S_{n}}.

(2) We check the defining property of the Ferrand homomorphism. First, let a ∈ A; we will see which endomorphism of ⋀^{n}A the element a ⊗ ⋯ ⊗ a ∈ (A^{⊗n})^{S_{n}} gives rise to. The (A^{⊗n})^{S_{n}}-module structure on ⋀^{n}A is the one inherited from the (A^{⊗n})^{S_{n}}-algebra A^{⊗n}, so the endomorphism of ⋀^{n}A corresponding to a ⊗ ⋯ ⊗ a is

b_{1} ∧ ⋯ ∧ b_{n} ↦ (ab_{1}) ∧ ⋯ ∧ (ab_{n}).

But by definition, this endomorphism is multiplication by Nm_{A/R}(a) ∈ R, as desired. Now we check that for each R-algebra R′, if we set A′ = R′ ⊗_{R} A then the base-changed homomorphism

(A′^{⊗_{R′} n})^{S_{n}} ≅ R′ ⊗_{R} (A^{⊗n})^{S_{n}} → R′ ⊗_{R} End_{R}(⋀^{n}A) ≅ End_{R′}(⋀^{n}A′)

has the same property, sending each a ⊗ ⋯ ⊗ a to multiplication by Nm_{A′/R′}(a) ∈ R′. But this follows because the base-changed homomorphism is again the one expressing the fact that ⋀^{n}A′ inherits an (A′^{⊗_{R′} n})^{S_{n}}-module structure from A^{′⊗_{R′} n}. □

Proof of Theorem 4.1(a). By Lemma 4.2(1), we obtain a short exact sequence of (A^{⊗n})^{S_{n}}-modules

0 → (A^{⊗n})^{S_{n}} → (A^{⊗n})^{A_{n}} → ⋀^{n}A → 0

where the right-hand map sends γ^{A_{n}}(a_{1}, …, a_{n}) ↦ a_{1} ∧ ⋯ ∧ a_{n}. Tensoring along the Ferrand homomorphism Φ_{A/R}: (A^{⊗n})^{S_{n}} → R is equivalent to quotienting by its kernel I = (x − Φ_{A}(x) : x ∈ (A^{⊗n})^{S_{n}}), obtaining an a priori merely right-exact sequence

(A^{⊗n})^{S_{n}}/I → (A^{⊗n})^{A_{n}}/(I · (A^{⊗n})^{A_{n}}) → ⋀^{n}A/(I · ⋀^{n}A) → 0.

From left to right, these modules are R, Δ_{A}, and ⋀^{n}A, in the latter case because I, being the kernel of the Ferrand homomorphism, is also the annihilator of the (A^{⊗n})^{S_{n}}-module ⋀^{n}A by Lemma 4.2(2). Thus we have a right exact sequence of R-modules

R → Δ_{A} → ⋀^{n}A → 0,

where the left-hand map sends 1 ↦ 1 and the right-hand map sends each γ̇^{A_{n}}(a_{1}, …, a_{n}) to a_{1} ∧ ⋯ ∧ a_{n}. But then R → Δ_{A} must be injective: if r ↦ 0, then r acts by zero on Δ_{A} and hence on its quotient ⋀^{n}A. But then r must be zero, since ⋀^{n}A is locally free of rank 1 and thus a faithful R-module. (Another way to see that R → Δ_{A} is injective is to note that I is the annihilator of the (A^{⊗n})^{S_{n}}-module (A^{⊗n})^{A_{n}}/(A^{⊗n})^{S_{n}}. It is thus the largest ideal of (A^{⊗n})^{S_{n}} whose product with (A^{⊗n})^{A_{n}} is contained in (A^{⊗n})^{S_{n}}; therefore I · (A^{⊗n})^{A_{n}} = I. Thus the map R → Δ_{A} is the map

(A^{⊗n})^{S_{n}}/I → (A^{⊗n})^{A_{n}}/(I · (A^{⊗n})^{A_{n}}) = (A^{⊗n})^{A_{n}}/I,

which is manifestly injective.) □

The remainder of the section is devoted to proving Theorem 4.1(b). Our first result relates the multiplication of some elements of Δ_{A} to the discriminant form of A.

Lemma 4.3. Let R be a ring. Let A be an R-algebra of rank n ≥ 2. Let a = (a_{1}, …, a_{n}) and b = (b_{1}, …, b_{n}) be in A^{n}. Then the equality

(γ̇^{A_{n}}(a) − γ̇^{\overline{A}^{n}}(a))(γ̇^{A_{n}}(b) − γ̇^{\overline{A}^{n}}(b)) = δ_{A}(a_{1} ∧ ⋯ ∧ a_{n}, b_{1} ∧ ⋯ ∧ b_{n})

holds in Δ_{A}.

Proof. Recall fromm Eq. (2) at the beginning of Section 3 the notation (·)^{(i)} for the ith embedding A → A^{⊗n}. Then we may write the difference γ^{A_{n}}(a)−γ^{\overline{A}^{n}}(a) in A^{⊗n} as

∑_{σ∈A_{n}} ∏_{i=1}^{n} a_{i}^{(σ(i))} − ∑_{σ∈\overline{A}_{n}} ∏_{i=1}^{n} a_{i}^{(σ(i))} = ∑_{σ∈S_{n}} sgn(σ) ∏_{i=1}^{n} a_{i}^{(σ(i))},

the determinant of the n × n-matrix whose ijth element is a_{i}^{(j)}, and similarly γ^{A_{n}}(b) − γ^{\overline{A}^{n}}(b) = det(b_{k}^{(j)})_{jk}. From this, we can write

(γ^{A_{n}}(a) − γ^{\overline{A}^{n}}(a))(γ^{A_{n}}(b) − γ^{\overline{A}^{n}}(b)) = det(a_{i}^{(j)})_{ij}det(b_{k}^{(j)})_{jk}

= det(∑_{j=1}^{n} a_{i}^{(j)}b_{k}^{(j)})_{ik}.

The symbol a_{i}^{(j)}b_{k}^{(j)} is equal to (a_{i}b_{k})^{(j)}, and the sum over all j gives by definition e_{1}(a_{i}b_{k}). This is in (A^{⊗n})^{S_{n}} and by Lemma 3.1 its image under the Ferrand homomorphism is s_{1}(a_{i}b_{k}) = Tr_{A}(a_{i}b_{k}). Hence in Δ_{A} we have

(γ̇^{A_{n}}(a) − γ̇^{\overline{A}^{n}}(a))(γ̇^{A_{n}}(b) − γ̇^{\overline{A}^{n}}(b)) = det(Tr_{A}(a_{i}b_{k}))_{ik}
= δ_{A}(a_{1} ∧ ⋯ ∧ a_{n}, b_{1} ∧ ⋯ ∧ b_{n})

as we wanted to show. □

For comparing the discriminant form of Δ_{A} with that of A, we will need to understand the trace map Tr_{Δ_{A}}: Δ_{A} → R. A helpful intermediate step is to understand the so-called standard involution on any rank-2 algebra:

Definition 4.4. Let R be a ring and D an R-algebra of rank 2. The function τ_{D}: D → D sending d ↦ Tr_{D}(d) − d is called the standard involution on D, and has the property that

d + τ_{D}(d) = Tr_{D}(d) and d · τ_{D}(d) = Nm_{D}(d)

for all d ∈ D. It is in fact an involution and an R-algebra homomorphism, and is even the unique R-algebra involution τ: D → D such that d · τ(d) ∈ R for all d ∈ D (see, for example, [18, Lemma 2.9]).

Corollary 4.5. Let R be a ring and A an R-algebra of rank n, with n ≥ 2. Let τ be the map Δ_{A} → Δ_{A} induced by the action of a transposition on the tensor factors in (A^{⊗n})^{A_{n}}. Then τ is the standard involution on Δ_{A}.

Proof. Since the action of a transposition on any S_{n}-invariant element is trivial, the map τ: Δ_{A} → Δ_{A} is a well-defined R-algebra involution on Δ_{A}. And for any element x of (A^{⊗n})^{A_{n}}, we have that x · τ(x) is S_{n}-invariant, and so is sent to an element of R in Δ_{A}. Therefore τ is indeed the standard involution on Δ_{A}. □

We can now prove that the exact sequence of Theorem 4.1 identifies the discriminant forms of A and Δ_{A}.

Proof of Theorem 4.1(b). The composite isomorphism ⋀^{n}A ≅ Δ_{A}/R ≅ ⋀^{2}Δ_{A} sends an element a_{1} ∧ ⋯ ∧ a_{n} of ⋀^{n}A to 1 ∧ γ̇^{A_{n}}(a_{1}, …, a_{n}) in ⋀^{2}Δ_{A}. Given a = (a_{1}, …, a_{n}) and b = (b_{1}, …, b_{n}) in A^{n} we have

δ_{Δ_{A}}(1 ∧ γ̇^{A_{n}}(a), 1 ∧ γ̇^{A_{n}}(b)) = det (\begin{matrix}Tr_{Δ_{A}}(1) & Tr_{Δ_{A}}(γ̇^{A_{n}}(b)) \\ Tr_{Δ_{A}}(γ̇^{A_{n}}(a)) & Tr_{Δ_{A}}(γ̇^{A_{n}}(a)γ̇^{A_{n}}(b)) \end{matrix})

which is equal to

2Tr_{Δ_{A}}(γ̇^{A_{n}}(a)γ̇^{A_{n}}(b)) − Tr_{Δ_{A}}(γ̇^{A_{n}}(a))Tr_{Δ_{A}}(γ̇^{A_{n}}(b)).

By Corollary 4.5, the standard involution of Δ_{A} is the one arising from the action of a transposition on the tensor factors of A^{⊗n}, so we may compute these traces as sums:

δ_{Δ_{A}}(1 ∧ γ̇^{A_{n}}(a), 1 ∧ γ̇^{A_{n}}(b)) = 2(γ̇^{A_{n}}(a)γ̇^{A_{n}}(b) + γ̇^{\overline{A}^{n}}(a)γ̇^{\overline{A}^{n}}(b))
− (γ̇^{A_{n}}(a) + γ̇^{\overline{A}^{n}}(a))(γ̇^{A_{n}}(b) + γ̇^{\overline{A}^{n}}(b))
= (γ̇^{A_{n}}(a) − γ̇^{\overline{A}^{n}}(a))(γ̇^{A_{n}}(b) − γ̇^{\overline{A}^{n}}(b))

which is equal to δ_{A}(a_{1} ∧ ⋯ ∧ a_{n}, b_{1} ∧ ⋯ ∧ b_{n}) by Lemma 4.3. So the given isomorphism identifies the discriminant forms of A and Δ_{A}, as we wanted to show. □

5. Examples of discriminant algebras

To show that computing Δ_{A} for A an algebra of rank n is a straightforward process, in this section we will exhibit some examples of discriminant algebras and how to compute them. The simplest case, the discriminant algebra of a quadratic (i.e. rank-2) algebra, is reassuring but not very illuminating.

