Witt Ring of a Commutative Ring

WITT RING OF A COMMUTATIVE RING

KAZIMIERZ SZYMICZEK

Introduction These notes give a detailed version of the construction of the Witt ring W (R) of a commutative ring R as presented in [7] and [11]. We also include some additional material on the Witt rings of Dedekind domains and the Witt rings of the product of rings. The prerequisites include the notion of the tensor product of modules.

1. Inner product spaces Let R be a commutative ring with identity element 1. We shall consider left R−modules. For an R−module M we write M ∗ for the dual module, that is, the R−module of linear functionals Hom(M,R). Given a symmetric bilinear form β : M × M → R the pair (M, β) is said to be a bilinear module. The module homomorphism βb : M → M ∗, βb(m)(n) = β(m, n) is said to be the adjoint homomorphism. The bilinear module (M, β) is said to be nonsingular if the adjoint homomorphism βb is an isomorphism. Then (M, β) is also said to be an inner product module. When the module M is ﬁnitely generated projective, the bilinear module (M, β) is said to be a bilinear space, and the inner product module (M, β) is said to be inner product space. All bilinear forms are assumed to be symmetric. Example 1.1. The ring R can be viewed as an R−module. Actually, we view it as a bilinear module with the symmetric bilinear form β : R × R → R, β(a, b) = ab. The dual module R∗ consists of R−linear maps ϕ : R → R. Thus for all x, y ∈ R, ϕ(x + y) = ϕ(x) + ϕ(y), ϕ(xy) = xϕ(y). Setting y = 1 we get ϕ(x) = xϕ(1). Thus the linear functional ϕ ∈ R∗ is just the multiplication by ϕ(1) on R. On the other hand, for all a, b ∈ R, βb(a)(b) = β(a, b) = ab and so βb(a) is the multiplication by a on R. It follows that for any ϕ ∈ R∗, βb(ϕ(1)) = ϕ. Thus βb : R → R∗ is surjective and since it is clearly injective, it is an isomorphism. Moreover, the R−module R is free, hence projective. This shows that the ring R viewed as an R−module is an inner product space. 1 2 KAZIMIERZ SZYMICZEK

2. Projective modules Recall that M is projective if there exists a module N such that M ⊕ N is a free module. The following lemma gives a nice property of f.g. projectives. Lemma 2.1. If M is f.g. projective, then there is a f.g. module N such that M ⊕N is free.

Proof. Let N be anyP module such that F := M ⊕ N is free. Let {ei}i∈I be a basis for F . Hence F = i∈I Rei. Let m1, . . . , mg be a set of generators for M. Write each mj as a linear combination of the ei’s. This involves only ﬁnitely many basis elements ei. Hence there is a ﬁnite subset J ⊂ I so that X M ⊂ Rei =: F0. i∈J

Observe that F0 = M + (N ∩ F0). Indeed each f0 ∈ F0 can be written as f0 = m + n, where m ∈ M and n ∈ N. But then n = f0 − m ∈ F0. This shows that F0 ⊂ M + (N ∩ F0), and the inverse inclusion is trivial. Because of the uniqueness of the representation f0 = m + n we actually have F0 = M ⊕ (N ∩ F0). Here F0 is ∼ free and f.g., hence also F0/M is f.g. But F0/M = N ∩ F0, hence the latter is also f.g. ¤ Corollary 2.2. A ﬁnitely generated module M is projective if and only if there is a module N and a natural number n such that M ⊕ N =∼ Rn. Notice that the complementary module N is also a f.g. projective module.

Remark 2.3. When F is a free module of rank n and {e1, . . . , en} is a basis for F , ∗ then there are linear functionals {ϕ1, . . . , ϕn} ⊂ F satisfying

ϕi(ej) = δij, 1 ≤ i, j ≤ n.

The functional ϕi can be viewed as the projection of f = x1e1 + ··· + xnen onto the i−th coordinate xi, that is ϕi(f) = xi. Hence we can write Xn (2.1) f = ϕi(f)ei i=1 and this is the unique representation of f as a linear combination of the basis elements ei. One can easily prove that the functionals ϕi form a basis for the dual ∗ ∗ module F . The bases {ei} and {ϕi} are said to be dual bases for F and F . It turns out that for projective modules there is a similar construction for which the representation of elements in the form (2.1) holds although we cannot expect any uniqueness of the representation. Lemma 2.4 (Dual bases lemma). Let M be a f.g. module. The following statements are equivalent. (a) M is projective. ∗ (b) For each set of generators {y1, . . . , yn} of M there is a set {ϕ1, . . . , ϕn} ⊂ M such that Xn (2.2) m = ϕi(m)yi for all m ∈ M. i=1 WITT RING OF A COMMUTATIVE RING 3

∗ (c) There exist sets {y1, . . . , yn} ⊂ M and {ϕ1, . . . , ϕn} ⊂ M such that (2.2) holds. Proof. (a) ⇒ (b) M is generated by n elements, hence M is a homomorphic image of a free module of rank n. So there is a surjective homomorphism σ : Rn → M and n we can choose σ so that σ(ei) = yi, where {e1, . . . , en} is the standard basis for R . Hence, by the fundamental property of projective modules there is a homomorphism λ : M → Rn so that the following diagram commutes.

λ id

+ ? Rn - M σ

n n Consider the projection πi : R → R which sends any n−tuple in R into its i−th n coordinate. Observe that for the standard basis {e1, . . . , en} of R and for m ∈ M we have Xn λ(m) = πi(λ(m))ei. i=1 ∗ The map ϕi = πi ◦ λ : M → R is a linear form (functional) in M and for any m ∈ M we get Xn Xn m = σλ(m) = σ( πi(λ(m))ei) = πi(λ(m))σ(ei) i=1 i=1 Xn = πi(λ(m))yi i=1 Xn = ϕi(m)yi i=1 (b) ⇒ (c) This is trivial. n (c) ⇒ (a) We again take surjective homomorphism σ : R → M so that σ(ei) = yi, n where {e1, . . . , en} is the standard basis for R . We will show that the exact sequence Rn −→σ M −→ 0 splits. For this we deﬁne the linear map Xn n τ : M −→ R , τ(m) = ϕi(m)ei i=1 and for any m ∈ M calculate Xn Xn Xn στ(m) = σ( ϕi(m)ei) = ϕi(m)σ(ei) = ϕi(m)yi = m. i=1 i=1 i=1 Hence τ splits σ and so M is a direct summand of Rn. This proves (a). ¤ 4 KAZIMIERZ SZYMICZEK

Corollary 2.5. Let M be a f.g. projective R−module. For each m ∈ M, m 6= 0, there is a linear functional ϕ ∈ M ∗ satisfying ϕ(m) 6= 0.

Proof. Since m 6= 0, one of the functionals ϕi satisfying (2.2) does not vanish at m. ¤

As an application we now prove a convenient condition for a bilinear space to be an inner product space. Lemma 2.6. Let (M, β) be a bilinear space. If the adjoint homomorphism βb : M → M ∗ is surjective, then it is also injective, and the bilinear space is an inner product space. Proof. Suppose βb(m) = 0 for some m ∈ M. Then βb(m)(n) = β(m, n) = 0 for all n ∈ M. By symmetry, β(n, m) = 0 for all n ∈ M, that is, βb(n)(m) = 0 for all n ∈ M. Since βb is surjective, it follows that all linear functionals on M vanish on the element m. Since M is f.g. projective, Corollary 2.5 implies that m = 0. This proves the injectivity of βb. Hence the adjoint homomorphism βb is an isomorphism, as desired. ¤

3. Direct sums Lemma 3.1. For any modules M and N, (M ⊕ N)∗ =∼ M ∗ ⊕ N ∗. Proof. For ϕ ∈ (M ⊕ N)∗ and all m ∈ M, n ∈ N we have ϕ(m, n) = ϕ((m, 0) + (0, n)) = ϕ(m, 0) + ϕ(0, n). We can interpret ϕ(m, 0) as the value at m of the functional ϕ( , 0) ∈ M ∗, which is the restriction ϕ|M of ϕ to M, and similarly interpret ϕ(0, n). Hence we can deﬁne the map ∗ ∗ ∗ Φ:(M ⊕ N) → M ⊕ N , Φ(ϕ) = (ϕ|M , ϕ|N ). Clearly Φ is a linear map. We now show that Φ is injective. For this we prove that ker Φ = 0. We have ∗ ∗ ϕ ∈ ker Φ ⇐⇒ ϕ|M = 0 ∈ M and ϕ|N = 0 ∈ N ⇐⇒ ϕ(m, 0) = 0 and ϕ(0, n) = 0 for all m ∈ M, n ∈ N ⇐⇒ ϕ(m, n) = 0 for all m ∈ M, n ∈ N ⇐⇒ ϕ = 0 ∈ (M ⊕ N)∗. ∗ ∗ It remains to establish the surjectivity of Φ. Let ϕ1 ∈ M , ϕ2 ∈ N . Then

ϕ : M ⊕ N → R, ϕ(m, n) = ϕ1(m) + ϕ2(n) ∗ is a linear map, that is, ϕ ∈ (M ⊕ N) . Moreover ϕ|M = ϕ1, ϕ|N = ϕ2. Hence

Φ(ϕ) = (ϕ|M , ϕ|N ) = (ϕ1, ϕ2). ¤

Corollary 3.2. If M is f.g. projective, then M ∗ is also f.g. projective. WITT RING OF A COMMUTATIVE RING 5

Proof. There is a module N and a natural number n so that M ⊕ N =∼ Rn. Then, by the Lemma, we have M ∗ ⊕ N ∗ =∼ (M ⊕ N)∗ =∼ (Rn)∗ =∼ Rn. The isomorphism (Rn)∗ =∼ Rn is easily established by using the concept of dual bases of the free module Rn and its dual (Rn)∗ (see Remark 2.3). This shows that M ∗ is f.g. projective. ¤

Corollary 3.3. If M and N are f.g. projective modules, then so are M ⊕ N and M ⊗ N.

