A new discriminant algebra construction

A discriminant algebra operation sends a commutative ring $R$ and an $R$-algebra $A$ of rank $n$ to an $R$-algebra $\Delta_{A/R}$ 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.


Introduction
Consider this fact from elementary Galois theory: Theorem 1.1.Let K ֒→ L be a degree-n separable field extension in characteristic other than 2, and let K s be a separable closure of K. Let N be the Galois closure of L over K, namely the sum of all the n images of L in K s , and let G = Gal(N/K) be the resulting Galois group.Then G is alternating if and only if the discriminant of L over K is a square in K.
Here, G being alternating means that it acts by even permutations on the K-algebra homomorphisms L → N .The discriminant of L over K, with respect to a basis (ℓ 1 , . . ., ℓ n ) for L as a K-vector space, is det(Tr(ℓ i ℓ j )) ij .This is only well-defined up to multiplying by the square of a unit in K, but this doesn't change whether the discriminant is a square in K.
Proof Outline.Let (ρ 1 , . . ., ρ n ) be the K-algebra maps L → N , on which G acts by postcomposition.If we assume that G is alternating, then the combination det(ρ i (ℓ j )) ij ∈ N is G-invariant, hence belongs to K.But this determinant is a square root of the discriminant of L over K with respect to the basis (ℓ 1 , . . ., ℓ n ).
The converse uses a little more Galois theory.Note that it needs that the characteristic of K is not 2, for otherwise, the discriminant's square root det(ρ i (ℓ j )) in L is actually S n -invariant, and hence belongs to K whether or not G is alternating.
Let ∆ L/K be the K-algebra K[x]/(x 2 − D), where D is the discriminant of L/K.We can rephrase Theorem 1.1 by saying that G is alternating if and only if there is a K-algebra homomorphism ∆ L/K → K.The Kalgebra ∆ L/K is called the naive discriminant algebra of L over K, and has the following properties.
• The dimension of ∆ L/K as a K-vector space is 2.
• The discriminant of ∆ L/K with respect to the natural K-basis (1, x) is 2 0 0 2D = 4D, which is equal to the discriminant of L over K, up to the square of the unit 2. (We could make the discriminants equal exactly if we set ∆ L/K = K[x]/(x 2 − 1 4 D) instead; this point will come up again later.) • If K s is any separable closure of K, the proof of Theorem 1.1 gives us a recipe for a square root of D in K s given an ordering of the n homomorphisms L → K s .This gives a natural map Bij {1, . . ., n}, Hom K (L, K s ) → Hom K (∆ L/K , K s ).
The square root of D resulting from an ordering of the n homomorphisms L → K is unchanged if we permute them with an even permutation, so we find that this map descends to a map out of the A n -orbits: Bij {1, . . ., n}, Hom K (L, K s ) /A n → Hom K (∆ L/K , K s ).
Each of these two sets has two elements, and in fact this map is a bijection between them.We can interpret this fact using the contravariant equivalence of categories {finite separable K-algebras} ∼ −→ {finite continuous Gal(K s /K)-sets} sending a finite separable K-algebra A (a finite product of finite separable field extensions of K) to Hom K (A, K s ).If we suppose that L corresponds to the finite set X, we find that ∆ L/K corresponds to the finite set Bij({1, . . ., n}, X)/A n .We call this the set of orientations of X, denoted Or(X).
In this paper, we will construct a discriminant algebra ∆ A/R for every pair (R, A) where R is an arbitrary (commutative) ring and A is an (associative, unital, commutative) R-algebra that is locally free of some rank n ≥ 2 as an R-module.This condition is analogous to the stipulation that L be a dimension-n vector space over K, and is a setting in which it still makes sense to talk about traces, and therefore discriminants, in the following sense: we have a discriminant bilinear form δ A/R on n A defined by Note, however, that we make no characteristic assumptions about R, and no separability assumptions about A.
In a 2005 letter to Manjul Bhargava and Markus Rost, Pierre Deligne suggested a list of four criteria a discriminant algebra operation (sending an R-algebra A of rank n to another R-algebra ∆ A/R of rank 2) should satisfy; see [3] for the original formulation.We prepend the following zeroth criterion to his list, as it justifies calling such a construction a discriminant algebra: Criterion 0. For each ring R and R-algebra A of rank n, the discriminant algebra ∆ A/R is an R-algebra of rank 2. Furthermore, there is a canonical isomorphism 2 ∆ A/R ∼ = n A that identifies the discriminant bilinear form of ∆ A/R with that of A.
Deligne's first criterion is that the discriminant algebra should commute with base change, in the following sense: Criterion 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 The second criterion gives an explicit description of the discriminant algebra in the case that 2 is invertible, and is analogous to the definition of the naive discriminant algebra of a field extension we gave at the beginning of the introduction.Criterion 2. If 2 is a unit in R and A is an R-algebra of rank n, then there is an R-algebra isomorphism ∆ 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 The third criterion concerns the case of R connected and A étale over R. In this situation, it is known that there is a contravariant equivalence between finite étale R-algebras and finite continuous π R -sets for a suitable profinite group π R .Criterion 3 relates the π R -set corresponding to an étale R-algebra and the π R -set corresponding to its discriminant algebra: Criterion 3. Let R be a connected ring and A an étale R-algebra of rank n corresponding to a π R -set X.Then ∆ A/R is also étale, and corresponds to the 2-element set Or(X) of orientations of X.
Deligne's fourth criterion mandated the compatibility of Criteria 2 and 3 when both apply, and we omit the statement here.In his letter, Deligne sketched a construction of a discriminant algebra operation satisfying his four criteria, by reducing to a similar list of properties for a discriminant algebra for quadratic forms and extending Criteria 2 and 3 across their codimension-2 complement in a universal case.This construction only works for odd-rank algebras, so he defined ∆ A/R to be ∆ (R×A)/R when A has even rank.It is not easy to make this construction explicit, in the general case.
Ottmar Loos later gave a different construction in [10] for the discriminant algebra of an even-rank algebra, extending to the odd-rank case again by defining ∆ A/R := ∆ (R×A)/R when A has odd rank.His construction uses a method, called shifting, that changes the multiplication on a rank-2 algebra to correct its discriminant form; Rost had previously used this technique to define a discriminant algebra for rank-3 algebras in [13].These constructions are explicit, but very complicated to compute in higher rank.
For a given R-algebra A, all three of these constructions proceed by considering a suitable quadratic form on the locally free module A/R, building a discriminant algebra for this quadratic form, and then suitably modifying it to work as a discriminant algebra for A itself.In this paper, we give a construction of a discriminant algebra operation (R, A) → ∆ A/R that does not appeal to a separate construction for quadratic forms, that does not split into cases based on whether the rank of A is odd or even, and that does not require first handling special cases (such as 2 being invertible or the algebra being étale).
Our definition of ∆ A/R is also both simple and explicit.Given a ring R and a rank-n R-algebra A, there is a canonical R-algebra homomorphism from the S n -invariant tensors of A ⊗n to R; this homomorphism (A ⊗n ) Sn → R is called the Ferrand homomorphism.Then we define the discriminant algebra of A over R to be the tensor product We begin in Section 2 by defining the Ferrand homomorphism and establishing its basic properties.Then in Section 3 we define our discriminant algebra as above; the main goal of the section is to prove Theorem 3.8, which implies that our construction satisfies Criterion 0. We also prove that our discriminant algebra satisfies Criteria 1 and 2. In fact, we show that satisfying Criterion 2 is a consequence of satisfying Criterion 0. Section 4 exhibits some examples of discriminant algebras, as well as a general module presentation.In Section 5, we analyze the discriminant algebra of product algebras; our main results are Theorem 5.7, that the discriminant algebra of a product can be computed from the discriminant algebras of the factors, and Theorem 5.12, that an added factor of the base ring does not change the discriminant algebra.Finally, Theorem 6.5, which proves that our construction satisfies Criterion 3, is the main result of Section 6.Now, on to the definitions and proofs of the theorems!