Proposition 5.1. Let R be a ring and A a quadratic R-algebra. Then the R-algebra homomorphism A → Δ_{A} sending a ↦ γ̇^{A_{2}}(1, a) is an isomorphism.

Proof. That this is an R-algebra homomorphism is evident from its factorization as

A → A^{⊗2} = (A^{⊗2})^{A_{2}} → Δ_{A}

sending a ↦ 1 ⊗ a = γ^{A_{2}}(1, a) ↦ γ̇^{A_{2}}(1, a). That it is an isomorphism then follows by the Five Lemma from the following commuting diagram:

[figure: 0; R; A; ⋀^{2}A; 0; 0; R; Δ_{A}; ⋀^{2}A; 0,]

where the bottom row is the short exact sequence of Theorem 4.1, and the map A → ⋀^{2}A in the top row sends a to 1 ∧ a. The left-hand square commutes because A → Δ_{A} is an R-algebra homomorphism, and the right-hand square commutes because the composite A → Δ_{A} → ⋀^{2}A also sends a ↦ γ̇^{A_{2}}(1, a) ↦ 1 ∧ a. □

The simplest non-trivial example of a discriminant algebra will therefore be that of a rank-3 algebra. The following proposition is quite useful for computing discriminant algebras of algebras that are free as modules.

Proposition 5.2. Let R be a ring, and suppose A is an R-algebra that is free of rank n ≥ 2 as an R-module, with R-basis θ = (θ_{1}, …, θ_{n}). Then Δ_{A} is also free as an R-module, with R-basis {1, γ̇^{A_{n}}(θ)}.

Proof. Note that if A is free with R-basis (θ_{1}, …, θ_{n}), then ⋀^{n}A is free of rank 1, with generator θ_{1} ∧ ⋯ ∧ θ_{n}. Then from the exact sequence of Theorem 4.1, we find that {1, γ̇^{A_{n}}(θ)} is an R-basis for Δ_{A}, because it is the disjoint union of an R-basis for R and a lifting of an R-basis for ⋀^{n}A. □

Remark 5.3. More generally, if Θ ⊆ A^{n} is a set of tuples such that

{θ_{1} ∧ ⋯ ∧ θ_{n} : (θ_{1}, …, θ_{n}) ∈ Θ}

generates ⋀^{n}A as an R-module, then Δ_{A} is generated as an R-module by 1 together with

{γ̇^{A_{n}}(θ_{1}, …, θ_{n}) : (θ_{1}, …, θ_{n}) ∈ Θ}.

Remark 5.4. In the setting of Proposition 5.2, note that since γ̇^{A_{n}}(θ) has trace γ̇^{A_{n}}(θ) + γ̇^{\overline{A}^{n}}(θ) and norm γ̇^{A_{n}}(θ)γ̇^{\overline{A}^{n}}(θ) by Corollary 4.5, we find that

Δ_{A} ≅ R[X]/(X^{2} − (γ̇^{A_{n}}(θ) + γ̇^{\overline{A}^{n}}(θ))X + (γ̇^{A_{n}}(θ)γ̇^{\overline{A}^{n}}(θ))).

Note that this quadratic polynomial in X has discriminant

(γ̇^{A_{n}}(θ) + γ̇^{\overline{A}^{n}}(θ))^{2} − 4(γ̇^{A_{n}}(θ)γ̇^{\overline{A}^{n}}(θ)) = (γ̇^{A_{n}}(θ) − γ̇^{\overline{A}^{n}}(θ))^{2}

= δ_{A}(θ_{1} ∧ ⋯ ∧ θ_{n}, θ_{1} ∧ ⋯ ∧ θ_{n}),

the discriminant of A with respect to the basis θ. As an immediate consequence, we find that the discriminant of A with respect to the R-basis θ is a square in R/(4). This generalizes Stickelberger’s theorem that the discriminant of a number field is always congruent to 0 or 1 modulo 4.

Example 5.5. Let R be a ring, and let E be a locally-free R-module of rank n ≥ 1. Then R ⊕ E can be given an R-algebra structure in which (1, 0) is the multiplicative identity and E is a square-zero ideal. Its discriminant algebra is then

Δ_{R⊕E} ≅ R ⊕ ⋀^{n}E,

again with multiplicative identity (1, 0) and ⋀^{n}E a square-zero ideal. We show this by proving that there is a unique R-module homomorphism Δ_{R⊕E} → R⊕⋀^{n}E sending 1 ↦ (1, 0) and γ̇^{A_{n+1}}(1, e) := γ̇^{A_{n+1}}(1, e_{1}, …, e_{n}) ↦ (0, e_{1} ∧ ⋯ ∧ e_{n}) for all tuples e = (e_{1}, …, e_{n}) ∈ E^{n}. We will then show that this homomorphism is an isomorphism, and that the induced multiplication on R ⊕ ⋀^{n}E is the desired one. Uniqueness is clear, as such elements generate Δ_{R⊕E} by Remark 5.3. For existence, then, it suffices to check locally, when E is free with basis θ = (θ_{1}, …, θ_{n}). Then Δ_{R⊕E} is free with basis (1, γ̇^{A_{n+1}}(1, θ)) and R ⊕ ⋀^{n}E is free with basis ((1, 0), (0, θ_{1} ∧ ⋯ ∧ θ_{n})), so we may define an R-module homomorphism Δ_{R⊕E} → R ⊕ ⋀^{n}E sending 1 ↦ (1, 0) and

γ̇^{A_{n+1}}(1, θ) ↦ (0, θ_{1} ∧ ⋯ ∧ θ_{n}).

We claim that this homomorphism has the desired property, that

γ̇^{A_{n+1}}(1, e) ↦ (0, e_{1} ∧ ⋯ ∧ e_{n})

for all e = (e_{1}, …, e_{n}) ∈ E^{n}. We check this first in case e = θ_{σ}, that is, the e_{i} are a permutation of the θ_{i}—see the discussion after Lemma 3.7. If σ is even, then γ̇^{A_{n+1}}(1, e) = γ̇^{A_{n+1}}(1, θ) and e_{1} ∧ ⋯ ∧ e_{n} = θ_{1} ∧ ⋯ ∧ θ_{n}, and we are done. If σ is odd, then γ̇^{A_{n+1}}(1, e) = γ̇^{\overline{A}^{n+1}}(1, θ), and what we instead must show is that

γ̇^{\overline{A}^{n+1}}(1, θ) ↦ (0, −θ_{1} ∧ ⋯ ∧ θ_{n}).

We show this by proving that the sum of γ̇^{A_{n+1}}(1, θ) and γ̇^{\overline{A}^{n+1}}(1, θ) is zero, and so the image of the latter is the negative of the image of the former. Now we can compute γ̇^{A_{n+1}}(1, θ)+γ̇^{\overline{A}^{n+1}}(1, θ) = γ̇^{S_{n+1}}(1, θ) = Φ_{A}(γ^{S_{n+1}}(1, θ)) as the coefficient of λ_{0}λ_{1} ⋯ λ_{n} in the norm of λ_{0} + ∑_{i=1}^{n} λ_{i}θ_{i}. This element acts via the following matrix:

(\begin{matrix}λ_{0} & 0 & 0 & … & 0 \\ λ_{1} & λ_{0} & 0 & … & 0 \\ λ_{2} & 0 & ... & ... & ⋮ \\ ⋮ & ⋮ & ... & λ_{0} & 0 \\ λ_{n} & 0 & … & 0 & λ_{0} \end{matrix})

This matrix is lower-triangular and has λ_{0}s on the diagonal, so the determinant is λ_{0}^{n+1}, in which the coefficient of λ_{0}λ_{1} ⋯ λ_{n} is zero, as desired. Thus the homomorphism we have defined does indeed send γ̇^{A_{n+1}}(1, e) to (0, e_{1} ∧⋯∧e_{n}) whenever the e_{i} are a permutation of the θ_{i}.

Now we check that this holds for general e. Write e_{i} = ∑_{j} m_{ij}θ_{j} for some m_{ij} ∈ R. Then we have

γ̇^{A_{n+1}}(1, e) = γ̇^{A_{n+1}}(1, ∑_{j=1}^{n} m_{1j}θ_{j}, …, ∑_{j=1}^{n} m_{nj}θ_{j})

= ∑_{f: {1,...,n}→{1,...,n}}(∏_{i=1}^{n} m_{if(i)})γ̇^{A_{n+1}}(1, θ_{f(1)}, …, θ_{f(n)}).

We claim that the terms with f not injective equal zero. Indeed, if f is not injective then γ^{A_{n+1}}(1, θ_{f(1)}, …, θ_{f(n)}) is S_{n+1}-invariant and equals γ^{α}(1, θ_{1}, …, θ_{n}) for some α ∈ {0, …, n}^{n+1}/S_{n+1}. But then its image in Δ_{A} is the coefficient of λ^{α} in the norm of λ_{0} + ∑_{i=1}^{n} λ_{i}θ_{i}, which vanishes unless α = {(0, 0, …, 0)}, but this case cannot arise. Therefore

γ̇^{A_{n+1}}(1, e) = ∑_{σ∈S_{n}}(∏_{i=1}^{n} m_{iσ(i)})γ̇^{A_{n+1}}(1, θ_{σ(1)}, …, θ_{σ(n)})

↦ (0, ∑_{σ∈S_{n}}(∏_{i=1}^{n} m_{iσ(i)})sgn(σ) θ_{1} ∧ ⋯ ∧ θ_{n})

= (0, det(m_{ij})_{ij} θ_{1} ∧ ⋯ ∧ θ_{n})

= (0, e_{1} ∧ ⋯ ∧ e_{n})

as desired. Thus we have an R-module homomorphism Δ_{R⊕E} → R ⊕ ⋀^{n}E sending γ̇^{A_{n+1}}(1, e) to (0, e_{1} ∧ ⋯ ∧ e_{n}), and since it is an isomorphism locally, it is an isomorphism of R-modules. Then all that remains to check is that the induced R-algebra structure on R ⊕ ⋀^{n}E is the indicated one. Now as long as n ≥ 2, every term in a product of the form

γ̇^{A_{n+1}}(1, e_{1}, …, e_{n})γ̇^{A_{n+1}}(1, f_{1}, …, f_{n})

vanishes because every product e_{i}f_{j} = 0. Thus the multiplication on R ⊕ ⋀^{n}E sets (e_{1} ∧⋯∧e_{n})(f_{1} ∧⋯∧f_{n}) = 0 as desired. If we are in the case n = 1, then R ⊕ E is a rank-2 algebra and we have Δ_{R⊕E} ≅ R ⊕ E ≅ R ⊕ ⋀^{1}E anyway.