Proof. By the hypothesis, there are modules M1,N1 and natural numbers k and ` so that ∼ k ∼ ` M ⊕ M1 = R ,N ⊕ N1 = R . Then ∼ k ` ∼ k+` (M ⊕ N) ⊕ (M1 ⊕ N1) = R ⊕ R = R showing that M ⊕ N is f.g. projective module. Similarly, we have ¡ ¢ ∼ k ` ∼ k` (M ⊗ N) ⊕ (M ⊗ N1) ⊕ (M1 ⊗ N) ⊕ (M1 ⊗ N1) = R ⊗ R = R showing that M ⊗ N is f.g. projective. ¤

Lemma 3.4. Let (M, β) and (N, γ) be bilinear modules. Then the following diagram is commutative:

M ⊕ N

β\⊕ γ βb ⊕ γb

Proof. For m0 ∈ M and n0 ∈ N we have (Φ ◦ (β\⊕ γ))(m0, n0) = Φ((β\⊕ γ)(m0, n0)) = ((β\⊕ γ)(m0, n0)( , 0), (β\⊕ γ)(m0, n0)(0, )) = (βb(m0), γb(n0)) = (β,b γb)(m0, n0) = (βb ⊕ γb)(m0, n0). Here the passage from the second to the third row is justiﬁed by the following calculation: (β\⊕ γ)(m0, n0)( , 0) = (β ⊕ γ)((m0, n0), ( , 0)) = β(m0, ) + γ(n0, 0) = β(m0, ) = βb(m0). ¤ 6 KAZIMIERZ SZYMICZEK

Theorem 3.5. For bilinear modules (M, β) and (N, γ), (M, β) ⊥ (N, γ) is nonsingular ⇐⇒ (M, β) and (N, γ) are nonsingular. Proof. We have (M, β) ⊥ (N, γ) = (M ⊕ N, β ⊕ γ) and so (M, β) ⊥ (N, γ) is nonsingular if and only if β\⊕ γ is an isomorphism. By Lemma 3.4 this happens exactly when βb ⊕ γb is an isomorphism. The latter is equivalent to saying that βb and γb are isomorphisms which means that (M, β) and (N, γ) are nonsingular. ¤

4. Tensor products We want to prove that the tensor product of two inner product spaces is an inner product space. This is a nontrivial theorem and we shall give two proofs for the key Lemma needed (Lemma 4.4). The ﬁrst proof requires the knowledge of all the basic tricks on tensor products of f.g. projective modules and their duals. These are available in [4] and [5]. The latter gives more detail and discusses the matter in full generality. The second proof is direct and uses only Lemma 2.4 and its corollary. We begin with gathering all the necessary results. Lemma 4.1. For any modules M,N,C there is a canonical isomorphism Hom(M ⊕ N,C) =∼ Hom(M,C) ⊕ Hom(N,C). Proof. For a proof see [4, Chapter III, §3]. The isomorphism sends a homomorphism f : M ⊕ N → C into the pair of restrictions (f|M , f|N ). Notice that we have given a detailed proof for the special case when C = R in Lemma 3.1. ¤ Lemma 4.2. If M is a f.g. projective module, then for any module C there is a canonical isomorphism M ∗ ⊗ C =∼ Hom(M,C). Proof. First observe that there is a natural homomorphism ζ : M ∗ ⊗ C −→ Hom(M,C), ζ(ϕ ⊗ c)(m) = ϕ(m) c. We shall prove that if M is f.g. projective, this is an isomorphism. We begin with the case when M is a f.g. free module. From this we will deduce ∼ n the result in the general© case. So letªM = R be a f.g. free module. Let {e1, . . . , en} be a basis for M and let e1, . . . , en be the dual basis for M ∗. This is characterized by the requirement that i i e (ej) = 0 when i 6= j and e (ei) = 1 for all 1 ≤ i, j ≤ n. Any element of M ∗ ⊗ C has a unique representation in the Pn i form i=1 e ⊗ ci, where ci ∈ C. Then ³Xn ´ ∗ i ζ : M ⊗ C −→ Hom(M,C), ζ e ⊗ ci = f, i=1 where f : M → C is the unique homomorphism satisfying f(ej) = cj, j = 1, . . . , n. Indeed, ³Xn ´ Xn Xn i i i f(ej) = ζ e ⊗ ci (ej) = ζ(e ⊗ ci)(ej) = e (ej)ci = cj. i=1 i=1 i=1 WITT RING OF A COMMUTATIVE RING 7

Since f is uniquely determined by its values on the basis elements ei, it follows that ζ is an isomorphism. Now we consider the general case. So let M be a f.g. projective module. Then there is a f.g. projective module P so that M ⊕ P =∼ Rn for some n. Since the result is known in the case of free modules, there is an isomorphism

ζ :(M ⊕ P )∗ ⊗ C −→ Hom(M ⊕ P,C) © ª 1 n and we recall how ζ acts. Let {e1, . . . , en} be a basis for M ⊕P and let e , . . . , e ∗ ∗ be the dual basis for (M ⊕ P ) . Any element of (M ⊕ P )³ ⊗ C has a´ unique Pn i Pn i representation in the form i=1 e ⊗ ci, where ci ∈ C and ζ i=1 e ⊗ ci = f, where f : M ⊕ P → C is the unique homomorphism satisfying f(ei) = ci for i = 1, . . . , n. Thus, using the distributivity of ⊗ with respect to ⊕, Lemma 3.1, the isomorphism ζ, and Lemma 4.1, we have the following chain of isomorphisms (M ∗ ⊗ C) ⊕ (P ∗ ⊗ C) =∼ (M ∗ ⊕ P ∗) ⊗ C =∼ (M ⊕ P )∗ ⊗ C =∼ Hom(M ⊕ P,C) =∼ Hom(M,C) ⊕ Hom(P,C).

The proof will be complete when we show that the composition of the isomorphisms from the left to the right hand side maps M ∗ ⊗ C onto Hom(M,C). The ﬁrst step is to show that M ∗ ⊗ C maps into Hom(M,C), and for this it is suﬃcient to show that each generator of M ∗ ⊗ C has the image in Hom(M,C). So ∗ ∗ ∗ take a generator f1 ⊗c in M ⊗C. Viewed as an element of (M ⊗C) ⊕ (P ⊗C) it ∗ ∗ should be written as (f1 ⊗c, 0⊗c) and then it goes onto (f1, 0)⊗c ∈ (M ⊕P )⊗C, ∗ and further, onto (f1, 0) ⊗ c ∈ (M ⊕ P ) ⊗ C. The linear form

ϕ = (f1, 0) : M ⊕ P → R has the characteristic property that its restriction to P is the zero map, ϕ|P = 0. Now we apply ζ to ϕ ⊗ c and, by deﬁnition, we get f ∈ Hom(M ⊕ P,C) described P i as follows. If ϕ = i aie , then X i ζ(ϕ ⊗ c) = ζ( e ⊗ aic) = f, i where f : M ⊕ P → C, f(ei) = aic. We now show that from ϕ|P = 0 itP follows that f|P = 0. Indeed, take any p ∈ P . Then we can write uniquely p = j yjej and then we have X X X i¡ ¢ 0 = ϕ(p) = aie yjej = aiyi. i j i

Hence for f(p) we have X X X f(p) = f( yjej) = yjf(ej) = yjajc = 0 · c = 0. j j j

The next isomorphism maps f onto (f|M , f|P ) ∈ Hom(M,C) ⊕ Hom(P,C) and since the restriction f|P is the zero map, the image is in Hom(M,C), as required. 8 KAZIMIERZ SZYMICZEK

We have proved that M ∗ ⊗ C maps into Hom(M,C), and, by symmetry, P ∗ ⊗ C maps into Hom(P,C). Since these are the restrictions of an isomorphism, both maps have to be surjective. This proves the Lemma. 1 ¤

Lemma 4.3. For any modules M,N,C there is a canonical isomorphism Hom(M ⊗ N,C) =∼ Hom(M, Hom(N,C)). Proof. The map η : Hom(M ⊗ N,C) −→ Hom(M, Hom(N,C)), η(f)(m)(n) = f(m ⊗ n) turns out to be an isomorphism. For a proof see [4, Chapter XVI, §2] and [5, Chapter V,§3]. We shall need the inverse isomorphism η−1. It is deﬁned as follows. Take a homomorphism h : M → Hom(N,C). Then h can be viewed as bilinear map h : M × N → C since h(m)(n) is bilinear in m, n. Thus by the universal property of the tensor product there is a unique homomorphism h0 : M ⊗ N → C satisfying h0(m ⊗ n) = h(m)(n). We deﬁne ρ : Hom(M, Hom(N,C)) −→ Hom(M ⊗ N,C), ρ(h) = h0. Thus ρ(h)(m ⊗ n) = h(m)(n) for all m ∈ M, n ∈ N. A direct check shows that η ◦ ρ and ρ ◦ η are identities and so ρ = η−1. ¤

Now we will apply the results discussed so far in this section to the proof of the main result stating that the tensor product of two inner product spaces is an inner product space. Let’s explain the notation we are going to use. Let M and N be R−modules. Let f ∈ M ∗ and g ∈ N ∗ be linear forms (functionals) on M and N, respectively. We consider the following three operations on the forms f, g. f × g : M × N −→ R, (f × g)(m, n) = f(m) · g(n). This is a bilinear form on M × N. f £ g : M ⊗ N −→ R is the linear form on M ⊗ N associated with f × g, that is the unique linear map making the following diagram commute.