The Ferrand homomorphism
The Ferrand homomorphism for an R-algebra A of rank n is a certain Ralgebra homomorphism Φ A/R : (A ⊗n ) Sn → R. We call this homomorphism the Ferrand homomorphism, because we took from D. Ferrand's paper [5] the idea to use it to give R the structure of a (A ⊗n ) Sn -algebra, thereby obtaining interesting constructions by tensoring other algebras with R over this map.The construction of the discriminant algebra in the next section will be an example.
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 ) Sn .The image of any S n -invariant tensor in N will be invariant under the Galois action on N , and thus belongs to K, giving us the Ferrand homomorphism Φ L/K : (L ⊗n ) Sn → K.
In this section, we will construct the Ferrand homomorphism Φ A/R for any base ring R and R-algebra A of rank n, by showing that the nth exterior power of A has a natural (A ⊗n ) Sn -module structure.Then we obtain the Ferrand homomorphism as the map (A ⊗n First, however, we have to make precise what it means for an R-algebra to have rank n: 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 the unit ideal of R is generated by the set of all r ∈ R for which R r ⊗ M is free (of rank n) as an R r -module.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, finitely generated, and the rank of each (necessarily free) R p -module M p is n for each prime ideal p ∈ M ; see [2, Thm. 2 on p. II.141].
To prove that n A inherits from A ⊗n the structure of an (A ⊗n ) Snmodule, we will need to show that the kernel of the canonical surjection A ⊗n ։ n A is not just an R-submodule, but an (A ⊗n ) Sn -submodule.The following definition and lemma provide module generators for the Ginvariants of tensor powers, and will be useful throughout this paper: Definition 2.3.Let R be a ring and M be an R-module.Given a permutation group G ⊆ S d , a family m = (m i ) i∈I of elements in M with index set I, and a G-orbit α ∈ I d /G of d-tuples from I, we define the element γ α (m) of (M ⊗d ) G by γ α (m) = If I = {1, . . ., n} for some natural number n, then the family m of elements of M is an ordinary n-tuple (m 1 , . . ., m n ), and in this case we will also write γ α (m) as γ α (m 1 , . . ., m n ).Note that even for general I and m ∈ M I , each γ α (m) ∈ (M ⊗d ) G is equal to an element of the form γ α ′ (m ′ 1 , . . ., m ′ d ), since α ∈ I d /G uses at most d elements from I. 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 Such a linear combination is Ginvariant 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 left 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.
We are now ready to prove that n A has a natural (A ⊗n ) Sn -module structure: Proposition 2.5.Let R be a ring.Let A be a rank-n R-algebra.The (A ⊗n ) Sn -module structure on A ⊗n descends to a (A ⊗n ) Sn -module structure on its quotient n A.
This fact was shown in case A is free as an R-module in [14, Prop.1.3], but does not hold in general: see [11,Ex. 4.6].
Proof.Let R be a ring and A an R-algebra of rank n.We need to show that the kernel of the canonical surjection A ⊗n → n A is not just an Rsubmodule of A ⊗n , but an (A ⊗n ) Sn -submodule.This kernel is by definition the R-submodule of A ⊗n generated by elements of the form x = x 1 ⊗• • •⊗x n such that x i = x j for some i = j.We would like to show that for each such x and for each s ∈ (A ⊗n ) Sn , the product s • x is still in the kernel of A ⊗n → n A. By Lemma 2.4, it is enough to show that s • x is sent to zero in n A for each s of the form γ α (a 1 , . . ., a n ) for some a 1 , . . ., a n in A and some α in A n /S n .
We have Now consider the transposition τ ∈ S n that interchanges i and j (the two indices for which x i = x j ).Since α is an S n -orbit of n-tuples, the transposition τ acts on the n-tuples in α, sending (k 1 , . . ., k n ) to (k τ (1) , . . ., k τ (n) ).We can partition α into τ -orbits of size 1 or 2, according as each (k 1 , . . ., k n ) ∈ α has k i = k j or k i = k j .We will break up the above sum into a sum over τ -orbits of α, and handle each orbit in turn.Suppose that (k 1 , . . ., k n ) is fixed by τ , so that k i = k j .Then a k i equals a k j , so then the corresponding pure tensor a k 1 x 1 ⊗ • • • ⊗ a kn x n has two equal tensor factors and is sent to zero in n A.
Now suppose that k i = k j .Then the sum over the τ -orbit of (k 1 , . . ., k n ) is the sum of two pure tensors, identical but for two transposed tensor factors.Such a sum is sent to zero in n A. Therefore the entire sum γ α (a 1 , . . ., a n )x is sent to zero in n A, so the kernel of A ⊗n ։ n A is an (A ⊗n ) Sn -submodule of A ⊗n .Therefore the quotient n A itself is an (A ⊗n ) Sn -submodule as well.