Example 5.6. Let R be a ring and A be an R-algebra of rank n that can be generated by a single element a. Then if p_{a}(λ) is the characteristic polynomial of a, we have A ≅ R[X]/(p_{a}(X)). In particular, {1, a, …, a^{n−1}} is an R-basis of A. So we find that {1, γ̇^{A_{n+1}}(1, a, …, a^{n−1})} is an R-basis of Δ_{A}. For example, if n = 3 and p_{a}(X) = X^{3} − sX^{2} + tX − u, then we have an Rbasis for Δ_{A} given by {1, γ̇^{A_{3}}(1, a, a^{2})}. We can compute the trace and norm of the generator γ̇^{A_{3}}(1, a, a^{2}) as follows: for the trace, we need only a small calculation

Tr_{Δ_{A}}(γ̇^{A_{3}}(1, a, a^{2})) = Φ_{A}(γ^{S_{3}}(1, a, a^{2}))
= Φ_{A}(e_{1}(a)e_{2}(a) − 3e_{3}(a))
= s_{1}(a)s_{2}(a) − 3s_{3}(a)
= st − 3u.

For the norm, we have

Nm_{Δ_{A}}(γ̇^{A_{3}}(1, a, a^{2})) = γ̇^{A_{3}}(1, a, a^{2})γ̇^{\overline{A}^{3}}(1, a, a^{2})
= γ̇^{A_{3}}(1, a^{3}, a^{3}) + γ̇^{A_{3}}(a^{2}, a^{2}, a^{2}) + γ̇^{A_{3}}(a, a, a^{4}).

(The reader may check this expansion by hand, or appeal to Lemma 3.7.) We have γ̇^{A_{3}}(a^{2}, a^{2}, a^{2}) = 3u^{2}, as this is just three times the norm of a^{2}. Moreover, we have that γ^{A_{3}}(a, a, a^{4}) is equal to e_{3}(a)γ^{A_{3}}(1, 1, a^{3}) so that γ̇^{A_{3}}(a, a, a^{4}) = uγ̇^{A_{3}}(1, 1, a^{3}). Moreover γ̇^{A_{3}}(1, 1, a^{3}) and γ̇^{A_{3}}(1, a^{3}, a^{3}) are equal to s_{1}(a^{3}) and s_{2}(a^{3}) respectively. Hence, we just need to compute s_{1}(a^{3}) and s_{2}(a^{3}). The action of a^{3} with respect to the basis (1, a, a^{2}) is given by the matrix

(\begin{matrix}0 & 0 & u \\ 1 & 0 & −t \\ 0 & 1 & s \end{matrix})^{3} = (\begin{matrix}u & su & s^{2}u − tu \\ −t & u − st & su − s^{2}t + t^{2} \\ s & s^{2} − t & u − 2st + s^{3} \end{matrix}) ,

which has trace s_{1}(a^{3}) = 3u − 3st + s^{3} and quadratic trace s_{2}(a^{3}) = 3u^{2} − 3stu + t^{3}. So we obtain

Nm_{Δ_{A}}(γ̇^{A_{3}}(1, a, a^{2})) = s_{2}(a^{3}) + 3u^{2} + us_{1}(a^{3})
= (3u^{2} − 3stu + t^{3}) + 3u^{2} + u(3u − 3st + s^{3})
= 9u^{2} − 6stu + t^{3} + s^{3}u.

Therefore we have the following algebra presentation:

Δ_{A} ≅ R[X]/(X^{2} − (st − 3u)X + (9u^{2} − 6stu + t^{3} + s^{3}u)).

Example 5.7. Consider the cyclotomic extension of rings of integers ℤ[ζ] ≅ ℤ[x]/(x^{4} + x^{3} + x^{2} + x + 1) over ℤ for ζ a primitive 5th root of unity. The discriminant of this ℤ-algebra is

|\begin{matrix}Tr_{K}(1) & Tr_{K}(ζ) & Tr_{K}(ζ^{2}) & Tr_{K}(ζ^{3}) \\ Tr_{K}(ζ) & Tr_{K}(ζ^{2}) & Tr_{K}(ζ^{3}) & Tr_{K}(ζ^{4}) \\ Tr_{K}(ζ^{2}) & Tr_{K}(ζ^{3}) & Tr_{K}(ζ^{4}) & Tr_{K}(1) \\ Tr_{K}(ζ^{3}) & Tr_{K}(ζ^{4}) & Tr_{K}(1) & Tr_{K}(ζ) \end{matrix}| = |\begin{matrix}4 & −1 & −1 & −1 \\ −1 & −1 & −1 & −1 \\ −1 & −1 & −1 & 4 \\ −1 & −1 & 4 & −1 \end{matrix}| = 5^{3}.

Adjoining a square root of the discriminant to ℤ, we obtain a naive discriminant algebra Δ_{O_{K}/ℤ}^{naive} = ℤ[x]/(x^{2} − 125). But the discriminant of this quadratic ℤalgebra is (0)^{2} − 4(−125) = 2^{2} · 5^{3}, which is not equivalent to that of O_{K}/ℤ, since 2^{2} is not the square of a unit in ℤ. Let us see instead what our construction Δ_{O_{K}/ℤ} gives. The ℤ-basis {1, ζ, ζ^{2}, ζ^{3}} for ℤ[ζ] gives a basis {1, γ̇^{A_{4}}(1, ζ, ζ^{2}, ζ^{3})} for the discriminant algebra. We compute the trace and norm of γ̇^{A_{4}}(1, ζ, ζ^{2}, ζ^{3}) to give a presentation for this algebra:

Tr_{Δ_{ℤ[ζ]}}(γ̇^{A_{4}}(1, ζ, ζ^{2}, ζ^{3})) = γ̇^{A_{4}}(1, ζ, ζ^{2}, ζ^{3}) + γ̇^{\overline{A}^{4}}(1, ζ, ζ^{2}, ζ^{3})
= Φ_{ℤ[ζ]}(γ^{S_{4}}(1, ζ, ζ^{2}, ζ^{3})),

which the algorithm in Lemma 3.6 expresses as

Φ_{ℤ[ζ]}(e_{1}(ζ)e_{1}(ζ^{2})e_{1}(ζ^{3}) − e_{1}(ζ)e_{1}(ζ^{5}) − e_{1}(ζ^{2})e_{1}(ζ^{4}) − 2e_{2}(ζ^{3}) + e_{1}(ζ^{6}))

= s_{1}(ζ)s_{1}(ζ^{2})s_{1}(ζ^{3}) − s_{1}(ζ)s_{1}(ζ^{5}) − s_{1}(ζ^{2})s_{1}(ζ^{4}) − 2s_{2}(ζ^{3}) + s_{1}(ζ^{6})

= (−1)(−1)(−1) − (−1)(4) − (−1)(−1) − 2(1) + (−1)

= −1 + 4 − 1 − 2 − 1 = −1.

The norm we compute with the help of Lemma 3.7:

Nm_{Δ_{ℤ[ζ]}}(γ̇^{A_{4}}(1, ζ, ζ^{2}, ζ^{3})) = γ̇^{A_{4}}(1, ζ, ζ^{2}, ζ^{3})γ̇^{\overline{A}^{4}}(1, ζ, ζ^{2}, ζ^{3})
=  γ̇^{A_{4}}(ζ, ζ, ζ^{4}, ζ)  + γ̇^{A_{4}}(ζ, ζ^{2}, 1, ζ^{4})
+ γ̇^{A_{4}}(ζ, ζ^{3}, ζ^{3}, 1) + γ̇^{A_{4}}(ζ^{2}, 1, 1, 1)
+ γ̇^{A_{4}}(ζ^{2}, ζ^{2}, ζ^{2}, ζ) + γ̇^{A_{4}}(ζ^{2}, ζ^{3}, ζ^{4}, ζ^{3})
+ γ̇^{A_{4}}(ζ^{3}, 1, ζ^{3}, ζ) + γ̇^{A_{4}}(ζ^{3}, ζ, 1, ζ^{3})
+ γ̇^{A_{4}}(ζ^{3}, ζ^{3}, ζ^{2}, ζ^{4}) + γ̇^{A_{4}}(ζ^{4}, 1, ζ^{4}, ζ^{4})
+ γ̇^{A_{4}}(ζ^{4}, ζ, ζ^{2}, 1) + γ̇^{A_{4}}(ζ^{4}, ζ^{2}, ζ^{3}, ζ^{3}).

Each term with a repeated entry is easy to expand in terms of the s_{k}(ζ^{j}); for example:

γ̇^{A_{4}}(ζ, ζ, ζ^{4}, ζ) = s_{4}(ζ)γ̇^{A_{4}}(1, 1, ζ^{3}, 1) = 3s_{4}(ζ)s_{1}(ζ^{3}) = 3(1)(−1) = −3

and

γ̇^{A_{4}}(ζ, ζ^{3}, ζ^{3}, 1) = s_{1}(ζ)s_{2}(ζ^{3}) − s_{1}(ζ^{3})s_{1}(ζ^{4}) + s_{1}(ζ^{7})
= (−1)(1) − (−1)(−1) + (−1) = −3.

In fact, all the terms equal −3 except for γ̇^{A_{4}}(ζ, ζ^{2}, 1, ζ^{4}) and γ̇^{A_{4}}(ζ^{4}, ζ, ζ^{2}, 1), whose sum when multiplied by s_{4}(ζ) = 1 gives Tr(γ̇^{A_{4}}(ζ^{2}, ζ^{3}, ζ, 1)) = −1. Thus the trace and norm of γ̇^{A_{4}}(1, ζ, ζ^{2}, ζ^{3}) are −1 and 10(−3) − 1 = −31, so

Δ_{O_{K}/ℤ} ≅ ℤ[x]/(x^{2} + x − 31),

which does have discriminant (1)^{2} − 4(−31) = 125 equal to that of O_{K}.