⊗ M × N - M ⊗ N

f × g f £ g j ? R

Hence (f £ g)(m ⊗ n) = f(m) · g(n). f ⊗g : M ⊗N −→ R⊗R =∼ R, (f ⊗g)(m⊗n) = f(m)⊗g(n) = (f(m)·g(n))(1⊗1). This is the tensor product of linear forms f and g and is an element of M ∗ ⊗ N ∗. Now we deﬁne Ψ: M ∗ ⊗ N ∗ −→ (M ⊗ N)∗

1The result is stated in [5, Chapter V, §4] but details are given only for the easier part of the proof when M is assumed to be a free module. WITT RING OF A COMMUTATIVE RING 9 to be the unique linear map such that Ψ(f ⊗ g) = f £ g. Its existence follows from the commutative diagram

⊗ M ∗ × N ∗ - M ∗ ⊗ N ∗

£ Ψ j ? (M ⊗ N)∗

The hardest part of the proof of nonsingularity of the tensor product of two nonsingular modules is to show that Ψ is an isomorphism. Now we can do that. Lemma 4.4. If M is ﬁnitely generated projective and N is any module, then Ψ is an isomorphism. Proof. We have the following isomorphisms of R−modules: M ∗ ⊗ N ∗ =∼ Hom(M,N ∗) by Lemma 4.2 = Hom(M, Hom(N,R)) =∼ Hom(M ⊗ N,R) by Lemma 4.3 = (M ⊗ N)∗. It remains to check that the resulting isomorphism coincides with Ψ. So take any simple tensor f ⊗ g ∈ M ∗ ⊗ N ∗. The isomorphism ζ of Lemma 4.2 sends f ⊗ g into the map h := f() · g in Hom(M, Hom(N,R)). This in turn goes under the isomorphism η−1 of Lemma 4.3 onto the map h0 satisfying h0(m ⊗ n) = h(m)(n) = f(m) · g(n) = (f £ g)(m ⊗ n) for all m ∈ M, n ∈ N. Thus the composition η−1 ◦ ζ acts exactly in the same way as Ψ. Both send f ⊗ g onto f £ g. This proves the Lemma. ¤

Remark 4.5 (Second proof). Here we give a direct proof that Ψ is an isomorphism in the case when M and N are f. g. projective modules. For this we use Lemma 2.4. First we show that Ψ is a surjective homomorphism. Let {mi} and {fi} be the sets of generators for the projective module M and linear forms satisfying (2.2), and let {nj} and {gj} be the corresponding sets for the projective module N. Thus we have X X m = fi(m)mi, n = gj(n)nj for all m ∈ M, n ∈ N. Then, by the bilinearity of the tensor product we get X X m ⊗ n = fi(m)gj(n) mi ⊗ nj. i j 10 KAZIMIERZ SZYMICZEK

Now, given any linear form ϕ ∈ (M ⊗ N)∗ we have X X ϕ(m ⊗ n) = fi(m)gj(n) ϕ(mi ⊗ nj) i j X X = ϕ(mi ⊗ nj) · (fi £ gj)(m ⊗ n). i j It follows that X X ³X X ´ ϕ = ϕ(mi ⊗ nj) · (fi £ gj) = Ψ ϕ(mi ⊗ nj) · (fi ⊗ gj) , i j i j showing the surjectivity of Ψ. And now we prove the injectivity of Ψ. For this we shall use the just proved surjectivity of Ψ for any f.g. projective modules M and N. In particular, M ∗ and N ∗ are f.g. projective, hence the module homomorphism Ψ∗ : M ∗∗ ⊗ N ∗∗ → (M ∗ ⊗ N ∗)∗ satisfying Ψ∗(m ⊗ n) = m £ n is surjective. Here we identify the bidual module M ∗∗ with M as we can for f.g. projective module M (for a proof see [5, Chapter V, Theorem 4.1]). To prove the injectivity of Ψ we show that its kernel equals zero. So let X ¡ ¢ ∗ ∗ Ψ fi ⊗ gi = 0 for some fi ∈ M , gi ∈ N . Hence for all m ∈ M and n ∈ N we have ¡X ¢ X X 0 = Ψ fi ⊗ gi (m ⊗ n) = Ψ(fi ⊗ gi)(m ⊗ n) = fi(m) · gi(n). Now observe that viewing m and n as elements of M ∗∗ and N ∗∗, respectively, we can write fi(m) · gi(n) = m(fi) · n(gi) = (m £ n)(fi ⊗ gi). P Thus if t := fi ⊗ gi is in the kernel of Ψ, then (m £ n)(t) = 0 for all m ∈ M, n ∈ N. Since Ψ∗ is surjective, it follows that for all linear functionals ψ ∈ (M ∗ ⊗ N ∗)∗ we have ψ(t) = 0. According to Corollary 2.5 we must have t = 0, as required. Lemma 4.6. Let (M, β) and (N, γ) be bilinear modules. Then the following diagram commutes.

M ⊗ N

βb ⊗ γb β\⊗ γ

Proof. It suﬃces to check that the maps involved act in the same way on the generators m ⊗ n of M ⊗ N. On the one hand we have (βb ⊗ γb)(m ⊗ n) = βb(m) ⊗ γb(n) WITT RING OF A COMMUTATIVE RING 11 and this is sent by Ψ to βb(m) £ γb(n). Thus for all m0 ∈ M, n0 ∈ N, Ψ((βb ⊗ γb)(m ⊗ n))(m0 ⊗ n0) = (βb(m) £ γb(n))(m0 ⊗ n0) = βb(m)(m0) · γb(n)(n0) = β(m, m0) · γ(n, n0). On the other hand, (β\⊗ γ)(m ⊗ n) is a linear form on M ⊗ N and we have (β\⊗ γ)(m ⊗ n)(m0 ⊗ n0) = (β ⊗ γ)(m ⊗ n, m0 ⊗ n0) = β(m, m0) · γ(n, n0) for all m0 ∈ M, n0 ∈ N. This shows the commutativity of the diagram. ¤ Theorem 4.7. For bilinear spaces (M, β) and (N, γ), (M, β) and (N, γ) are nonsingular ⇒ (M, β) ⊗ (N, γ) is nonsingular. Proof. By Lemma 4.4 and Lemma 4.6 it is suﬃcient to show that βb ⊗ γb is an isomorphism. For then β\⊗ γ is also an isomorphism and so (M, β) ⊗ (N, γ) = (M ⊗ N, β ⊗ γ) is nonsingular. By the hypothesis, βb and γb are isomorphisms. It is easy to show that βb−1 ⊗ γb−1 : M ∗ ⊗ N ∗ −→ M ⊗ N is the inverse map of βb ⊗ γb. Indeed, for any f ∈ M ∗ and g ∈ N ∗ there are m ∈ M and n ∈ N so that f = βb(m) and g = γb(n) and so we have ((βb ⊗ γb) ◦ (βb−1 ⊗ γb−1))(f ⊗ g) = (βb ⊗ γb)(m ⊗ n) = f ⊗ g and similarly, for all m ∈ M and n ∈ N and f = βb(m) and g = γb(n), ((βb−1 ⊗ γb−1) ◦ (βb ⊗ γb))(m ⊗ n) = (βb−1 ⊗ γb−1)(f ⊗ g) = m ⊗ n. Thus if βb and γb are isomorphisms, βb ⊗ γb is also an isomorphism. ¤

5. Metabolic spaces For a bilinear module (S, β) and for a submodule N of S we consider the orthogonal complement of N in S. This is the submodule N ⊥ := {s ∈ S : β(s, N) = 0} . The submodule N is said to be totally isotropic (or self-orthogonal) if β(N,N) = 0. For this it is necessary and suﬃcient that N ⊆ N ⊥. Proposition 5.1. Let (S, β) be an inner product space and let M and N be submodules of S. The following are equivalent. (a) S = M ⊕ N and N = N ⊥. (b) S = M ⊕ N, βb(N) = M ∗ and β(N,N) = 0. The condition βb(N) = M ∗ is to be understood as follows. We have the isomorphisms βb M ⊕ N −→ (M ⊕ N)∗ −→Φ M ∗ ⊕ N ∗. We use Φ to identify (M ⊕ N)∗ with M ∗ ⊕ N ∗ so that βb(N) = M ∗ is to be understood as Φ(βb(N)) = M ∗. 12 KAZIMIERZ SZYMICZEK

Proof. (a) ⇒ (b) We have to show that βb(N) = M ∗. First notice that β(N,N) = 0 follows from N = N ⊥. Further, for n, n0 ∈ N we have βb(0, n)(0, n0) = β((0, n), (0, n0)) = 0 since β(N,N) = 0. Hence in the identification b b b ∗ ∗ Φ(β(0, n)) = (β(0, n)|M , β(0, n)|N ) ∈ M ⊕ N the second component vanishes. Hence b b ∗ Φ(β(0, n)) = (β(0, n)|M , 0) ∈ M showing that βb(N) ⊆ M ∗. To prove the inverse inclusion take ϕ ∈ M ∗. By nonsingularity there is (m, n) ∈ S such that βb(m, n) = ϕ ∈ M ∗. It follows that in b b b Φ(β(m, n)) = (β(m, n)|M , β(m, n)|N ) the second component vanishes. Hence βb(m, n)(0, n0) = 0 for all n0 ∈ N, that is, β((m, n), (0, n0)) = 0 for all n0 ∈ N. This means (m, n) ∈ N ⊥ = N, the latter by (a). Hence m = 0. Thus any ϕ ∈ M ∗ is of the shape ϕ = βb(0, n) for some n ∈ N, that is, ϕ ∈ βb(N). Thus M ∗ ⊆ βb(N), as required. (b) ⇒ (a) We must show that N = N ⊥. From β(N,N) = 0 we get N ⊆ N ⊥ so we must prove the inverse inclusion. Let s = (m, n) ∈ N ⊥. Then β((m, n), (0, n0)) = 0 for all n0 ∈ N. This says that in the identification ∗ ∗ ∗ b b b Φ:(M ⊕ N) → M ⊕ N , β(m, n) 7→ (β(m, n)|M , β(m, n)|N ) the second component vanishes and hence βb(m, n) ∈ M ∗. By (b) we get βb(m, n) ∈ βb(N), that is, βb(m, n) = βb(0, n0) for some n0 ∈ N. It follows βb(m, n − n0) = 0 and since βb is an isomorphism we get m = 0 and n − n0 = 0. Thus we have proved that if s = (m, n) ∈ N ⊥, then m = 0, that is, s = (0, n) ∈ N. This proves that N ⊥ ⊆ N, and so N = N ⊥, as desired. ¤ Definition 5.2. The inner product space (S, β) is said to be metabolic if it satisfies one (hence both) of the conditions (a) or (b). Remark 5.3. Here we try to clarify the definition of metabolic spaces given by Milnor and Husemoller in [7]. For this we introduce two more conditions related to (a) and (b) above. So (S, β) is an inner product space and M and N are submodules of S. (c) S = M ⊕ N, βb(N) = M ∗, βb(M) = N ∗ and β(N,N) = 0. (d) S = M ⊕ N, βb(N) = M ∗, βb(M) = N ∗, β(N,N) = 0 and β(M,M) = 0. First observe that (c) and (d) are equivalent. We only have to check that (c) ⇒ (d). So assuming (c) we must prove that β(M,M) = 0. For m ∈ M we have b b b ∗ ∗ Φ(β(m, 0)) = (β(m, 0)|M , β(m)|N ) ∈ M ⊕ N . WITT RING OF A COMMUTATIVE RING 13