Remark 2.6.Note that the (A ⊗n ) Sn -action on n A has the following description: the application of an S n -invariant tensor It therefore appears to be the restriction to (A ⊗n ) Sn of an action of A ⊗n on n A given by However, the action ( * ) is not well-defined, because the right-hand side does not necessarily vanish when two of the b i are equal.This is why care must be taken to ensure well-definedness of the (A ⊗n ) Sn -module structure on n A.
We can now define the Ferrand homomorphism as follows: Definition 2.7.Let R be a ring and A an R-algebra of rank n.By Proposition 2.5, the R-module structure on n A extends to an (A ⊗n ) Snmodule structure.Hence we have an induced R-algebra homomorphism (A ⊗n ) Sn → End R ( n A).But since n A is a locally free R-module of rank 1, we have a canonical isomorphism End R ( n A) ∼ = R.The resulting Ralgebra homomorphism (A ⊗n ) Sn → R is called the Ferrand homomorphism of A over R and denoted Φ A/R .
We will now demonstrate some useful features of the Ferrand homomorphism.One such property is that the Ferrand homomorphism commutes with base change.To show this, we need to first show that the operation of taking invariant tensors commutes with base change for locally free modules: Proposition 2.8.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 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, [7, 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 in the sense that the following diagram commutes: We say that the Ferrand homomorphism commutes with base change.
Proof.The map Φ A/R is the map representing the action of (A ⊗n ) Sn on n A, which is induced by the multiplication in A ⊗n (see Proposition 2.5).Since multiplication and exterior powers commute with base change, the statement follows.
To better understand the Ferrand homomorphism, and to connect it to its original definition by Ferrand, we make the following definition: Definition 2.10.Let R be a ring, and let A be an R-module 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.
Remark 2.11.Note that Nm A/R (a) is the unique element of R such that the action of a ⊗ • • • ⊗ a on n A is the same as multiplication by Nm A/R (a).According to the definition of the Ferrand homomorphism, then, we see that This property, together with the fact that the Ferrand homomorphism commutes with base change, was Ferrand's original characterization of the Ferrand homomorphism in [5, 3.1.2].
More specifically, the norm function defines a homogeneous degree-n multiplicative polynomial law from A to R, and therefore corresponds to an R-algebra homomorphism from Γ n R (A) → R by [12].Here, Γ n R (A) is the degree-n component of A's divided powers algebra, and since R → A is flat, it is canonically isomorphic to (A ⊗n ) Sn by [4, 5.5.2.5 on p. 123].The resulting R-algebra homomorphism (A ⊗n ) Sn → R is Φ A/R .We will not explain here what a polynomial law is, as we will not use this theory.
A slightly more general statement than Remark 2.11 will also be useful.
To that end, we make the following pair of definitions: Definition 2.12.Let R be a ring and A an R-algebra of rank n.
The characteristic polynomial of a (with indeterminate λ) is the norm of λ − a and is always a monic degree-n polynomial in λ with coefficients from R. Similarly, the unsigned characteristic polynomial of a is the norm of λ + a.The coefficient of λ n−k in the unsigned characteristic polynomial of a is denoted s k (a).A special case is that s n (a) is the norm Nm A/R (a) of a; in addition, s 1 (a) is called the trace of a and will sometimes be denoted Tr A/R (a).When A is free, then s 1 coincides with the classical definition of the trace.Definition 2.13.Let R be a ring and A an R-algebra.Consider the poly- , where for a ∈ A the symbol a with a in the ith position.We will denote the coefficient of λ n−k in this polynomial by e k (a).Note that e k (a) is in (A ⊗n ) Sn for all a and k, and it is the kth elementary symmetric function in a (1) , . . ., a (n) .
Lemma 2.14.Let R be a ring and A an R-algebra of rank n, with Ferrand homomorphism Φ A/R : (A ⊗n ) Sn → R. Then for every a ∈ A and k ∈ {1, . . ., n}, we have This property extends the well-known fact that, given an element a of a degree-n separable field extension L of K, the sum of the n conjugates of a in L's Galois closure is equal to Tr L/K (a), and the product of the n conjugates is equal to Nm L/K (a).