While theoretically well-behaved, sometimes the coefficients in a presentation of the discriminant algebra are hard to interpret.

Example 5.8. Let A be the rank-n ℤ-algebra ℤ[x]/(x^{n} − 1) for n ≥ 2. Then as in Example 5.6 the discriminant algebra Δ_{A/ℤ} has basis {1, γ̇^{A_{n}}(1, x, …, x^{n−1})}, and we can present the algebra structure of Δ_{A/ℤ} if we know the trace and norm of γ̇^{A_{n}}(1, x, …, x^{n−1}). The trace is equal to Φ_{A/ℤ}(γ^{S_{n}}(1, x, …, x^{n−1})), which as in Example 3.3 we can compute as the coefficient of λ_{1} … λ_{n} in the norm of λ_{1} + λ_{2}x + ⋯ + λ_{n}x^{n−1}, which equals the determinant

|\begin{matrix}λ_{1} & λ_{n} & λ_{n−1} & … & λ_{2} \\ λ_{2} & λ_{1} & λ_{n} & ... & ⋮ \\ λ_{3} & λ_{2} & ... & ... & λ_{n−1} \\ ⋮ & ... & ... & λ_{1} & λ_{n} \\ λ_{n} & … & λ_{3} & λ_{2} & λ_{1} \end{matrix}| .

This coefficient is equal to the permanent of Schur’s matrix (ζ^{ij})_{i,j=1}^{n} for ζ a primitive nth root of unity; see [7] for this and the following facts. For small values of n, the trace of γ̇^{A_{n}}(1, x, …, x^{n−1}) is shown in the following table:

\begin{matrix}n & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 … \\ Tr & 0 & −3 & 0 & −5 & 0 & 105 & 0 & 81 & 0 & 6765 … \end{matrix}

For even n, the trace is always zero; thus γ̇^{A_{n}}(1, x, …, x^{n−1})^{2} is in ℤ and equals one fourth of the discriminant, so

Δ_{A/ℤ} ≅ ℤ[x]/(x^{2} − \frac{1}{4}(−1)^{n/2}n^{n}),

the “naive” discriminant algebra if one remembers the factor of 1/4. But if n is odd, then the trace of γ̇^{A_{n}}(1, x, …, x^{n−1}) is nonzero.

6. Proofs of Theorems 2 to 4

In this section we demonstrate proofs of the other three main theorems from the introduction. First, we show that Theorem 2 holds, so that the operation (R, A) ↦ Δ_{A/R} commutes with base change.

Theorem 6.1. Let R be a ring, and let A be an R-algebra of rank n with n ≥ 2. Let R′ be an R-algebra, and let A′ = R′ ⊗A, so that A′ is an R′-algebra of rank n. The natural R′-algebra isomorphism R′ ⊗ (A^{⊗n})^{A_{n}} → (A′^{⊗_{R′} n})^{A_{n}} descends to an R′-algebra isomorphism R′ ⊗ Δ_{A/R} → Δ_{A′/R′}.

Proof. We proved that the canonical homomorphism R′ ⊗ (A^{⊗n})^{A_{n}} → (A′^{⊗_{R′} n})^{A_{n}} is an isomorphism in Proposition 3.5. Then since the Ferrand homomorphism also commutes with base change, we obtain an isomorphism R′ ⊗ Δ_{A/R} ≅ Δ_{A′/R′} as desired. □

Second, we can immediately show that satisfying Theorem 3, i.e. the discriminant algebra having the right description when 2 is a unit in the base ring, is a consequence of satisfying Theorem 1, as a corollary of the following proposition:

Proposition 6.2. Let R be a ring in which 2 is invertible. Let D be an Ralgebra of rank 2. Give the rank-2 R-module R⊕⋀^{2}D an R-algebra stucture by making (1, 0) the multiplicative identity and setting (0, ξ) · (0, ψ) = \frac{1}{4}δ_{D}(ξ, ψ).

Then there is a unique isomorphism of R-algebras D ⟶^{∼} R ⊕ ⋀^{2}D inducing the identity map on top exterior powers

⋀^{2}D ⟶^{∼} ⋀^{2}(R ⊕ ⋀^{2}D) ≅ (R ⊕ ⋀^{2}D)/R = ⋀^{2}D;

the isomorphism is d ↦ (\frac{1}{2}Tr_{D}(d), 1 ∧ d).

Proof. First we establish existence. Note that the isomorphism D/R ⟶^{∼} ⋀^{2}D sending the class of d to 1 ∧ d gives us an exact sequence

0 → R → D → ⋀^{2}D → 0,

which is split by the homomorphism D → R: d ↦ \frac{1}{2}Tr_{D}(d). This gives us the desired isomorphism of R-modules D ≅ R⊕⋀^{2}D sending d to (\frac{1}{2}Tr_{D}(d), 1∧d). Next we show that transporting the R-algebra structure from D to R ⊕ ⋀^{2}D gives the desired multiplication on R ⊕ ⋀^{2}D. The image of the multiplicative identity 1 is (\frac{1}{2}Tr_{D}(1), 1 ∧ 1) = (1, 0), as desired. All that remains, then, is to check the multiplication on elements of the form (0, ξ), that is, the images of trace-zero elements of D. Then let d, e ∈ D have trace zero; we will show that de = \frac{1}{4}δ_{D}(1 ∧ d, 1 ∧ e) in D. First, note that the squares of d, e, and d + e are all in R: they are roots of their characteristic polynomials, which have no linear term. Therefore de = \frac{1}{2}((d + e)^{2} − d^{2} − e^{2}) is in R as well. Then we may compute δ_{D}(1 ∧ d, 1 ∧ e) as

δ_{D}(1 ∧ d, 1 ∧ e) = det (\begin{matrix}Tr_{D}(1) & Tr_{D}(d) \\ Tr_{D}(e) & Tr_{D}(de) \end{matrix}) = det (\begin{matrix}2 & 0 \\ 0 & 2de \end{matrix}) = 4de,

so we have de = \frac{1}{4}δ_{D}(1 ∧ d, 1 ∧ e) as we wanted to show. Thus the induced product on R ⊕ ⋀^{2}D has (0, ξ) · (0, ψ) = \frac{1}{4}δ_{D}(ξ, ψ), as claimed, and we have demonstrated existence of the desired algebra isomorphism. Now we show that there is only one isomorphism with these properties. It is sufficient to show that the only algebra automorphism of D inducing the identity on ⋀^{2}D is itself the identity. For this we work locally, so assume D has a basis of the form {1, x}. An automorphism of D sends x to an element ux + r for some r ∈ R and u a unit of R. If this is to descend to the identity on ⋀^{2}D, we must have u = 1. And if this is an automorphism of D, we must have Tr_{D}(x) = Tr_{D}(ux + r) = Tr_{D}(x + r) = Tr_{D}(x) + 2r, so 2r = 0. Since 2 is a unit of R, we have r = 0 and the automorphism is id_{D}. □

Corollary 6.3. Let R be a ring in which 2 is a unit, and let A be a rank-n R-algebra. Put an R-algebra structure on R ⊕ ⋀^{n}A by making (1, 0) the multiplicative identity, and setting (0, ξ) · (0, ψ) = \frac{1}{4}δ_{A}(ξ, ψ). Then there is a unique R-algebra isomorphism Δ_{A} ≅ R ⊕ ⋀^{n}A such that the resulting isomorphism ⋀^{2}Δ_{A} ≅ ⋀^{2}(R ⊕ ⋀^{n}A) ≅ ⋀^{n}A is the one from Theorem 4.1.

Note the extra factor of \frac{1}{4} in the definition of the multiplication, as compared with the statement of Theorem 3 on page 1053. Since 2 is a unit, the two multiplications yield isomorphic algebra structures on R⊕⋀^{n}A, but we see that the one given here is the one which arises naturally from the identification of the discriminant forms. This is analogous to the factor of \frac{1}{4} one could introduce to the naive discriminant algebra K[x]/(x^{2} − \frac{1}{4}D), to make its discriminant D instead of 4D. For the remainder of this section, we will suppose that R is connected, i.e. that it has exactly two idempotents 0 and 1, so that we are working in the setting of Theorem 4. A finite étale algebra over a connected ring R is automatically an R-algebra of rank n for some natural number n, and a R-algebra A of rank n is finite étale if and only if its discriminant form δ_{A}: ⋀^{n}A × ⋀^{n}A → R is nondegenerate, in the sense that the induced R-module homomorphism

⋀^{n}A → Hom(⋀^{n}A, R)

is an isomorphism. It is clear from Theorem 1, then, that A is étale if and only if Δ_{A/R} is étale, since their discriminant forms are isomorphic. The significance of finite étale R-algebras is clarified by the following theorem, due to Grothendieck and proven in [8, Ch. V, §7]:

Theorem 6.4. Let R be a connected ring equipped with a ring homomorphism to a separably closed field K. Then there is a profinite group π_{R}, called the fundamental group of R (at K), such that for each finite étale R-algebra A, the finite set F (A) = Hom_{R -Alg}(A, K) is naturally equipped with a continuous π_{R}-action. Furthermore, the assignment A ↦ F (A) defines a contravariant equivalence of categories

F: R -ét → π_{R} -set

from the category of finite étale R-algebras to the category of π_{R}-sets, i.e. finite sets equipped with a continuous π_{R}-action.

Note: Grothendieck’s original formulation was for the fundamental group of locally Noetherian schemes, but the Noetherian hypothesis is not necessary; an excellent reference is [11, Section 5]. Note also that the fundamental group π_{R} implicitly depends on the choice of K; different choices of K yield fundamental groups that are isomorphic but not canonically so, a behavior which is analogous to the dependence of the topological fundamental group on a choice of base point. In the following, we will suppress the dependence of π_{R} on K whenever possible without causing confusion.

Example 6.5. If R = K is a field and K^{s} is its separable closure, then π_{K} is naturally identified with the absolute Galois group of K, which acts continuously on Hom_{K -Alg}(A, K^{s}) for each finite separable K-algebra A.

Remark 6.6. In general, Theorem 6.4 tells us quite a lot about the finite π_{R}-set corresponding to a given finite étale R-algebra:

(1) For each finite étale R-algebra A, the rank of A is the cardinality of F (A). Indeed, suppose that A is locally free of rank n as an R-module. Then as sets, F (A) = Hom_{R -Alg}(A, K) ≅ Hom_{K -Alg}(K ⊗_{R} A, K), but since K ⊗_{R} A is a finite separable K-algebra of rank n, it is isomorphic to K^{n} and there are exactly n K-algebra homomorphisms K^{n} → K. (In fact, the choice of isomorphism K ⊗_{R} A ≅ K^{n} amounts to a choice of bijection {1, …, n} ⟶^{∼} F (A), so K ⊗_{R} A is canonically isomorphic to K^{F (A)} as a K-algebra.)