b ∗ b ∗ From β(M) = N we conclude that β(m, 0)|M = 0 ∈ N . Hence β((m, 0), (m0, 0)) = βb(m, 0)(m0, 0) = 0 for all m0 ∈ M, as required. Thus we have (a) ⇐⇒ (b) ⇐= (c) ⇐⇒ (d). Milnor and Husemoller assert that (a) and (c) are equivalent and so one can take anyone of them for definition of metabolic spaces. This is, however, not true. We give here an example showing that (a) and (c) are not equivalent sentences. Example 5.4. Let R be a field of characteristic 2 and let S = Ru + Rv be a 2- dimensional vector space endowed with the symmetric bilinear form β satisfying β(u, u) = 0, β(u, v) = β(v, v) = 1. In other words, S is an isotropic but not hyperbolic plane and β has in the basis {u, v} the matrix · ¸ 0 1 (5.1) . 1 1 Then the condition (a) is satisfied with N = Ru and M = Rv. On the other hand, S does not satisfy (d). The point is that N = Ru is the unique totally isotropic subspace of dimension 1 in S. For, if R(xu + yv) is totally isotropic, then (5.2) 0 = β(xu + yv, xu + yv) = xy + xy + y2 = y2 and so y = 0 and R(xu + yv) = Ru. Thus S satisfies (a) but cannot be written as the direct sum of two totally isotropic subspaces. Hence S does not satisfy (d). Definition 5.5. The inner product space (S, β) is said to be hyperbolic if it satisfies one (hence both) of the conditions (c) or (d). Thus every hyperbolic space is metabolic but there are non-hyperbolic metabolic spaces. Example 5.4 may be viewed as a very special counter-example but in fact the obstruction is of much more general nature. Here we only indicate that essentially the same counterexample works over the ring of integers Z. Example 5.6. We consider the free Z-module S = Zu + Zv with bilinear form β defined by the matrix (5.1). Then S is metabolic. As in Example 5.4 we show that Zu is the unique totally isotropic subspace of rank 1 in S. This time the equation (5.2) becomes 0 = 2xy + y2 and it is to be solved in integers. We claim that only y = 0 is admissible. For in the other solution we have y = −2x and we have S = Zu ⊕ Z(xu + yv) = Zu ⊕ Z(xu − 2xv). It follows that every element of S has a unique representation in the form au + b(xu − 2xv) = (a + bx)u − 2bxv, a, b ∈ Z which is not the case, since here the second coefficient can take only even values. This shows that Zu is the unique totally isotropic submodule of rank 1 and so S is not hyperbolic. Now we prove the properties of metabolic spaces needed in the construction of the Witt ring. 14 KAZIMIERZ SZYMICZEK

Lemma 5.7. Let (S, α) and (T, β) be metabolic spaces. Then the orthogonal sum (S ⊕ T, α ⊕ β) is a metabolic space. Proof. By the hypothesis, we have decompositions

S = M ⊕ N, α(N,N) = 0,T = M1 ⊕ N1, β(N1,N1) = 0 and ∗ b ∗ (5.3) αb(N) = M , β(N1) = M1 . Then

S ⊕ T = (M ⊕ M1) ⊕ (N ⊕ N1) and 0 0 0 0 (α ⊕ β)((n, n1), (n , n1)) = α(n, n ) + β(n1, n1) = 0 0 0 for all n, n ∈ N and n1, n1 ∈ N1. Hence

(α ⊕ β)(N ⊕ N1,N ⊕ N1) = 0, that is, N ⊕ N1 is a totally isotropic subspace of S ⊕ T . Moreover, \ b ∗ ∗ ∗ (α ⊕ β)(N ⊕ N1) = αb(N) ⊕ β(N1) = M ⊕ M1 = (M ⊕ M1) , where we have used Lemma 3.4, the condition (5.3), and Lemma 3.1. Hence S ⊕ T satisﬁes the condition (b) in Prop. 5.1 and so is a metabolic space. ¤

Lemma 5.8. Let (S, α) be a metabolic space and let (X, β) be an arbitrary inner product space. Then the tensor product (S ⊗ X, α ⊗ β) is a metabolic space. Proof. We have S = M ⊕ N with αb(N) = M ∗ and α(N,N) = 0. Then S ⊗ X = (M ⊗ X) ⊕ (N ⊗ X). First notice that N ⊗ X is totally isotropic subspace of S ⊗ X. This follows from the fact that N is a totally isotropic subspace of S, for we have (α ⊗ β)(n ⊗ x, n0 ⊗ x0) = α(n, n0) · β(x, x0) = 0 · β(x, x0) = 0 for all n, n0 ∈ N and x, x0 ∈ X. Moreover, (α\⊗ β)(N ⊗ X) = αb(N) ⊗ βb(X) = M ∗ ⊗ X∗ = (M ⊗ X)∗, where we have used Lemma 4.6, αb(N) = M ∗ and nonsingularity of (X, β), and Lemma 4.4. Hence S ⊗ X satisﬁes the condition (b) in Prop. 5.1 and so is a metabolic space. ¤

Lemma 5.9. Let (M, β) be an inner product space. Then the orthogonal sum (M, β) ⊥ (M, −β) is a metabolic space. Proof. Let S = Ru ⊕ Rv be the orthogonal sum of two free submodules of rank 1 with basis elements u and v and bilinear forms α and α0 deﬁned by α(au, bu) = ab, α0(av, bv) = −ab. We denote (S, α ⊕ α0) by S = (1) ⊕ (−1). WITT RING OF A COMMUTATIVE RING 15

This is known to be a metabolic space (see [7, p.13]). Then we have (S, α) ⊗ (M, β) = ((1) ⊕ (−1)) ⊗ (M, β) ¡ ¢ ¡ ¢ = (1) ⊗ (M, β) ⊕ (−1) ⊗ (M, β) = (M, β) ⊥ (M, −β). This is metabolic by Lemma 5.8. ¤

6. The Witt ring Witt ring of similarity classes. Deﬁnition 6.1. Two inner product spaces (M, β) and (N, γ) are said to be isometric, written M =∼ N, if there exists a module isomorphism f : M → N satisfying γ(f(m), f(m0)) = β(m, m0) for all m, m0 ∈ M. Lemma 6.2. The isometry relation is an equivalence relation. It preserves orthogonal sums and tensor products, that is, for inner product spaces M,N,M 0,N 0, M =∼ M 0 and N =∼ N 0 =⇒ M ⊥ N =∼ M 0 ⊥ N 0 and M ⊗ N =∼ M 0 ⊗ N 0. Proof. ¤

Definition 6.3. Two inner product spaces M and N are said to be similar, written M ∼ N, if there exist metabolic spaces S and T so that M ⊥ S is isometric to N ⊥ T . Lemma 6.4. Similarity is an equivalence relation. It preserves orthogonal sums and tensor products, that is, for inner product spaces M,N,M 0,N 0, M ∼ M 0 and N ∼ N 0 =⇒ M ⊥ N ∼ M 0 ⊥ N 0 and M ⊗ N ∼ M 0 ⊗ N 0. Proof. If M ∼ M 0 and N ∼ N 0, then there are metabolic spaces S, S0,T,T 0 so that M ⊥ S =∼ M 0 ⊥ S0 and N ⊥ T =∼ N 0 ⊥ T 0. From this, by the extension of isometries (Lemma 6.2) we get (M ⊥ N) ⊥ (S ⊥ T ) =∼ (M 0 ⊥ N 0) ⊥ (S0 ⊥ T 0). By the Lemma 5.7 the direct sums S ⊥ T and S0 ⊥ T 0 are metabolic spaces, hence, by definition, M ⊥ N ∼ M 0 ⊥ N 0. The proof of the remaining part is similar. From the assumptions and Lemma 6.2 it follows that (M ⊥ S) ⊗ (N ⊥ T ) =∼ (M 0 ⊥ S0) ⊗ (N 0 ⊥ T 0). Hence M ⊗ N ⊥ M ⊗ T ⊥ S ⊗ N ⊥ S ⊗ T =∼ M 0 ⊗ N 0 ⊥ M 0 ⊗ T 0 ⊥ S0 ⊗ N 0 ⊥ S0 ⊗ T 0. Now from Lemma 5.8 we infer that all the tensor products except possibly M ⊗ N and M 0 ⊗ N 0 are metabolic, hence their corresponding direct sums are metabolic by Lemma 5.7, and so M ⊗ N ∼ M 0 ⊗ N 0, by definition. ¤ 16 KAZIMIERZ SZYMICZEK

We write W (R) for the collection of all similarity classes (called also Witt classes) of inner product spaces over the commutative ring R. The class containing the space (M, β) is written hM, βi or simply hMi. We deﬁne two operations in W (R), the addition and the multiplication of classes by setting hMi + hNi = hM ⊥ Ni and hMi · hNi = hM ⊗ Ni. According to Lemma 6.4, these operations do not depend on the choice of representatives of the similarity classes. But we also have to check that the sum and the product of similarity classes of two inner product spaces are the similarity classes of inner product spaces. This follows from Theorem 3.5 for addition, and from Theorem 4.7 for the multiplication operation. Thus we have well deﬁned addition and multiplication operations in W (R). Theorem 6.5. The collection W (R) of all Witt classes of symmetric inner product spaces over R forms a commutative ring with 1, using the orthogonal sum as addition operation and the tensor product as multiplication operation. Proof. Commutativity and associativity of both operations and the distributivity of multiplication with respect to addition follow easily from the standard properties of direct sums and tensor products of modules. The metabolic spaces form a similarity class which will be denoted h0i. Clearly, hMi + h0i = hMi for any class hMi. Moreover, from Lemma 5.9 we get hM, βi + hM, −βi = h0i ∈ W (R). This proves that W (R) is an additive abelian group. As to multiplicative properties it remains to show that there is a unit element in W (R). We consider the class h1i of the free rank 1 inner product space U = Ru, where the bilinear form α on U is given by α(au, bu) = ab. Then U ⊗ M =∼ M for any inner product space M, and so h1i · hMi = hMi for any class hMi ∈ W (R). This proves the theorem. ¤

Witt ring via Grothendieck ring. There is another approach to Witt rings which has less geometric and more algebraic flavor. The construction starts with the semiring S(R) of the equivalence classes of stable isometry relation and uses the construction of the Grothendieck ring G(R) for that semiring. A suitable factor ring of the Grothendieck ring G(R) is then the Witt ring of R. We proceed to explaining the construction in some detail. We adopt the notion of the Grothendieck ring from [11, pp. 30–31]. According to that approach we start from a commutative semigroup R and define the following equivalence relation on the set of pairs of elements of R: (a, b) ≈ (a0, b0) ⇐⇒ a + b0 + c = a0 + b + c for some element c ∈ R. The equivalence class of the pair (a, b) is written [a, b] and the addition of classes is defined by the formula [a, b] + [a0, b0] = [a + a0, b + b0]. This defines a group structure on the set G of equivalence classes with the zero element [0, 0] (which equals [a, a] for any a ∈ R), and with the additive inverse of [a, b] being [b, a]. The map ı : R → G, a 7→ [a, 0] is a semigroup homomorphism. It is injective, if R is a cancellative semigroup (cancellation law holds for addition). In that case we can identify the classes [a, 0] ∈ WITT RING OF A COMMUTATIVE RING 17