The discriminant algebra: its definition and basic properties
Using the Ferrand homomorphism we can define the discriminant algebra right away.
Definition 3.1.Let R be a ring and let A be an R-algebra of rank n with n ≥ 2. Then the discriminant algebra ∆ A/R of A over R is the tensor product of (A ⊗n ) Sn -algebras defined by the inclusion (A ⊗n ) Sn ֒→ (A ⊗n ) An and the Ferrand homomorphism Φ A/R : (A ⊗n ) Sn → R.
(If the base ring R is understood, it may be omitted and the discriminant algebra of A denoted ∆ A .We will adopt similar conventions for the trace and norm maps, the Ferrand homomorphism, and the discriminant bilinear form.) This construction is very simple, but it is not immediately clear whether this algebra has the desired properties.In this section, we show that this definition of the discriminant algebra does satisfy Criteria 0 to 2 from the introduction-Criterion 3 we will leave to Section 6.
The main result of this section is that the discriminant algebra operation satisfies Criterion 0: that for any R-algebra A of rank n ≥ 2, its discriminant algebra ∆ A is always an R-algebra of rank 2; and furthermore that its discriminant form can be identified with the one of A. We prove it by exhibiting in Theorem 3.8 a certain short exact sequence of R-modules It is clear that the satisfying of Criterion 0 is equivalent to the existence of such an exact sequence.Describing the map ∆ A → n A, and building up the ancillary facts that support the proof of Theorem 3.8, will make up the bulk of this section.But first, we show that the operation (R, A) → ∆ A/R satisfies Criterion 1, commuting with base change.Theorem 3.2.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 Proof.This follows since the Ferrand homomorphism and taking invariants commutes with base change, as we proved in Propositions 2.8 and 2.9.
Second, we can immediately show that satisfying Criterion 2, i.e. the discriminant algebra having the right description when 2 is a unit in the base ring, is a consequence of satisfying Criterion 0: Proposition 3.3.Let R be a ring, and suppose 2 is invertible in R. Let A be an R-algebra of rank n.Let D be an R-algebra of rank 2. Suppose we have an exact sequence Then there is an R-algebra isomorphism where the multiplication on R⊕ n A has identity (1, 0) and on n A is given by This isomorphism is given by . (Note the extra factor of 1  4 in the definition of the multiplication, as compared with the statement of Criterion 2. Since 2 is a unit, the two multiplications yield isomorphic algebras, but we will see that the one listed here arises naturally from the identification of the discriminant forms.This is analogous to the factor of 1  4 one could introduce to the naive discriminant algebra K[x]/(x 2 − 1 4 D), to make its discriminant D instead of 4D.) Proof.The map D → R sending d to 1 2 Tr D (d) splits the exact sequence.So we have an R-module isomorphism We want to show that when the R-algebra structure on D is transported along this isomorphism, it gives the indicated algebra structure on R⊕ n A.
The multiplicative identity will be the image of 1, namely ( 1 2 Tr D (1), f (1)) = (1, 0), as described.Then to check that the rest of the multiplication is as desired, it is enough to show that if d and e are two trace-zero elements of D, then their product in D is 1  4 δ D (f (d), f (e)).Notice that d 2 , e 2 and (d + e) 2 all belong to R, because d, e and d + e are roots of their characteristic polynomials, which have no linear term.Then also Since the discriminant forms of A and D are identified by the isomor- And since the traces of d and e are zero, and de is in R, we have that In the rest of the section we will work towards establishing a short exact sequence 0 → R → ∆ A → n A → 0. We will start by looking for a convenient set of generators for the discriminant algebra.First note that ∆ A is a quotient of the A n -invariants of A ⊗n : Then by Lemma 2.4, the images of elements of the form γ α (a) (where a ∈ A I is some tuple and α ∈ I n /A n is some A n -orbit) generate ∆ A as an R-module.We will refer to the image of such an element γ α (a) by γα (a) ∈ ∆ A .It is inconvenient, however, to work with arbitrary indexing sets I; in fact, a smaller collection of the γ α (a) will work.
The set of odd permutations A d is also an A d -orbit in {1, . . ., d} d : the orbit of (2, 1, . . ., d).Thus we may also consider the element γ A d (m 1 , . . ., m d ) of (M ⊗d ) A d defined by We will show that 1 and the set of the γAn generate the discriminant algebra.For this proof, we need some more results.Lemma 3.6.Let R be a ring and M a projective R-module.Let θ = {θ i : i ∈ I} be a generating family for M , with I totally ordered.Then for each natural number d ≥ 2, we have that (M ⊗d ) A d is generated by (M ⊗d ) S d together with the family Proof.We know from Lemma 2.4 that (M ⊗d ) A d is generated by elements of the form γ α (θ), with α ranging over I d /A d .Now note that if α uses any index from I more than once, it is in fact S d -invariant, and hence γ α (θ) is in (M ⊗d ) S d .Otherwise, α uses d distinct indices from I; let i 1 < • • • < i d be those indices in order.Then either γ α (θ) = γ A d (θ i 1 , . . ., θ i d ), in which case it belongs to our purported generating set, or , in which case it is a linear combination of such elements.