(2) If A_{1}, …, A_{n} are finite étale R-algebras, then F (∏_{i=1}^{n} A_{i}) is isomorphic to the disjoint union ∐_{i=1}^{n} F (A_{i}), and F (⨂_{i=1}^{n} A_{i}) to the product ∏_{i=1}^{n} F (A_{i}), each with the induced π_{R}-action. This is just because a contravariant equivalence sends limits to colimits and vice versa, and products and tensor products of étale algebras are étale.

(3) In particular, the zero R-algebra corresponds to the empty π_{R}-set, and R itself to a singleton {∗} with the trivial π_{R}-action. Then for any finite set S, the trivial étale R-algebra R^{S} = ∏_{s∈S} R corresponds to a π_{R}-set isomorphic to the set S equipped with the trivial π_{R}-action.

(4) If G is a finite group acting via R-algebra isomorphisms on a finite étale R-algebra A, then G also acts naturally on the corresponding π_{R}-set F (A). Furthermore, the R-algebra of G-invariants A^{G} is also finite étale, and the corresponding π_{R}-set F (A^{G}) is isomorphic to F (A)/G, the set of G-orbits of F (A). (See [11, Prop. 5.20].)

With these remarks, we will see that the Ferrand homomorphism has an especially nice interpretation in the case of étale algebras:

Proposition 6.7. Let R be a connected ring equipped with a ring homomorphism to a separably closed field K. Let π_{R} be the fundamental group. Let A be a finite étale R-algebra corresponding to an n-element π_{R}-set X under the equivalence of Theorem 6.4. Then the Ferrand homomorphism Φ_{A}: (A^{⊗n})^{S_{n}} → R corresponds to the map of π_{R}-sets {∗} → X^{n}/S_{n} sending ∗ to the S_{n}-orbit of bijections {1, …, n} ⟶^{∼} X.

Proof. By changing base (to K if necessary) we may assume that A is the trivial étale algebra R^{X}. Then Ferrand shows in [6, Ex. 3.1.3(b)] that the Ferrand homomorphism ((R^{X})^{⊗n})^{S_{n}} ≅ R^{X^{n}/S_{n}} → R is projection onto the coordinate indexed by the S_{n}-orbit of bijections {1, …, n} ⟶^{∼} X. □

Finally, we prove the main theorem of the section.

Theorem 6.8. Let R be a connected ring with fundamental group π_{R}, and let A be a finite étale R-algebra corresponding to an n-element π_{R}-set X, with n ≥ 2. Then the discriminant algebra Δ_{A} is a finite étale R-algebra corresponding to the 2-element π_{R}-set

Or(X) := Bij({1, …, n}, X)/A_{n}

of orientations on X.

Proof. Note that Δ_{A} is defined so that the following square is a tensor product diagram:

\begin{matrix}Δ_{A} & R \\ (A^{⊗n})^{A_{n}} & (A^{⊗n})^{S_{n}} \end{matrix}

Since A is étale, this square is in fact a cofibered product in the category of finite étale R-algebras. Hence under the contravariant equivalence F: R -ét → π_{R} -set, this diagram becomes a fiber product diagram of π_{R}-sets:

\begin{matrix}F (Δ_{A}) & \{∗\} \\ X^{n}/A_{n} & X^{n}/S_{n} \end{matrix}

This pullback can be computed in the category of sets, and then equipped canonically with a π_{R}-action. By Proposition 6.7 the image of ∗ in X^{n}/S_{n} is the class of bijections Bij({1, …, n}, X). So the elements of F (Δ_{A}) are A_{n}- equivalence classes of maps {1, …, n} → X, whose underlying S_{n}-equivalence class is Bij({1, …, n}, X). In other words, this is the set of A_{n}-equivalence classes of bijections {1, …, n} ⟶^{∼} X, namely the two-element set Or(X). □

7. Functoriality

Given an isomorphism of rank-n R-algebras f: A → B, we obtain isomorphisms (f^{⊗n})^{A_{n}}: (A^{⊗n})^{A_{n}} → (B^{⊗n})^{A_{n}} and (f^{⊗n})^{S_{n}}: (A^{⊗n})^{S_{n}} → (B^{⊗n})^{S_{n}}, and thus an isomorphism of tensor products Δ_{f}: Δ_{A/R} → Δ_{B/R}. The key here is that the following triangle of R-algebra homomorphisms commutes if f is an isomorphism:

(3)  \begin{matrix}(A^{⊗n})^{S_{n}} & (B^{⊗n})^{S_{n}} \\ R \end{matrix} Φ_{A/R} (f^{⊗n})^{Sn} Φ_{B/R}

For a general homomorphism f: A → B, if the triangle (3) commutes, then we obtain a homomorphism of discriminant algebras Δ_{f}: Δ_{A} → Δ_{B}.

Proposition 7.1. Let R be a ring, and A and B two R-algebras of rank n. Let f: A → B be an R-algebra homomorphism, and let Ω ⊆ A be a set of elements of A whose powers generate A as an R-module. Then the following are equivalent:

(i) The triangle (3) of R-algebra homomorphisms commutes.

(ii) For all a ∈ A, we have Nm_{A/R}(a) = Nm_{B/R}(f(a)), and the same holds

after base change to any R-algebra R′. (iii) For all a ∈ Ω, the characteristic polynomial of a equals the character-

istic polynomial of f(a).

Proof. (i ⇒ ii) After base changing to R′, we obtain a commuting triangle

\begin{matrix}(A′^{⊗_{R′} n})^{S_{n}} & (B′^{⊗_{R′} n})^{S_{n}} \\ R′ \end{matrix} Φ_{A′/R′} (f′^{⊗n})^{Sn} Φ_{B′/R′}

One path sends a⊗⋯⊗a to Nm_{A′/R′}(a), and other sends it to Nm_{B′/R′}(f′(a)), so these must be equal. (ii ⇒ iii) Given any a ∈ A, we have from (ii) applied to R′ = R[λ] the equation Nm_{A[λ]/R[λ]}(λ − a) = Nm_{B[λ]/R[λ]}(λ − f(a)), but these are exactly the characteristic polynomials of a and f(a). (iii ⇒ i) The composite Φ_{B/R} ∘ (f^{⊗n})^{S_{n}} sends e_{k}(a) ↦ e_{k}(f(a)) ↦ s_{k}(f(a)) = s_{k}(a) for each a ∈ Ω, and therefore equals Φ_{A/R} by Lemma 3.6. □

We call an R-algebra homomorphism universally norm-preserving if it satisfies the equivalent conditions of Proposition 7.1. Isomorphisms are always universally norm-preserving, but the following example shows that universally norm-preserving homomorphisms are in general neither injective nor surjective.

Example 7.2. Let A be any rank-n R-algebra and a ∈ A any element, with characteristic polynomial p_{a}(λ) = Nm_{A[λ]/R[λ]}(λ − a). Then the R-algebra homomorphism R[x]/(p_{a}(x)) → A sending x to a is universally norm-preserving, since Ω = {x} ⊆ R[x]/(p_{a}(x)) is a set whose powers {1, x, x^{2}, … } generate R[x]/(p_{a}(x)) as an R-module, and the characteristic polynomial of x in R[x]/(p_{a}(x)) is again p_{a}(λ). For example, let r and s be two elements of any ring R. Then we have a universally norm-preserving homomorphism R[x]/(x − r)(x − s) → R^{2} sending x ↦ (r, s). This map is injective if and only if r − s is not a zerodivisor in R, and surjective if and only if r − s is a unit. Note that there are also well-defined homomorphisms R[x]/(x−r)(x−s) → R^{2} sending x to (r, r) or (s, s), but these are not universally norm-preserving unless r = s.

Remark 7.3. The S_{n}-closure operation defined by Bhargava and Satriano in [1] has a universal property related to universally norm-preserving homomorphisms. Namely, a full set of sections for a rank-n algebra A is a family of n algebra homomorphisms A → R such that the resulting homomorphism of rank-n algebras A → R^{n} is universally norm-preserving, and the S_{n}-closure of A over R is the universal R-algebra over which A is equipped with a full set of sections.

8. The discriminant algebra of a product

If A and B are R-algebras of ranks m and n, respectively, then their product A × B is an R-algebra of rank m + n. It is well-known that ⋀^{m+n}(A × B) is canonically isomorphic to ⋀^{m}A⊗⋀^{n}B, and that if we identify these two locallyfree R-modules we find that the discriminant quadratic form δ_{A×B} is just δ_{A} ⊗ δ_{B}. In this sense, the construction of the discriminant form is multiplicative. We extend this multiplicativity to the construction of the discriminant algebra Δ_{A×B}. Namely, for each ring R there is a commutative monoid structure ∗_{R} on Quad_{R}, the set of isomorphism classes of quadratic R-algebras; these monoid structures have recently been characterized by Voight in [19] as the unique family of functions ∗_{R}: Quad_{R} × Quad_{R} → Quad_{R} such that

• For each R, the set Quad_{R} is a commutative monoid with multiplication ∗_{R} and unit the class of R^{2},

• For each R-algebra R′, the base-change operation Quad_{R} → Quad_{R′} is a homomorphism of commutative monoids, and

• If S and T are quadratic étale R-algebras with standard involutions σ and τ, then S ∗_{R} T is the class of the subring of S ⊗ T fixed by σ ⊗ τ.

In [5], Deligne asserted the existence of such a binary operation over which the discriminant algebra should distribute; an explicit construction of this binary operation on quadratic algebras is due to Loos and can be found in [12]. Our goal in this section is to show that Δ is multiplicative in the sense that Δ_{A×B} ≅ Δ_{A} ∗ Δ_{B}. Our approach will be to show first that the operation on quadratic algebras (S, T) ↦ Δ_{S×T} satisfies the three properties of the operations ∗_{R} listed above, and then to exhibit an isomorphism Δ_{A×B} ≅ Δ_{Δ_{A}×Δ_{B}}. Then we will have

Δ_{A×B} ≅ Δ_{Δ_{A}×Δ_{B}} ≅ Δ_{A} ∗ Δ_{B}.

Remark 8.1. The easiest properties to check are base-change and commutativity, and the description in the étale case is also straightforward to verify. Indeed, if S and T are quadratic R-algebras, then S × T ≅ T × S so by functoriality under isomorphisms we find that

Δ_{S×T} ≅ Δ_{T ×S}.