G with the elements a ∈ R and view R as a subset of G. Then any element of G can be written as [a, b] = [a, 0] + [0, b] = [a, 0] − [b, 0] = a − b. Thus when R is cancellative semigroup, any element of its Grothendieck group can be written as the difference of two elements of R. To round off the construction it remains to consider the case when R has the structure of a semiring, that is, on the additive semigroup R there is another operation, the multiplication, which is commutative, associative and distributive over the addition. Then we can define the multiplication operation in G, the Grothendieck group of the additive semigroup R, by setting [a, b] · [a0, b0] = [aa0 + bb0, ab0 + ba0]. This makes the group G into a commutative ring called the Grothendieck ring of the semiring R. Now we apply the above construction in the case when R is the semiring of stable isometry classes of inner product spaces over a ring. We recall that two inner product spaces M and N are stably isometric when there is another inner product space P so that there is an isometry M ⊥ P =∼ N ⊥ P. Stable isometry is an equivalence relation. Write [(M, β)] or simply [M] for the stable isometry class of the inner product space (M, β). The set of all stable isometry classes [M] of inner product spaces over the commutative ring R will be denoted S(R). This set can be equipped with two operations, the addition and multiplication by the rules [M] + [N] = [M ⊥ N] and [M] · [N] = [M ⊗ N]. Checking that these definitions are independent of the choice of representatives in the classes is straightforward. It is also easy to check that (S(R), +, · ) is a commutative semi-ring. This means that it satisfies all the ring axioms except for the existence of additive inverses. The zero element of the semiring is the isometry class [0] of the zero inner product space (the module is the zero module over R and the bilinear form is the zero form) and the multiplicative identity is the class [1] of the free inner product space (R, · ) of rank one, where the bilinear form on R is just the ring multiplication in R. In the Grothendieck group construction the pairs ([M], [N]) and ([M 0], [N 0]) are equivalent when there is an inner product space P so that M ⊥ N 0 ⊥ P =∼ M 0 ⊥ N ⊥ P. It is important to notice that for stable isometry the cancellation law holds. For suppose M ⊥ P and N ⊥ P are stably isometric. Then there is a space Q and an isometry (M ⊥ P ) ⊥ Q =∼ (N ⊥ P ) ⊥ Q. But then we have the isometry M ⊥ (P ⊥ Q) =∼ N ⊥ (P ⊥ Q) showing that M and N are stably isometric. It follows that the semiring S(R) of stable isometry classes of inner product spaces over the ring R imbeds injectively into its Grothendieck ring denoted G(R). 18 KAZIMIERZ SZYMICZEK

Hence the elements of the Grothendieck ring G(R) can be viewed as diﬀerences [M]−[N] of equivalence classes of spaces stably isometric to M and N. The equality relation in G(R) is described as follows:

[M] − [N] = [M 0] − [N 0] ⇐⇒ M ⊥ N 0 ⊥ P =∼ M 0 ⊥ N ⊥ P for some inner product space P . Addition and multiplication in G(R) are expressed in terms of orthogonal sums and tensor products as follows

([M] − [N]) + ([M 0] − [N 0]) = [M ⊥ M 0] − [N ⊥ N 0],

([M] − [N]) · ([M 0] − [N 0]) = [(M ⊗ M 0) ⊥ (N ⊗ N 0)] − [(M ⊗ N 0) ⊥ (N ⊗ M 0)]. Now in the Grothendieck ring G(R) we consider the ideal M(R) generated by the set of all classes [S], where S is a metabolic space. The factor ring G(R)/M(R) is said to be the Witt ring of the commutative ring R. Thus we have two candidates for the Witt ring and we now show that the new one is just the former in a disguise.

Theorem 6.6. For any commutative ring R the following map

G(R) → W (R), [M] − [N] 7→ hMi − hNi is a surjective ring homomorphism with the kernel M(R). Hence we have the isomorphism of rings G(R)/M(R) =∼ W (R). Proof. Clearly the map is well deﬁned on the elements of G(R) (does not depend on the representation of the element as the diﬀerence of two stable isometry classes). Further, easy calculation shows that the map is a ring homomorphism. Surjectivity does not present any problem. So the main thing is to determine the kernel. This is very simple. On the one hand, M(R) is contained in the kernel, since it is generated as an ideal by the classes [S], where S is a metabolic space, and [S] 7→ hSi = 0 ∈ W (R). On the other hand, given any element [M] − [N] in the kernel, we have hMi = hNi. Hence, there are metabolic spaces S and T so that M⊥S =∼ N⊥T , that is [M] + [S] = [N] + [T ]. Then we have

[M] − [N] = [T ] − [S] ∈ M(R), as required. ¤

Explicit determination of the Witt ring W (R) for a given ring R which is not a ﬁeld is known only for a few rings R. These include the ring of rational integers, where W (Z) =∼ Z (see [7, p. 90]), and the ring of polynomials K[X] over a ﬁeld K of characteristic not 2, where W (K[X]) =∼ W (K) (see [11, p. 211]). For determination of the additive group of the Witt ring W (R) for several rings R including R[X,Y ]/(X2 + Y 2 − 1) see [3, pp. 495–499]. WITT RING OF A COMMUTATIVE RING 19

7. Change of rings Extending the module. Let f : R → R0 be a ring homomorphism. Then R0 can be viewed as an R−module with multiplication by scalars deﬁned as follows R × R0 → R0, (a, a0) 7→ f(a) · a0. Let M be a module over the ring R. We consider the tensor product of R−modules 0 0 R ⊗R M =: M . Observe that M 0 can be viewed as R0−module with multiplication by scalars from R0 deﬁned on simple tensors by (7.1) a0 · (b0 ⊗ m) = a0b0 ⊗ m. Here is a formal proof that such multiplication of simple tensors extends onto the whole product R0 ⊗ M. We consider the 3-linear map α : R0 × R0 × M −→ R0 ⊗ M, (a0, b0, m) 7→ a0b0 ⊗ m. By the universal property of the tensor product there is a unique R−linear map h : R0 ⊗ (R0 ⊗ M) → R0 ⊗ M such that h(a0 ⊗ (b0 ⊗ m)) = a0b0 ⊗ m. This map h composed with the bilinear map ⊗ : R0 × (R0 ⊗ M) → R0 ⊗ (R0 ⊗ M) gives the R−bilinear map µ : R0 × (R0 ⊗ M) → R0 ⊗ M, µ(a0, (b0 ⊗ m)) = a0b0 ⊗ m. All the relevant maps are visualized in the following commutative diagram.

R0 ⊗ (R0 ⊗ M) * Y ⊗ ⊗ R0 × R0 × M h R0 × (R0 ⊗ M)

α µ j ? R0 ⊗ M

It is now easy to check that the map µ makes M 0 = R0 ⊗ M into an R0−module. Each element of M 0 can be written as the ﬁnite sum of simple tensors X 0 0 0 ai ⊗ mi, ai ∈ R , mi ∈ M. When M 0 is viewed as R0−module this can be rewritten as X 0 0 0 ai(1 ⊗ mi), ai ∈ R , mi ∈ M. Hence M 0 is generated as an R0−module by the set {1 ⊗ m : m ∈ M}. The following lemma shows that under certain circumstances we can identify M with a submodule of M 0. For an application see Lemma 7.8. Lemma 7.1. Let M be an R−module, and let f : R → R0 be a ring homomorphism. Suppose the following condition is satisﬁed: ∀ m ∈ M, m 6= 0 ∃ ϕ ∈ M ∗ f(ϕ(m)) 6= 0. 20 KAZIMIERZ SZYMICZEK

Then the R−linear map

0 τ : M −→ R ⊗R M, m 7→ 1 ⊗ m is injective.

Proof. We must show that the kernel of the linear map τ is zero. For this it suﬃces to show that if 0 6= m ∈ M, then 0 6= 1 ⊗ m ∈ R0 ⊗ M. So we ﬁx m ∈ M, m 6= 0. From the assumption we know that there is a linear functional ϕ ∈ M ∗ such that f(ϕ(m)) 6= 0. We consider the R−bilinear map R0 × M −→ R0, (a0, m) 7→ a0f(ϕ(m)). For this there is a unique R−linear map ϕ0 : R0 ⊗ M → R0 so that ϕ0(a0 ⊗ m) = a0f(ϕ(m)). It follows that ϕ0(1 ⊗ m) = f(ϕ(m)) 6= 0. Hence also 1 ⊗ m 6= 0. ¤

In the lemma below we use the notation of the above proof. Thus for any linear functional ϕ : M → R we write ϕ0 for the R−linear map M 0 → R0 deﬁned by the rule ϕ0(a0 ⊗ m) = a0f(ϕ(m)) for all m ∈ M. Observe that ϕ0 is obviously R0−linear and so ϕ0 ∈ M 0∗ and that ϕ0 makes the following diagram commute: ϕ M - R

τ f

? ϕ0 ? M 0 - R0

Lemma 7.2. Let M be a ﬁnitely generated projective R−module, and let

∗ {y1, . . . , yn} ⊂ M and {ϕ1, . . . , ϕn} ⊂ M be dual bases for M and M ∗. Then

0 0 0 0∗ {1 ⊗ y1,..., 1 ⊗ yn} ⊂ M and {ϕ1, . . . , ϕn} ⊂ M are dual bases for M 0 and M 0∗. Hence, if M is a finitely generated projective R−module, then M 0 is a finitely generated projective R0−module. P Proof. Take any m ∈ M. Then m = ϕi(m)yi. Hence X X X 0 1 ⊗ m = 1 ⊗ ϕi(m)yi = fϕi(m)(1 ⊗ yi) = ϕi(1 ⊗ m)(1 ⊗ yi). Thus any generator 1 ⊗ m of the R0−module M 0 has the required representation. By linearity this holds for all elements of M 0 and so according to Lemma 2.4 this finishes the proof. ¤ WITT RING OF A COMMUTATIVE RING 21

Extending bilinear forms. Now suppose M has the structure of a bilinear module with the bilinear form β : M × M → R. We want to extend the R−bilinear form β on M to an R0−bilinear form β0 : M 0 × M 0 −→ R0 satisfying

β0(a0 ⊗ m, b0 ⊗ n) = a0b0f(β(m, n)), a0, b0 ∈ R0, m, n, ∈ M.