We thus find that the classes of the γ An (θ i 1 , . . ., But if θ is an R-basis for M , then no nonzero R-linear combination of these elements can be S d -invariant, since one can only produce R-linear combinations of pure tensors θ i 1 ⊗• • •⊗θ i d that have an even number of reversals (pairs j < k in {1, . . ., d} with i j > i k ).Hence the family is a linearly independent generating set for (M ⊗d ) A d /(M ⊗d ) S d .Corollary 3.7.Let R be a ring.Let A be an R-algebra of rank n, with n ≥ 2. If θ = {θ i : i ∈ I} is a set of generators for A, then ∆ A is generated as an R-module by {1} ∪ { γAn (θ i 1 , . . ., θ in ) : i j ∈ I}.
Proof.By Lemma 3.6 we have that (A ⊗n ) An is generated by Since the map (A ⊗n ) An → ∆ A is surjective, the discriminant algebra is generated by the image of the above set under the natural map.Every element in (A ⊗n ) Sn is sent to an R-multiple of 1 in ∆ A , establishing the claim.
Using this generating set for ∆ A , we can finally describe the desired short exact sequence needed to satisfy Criterion 0. Theorem 3.8.Let R be a ring, and let A be an R-algebra of rank n, with n ≥ 2.
(a) Its discriminant algebra ∆ A fits naturally into a short exact sequence of R-modules where the map R → ∆ A sends 1 → 1, and the map . ., a n ) identifies the discriminant bilinear forms of A and ∆ A .
All we need to prove Theorem 3.8(a) is a strengthening of Lemma 3.6: Lemma 3.9.Let R be a ring and M a locally free R-module.Let d ≥ 2 be a natural number.Then there is an isomorphism of R-modules Both n M and the quotient on the right hand side commute with localization.Hence we can reduce to proving that the map is an isomorphism if M is free.Let {θ i : i ∈ I} be a basis for M , with I totally ordered.Then d M has an R-basis of elements of the form By Lemma 3.6 these form an R-basis for (M ⊗d ) A d /(M ⊗d ) S d , so the lemma is proved.
Proof of Theorem 3.8(a).Lemma 3.9 with A and n in place of M and d gives us a short exact sequence where the first map is the natural inclusion, and the second sends each element of the form γ An (a 1 , . . ., a n ) to a 1 ∧ • • • ∧ a n .Tensoring with R over (A ⊗n ) Sn , we get an exact sequence Since the Ferrand homomorphism is an R-algebra map it is surjective.So R is isomorphic to the quotient of (A ⊗n ) Sn by the ideal generated by x−Φ A (x).
But this ideal acts trivially on n A, So we are left to show that the natural map R → ∆ A is injective.Suppose r ∈ R is sent to zero in ∆ A .Then multiplication by r on ∆ A is the zero map.Then multiplication by r is the zero map also on n A. Since n A is locally free, this implies r = 0. Thus we have the desired short exact sequence of R-modules Since R and n A are locally free R-modules of rank 1, the exact sequence implies that ∆ A is an R-algebra of rank 2.
The remainder of the section is devoted to building up enough machinery to prove Theorem 3.8(b).Our first result relates the multiplication of some elements of ∆ A to the discriminant form of A.
Proof.First note that in A ⊗n we find that k jk .From this, we can write , and the sum over all j gives by definition e 1 (a i b k ).This is in (A ⊗n ) Sn and by Lemma 2.14 its image under the Ferrand homomorphism is s as we wanted to show.
For comparing the discriminant form of ∆ A with that of A, we will need to understand the trace map Tr   So it remains only to show that d → Tr D (d) − d is an R-algebra homomorphism.For this, note that we may check locally, so that we may assume D is of the form R[x]/(x 2 −sx+t) for some s, t ∈ R (which are then the trace and norm of x ∈ D, respectively).Then the trace of an element ax + b ∈ D is as+2b, so we get Tr D (ax+b)−(ax+b) = (as+2b)−(ax+b) = a(s−x)+b.But this is just the R-algebra endomorphism of R[x]/(x 2 − sx + t) sending x to s − x, which is well-defined because (s − x) 2 − s(s − x) + t = 0.
We will use the following lemma to characterize the standard involution on ∆ A , and so obtain a description of its trace and norm maps.Proof.It is enough to show that τ and the standard involution agree locally.Since τ retains its properties under localization, the problem reduces to the case that D again is of the form D = R[x]/(x 2 − sx + t) for some s, t ∈ R.
By assumption, we know that x + τ (x) and x • τ (x) are both in R, so we get an equation 0 = (x − x)(x − τ (x)) = x 2 − (x + τ (x))x + (x • τ (x)) = 0, which we can write as Since x and 1 are an R-basis for D, we have x + τ (x) = s = Tr D (x) (and also x • τ (x) = t = Nm D (x)).We also have assumed that τ (1) = 1, so by R-bilinearity of τ we have τ (ax + b) = a(s − x) + b, which is indeed the standard R-algebra involution x → s − x on D. Corollary 3.13.Let R be a ring and A an R-algebra of rank n, with n ≥ 2. Let τ be the map ∆ A → ∆ A induced by the action of a transposition on the tensor factors in (A ⊗n ) An .Then τ is the standard involution on ∆ A .In particular, given an n-tuple θ = (θ 1 , . . ., θ n ) of elements of A, the trace and norm of γ An (θ) ∈ ∆ A are given by Proof.Since the action of a transposition on any S n -invariant element is trivial, the map τ : ∆ A → ∆ A is well-defined.It also is an R-algebra homomorphism, so is R-linear and sends 1 → 1.And for any element a of (A ⊗n ) An , we have that a + τ (a) and a • τ (a) are both S n -invariant, and so are sent to elements of R in ∆ A .Then Lemma 3.12 tells us that τ is indeed the standard involution τ .
We can now prove that the exact sequence of Theorem 3.8 identifies the discriminant forms of A and ∆ A .