Furthermore, if R′ is any R-algebra, then

R′ ⊗_{R} Δ_{(S×T)/R} ≅ Δ_{R′⊗(S×T)/R′} ≅ Δ_{(S′×T′)/R′},

so the operation sending (S, T) to the isomorphism class of Δ_{S×T} commutes with base change. For the étale case, it is sufficient to check in case the base ring R is connected. (Indeed, in the proof of uniqueness in [19] the étale description is only used in a single universal case for which the base ring is even a domain.) Then if we suppose S and T to correspond to π_{R}-sets X and Y, to check that (S ⊗ T)^{S_{2}} is isomorphic to Δ_{S×T} we need only find an isomorphism of π_{R}-sets

(X × Y)/S_{2} ≅ Bij({1, …, 4}, X ⊔ Y )/A_{4}.

To wit, we have isomorphisms X ≅ Bij({1, 2}, X) and Y ≅ Bij({3, 4}, Y), and an inclusion

Bij({1, 2}, X) × Bij({3, 4}, Y) ↪ Bij({1, …, 4}, X ⊔ Y ).

It is easy to check that after projecting to Bij({1, …, 4}, X ⊔Y )/A_{4}, we obtain a well-defined bijection from (Bij({1, 2}, X) × Bij({3, 4}, Y))/S_{2}.

Checking that the operation (S, T) ↦ Δ_{S×T} is associative and has unit R^{2}, and that there is an isomorphism Δ_{A×B} ≅ Δ_{Δ_{A}×Δ_{B}} for general A and B, will occupy the rest of this section. To check these statements, we will have to dig into explicit computations with elements. We use the following abuse of notation in the hope that it will prevent the reader from being buried by parentheses. In product algebras of the form A × B, we will often need to refer to elements of the form (a, 0) or (0, b). We will usually denote such elements simply by a or b, thus implicitly identifying the rings A and B with their corresponding ideals A × 0 and 0 × B in A × B. Since each of A and B naturally contains an image of R, for each r ∈ R we will write r_{A} for (r, 0) and r_{B} for (0, r); thus the two idempotents (1, 0) and (0, 1) in A × B are 1_{A} and 1_{B}, respectively, and their sum is the unit 1.

Remark 8.2. For example, in this notation the isomorphism

⋀^{m}A ⊗ ⋀^{n}B ≅ ⋀^{(m+n)}(A × B)

is the one sending (a_{1} ∧ ⋯ ∧ a_{m}) ⊗ (b_{1} ∧ ⋯ ∧ b_{n}) ↦ a_{1} ∧ ⋯ ∧ a_{m} ∧ b_{1} ∧ ⋯ ∧ b_{n} for all a_{1}, …, a_{m} ∈ A and b_{1}, …, b_{n} ∈ B (see [2, Prop. 10 on p. III.84]).

Remark 8.3. As a consequence, Δ_{A×B} is generated as an R-module by 1 together with elements of the form γ̇^{A_{m+n}}(a_{1}, …, a_{m}, b_{1}, …, b_{n}) with a_{1}, …, a_{m} ∈ A and b_{1}, …, b_{n} ∈ B by Remark 5.3. We will usually write such an element as γ̇^{A_{m+n}}(a, b), where a = (a_{1}, …, a_{m}) ∈ A^{m} and b = (b_{1}, …, b_{n}) ∈ B^{n}.

We will also often use the following description of the Ferrand homomorphism for A × B:

Lemma 8.4. Let R be a ring, and let A and B be R-algebras of ranks m and n, respectively. Identify S_{m} and S_{n} with subgroups of S_{m+n} permuting the first m and last n elements of {1, …, m + n}, respectively.

(1) For each subgroup G ⊆ S_{m+n}, the R-algebra projection

π: (A × B)^{⊗m+n} → A^{⊗m} ⊗ B^{⊗n}

sending (a_{1}, b_{1}) ⊗ ⋯ ⊗ (a_{m+n}, b_{m+n}) to (a_{1} ⊗ ⋯ ⊗ a_{m}) ⊗ (b_{m+1} ⊗ ⋯ ⊗ b_{m+n}) restricts to an R-algebra homomorphism

((A × B)^{⊗(m+n)})^{G} → (A^{⊗m})^{G∩S_{m}} ⊗ (B^{⊗n})^{G∩S_{n}}.

(2) In case G = S_{m+n}, the resulting composite

\begin{matrix}((A × B)^{⊗(m+n)})^{S_{m+n}} → (A^{⊗m})^{S_{m}} ⊗ (B^{⊗n})^{S_{n}} \\ R ⊗ R ≅ R \end{matrix}_{Φ_{A}⊗Φ_{B}}

is the Ferrand homomorphism Φ_{A×B}.

Proof. (1) Set H = G ∩ S_{m} and K = G ∩ S_{n}. Then H × K is naturally a subgroup of G, and we have an inclusion

((A × B)^{⊗(m+n)})^{G} ↪ ((A × B)^{⊗(m+n)})^{H×K}.

Thus it is enough to show that π: (A × B)^{⊗(m+n)} → A^{⊗m} ⊗ B^{⊗n} restricts to a homomorphism ((A × B)^{⊗(m+n)})^{H×K} → (A^{⊗m})^{H} ⊗ (B^{⊗n})^{K}. Now S_{m} × S_{n} acts on both (A × B)^{⊗m+n} and A^{⊗m} ⊗ B^{⊗n} by permuting the first m and last n tensor factors separately, and the projection π is equivariant with respect to this action. Thus we obtain a map of H × K-invariants

((A × B)^{⊗(m+n)})^{H×K} → (A^{⊗m} ⊗ B^{⊗n})^{H×K} .

Finally, observe that (A^{⊗m} ⊗ B^{⊗n})^{H×K} = (A^{⊗m})^{H} ⊗ (B^{⊗n})^{K} as subalgebras of A^{⊗m} ⊗ B^{⊗n}. (By Proposition 3.5, this can be checked after changing base to a localization in which both A and B are free R-modules, in which case the isomorphism is elementary.) Thus we have obtained the desired homomorphism as the composite

((A × B)^{⊗(m+n)})^{G} ↪ ((A × B)^{⊗(m+n)})^{H×K} → (A^{⊗m})^{H} ⊗ (B^{⊗n})^{K}.

(2) We check the defining property of Φ_{A×B}. Let (a, b) be an arbitrary element of A×B; is (a, b)⊗⋯⊗(a, b) sent to Nm_{A×B}(a, b) = Nm_{A}(a)Nm_{B}(b)? Indeed so: we have

(a, b) ⊗ ⋯ ⊗ (a, b) ↦ (a ⊗ ⋯ ⊗ a) ⊗ (b ⊗ ⋯ ⊗ b)
↦ (Nm_{A}(a)) ⊗ (Nm_{B}(b))
↦ Nm_{A}(a)Nm_{B}(b)

as desired. The same property holds also after base change, so the indicated homomorphism is Φ_{A×B}. □

Our first application is to check that R^{2} is a unit for the operation (S, T) ↦ Δ_{S×T}. More generally, if A is any rank-n R-algebra, then we have the following R-algebra isomorphism Δ_{R×A} ≅ Δ_{A}:

Theorem 8.5. Let R be a ring, and let A be an R-algebra of rank n ≥ 2. The R-algebra homomorphism (R × A)^{⊗(n+1)} → A^{⊗n} sending

(r_{0}, a_{0}) ⊗ (r_{1}, a_{1}) ⊗ ⋯ ⊗ (r_{n}, a_{n}) ↦ r_{0} · (a_{1} ⊗ ⋯ ⊗ a_{n})

restricts to an R-algebra homomorphism ((R × A)^{⊗(n+1)})^{A_{n+1}} → (A^{⊗n})^{A_{n}} that descends to an R-algebra isomorphism Δ_{R×A} → Δ_{A}.

Proof. The homomorphism (R × A)^{⊗n+1} → A^{⊗n} ≅ R^{⊗1} ⊗ A^{⊗n} is of the form considered in Lemma 8.4(1), with R and A in place of A and B. Using the subgroup A_{n+1} ⊆ S_{n+1}, whose intersection with S_{n} is A_{n}, Lemma 8.4(1) tells us that the homomorphism restricts to one

((R × A)^{⊗(n+1)})^{A_{n+1}} → R ⊗ (A^{⊗n})^{A_{n}} ≅ (A^{⊗n})^{A_{n}}, as desired. In addition, we know from Lemma 8.4(2) that the further restriction ((R × A)^{⊗(n+1)})^{S_{n+1}} → (A^{⊗n})^{S_{n}} commutes with the Ferrand homomorphisms to R; thus we obtain a morphism of tensor products

((R × A)^{⊗(n+1)})^{A_{n+1}} ⨂ R_{((R×A)^{⊗(n+1)})^{S_{n+1}}} → (A^{⊗n})^{A_{n}} ⨂ R_{(A^{⊗n})^{Sn}},

i.e. an R-algebra homomorphism Δ_{R×A} → Δ_{A}. To show that this homomorphism is an isomorphism, we show that it fits into a map of short exact sequences

[figure: 0; R; Δ_{R×A}; ⋀^{n+1}(R × A); 0; 0; R; Δ_{A}; ⋀^{n}A; ∼; 0]

where the right-hand isomorphism is the one ⋀^{n}A ≅ ⋀^{1}R ⊗ ⋀^{n}A ≅ ⋀^{n+1}(R × A) from Remark 8.2. The left-hand square commutes because Δ_{R×A} → Δ_{A} is an R-algebra homomorphism. To show that the right-hand square commutes, follow an element γ̇^{A_{n+1}}(1_{R}, a_{1}, …, a_{n}) of Δ_{R×A}; such elements along with 1 generate Δ_{R×A} by Remark 8.3. Its image in Δ_{A} is γ̇^{A_{n}}(a_{1}, …, a_{n}), giving a_{1} ∧ ⋯ ∧ a_{n} in ⋀^{n}A, thence 1_{R} ∧ a_{1} ∧ ⋯ ∧ a_{n} in ⋀^{n+1}(R × A). But this corresponds exactly to the action of the homomorphism Δ_{R×A} → ⋀^{n+1}(R×A). Thus the right-hand square commutes as well, so by the Five Lemma, the homomorphism Δ_{R×A} → Δ_{A} is an isomorphism of R-algebras. □

Remark 8.6. Certain authors (such as Deligne in [5] and Loos in [13]) construct a discriminant algebra for only even-rank or only odd-rank algebras, defining the discriminant algebra of an algebra A whose rank is of the wrong parity to be that of R×A. The fact that our construction is invariant under adding a factor of R will be useful in future work comparing these different constructions.