This can be done by using again the universal property of the tensor product. We consider the 4-linear map

α : R0 × M × R0 × M −→ R0, (a0, m, b0, n) 7→ a0b0f(β(m, n)).

By the universal property of the tensor product there is a unique R−linear map h : (R0⊗M)⊗(R0⊗M) → R0 such that h((a0⊗m)⊗(b0⊗n)) = a0b0f(β(m, n)). This map h composed with the bilinear map ⊗ :(R0 ⊗M)×(R0 ⊗M) → (R0 ⊗M)⊗(R0 ⊗M) gives the R−bilinear map β0 :(R0 ⊗ M) × (R0 ⊗ M) → R0 with the property that β0(a0 ⊗ m, b0 ⊗ n) = a0b0f(β(m, n)) for all a0, b0 ∈ R0, m, n ∈ M. The maps form the following commutative diagram.

(R0 ⊗ M) ⊗ (R0 ⊗ M) * Y ⊗ ⊗ R0 × M × R0 × M h (R0 ⊗ M) × (R0 ⊗ M)

α β0 j ? R0

(M 0, β0) is said to be the bilinear R0−module obtained from the bilinear R−module (M, β) by change of rings from R to R0 (or by extension of scalars from R to R0). 0 Since the ring homomorphism f : R → R plays a role, we also write f#(M, β) for (M 0, β0).

0 0 Lemma 7.3. If (M, β) is an inner product space over R, then f#(M, β) = (M , β ) is an inner product space over R0.

Proof. By Lemma 7.2 we know that M 0 is a ﬁnitely generated projective R0−module. By Lemma 2.6 it suﬃces to show that the adjoint homomorphism βb0 : M 0 → M 0∗ is surjective. First we take dual bases

∗ {y1, . . . , yn} ⊂ M and {ϕ1, . . . , ϕn} ⊂ M

∗ b for M and M . Since β is assumed to be an isomorphism, there are xi ∈ M so that b b0 ϕi = β(xi), i = 1, . . . , n. To prove the surjectivity of the adjoint homomorphism β 22 KAZIMIERZ SZYMICZEK take any ψ ∈ M 0∗. Then for any generator 1 ⊗ m of M 0 we have X X ψ(1 ⊗ m) = ψ(1 ⊗ ϕi(m)yi) = ψ(1 ⊗ ϕi(m)yi) X X b = fϕi(m)ψ(1 ⊗ yi) = fβ(xi)(m) · ψ(1 ⊗ yi) X = fβ(xi, m) · ψ(1 ⊗ yi) X 0 = β (1 ⊗ xi, 1 ⊗ m) · ψ(1 ⊗ yi) X 0 = β (ψ(1 ⊗ yi) ⊗ xi, 1 ⊗ m) X 0¡ ¢ = β ψ(1 ⊗ yi) ⊗ xi, 1 ⊗ m) ¡X ¢ b0 = β ψ(1 ⊗ yi) ⊗ xi (1 ⊗ m) ¡P ¢ b0 By linearity argument it follows that ψ = β ψ(1 ⊗ yi) ⊗ xi . This shows that βb0 is surjective, as desired. ¤ Lemma 7.4. For two bilinear R−modules (M, β) and (N, γ), ∼ ∼ f#(M ⊥ N) = f#(M) ⊥ f#(N) and f#(M ⊗ N) = f#(M) ⊗ f#(N). Proof. ¤

0 0 Lemma 7.5. If (M, β) is a metabolic space over R, then f#(M, β) = (M , β ) is a metabolic space over R0. Proof. ¤ Lemma 7.6. For two bilinear R−modules (M, β) and (N, γ),

M ∼ N ⇒ f#(M) ∼ f#(N). Proof. ¤ Theorem 7.7. Any ring homomorphism f : R → R0 induces a ring homomorphism 0 of Witt rings W (R) → W (R ) by sending the similarity class hMi onto hf#(M)i. Proof. By Lemmas 7.3 and 7.6 we have a well deﬁned map 0 f\ : W (R) → W (R ), f\hMi = hf#(M)i.

By Lemma 7.4 the map preserves the ring operations. Hence f\ is a ring homomorphism. ¤ Extending Witt rings. A natural question arises to find the kernel of the induced homomorphism of Witt rings. When R is a domain (i.e. a commutative ring without zero divisors) then it has its field of fractions K and we have the natural imbedding R,→ K. Hence there is a homomorphism of the Witt ring W (R) into the Witt ring W (K) of the field K. In some important cases this homomorphism is injective, and then W (R) can be viewed as a subring of W (K). The most general situation known where this happens is the case when R is a regular domain of dimension ≤ 3 (theorem of Pardon (1984) and Ojanguren (1982), see [3, p. 472]). We explain the terminology. A ring R is said to be regular if it is noetherian and every f.g. R−module M has a finite projective resolution, that is, there is a finite exact sequence 0 → Pn → · · · → P0 → M → 0 WITT RING OF A COMMUTATIVE RING 23 where the Pi are projective modules. And the dimension (or, the Krull dimension) of the ring R is the largest length ` of a chain of prime ideals of the ring

0 ⊂ p1 ⊂ · · · ⊂ p`. For example, Dedekind ring is regular and has dimension 1 (every nonzero prime ideal is maximal). We reproduce here the proof of the Pardon-Ojanguren theorem in the special case of Dedekind rings. This has been proved much earlier by Knebusch. Before entering the proof we state the following lemma which is a special case of Lemma 7.1. Given an R−module M, the extended module M 0 is a K−module, that is, a vector space over the ﬁeld K. We want to view M as a subset of M 0. For this we prove the following lemma. Lemma 7.8. Let M be a f.g. projective R−module, where R is a domain, and let K be the ﬁeld of fractions of R. Then the R−linear map

M −→ K ⊗R M, m 7→ 1 ⊗ m is injective. Proof. We view R as a subring of the ﬁeld K and so we consider the injective homomorphism f : R → K, where f(a) = a. From Corollary 2.5 we know that for each m ∈ M, m 6= 0, there is a linear functional ϕ : M → R such that ϕ(m) 6= 0. Thus the hypothesis in Lemma 7.1 is satisﬁed and the result follows. ¤

We use Lemma 7.8 to identify the simple tensors 1 ⊗ m ∈ K ⊗ M with elements m ∈ M. Then for any simple tensor a ⊗ m ∈ K ⊗ M we have a ⊗ m = a · (1 ⊗ m) = am, where we have used the definition (7.1) of multiplication of tensors by scalars, and the identification implied by the Lemma. Thus a simple tensor a ⊗ m = am belongs to M when a ∈ R, and all the other simple tensors are of the form am, where a ∈ K and m ∈ M. The tensor product K ⊗ M appears to be the set of all linear combinations of elements of M with coefficients in K, and elements of M are recognized as those with coefficients in R. Yet another formulation is that K ⊗ M is a vector space over the field K generated by the elements of the module M. Now if M is a bilinear module with the bilinear form β, then the extended bilinear form β0 on K ⊗ M acts on simple tensors as follows: β0(am, bn) = β0(a ⊗ m, b ⊗ n) = abβ(m, n). We shall use this simplified description of the product K ⊗ M in the proof of the following theorem. Theorem 7.9. Let R be a Dedekind domain and K its field of fractions. Then there is an exact sequence 0 → W (R) → W (K). In other words, the homomorphism W (R) → W (K), induced by the imbedding f : R,→ K, is injective. 24 KAZIMIERZ SZYMICZEK

Proof. We must show that given an inner product space (M, β) over R,

if f#(M, β) is metabolic, then (M, β) is metabolic.

So suppose f#(M, β) is metabolic and V is a direct summand of the space with ⊥ 0 V = V . Recall that f#(M, β) = (K ⊗ M, β ), and K ⊗ M is a vector space over the field K. We view M as a subset generating the vector space K ⊗ M. Hence we can set N := M ∩ V ⊂ M. We want to prove N is a direct summand of M and N = N ⊥. First observe that N ⊆ N ⊥. For N ⊂ V = V ⊥, hence for any n, n0 ∈ N we have n, n0 ∈ V and since V is totally isotropic we have β(n, n0) = β0(n, n0) = 0. Conversely, assume n ∈ N ⊥, that is, β(n, N) = 0. Take any v ∈ V . Then there exists a nonzero a ∈ R so that av ∈ M. Hence av ∈ N and so β(n, av) = 0. Thus 0 = β(n, av) = a · β0(n, v) and since a 6= 0 it follows β0(n, v) = 0. Hence n is orthogonal to V , that is, n ∈ V ⊥. Thus we finally get n ∈ M ∩ V ⊥ = M ∩ V = N. This proves that N = N ⊥. It remains to show that N is a direct summand of M. For this we observe that M/N is a finitely generated torsion free module. It is finitely generated, since it is the homomorphic image of the f.g. module M. Moreover, since N ⊂ V , there is an injective module homomorphism M/N → (K ⊗ M)/V into torsion free R−module ((K ⊗ M)/V is actually a vector space over K, hence certainly torsion free as a module over R). Hence M/N is finitely generated and torsion free module over a Dedekind domain. Such a module is known to be projective ([2, Prop.4.1, p.133]). Hence the exact sequence 0 → N → M → M/N → 0 splits, and so N is a direct summand of M. This proves the theorem. ¤

8. Products of rings We consider in this section ﬁnite direct product of rings and establish a natural result for the structure of the Witt ring of the product. So let

R = R1 × · · · × R`, ` ≥ 2, be the direct product of rings Ri. Recall that the ring operations are deﬁned componentwise. We write ei for the element which has i−th coordinate 1 and the remaining coordinates 0. These elements form a system of orthogonal idempotents, ei · ej = 0 for i 6= j, and ei · ei = ei. The key property is that the ring identity 1 ∈ R satisﬁes

(8.1) 1 = e1 + ··· + e`. From this we get the decomposition

R = Re1 ⊕ · · · ⊕ Re`. This will be interpreted in two ways. First, when we view R as R−module, then Rei is a submodule (actually, this is the principal ideal in the ring R generated by ei). These submodules serve as simplest examples of projective R−modules which are not free R−modules. Indeed they are all projective R−modules (they are direct WITT RING OF A COMMUTATIVE RING 25 summands of the free module R) but they are not free (none of them has a basis since any linear combination of elements of Rei is annihilated by any ej, j 6= i). The second interpretation takes into account the fact that each Rei is, in fact, a ring on its own, with the multiplicative identity ei. After all,

Rei = 0 × · · · × 0 × Ri × 0 × · · · × 0.