Proof of Theorem 3.8(b). The composite isomorphism
By Corollary 3.13, the standard involution of ∆ 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: Lemma 3.10.So the given isomorphism identifies the discriminants on A and on ∆ A , as we wanted to show.

Presentations and examples
To show that computing ∆ 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 rank-2 algebra, is reassuring but not very illuminating.
Example 4.3.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 (X) 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, γAn (1, a, . . ., a n−1 )} is an R-basis of ∆ A .
For example, if n = 3 and p a (X) = X 3 − sX 2 + tX − u, then we have an R-basis for ∆ A given by {1, γA 3 (1, a, a 2 )}.We can compute the trace and norm of the generator γA 3 (1, a, a 2 ) as follows: for the trace, we need only a small calculation For the norm, we have (The reader may check this expansion by hand, or appeal to Corollary 4.9.)We have γA 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 γA 3 (a, a, a 4 ) = u γA 3 (1, 1, a 3 ).Moreover γA 3 (1, 1, a 3 ) and γA 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 Therefore we have the following algebra presentation: where we have used the facts that x 5 = 1 and that x, x 2 , x 3 , and x 4 are related by automorphisms of O K .If we use the naive construction in Criterion 2 to define a discriminant algebra, we obtain Z[x]/(x 2 − 125).However, this has discriminant (0) 2 − 4(−125) = 2 2 • 5 3 , which does not reflect the fact that K ramifies at 5 and no other primes.
On the other hand, a computation similar to that of Example 4.3 shows that ∆ O K /Z has the following presentation: In the remainder of this section, we will generalize Proposition 4.2 by giving a more general module presentation of the discriminant algebra of an arbitrary rank-n algebra.We will first look for relations among the γ An in ∆ A , and to that end we introduce some notation.Definition 4.5.Let n be a natural number.For each pair of distinct elements i and j of {1, . . ., n}, denote by α ij ⊆ {1, . . ., n} n the set of tuples that contain two entries equal to i, none equal to j, and one entry equal to every other integer in {1, . . ., n}.Under the action of S n on {1, . . ., n} n permuting the tuples, this subset constitutes an orbit α ij ∈ {1, . . ., n} n /S n .Lemma 4.6.Let R be a ring and let M be an R-module.Let n ≥ 2 be a natural number.Let (m 1 , . . ., m n ) be in M n .Suppose m i = m j for some i = j.
Proof.Note that if m i = m j for some i = j, the transposition (ij) acts trivially on γ An (m 1 , . . ., m n ), which is therefore S n -invariant.Moreover γ An (m 1 , . . ., m n ) is equal to γ α ij (m 1 , . . ., m n ) because m i = m j and the S n -orbit of a tuple with an index repeated twice is equal to its A n -orbit.
We prove that these relations, together with multilinearity, generate all the relations among the γAn in ∆ A .Here is the precise statement.
Theorem 4.7.Let R be a ring.Let A be an R-algebra of rank n ≥ 2. We denote by D A the quotient of the free R-module with basis by the submodule generated by and the relations making the map g : A n → D A defined by (a 1 , . . ., a n ) → g(a 1 , . . ., a n ) multilinear.Then the morphism f : Proof.First note that f is well defined.In fact g is multilinear, and by Lemma 4.6 we have that if (a 1 , . . ., a n ) in A n is such that a i = a j for some i = j, then γ An (a 1 , . . ., a n ) is equal to γ α ij (a 1 , . . ., a n ) ∈ (A ⊗n ) Sn .Therefore we have γAn (a 1 , . . ., a n ) = Φ A (γ α ij (a 1 , . . ., a n )) in ∆ A as desired.
Note that D A /R is isomorphic to n A. In fact, the symbols g(a 1 , . . ., a n ) in D A are multilinear and if a i = a j for some i = j, then by the relations in D A we have that g(a 1 , . . ., a n ) is in R • 1.So it is zero in the quotient D A /R.By definition these are the only relations we have in D A /R.So the map sending g(a 1 , . . ., a n ) in We have then the following diagram of R-modules, where the rows are exact, and the squares commute.
By the Five Lemma f is then an isomorphism of R-modules.
To conclude our description of ∆ A , we write the multiplication of two elements of the form γ An (a 1 , . . ., a n ) in terms of more elements of this form. .
Expanding the product we get since for fixed σ ′ the composite σ ′ τ runs over G as τ does.Now a