Remark 8.7. Note that we thus have an isomorphism Δ_{A} ≅ Δ_{R^{2}×A} for all R-algebras of rank at least 2, but the right-hand side is also well-defined for R-algebras of rank 0 or 1, and is then isomorphic to R^{2}. In this way we can extend the domain of the discriminant algebra operation to all constant rank (and even locally constant rank) algebras.

All that remains, then, to show that Δ_{S×T} ≅ S ∗ T is to check that the operation (S, T) ↦ Δ_{S×T} is associative. This will follow from our exhibiting an isomorphism Δ_{A×B} ≅ Δ_{Δ_{A}×Δ_{B}}, for then

Δ_{S×Δ_{T ×U}} ≅ Δ_{Δ_{S}×Δ_{T ×U}} ≅ Δ_{S×(T ×U)} ≅ ≅ Δ_{(S×T)×U} ≅ Δ_{Δ_{S×T} ×Δ_{U}} ≅ Δ_{Δ_{S×T} ×U},

where we have used the isomorphisms S ≅ Δ_{S} and U ≅ Δ_{U} from Proposition 5.1. So without any more ado, here is the promised isomorphism:

Theorem 8.8. Let R be a ring, and let A and B be R-algebras of ranks m and n, respectively, with both m and n at least 2. Then there is a unique Ralgebra isomorphism Δ_{A×B} ⟶^{∼} Δ_{Δ_{A}×Δ_{B}} that sends 1 ↦ 1 and γ̇^{A_{m+n}}(a, b) to γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(a), 1_{Δ_{B}}, γ̇^{A_{n}}(b)) for each a ∈ A^{m} and b ∈ B^{n}.

The proof that this assignment describes a well-defined algebra isomorphism is long and unenlightening. The authors hope that future work will reveal a simpler construction of this isomorphism that avoids the elementary slog to follow. Indeed, for the proof we use the following lemma several times:

Lemma 8.9. Let a = (a_{1}, …, a_{m}) ∈ A^{m} and b = (b_{1}, …, b_{n}) ∈ B^{n}. Then

Φ_{A×B}(γ^{S_{m+n}}(a, b)) = Φ_{A}(γ^{S_{m}}(a)) · Φ_{B}(γ^{S_{n}}(b)).

If two of the a_{i} are equal, then γ^{A_{m+n}}(a, b) is S_{m+n}-invariant, and

Φ_{A×B}(γ^{A_{m+n}}(a, b)) = Φ_{A}(γ^{A_{m}}(a)) · Φ_{B}(γ^{S_{n}}(b)).

Proof. By Lemma 8.4(2), the Ferrand homomorphism for A × B may be computed by first projecting to (A^{⊗m})^{S_{m}} ⊗(B^{⊗n})^{S_{n}} and then applying Φ_{A}⊗Φ_{B}. In our case, the terms of γ^{S_{m+n}}(a, b) that survive after the projection are exactly those with the a_{i} among the first m tensor factors and the b_{i} among the last n. Thus the image of γ^{S_{m+n}}(a, b) in (A^{⊗m})^{S_{m}} ⊗ (B^{⊗n})^{S_{n}} is γ^{S_{m}}(a) ⊗ γ^{S_{n}}(b). Therefore

Φ_{A×B}(γ^{S_{m+n}}(a, b)) = Φ_{A}(γ^{S_{m}}(a)) · Φ_{B}(γ^{S_{n}}(b))

as desired. If two of the a_{i} are equal, then the image of γ^{A_{m+n}}(a, b) after projecting to A^{⊗m} ⊗ B^{⊗n} is the sum γ^{A_{m}}(a) ⊗ γ^{A_{n}}(b) + γ^{\overline{A}^{m}}(a) ⊗ γ^{\overline{A}^{n}}(b), which equals γ^{A_{m}}(a) ⊗ γ^{S_{n}}(b) since γ^{A_{m}}(a) = γ^{\overline{A}^{m}}(a). Thus we obtain

Φ_{A×B}(γ^{A_{m+n}}(a, b)) = Φ_{A}(γ^{A_{m}}(a)) · Φ_{B}(γ^{S_{n}}(b))

as claimed. □

Example 8.10. As a special case, consider quadratic R-algebras S and T with s ∈ S and t ∈ T. We have

(4)

Φ_{S×T} (γ^{A_{4}}(s, s, 1_{T} , t)) = Φ_{S}(γ^{A_{2}}(s, s)) · Φ_{T} (γ^{S_{2}}(1, t))
= Φ_{S}(s ⊗ s) · Φ_{T} (1 ⊗ t + t ⊗ 1)
= Nm_{S}(s) · Tr_{T} (t).

Similarly, we have

(5)

Φ_{S×T} (γ^{A_{4}}(s, s, t, t)) = Φ_{S}(γ^{A_{2}}(s, s)) · Φ_{T} (γ^{S_{2}}(t, t))
= Φ_{S}(s ⊗ s) · Φ_{T} (t ⊗ t + t ⊗ t)
= Nm_{S}(s) · 2Nm_{T} (t)
= 2Nm_{S}(s)Nm_{T} (t).

These two identities will come up again in the proof of Theorem 8.8.

Proof of Theorem 8.8. For uniqueness, note that Δ_{A×B} is generated as an Rmodule by 1 and elements of the form γ̇^{A_{m+n}}(a_{1}, …, a_{m}, b_{1}, …, b_{n}) by Remark 8.3. We demonstrate the existence of such an isomorphism on each localization of R under which A and B become free R-modules. By uniqueness, then, these isomorphisms on the localizations will glue to an R-algebra isomorphism between the two original discriminant algebras. Thus it suffices to assume A and B are free, say with R-bases θ = (θ_{1}, …, θ_{m}) and ϕ = (ϕ_{1}, …, ϕ_{n}), respectively. In that case, Δ_{A×B} is freely generated as an R-module by 1 and γ̇^{A_{m+n}}(θ_{1}, …, θ_{m}, ϕ_{1}, …, ϕ_{n}); we abbreviate the latter as γ̇^{A_{m+n}}(θ, ϕ). Similarly, Δ_{Δ_{A}×Δ_{B}} is freely generated as an R-module by 1 and γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ)). Then we can naively define an R-module isomorphism

f: Δ_{A×B} → Δ_{Δ_{A}×Δ_{B}}

sending 1 ↦ 1 and γ̇^{A_{m+n}}(θ, ϕ) ↦ γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ)). We claim that in fact

f: γ̇^{A_{m+n}}(a, b) ↦ γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(a), 1_{Δ_{B}}, γ̇^{A_{n}}(b))

for all a = (a_{1}, …, a_{m}) ∈ A^{m} and b = (b_{1}, …, b_{n}) ∈ B^{n}, so that f acts elementwise as desired. Then we will show that f is multiplicative, making it an R-algebra isomorphism. Since the expressions γ̇^{A_{m+n}}(a, b) and γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(a), 1_{Δ_{B}}, γ̇^{A_{n}}(b)) are each multilinear in the a_{i} and b_{j}, we can reduce to proving the claim in case each of the a_{i} and b_{j} are among the basis elements for A and B. The first possibility is that two of the a_{i} or two of the b_{j} are equal. Without loss of generality, assume that the a_{i} are not all distinct. Then γ^{A_{m+n}}(a, b) is S_{m+n}-invariant and γ̇^{A_{m+n}}(a, b) is equal to Φ_{A×B}(γ^{A_{m+n}}(a, b)) = Φ_{A}(γ^{A_{m}}(a))Φ_{B}(γ^{S_{n}}(b)) by Lemma 8.9. On the other hand, since γ^{A_{m}}(a) ∈ (A^{⊗m})^{A_{m}} is also S_{m}-invariant, we can also express γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(a), 1_{Δ_{B}}, γ̇^{A_{n}}(b)) in a form amenable to the Example 8.10 identities:

γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(a), 1_{Δ_{B}}, γ̇^{A_{n}}(b)) = γ̇^{A_{4}}(1_{Δ_{A}}, Φ_{A}(γ^{A_{m}}(a))_{Δ_{A}}, 1_{Δ_{B}}, γ̇^{A_{n}}(b))
= Φ_{A}(γ^{A_{m}}(a))γ̇^{A_{4}}(1_{Δ_{A}}, 1_{Δ_{A}}, 1_{Δ_{B}}, γ̇^{A_{n}}(b))
= Φ_{A}(γ^{A_{m}}(a))Nm_{Δ_{A}}(1)Tr_{Δ_{B}}(γ̇^{A_{n}}(b))
= Φ_{A}(γ^{A_{m}}(a))Φ_{B}(γ^{S_{n}}(b)).

Therefore in this case f sends γ̇^{A_{m+n}}(a, b) to γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(a), 1_{Δ_{B}}, γ̇^{A_{n}}(b)) because these two elements are the same R-multiple of 1. The second possibility is that the a_{i} are a permutation of the θ_{i}, and the b_{j} are a permutation of the ϕ_{j}. Write a = θ_{σ} and b = ϕ_{τ} for appropriate permutations σ ∈ S_{m} and τ ∈ S_{n}, recalling the notation from Lemma 3.7 for the action of the symmetric group on the set of tuples. We must show that

f: γ̇^{A_{m+n}}(θ_{σ}, ϕ_{τ}) ↦ γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ_{σ}), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ_{τ})).

There are four cases, according to the signs of σ and τ, which determine the values of γ̇^{A_{m+n}}(θ_{σ}, ϕ_{τ}), γ̇^{A_{m}}(θ_{σ}), and γ̇^{A_{n}}(ϕ_{τ}). So we must show that the following four assignments hold:

f: γ̇^{A_{m+n}}(θ, ϕ) ↦ γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ)), from σ and τ even,
f: γ̇^{\overline{A}^{m+n}}(θ, ϕ) ↦ γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{\overline{A}^{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ)), from σ odd and τ even,
f: γ̇^{A_{m+n}}(θ, ϕ) ↦ γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{\overline{A}^{m}}(θ), 1_{Δ_{B}}, γ̇^{\overline{A}^{n}}(ϕ)), from σ and τ odd, and
f: γ̇^{\overline{A}^{m+n}}(θ, ϕ) ↦ γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ), 1_{Δ_{B}}, γ̇^{\overline{A}^{n}}(ϕ)), from σ even and τ odd.