We will often identify the rings Ri and Rei. Now let M be an R−module. For each principal ideal J = Re of R the set J M = {xm ∈ M : x ∈ J , m ∈ M} = {em ∈ M : m ∈ M} is a submodule of M. Hence each ReiM = eiM is an R−submodule of M. More- over, we have the direct sum decomposition

M = e1M ⊕ · · · ⊕ e`M.

As above, there is a second interpretation for eiM, and this is to view eiM as an Ri−module. To facilitate notation, we shall write Mi for eiM. And now we add one more structure component, the bilinear form. Suppose β : M × M → R is a bilinear form. Then from (8.1) we get

β(m, n) = e1β(m, n) + ··· + e`β(m, n) for all m, n ∈ M. Here eiβ(m, n) ∈ Ri. We deﬁne

βi : Mi × Mi → Ri, βi(m, n) = eiβ(m, n).

Then βi is a bilinear form on the Ri−module Mi. Observe that when we view Mi as an R−module, then the direct sum M = M1 ⊕ · · · ⊕ M` is in fact the direct orthogonal sum. Indeed, any two distinct direct summands are pairwise orthogonal since for i 6= j we have 0 0 0 β(eim, ejm ) = eiejβ(m, m ) = 0 for m, m ∈ M. Now let f : M → N be a homomorphism of R−modules. By linearity we have f(eim) = eif(m) for all m ∈ M. Hence f(eiM) ⊆ eiN. In other words the restriction fi := f|eiM of f to the R−submodule eiM is an R−module homomorphism

fi : Mi → Ni.

Observe that fi can be viewed as a homomorphism of Ri−modules as well. After all it is additive and for any a ∈ R the element eia ∈ Ri satisﬁes

fi(eia · eim) = f(eia · eim) = eia f(eim) = eia fi(eim) for all eim ∈ Mi. Lemma 8.1. Let f : M → N be a homomorphism of R−modules. Then the restriction fi of f to eiM maps eiM into eiN and is a homomorphism of Ri−modules. Moreover, f is an isomorphism if and only if each restriction fi : eiM → eiN is an isomorphism. Proof. We have the direct sum decompositions

M = e1M ⊕ · · · ⊕ e`M,N = e1N ⊕ · · · ⊕ e`N, and any homomorphism f maps eiM into eiN. So it is clear that f is an isomorphism if and only if its restrictions fi to eiM are isomorphisms. ¤

Lemma 8.2. (M, β) is an inner product space if and only if each (Mi, βi) is an inner product space. 26 KAZIMIERZ SZYMICZEK

Proof. If (M, β) is nonsingular, then βb : M → M ∗ is an isomorphism. Hence, by b b Lemma 8.1, the restriction (β)i of β to Mi = eiM is an isomorphism between Mi ∗ ∗ ∗ and Mi = eiM . To show that (Mi, βi) is nonsingular we must verify that eiM b is the dual module of eiM and also that the isomorphism (β)i is induced by the bilinear form βi on eiM. In other words we have to check that ∗ ∗ b d eiM = (eiM) and (β)i = (βi). ∗ ∗ So take a linear functional ϕ ∈ M and the corresponding ϕi := eiϕ ∈ eiM . Then ϕ : M → R is a homomorphism and so from Lemma 8.1 we know that ϕi maps eiM ∗ to eiR = Ri. Hence ϕi ∈ (eiM) . On the other hand, given any linear functional ∗ ϕi ∈ (eiM) we can view ϕi as the restriction of a linear functional ϕ on M. We can take for ϕ the functional which is zero on all direct summand ejM except for ∗ eiM where it is equal to ϕi. Hence ϕi ∈ eiM , as required. b It remains to verify that the isomorphism (β)i is induced by the bilinear form βi on eiM. This follows from the following calculation: b 0 0 0 d 0 (β)i(eim)(eim ) = eiβ(eim, eim ) = βi(eim, eim ) = (βi)(eim)(eim ).

This proves one part of the lemma. For the other we assume that each (Mi, βi) is ∗ ∗ ∗ an inner product space. We know already that (M )i = (Mi) =: Mi , hence ∗ ∗ ∗ ∗ M = (M1 ⊕ · · · ⊕ M`) = M1 ⊕ · · · ⊕ M` . d ∗ Further, each (βi): Mi → Mi is an isomorphism, hence M M M d ∗ (βi): Mi → Mi is an isomorphism. It remains to check that this isomorphism is induced by the bilinear form β, and we leave this to the patient reader as an exercise. ¤ Lemma 8.3. (M, β) and (N, γ) are isometric bilinear modules over R if and only if (Mi, βi) and (Ni, γi) are isometric bilinear modules over Ri for each i = 1, . . . , `.

Proof. Let ϕ : M → N be an isometry. By Lemma 8.1 ϕ maps eiM onto eiN. Moreover, we have β(m, m0) = γ(ϕ(m), ϕ(m0)) for all m, m0 ∈ M. Observe that by bilinearity, 0 2 0 0 eiβ(eim, eim ) = ei · ei · β(m, m ) = eiβ(m, m ). Hence we have X X 0 0 eiβ(eim, eim ) = eiβ(m, m ) = β(m, n) = γ(ϕ(m), ϕ(m0)) X 0 = eiγ(ϕ(m), ϕ(m )) X 0 = eiγ(ϕ(eim), ϕ(eim )) It follows that 0 0 0 eiβ(eim, eim ) = eiγ(ϕ(eim), ϕ(eim )), ∀ m, m ∈ M, i = 1, . . . , `.

Thus the restriction of ϕ to Mi = eiM is an isometry between (Mi, βi) and (Ni, γi). Conversely, if ϕi : Mi → Ni are isometries, then

ϕ := ϕ1 ⊕ · · · ⊕ ϕ` : e1M ⊕ · · · ⊕ e`M → e1N ⊕ · · · ⊕ e`N is an isometry between M and N. ¤ WITT RING OF A COMMUTATIVE RING 27

Lemma 8.4. (S, β) is metabolic if and only if each (Si, βi) is metabolic. Proof. Suppose S is metabolic. Then S = N ⊕P where N = N ⊥. On the one hand we have

S = e1(N ⊕ P ) ⊕ · · · ⊕ e`(N ⊕ P )

= (e1N ⊕ e1P ) ⊕ · · · ⊕ (e`N ⊕ e`P ).

⊥ So each Si splits into direct sum eiN ⊕eiP and we want to show that eiN = (eiN) . First we have 2 eiβ(ein, ein) = ei · ei β(n, n) = eiβ(n, n) = 0, ⊥ ⊥ so that eiN ⊆ (eiN) . Second, if eim ∈ (eiN) , then

eiβ(eim, ein) = 0 ⊥ for all n ∈ N. Hence β(eim, n) = 0 for all n ∈ N, and so eim ∈ N = N. But 2 ⊥ eim ∈ N implies ei m ∈ eiN, hence eim ∈ eiN. This proves (eiN) ⊆ eiN, as required. Conversely, assume all (Si, βi) are metabolic. Then Si = Ni ⊕ Pi, where Ni = ⊥ Ni . Since S is the direct sum of Si’s, we can write

S = N ⊕ P, where N = N1 ⊕ · · · ⊕ N`,P = P1 ⊕ · · · ⊕ P`. We prove that N = N ⊥. First take n, n0 ∈ N. Then we have X X 0 0 0 β(n, n ) = eiβ(n, n ) = eiβ(ein, ein ) = 0

⊥ since eiβ(Ni,Ni) = 0. This shows that β(N,N) = 0 and so N ⊆ N . ⊥ Now take any m ∈ N . Then β(m, n) = 0 for all n ∈ N, hence also eiβ(m, n) = 0 for all n ∈ N and for all i. It follows that eiβ(eim, ein) = 0 for all ein ∈ eiN = Ni ⊥ and so eim ∈ Ni = Ni. Thus X X m = eimi ∈ Ni = N, as required. ¤

Lemma 8.5. M and N are similar inner product spaces over R if and only if Mi and Ni are similar inner product spaces over Ri for each i = 1, . . . , `. Proof. Inner product spaces M and N are similar when there are metabolic spaces S, T so that M ⊥ S =∼ N ⊥ T . Now we have ∼ ∼ M ⊥ S = N ⊥ T ⇐⇒ (M ⊥ S)i = (N ⊥ T )i ∀ i by Lemma 8.3 ∼ ⇐⇒ Mi ⊥ Si = Ni ⊥ Ti ∀ i

⇐⇒ Mi ∼ Ni ∀ i. The last conclusion follows from Lemma 8.4 since S, T are metabolic if and only if all Si,Ti are metabolic. ¤

Theorem 8.6. The map ¡ ¢ W (R) → W (R1) × · · · × W (R`), hMi 7→ hM1i,..., hM`i is an isomorphism of rings. 28 KAZIMIERZ SZYMICZEK

Proof. By Lemma 8.5 the map is well deﬁned. And it is clearly a ring homomorphism. If hMi is in the kernel of the homomorphism, then each Mi is metabolic over Ri, hence by Lemma 8.4 the space M over R is metabolic, hence hMi = 0. It follows the homomorphism is injective. It remains to establish surjectivity. Given any `−tuple of inner product spaces (M1,...,M`), where Mi is Ri−module, we consider the cartesian product

M = M1 × · · · × M` and make it naturally into an R−module. That is, given (a1, . . . , a`) ∈ R and (m1, . . . , m`) ∈ M we deﬁne the multiplication by scalars R × M → M by setting

(a1, . . . , a`) · (m1, . . . , m`) = (a1m1, . . . , a`m`). Then we deﬁne the bilinear form β on M by setting

β(m, n) := e1β1(m, n) + ··· + e`β`(m, n).