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.We would like to relate the discriminant algebra of A×B to the discriminant algebras of A and B. We will use the fact that m+n (A× B) ∼ = m A⊗ n B to show that the Ferrand homomorphism Φ A×B factors through Φ A ⊗ Φ B ; see Lemma 5.3 for the precise statement.
Then we will show that the discriminant algebra of A × B depends only on ∆ A and ∆ B : We can rephrase this by defining a binary operation * on quadratic (i.e.rank-2) algebras by A * B := ∆ A×B , so that for A and B of arbitrary ranks we have ∆ A×B ∼ = ∆ A * ∆ B .We will show that this operation is associative and commutative, and in Section 6 we will give another description of this operation when the quadratic algebras are étale.In [3], Deligne posited the existence of such a binary operation over which the discriminant algebra should distribute; another explicit construction of this binary operation on quadratic algebras is due to Loos and can be found in [9].Finally, we note that the discriminant algebra is unchanged when extra factors of R are introduced: Certain authors (such as Deligne in [3] and Loos in [10]) 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.
In this section, we will use the following abuse of notation in the hope that it will prevent the reader from being bogged down by a deluge of 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.
Lemma 5.1.Let R be a ring, and let A and B be degree-m and n extensions of R. Then there is an R-module isomorphism Proof.By [1, Prop. 10 on p. III.84], there is an R-module isomorphism where the (k, ℓ)-th component of the isomorphism sends But unless k ≤ m and ℓ ≤ n, one of k A and ℓ B vanishes, so the only nonzero term in the direct sum is m A ⊗ n B.
In order to show that Φ A×B : ((A × B) ⊗(m+n) ) S m+n → R factors via the tensor product of Φ A : (A ⊗m ) Sm → R and Φ B : (B ⊗n ) Sn → R, we must first exhibit a canonical R-algebra homomorphism ((A × B) ⊗(m+n) ) S m+n → (A ⊗m ) Sm ⊗ (B ⊗n ) Sn .Indeed, the desired homomorphism is just a restriction of a projection from (A × B) ⊗(m+n) to A ⊗m ⊗ B ⊗n , as we prove in greater generality below: Lemma 5.2.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.Then for each subgroup G ⊆ S m+n , the R-algebra projection Proof.Set H = G ∩ S m and K = G ∩ S n .Then H × K is naturally a subgroup of G, and we have an inclusion 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 Finally, observe that (A ⊗m ⊗ B ⊗n ) H×K ∼ = (A ⊗m ) H ⊗ (B ⊗n ) K as subalgebras of A ⊗m ⊗ B ⊗n .(By Lemma 2.8, 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 Lemma 5.3.Let R be a ring, and let A and B be R-algebras of ranks m and n, respectively, so that A × B is an R-algebra of rank m + n.Then the Ferrand homomorphism Φ A×B factors as shown in the following commutative diagram: where the map ((A × B) ⊗(m+n) ) S m+n → (A ⊗m ) Sm ⊗ (B ⊗n ) Sn is that given by Lemma 5.2 with G = S m+n .
Proof.Recalling the definitions of the Ferrand homomorphisms Φ A×B , Φ A , and Φ B , we find that we must verify the commutativity of the following diagram: This can be seen to hold by inspection of the action of an element s of ((A × B) ⊗(m+n) ) S m+n on an element ξ of m+n (A × B): By Lemma 5.1, we may assume the element of Or if two of the a i are equal, then even γ A m+n (a, b) is S m+n -invariant.Its image after projecting to A ⊗m ⊗ B ⊗n is the sum γ Am As a special case, consider quadratic R-algebras A and B with a ∈ A and b ∈ B. We have Similarly, we have These two identities will come up again in the proof of Theorem 5.7.
Remark 5.5.Consider the idempotent element η of (A × B) ⊗m+n obtained by summing over all pure tensors with m tensor factors equal to 1 A and n tensor factors equal to 1 B .This idempotent η is S m+n -invariant, and its image under Φ A×B is given by Φ In particular, the image of any S m+n -invariant tensor s under the Ferrand homomorphism is equal to the image of η • s: So unless s has a term with at least m factors having nonzero A-components and at least n factors with nonzero B-components, its product with η will vanish, and hence so will its image under the Ferrand homomorphism.Proof.For uniqueness, note that ∆ A×B is generated as an R-module by 1 and elements of the form γA m+n (a 1 , . . ., a m , b 1 , . . ., b n ) by Corollary 5.6.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.
Since the expressions γA m+n (a, b) and γA 4 1 ∆ A , γAm (a), 1 ∆ B , γAn (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 or two of the b j are equal.Without loss of generality, assume that the a i are not all distinct.Then γA m+n (a, b) is S m+n -invariant and was shown in Example 5.4 to be equal to Φ A γ Am (a) Φ B γ Sn (b) .On the other hand, since γAm (a) ∈ ∆ A is also S m -invariant, we can also express γA 4 1 ∆ A , γAm (a), 1 ∆ B , γAn (b) in a form amenable to the Example 5.4 identities: Therefore in this case f sends γA m+n (a, b) to γA 4 1 ∆ A , γAm (a), 1 ∆ B , γAn (b) because these two elements are the same R-multiple of 1.
The first of the four assignments holds by the definition of f .As for the others, note that γA m+n (θ, ψ) + γAm+n (θ, ψ) = γS m+n (θ, ψ) so it is enough to show that each successive pair of outputs also sums to Φ A γ Sm (θ) Φ B γ Sn (ψ) .And this indeed holds: for example, The other two sums can be evaluated similarly.So indeed, we have shown that f must send γA m+n (θ σ , ψ τ ) to γA 4 1 ∆ A , γAm (θ σ ), 1 ∆ B , γAn (ψ τ ) , and established the claim that for all a ∈ A m and b ∈ B n .Now we show that the R-module isomorphism f : ∆ A×B → ∆ ∆ A ×∆ B is in fact an R-algebra isomorphism.Since f is R-linear and {1, γA m+n (θ, ψ)} Proof.Commutativity of * up to isomorphism just follows from commutativity of × up to isomorphism: For associativity, we have (Note that we are using the fact that C, being quadratic, is isomorphic to its own discriminant algebra.)Similarly, we have A * (B * C) ∼ = ∆ A×(B×C) , so associativity of * again follows from associativity of ×.
Theorem 5.7 itself can also be restated in terms of the operation * : is the identity if the rank of A or B is even, but is the standard involution on ∆ A×B if the ranks of A and B are both odd.This is precisely the behavior described by Deligne at the end of [3].
The last case of the discriminant algebra of a product is the case of one factor being R. First, we need a lemma on exterior powers: element γ = γA n+1 (1 R , a 1 , . . ., a n ) of ∆ R×A ; such elements along with 1 generate ∆ R×A by Corollary 5.6.Then the image of γ in n+1 (R × A) is 1 R ∧ a 1 ∧ • • • ∧ a n , corresponding to a 1 ∧ • • • ∧ a n in n A under the isomorphism of Lemma 5.11.On the other hand, γ is sent to γAn (a 1 , . . ., a n ) in ∆ A , and thence again to a 1 ∧ • • • ∧ a n in n A. Thus the right-hand square commutes as well, so by the Five Lemma, the homomorphism ∆ R×A → ∆ A is an isomorphism of R-algebras.6 The discriminant algebra of a finite étale algebra In this section, we will suppose that R is connected, i.e. that it has exactly two idempotents 0 and 1.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 Rmodule homomorphism n A → Hom( n A, R) is an isomorphism.It is clear, then, that any discriminant algebra operation (R, A) → ∆ A/R satisfying Criterion 0 will have the property that A is étale if and only if ∆ 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 [6, Ch.V, §7]: Theorem 6.1.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 Ralgebra 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 [8,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 choice of K except where the choice is relevant.Example 6.2.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 6.3.In general, Theorem 6.1 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 ( n i=1 A i ) is isomorphic to the disjoint union n i=1 F (A i ), and F ( n i=1 A i ) to the product n i=1 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 [8,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 6.4.Let R be a connected ring equipped with a ring homomorphism to a separably closed field K. Let π R be the fundamental group.