The first of the four assignments holds by the definition of f. As for the others, note that

γ̇^{A_{m+n}}(θ, ϕ) + γ̇^{\overline{A}^{m+n}}(θ, ϕ) = γ̇^{S_{m+n}}(θ, ϕ)
= Φ_{A×B} (γ^{S_{m+n}}(θ, ϕ))
= Φ_{A} (γ^{S_{m}}(θ)) Φ_{B} (γ^{S_{n}}(ϕ)) ,

so it is enough to show that the sum of each successive pair of outputs is Φ_{A} (γ^{S_{m}}(θ)) Φ_{B} (γ^{S_{n}}(ϕ)). And this indeed holds: for example,

γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ)) + γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{\overline{A}^{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ))
= γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ) + γ̇^{\overline{A}^{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ))
= γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{S_{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ))
= γ̇^{A_{4}} (1_{Δ_{A}}, Φ_{A} (γ^{S_{m}}(θ))_{Δ_{A}} , 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ))
= Φ_{A} (γ^{S_{m}}(θ)) γ̇^{A_{4}} (1_{Δ_{A}}, 1_{Δ_{A}}, 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ))
= Φ_{A} (γ^{S_{m}}(θ)) Nm_{Δ_{A}}(1) Tr_{Δ_{B}} (γ̇^{A_{n}}(ϕ))
= Φ_{A} (γ^{S_{m}}(θ)) Φ_{B} (γ^{S_{n}}(ϕ)) .

The other two sums can be evaluated similarly. So indeed, we have shown that f must send γ̇^{A_{m+n}}(θ_{σ}, ϕ_{τ}) to γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ_{σ}), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ_{τ})), and established the claim that

f: γ̇^{A_{m+n}}(a, b) ↦ γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(a), 1_{Δ_{B}}, γ̇^{A_{n}}(b))

for all a ∈ A^{m} and b ∈ B^{n}. Now we show that the R-module isomorphism f: Δ_{A×B} → Δ_{Δ_{A}×Δ_{B}} is in fact an R-algebra isomorphism. Since f is R-linear and {1, γ̇^{A_{m+n}}(θ, ϕ)} forms an R-basis for Δ_{A×B}, all that we need to check is whether

f(1 · 1) = f(1) · f(1),
f(1 · γ̇^{A_{m+n}}(θ, ϕ)) = f(1) · f(γ̇^{A_{m+n}}(θ, ϕ)), and
f(γ̇^{A_{m+n}}(θ, ϕ) · γ̇^{A_{m+n}}(θ, ϕ)) = f(γ̇^{A_{m+n}}(θ, ϕ)) · f(γ̇^{A_{m+n}}(θ, ϕ)).

The first two hold because f(1) = 1 by definition, so all that is left to check is that f (γ̇^{A_{m+n}}(θ, ϕ))^{2} = f (γ̇^{A_{m+n}}(θ, ϕ)^{2}) . Now we can use Lemma 3.7 to expand this product:

γ̇^{A_{m+n}}(θ, ϕ)^{2} = ∑_{σ∈A_{m+n}} γ̇^{A_{m+n}}((θ, ϕ)(θ, ϕ)_{σ})

= ∑_{\begin{matrix}(σ,τ)∈ \\ (A_{m}×A_{n})∪(\overline{A}_{m}×\overline{A}_{n}) \end{matrix}} γ̇^{A_{m+n}}(θθ_{σ}, ϕϕ_{τ}),

since (θ, ϕ)(θ, ϕ)_{σ} has a zero entry unless σ belongs to A_{m+n} ∩ (S_{m} × S_{n}) = (A_{m} × A_{n}) ∪ (\overline{A}_{m} × \overline{A}_{n}). Therefore

f (γ̇^{A_{m+n}}(θ, ϕ)^{2}) = ∑_{\begin{matrix}(σ,τ)∈ \\ (A_{m}×A_{n})∪(\overline{A}_{m}×\overline{A}_{n}) \end{matrix}} γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θθ_{σ}), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕϕ_{τ}))

= γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ)^{2}, 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ)^{2})
+ γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ)γ̇^{\overline{A}^{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ)γ̇^{\overline{A}^{n}}(ϕ)).

On the other hand,

f (γ̇^{A_{m+n}}(θ, ϕ))^{2} = γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ))^{2}
= γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ)^{2}, 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ)^{2})
+ γ̇^{A_{4}}(γ̇^{A_{m}}(θ), γ̇^{A_{m}}(θ), γ̇^{A_{n}}(ϕ), γ̇^{A_{n}}(ϕ)),

by a similar application of Lemma 3.7. One of each of the two terms on the right-hand sides matches immediately, but the others are also equal:

γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(θ)γ̇^{\overline{A}^{m}}(θ), 1_{Δ_{B}}, γ̇^{A_{n}}(ϕ)γ̇^{\overline{A}^{n}}(ϕ))
= γ̇^{A_{4}} (1_{Δ_{A}}, Nm_{Δ_{A}}(γ̇^{A_{m}}(θ))_{Δ_{A}}, 1_{Δ_{B}}, Nm_{Δ_{B}}(γ̇^{A_{n}}(ϕ))_{Δ_{B}})
= Nm_{Δ_{A}}(γ̇^{A_{m}}(θ)) Nm_{Δ_{B}}(γ̇^{A_{n}}(ϕ)) γ̇^{A_{4}} (1_{Δ_{A}}, 1_{Δ_{A}}, 1_{Δ_{B}}, 1_{Δ_{B}})
= 2 Nm_{Δ_{A}}(γ̇^{A_{m}}(θ)) Nm_{Δ_{B}}(γ̇^{A_{n}}(ϕ))
= γ̇^{A_{4}}(γ̇^{A_{m}}(θ), γ̇^{A_{m}}(θ), γ̇^{A_{n}}(ϕ), γ̇^{A_{n}}(ϕ)).

Thus f is an R-algebra isomorphism Δ_{A×B} → Δ_{Δ_{A}×Δ_{B}}, as desired. □

Remark 8.11. The isomorphism of Theorem 8.8 does not entirely commute with the isomorphisms interchanging A and B. While it is always the case that

γ̇^{A_{4}}(1_{Δ_{A}}, γ̇^{A_{m}}(a_{1}, …, a_{m}), 1_{Δ_{B}}, γ̇^{A_{n}}(b_{1}, …, b_{n}))

equals

γ̇^{A_{4}}(1_{Δ_{B}}, γ̇^{A_{n}}(b_{1}, …, b_{n}), 1_{Δ_{A}}, γ̇^{A_{m}}(a_{1}, …, a_{m})),

the elements γ̇^{A_{m+n}}(a_{1}, …, a_{m}, b_{1}, …, b_{n}) and γ̇^{A_{m+n}}(b_{1}, …, b_{n}, a_{1}, …, a_{m}) are generally only equal if m or n is even; otherwise they are conjugates. Thus the chain of isomorphisms

Δ_{A×B} ≅ Δ_{Δ_{A}×Δ_{B}} ≅ Δ_{Δ_{B}×Δ_{A}} ≅ Δ_{B×A} ≅ Δ_{A×B} 1088 Owen Biesel, Alberto Gioia

is the identity if the rank of A or B is even, but is the standard involution on Δ_{A×B} if the ranks of A and B are both odd. This is precisely the behavior described by Deligne at the end of [5].

References

[1] Bhargava, M., and Satriano, M. On a notion of “Galois closure” for extensions of rings. Journal of the European Mathematical Society 16, 9 (2014), 1881–1913.

[2] Bourbaki, N. Éléments de mathématique. Algèbre. Chapitres 1 à 3. Hermann, Paris, 1970.

[3] Bourbaki, N. Éléments de mathématique. Algèbre commutative. Chapitres 1 à 4, vol. 1. Springer, 1998.

[4] Deligne, P. Cohomologie a supports propres. Springer, 1973.

[5] Deligne, P. Letter to M. Rost and M. Bhargava. Available at

http://pub.math.leidenuniv.nl/~bieselod/references/deligneletter.pdf, 2005.

[6] Ferrand, D. Un foncteur norme. Bull. Soc. Math. France 126, 1 (1998), 1–49.

[7] Graham, R., and Lehmer, D. On the permanent of Schur’s matrix. Journal of the Australian Mathematical Society (Series A) 21, 04 (1976), 487–497.

[8] Grothendieck, A. Revêtements étales et groupe fondamental (SGA 1). Documents Mathématiques (Paris), 3. Société Mathématique de France, Paris, 2003.

[9] Katz, N. M., and Mazur, B. Arithmetic moduli of elliptic curves, vol. 108 of Annals of Mathematics Studies. Princeton University Press, Princeton, NJ, 1985.

[10] Knus, M.-A., Merkurjev, A., Rost, M., and Tignol, J.-P. The Book of Involutions, vol. 44. Amer. Math. Soc., 1998.

[11] Lenstra, H. W. Galois theory for schemes. Available at

http://websites.math.leidenuniv.nl/algebra/GSchemes.pdf, 2008.

[12] Loos, O. Tensor products and discriminants of unital quadratic forms over commutative rings. Monatshefte für Mathematik 122, 1 (1996), 45–98.

[13] Loos, O. Discriminant algebras of finite rank algebras and quadratic trace modules. Math. Z. 257, 3 (2007), 467–523.

[14] Roby, N. Lois polynomes et lois formelles en théorie des modules. Ann. Sci. École Norm. Sup. (3) 80 (1963), 213–348.

[15] Roby, N. Lois polynômes multiplicatives universelles. C. R. Acad. Sci. Paris Sér. A-B 290, 19 (1980), A869–A871.

[16] Rost, M. The discriminant algebra of a cubic algebra. Available at

http://www.math.uni-bielefeld.de/~rost/data/cub-disc.pdf, 2002.

[17] Vaccarino, F. Generalized symmetric functions and invariants of matrices. Mathematische Zeitschrift 260, 3 (2008), 509–526.

[18] Voight, J. Rings of low rank with a standard involution. Illinois Journal of Mathematics 55, 3 (2011), 1135–1154.

[19] Voight, J. Discriminants and the monoid of quadratic rings. ArXiv e-prints (Apr. 2015).

[20] Waterhouse, W. C. Discriminants of etale algebras and related structures. J. reine angew. Math 379 (1987), 209–220.

Owen Biesel Mathematisch Instituut Niels Bohrweg 1 Leiden University The Netherlands. bieselod@math.leidenuniv.nl Alberto Gioia alberto.gioia@univie.ac.at