Lemma 2 . 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 .

Definition 3 . 4 .Definition 3 . 5 .
Let R be a ring and M an R-module.Let d be a natural number, and let G be a subgroup of S d .For each d-tuple (m 1 , . . ., m d ) of elements of M , set γ G (m 1 , . . ., m d ) := σ∈Gm σ(1) ⊗ • • • ⊗ m σ(d) ∈ (M ⊗d ) G .This function γ G : M d → (M ⊗d ) G is multilinear, since each m i appears as exactly one tensor factor of each term in the sum.(Note that this notation is consistent with our previous γ notation: an d-tuple is an element of M I with I = {1, . . ., d}, andG ⊆ S d ⊆ Map({1, . . ., d}, {1, . . ., d}) = I d is itself a G-orbit.The resulting γ G (m 1 , . .., m d) is then given by the above formula.)As a special case, if d ≥ 2, then for each tuple (m 1 , . . ., m d ) in M d , we have an element γ A d (m 1 , . . ., m d ) of (M ⊗d ) A d defined by

γ
An (a) − γ An (aof the n × n-matrix whose ijth element is a (j) i , and similarly γ An (b) − γ A n (b) = det b (j)

1 .
∆ A : ∆ A → R. A helpful intermediate step is to understand the so-called standard involution on any rank-2 algebra: Proposition 3.11.Let R be a ring and D an R-algebra of rank 2. There is a unique R-algebra homomorphism τ : D → D with the following three properties: For all d ∈ D, we have d + τ (d) = Tr D (d).

3 .
The composite τ • τ is the identity on D. This homomorphism τ is called the standard involution on D. Proof.Note that property 1 determines that τ (d) must equal Tr D (d) − d, so a map with the above properties is necessarily unique.Moreover d → Tr D (d) − d satisfies properties 2 and 3.In fact we obtain immediately d • (Tr D (d) − d) = Nm D (d), since d is a root of its characteristic polynomial X 2 − Tr D (d)X + Nm D (d).And we also find that Tr D (Tr D (d) − d) = 2Tr D (d) − Tr D (d) = Tr D (d), so d and Tr D (d) − d have the same trace.Therefore this map is an involution since Tr D (d) − (Tr D (d) − d) = d.

Lemma 3 . 12 .
Let R be a ring and D an R-algebra of rank 2. Let τ be an R-module homomorphism D → D. Suppose that τ (1) is 1, and that for all d ∈ D both d + τ (d) and d • τ (d) are elements of R. Then τ is the standard involution on D.

Lemma 4 . 8 .
Let R be a ring and A an R-algebra.Let d be a natural number and G a subgroup of S d .Then for each pair of tuples (a 1 , . . ., a d ) and (b 1 , . . ., b d ) in A d , the equation γ G (a)γ G (b) = σ∈G γ G (a 1 b σ(1) , . . ., a d b σ(d) ).holds in (A ⊗d ) G .If we denote the tuple (b σ(1) , . . ., b σ(d) ) by b σ ∈ A d , and give A d the product algebra structure, then the product ab σ is (a 1 b σ(1) , . . ., a d b σ(d) ), and we may write the above identity as

some a 1 Example 5 . 4 .
, . . ., a n ∈ A and some b 1 . . ., b n ∈ B. Then to compute s • ξ, we would first consider the image of s under the projection π : (A×B) ⊗m+n → A ⊗m ⊗B ⊗n , since that is how s acts on ξ's representativea 1 ⊗ • • •⊗ a m ⊗ b 1 ⊗ • • • ⊗ b n .Then by Lemma 5.2, π(s) belongs to (A ⊗m ) Sm ⊗ (B ⊗n) Sn , and by Lemma 5.1 we may consider it to be acting on m A⊗ n B. And this action is exactly via Φ A ⊗ Φ B , as claimed.For example, let a 1 , . . ., a m ∈ A and b 1 , . . ., b n ∈ B, and consider the image of the S n -invariant element γ S m+n (a 1 , . . ., a m , b 1 , . . ., b n ) of (A × B) ⊗(m+n) under the Ferrand homomorphism; we will abbreviate this element as γ S m+n (a, b) with a = (a 1 , . . ., a m ) and b = (b 1 , . . ., b n ).Its image after projecting to A ⊗m ⊗ B ⊗n is γ Sm (a) ⊗ γ Sn (b), and therefore its image in R under the Ferrand homomorphism Φ A×B is (a) ⊗ γ An (b) + γ A m (a) ⊗ γ A n (b), which equals γ An (a) ⊗ γ Sn (b) since γ Am (a) = γ A m (a).Thus we obtain Φ A×B γ A m+n (a, b) = Φ A γ Am (a) • Φ B γ Sn (b) .

Corollary 5 . 6 .
Let R be a ring, and let A and B be degree-m and n extensions of R, respectively, with m + n ≥ 2. Then ∆ A×B is generated as an R-module by {1} ∪ { γA m+n (a 1 , . . ., a m , b 1 , . . ., b n ) | each a i ∈ A and b j ∈ B}.Proof.Since (A × {0}) ∪ ({0} × B) generates A × B as an R-module, from Corollary 3.7 we find that ∆ A×B is generated by 1 and elements γ of the form γ = γA m+n (c 1 , . . ., c m+n ) with each c i in either A × {0} or {0} × B. Since we may permute the c i by any even permutation without changing γ, and permuting the c i by an odd permutation only changes γ up to sign and the addition of an element of R, we find that 1, together with elements of the form γA m+n (a 1 , . . ., a k , b 1 , . . ., b ℓ ), with each a i in A and each b j in B, generate ∆ A×B .But by Remark 5.5, only those generators with k = m and ℓ = n are nonzero in ∆ A×B .Theorem 5.7.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 ∆ A×B ∼ −→ ∆ ∆ A ×∆ B that sends 1 → 1 and each element of the form γA m+n a 1 , . . ., a m , b 1 , . . ., b n to γA 4 1 ∆ A , γAm (a 1 , . . ., a m ), 1 ∆ B , γAn (b 1 , . . ., b n ) .