¡Then (M, β) is an¢ inner product space over R (by Lemma 8.2), and hMi maps to hM1i,..., hM`i . ¤

9. Matrices of module homomorphisms In this preparatory section we compile the fundamentals on representing module homomorphisms by matrices. This technique is successfully applied in Witt ring theory but in the existing texts like [1], [11] it is usually only sketched very briefly and relies on analogies with the well known elementary facts from linear algebra. For the orientation of the reader we mention that Baeza does not explain anything, Scharlau (cf. [11, p. 22]) says only that usual rules apply, and then (cf. [11, pp. 241–242]) gives the definitions. Also Knus gives some details in [3, pp. 12–13]. We follow the latter exposition. Assume the modules M and N are finite direct sums of the same number of submodules, M` M` M = Mi,N = Ni. i=1 i=1 L` L` We will write ρi : Mi → i=1 Mi and πi : i=1 Mi → Mi for the canonical injections and canonical projections, respectively. And for the corresponding maps 0 0 for N we write ρi : Ni → N and πi : N → Ni. It is important to notice that for i, j ∈ {1, . . . , `}, X`

πiρj = δij1Mi , and ρiπi = 1M , i=1 0 0 and similarly for the injections and projections ρi and πi. 2 Now let ϕ : M → N be a module homomorphism. We deﬁne ` maps aij as follows: 0 aij : Mj → Ni, aij = πiϕρj, i, j = 1, . . . , `. The following diagram visualizes the situation WITT RING OF A COMMUTATIVE RING 29

ρj Mj - M

? 0 πi - N Ni

The matrix [aij] of maps is said to be the matrix of the homomorphism ϕ : M → N with respect to the given direct sum decompositions of the modules M and N. With ﬁxed direct sum decompositions we will write Mat(ϕ) for the matrix [aij] of ϕ. Thus, given the direct sum decompositions of M and N, each module homomorphism ϕ : M → N determines its matrix Mat(ϕ). The signiﬁcance of the matrix comes from the fact that the matrix of ϕ determines completely the map ϕ itself. This follows from the following computation X` X` X X 0 0 0 0 0 (9.1) ϕ = 1N ◦ ϕ ◦ 1M = ρiπi ◦ ϕ ◦ ρjπj = ρiπiϕρjπj = ρiaijπj. i=1 j=1 i,j i,j

Thus given the maps aij we can recover the map ϕ itself.

Example 9.1. Consider the identity map 1M : M → M. If Mat(1M ) = [eij] is its matrix with respect to the given direct sum decomposition of M then

eij = πi1M ρj = πiρj = δij1Mi .

So the matrix of 1M is the diagonal matrix with the i−th diagonal entry equal to the identity map on Mi. Example 9.2. Consider the special case when M = N is a finite dimensional vector space over a field K and Mi = Ni = Kvi are 1−dimensional subspaces generated by the vectors vi. Since M is assumed to be the direct sum of subspaces Mi, the vectors vi form a basis for the space M. Given a vector space homomorphism ϕ : M → M we can consider the classical construction of the matrix [bij ] of ϕ with respect to the basis {v1, . . . , v`}. According to the definition,

πiϕ(vj) = bijvi.

On the other hand, ϕ has the matrix aij with respect to the given direct sum decomposition. Here aij : Kvj → Kvi is a linear map between one-dimensional subspaces so that it is determined completely by the image of vj. By deﬁnition we have aij (vj) = (πiϕρj)(vj) = πi(ϕ(vj)).

Thus aij (vj) = bijvi. This shows the relation between the two approaches to the concept of the matrix of ϕ. In the new approach aij is a linear map Kvj → Kvi and in the old approach bij is just the scalar determining the action of aij . If we identify the scalars with the multiplication-by-scalars endomorphisms, then the distinction between the matrices [aij ] and [bij] disappears. Now we analyze the matrix of a composition of two module homomorphisms. So let ϕ : M → N and ψ : N → L be two module homomorphisms. We assume that L` M and N have direct sum decompositions as above and that L = i=1 Li. We 00 00 write ρi : Li → L and πi : L → Li for the canonical injections and projections, 30 KAZIMIERZ SZYMICZEK respectively. Let Mat(ϕ) = [aij ] be the matrix of ϕ and Mat(ψ) = [bij ] be the matrix of ψ with respect to the given direct sum decompositions. Thus we have 0 00 0 aij = πiϕρj : Mj → Ni and bij = πi ψρj : Nj → Li.

Now we can also consider the matrix Mat(ψϕ) = [cij] of the composition ψϕ : M → L with respect to the given direct sum decompositions of M and L. Then we have ³X ´ 00 00 0 0 cij = πi ψϕρj = πi ψ ρkπk ϕρj k X 00 0 0 = πi ψρk ◦ πkϕρj k X = bikakj. k Thus (no surprise!) to get the matrix of ψϕ we formally multiply (row times column) the matrix of ψ by the matrix of ϕ: Mat(ψϕ) = Mat(ψ) · Mat(ϕ). The upshot of these remarks is that ϕ : M → N is an isomorphism if and only if its matrix is invertible. Indeed, if ψ : N → M is the inverse map then ψϕ = 1M and ϕψ = 1N and so we have

Mat(ψ) · Mat(ϕ) = Mat(1M ) and Mat(ϕ) · Mat(ψ) = Mat(1N ), where the Mat(1M ) and Mat(1N ) are the unit matrices (in the sense of Example 9.1). On the other hand, if the matrix Mat(ϕ) is invertible in the above sense and [bij] is the inverse matrix, then we take the homomorphism ψ determined by the matrix according to (9.1), X 0 ψ = ρibij πj, i,j −1 and this is the inverse map ϕ . For, one veriﬁes directly that if [bij] · [aij] is the unit matrix, then X X 0 0 ρibij πj ◦ ρsastπt = 1M i,j s,t and similarly for the other order of the involved matrices and maps.

10. Metabolic and hyperbolic spaces revisited We return to the discussion of metabolic spaces. Our aim is to give a new characterization of metabolic spaces and then to give a parallel characterization of hyperbolic spaces. By Definition 5.2 an inner product space (S, β) is metabolic if S = M ⊕ N and N = N ⊥. We will need the following construction. For any bilinear space (P, ρ) we consider the associated bilinear space (M(P ), µ) := (P ⊕ P ∗, µ), µ((p, f), (p0, f 0)) = f(p0) + f 0(p) + ρ(p, p0) for all p, p0 ∈ P , f, f 0 ∈ P ∗. It is fairly obvious that µ is a bilinear form. Recall that saying bilinear space (P, ρ) we assume that P is finitely generated projective module. In the proof below we will use the bi-dual module P ∗∗ and so we start with recalling that for a finitely generated projective module P the bidual module WITT RING OF A COMMUTATIVE RING 31

P ∗∗ is canonically isomorphic to P (for a proof see [5, Chapter V, Theorem 4.1]). The isomorphism is given by P → P ∗∗, p 7→ p,¯ wherep ¯(f) = f(p) for all f ∈ P ∗. It is customary to identify the mapp ¯ with p ∈ P and thus to identify P ∗∗ with P . Now we can prove the following somewhat surprising result. Lemma 10.1. For any bilinear space (P, ρ) the bilinear space (M(P ), µ) is an inner product space. Proof. We must show that the adjoint homomorphism µb : M(P ) → M(P )∗ is an isomorphism and for this we use the matrix of the homomorphism µb with respect to the direct sum decompositions M(P ) = P ⊕P ∗ and M(P )∗ = (P ⊕P ∗)∗ = P ∗ ⊕P ∗∗. First recall that for µb : P ⊕ P ∗ → (P ⊕ P ∗)∗ = P ∗ ⊕ P ∗∗ and for (p, f) ∈ P ⊕ P ∗ the map µb(p, f) act as follows ¡ ¢ µb(p, f)(p0, f 0) = µ (p, f), (p0, f 0) = f(p0)+f 0(p)+ρ(p, p0) = f(p0)+ρb(p)(p0)+p(f 0). Here we have written p(f 0) instead of f 0(p) as we identify P ∗∗ with P . Hence ¡ ¢ (10.1) µb(p, f) = f( ) + ρb(p) ⊕ p() ∈ P ∗ ⊕ P ∗∗.

Now let Mat(µb) = [aij ] be the matrix of µb with respect to the given direct sum decompositions. To determine a11 we recall that it is the map

∗ µb ∗ ∗∗ ∗ a11 : P → P ⊕ P −→ P ⊕ P → P , where the ﬁrst arrow is the canonical injection and the last is he canonical projection. Hence, in view of (10.1), we have

a11 : p 7→ (p, 0) 7→ ρb(p) + p() 7→ ρb(p).

Thus a11 = ρb. Similarly we determine the other matrix elements:

∗ ∗ µb ∗ ∗∗ ∗∗ a22 : P → P ⊕ P −→ P ⊕ P → P , f 7→ (0, f) 7→ f 7→ 0, hence a22 = 0 is the zero map. Further,

∗ ∗ µb ∗ ∗∗ ∗ a12 : P → P ⊕ P −→ P ⊕ P → P , f 7→ (0, f) 7→ f 7→ f,

∗ µb ∗ ∗∗ ∗∗ a21 : P → P ⊕ P −→ P ⊕ P → P , p 7→ (p, 0) 7→ ρb(p) + p() 7→ p.

Hence a12 = 1P ∗ and a21 = 1P . Thus we have · ¸ ρb 1 ∗ Mat(µb) = P . 1P 0 To prove that µb is an isomorphism we must show that this matrix is invertible. This is easy since we can point out explicitly the inverse matrix: · ¸ 0 1 Mat(µb)−1 = P . 1P ∗ −ρb ¤ Lemma 10.2. For any bilinear space (P, ρ) the inner product space (M(P ), µ) is metabolic. 32 KAZIMIERZ SZYMICZEK

Proof. (M(P ), µ) is an inner product space by Lemma 10.1. To show that M(P ) = P ⊕ P ∗ is metabolic it suffices to show that P ∗⊥ = P ∗. On the one hand µ((0, f), (0, f 0)) = f(0) + f 0(0) + β(0, 0) = 0 for all f, f 0 ∈ P ∗ and so P ∗ ⊆ P ∗⊥. On the other hand if (p, f) ∈ P ∗⊥, then µ((p, f), (0, f 0)) = f(0) + f 0(p) + ρ(p, 0) = f 0(p) = 0 for all f 0 ∈ P ∗. By Corollary 2.5 we have p = 0 and so (p, f) = (0, f) ∈ P ∗, as desired. ¤ Now we are ready to give a new characterization of metabolic spaces. Proposition 10.3. An inner product space (S, β) is metabolic if and only if there exists a bilinear space (P, ρ) so that (S, β) is isometric to (M(P ), µ). Proof. Suppose first that S = M ⊕ N and N = N ⊥. We will show that one can take P = M with the restriction of β to M as the bilinear form on M. We know from Proposition 5.1 that the isomorphism βb sends the module N onto M ∗, hence these two modules are isomorphic. We will show that S and M(M) = M ⊕ M ∗ are isometric. We have the isomorphism M ⊕ N → M ⊕ M ∗, (m, n) 7→ ((m, 0), βb(0, n)) and this is an isometry since β((m, n), (m0, n0)) = β((m, 0), (0, n0)) + β((0, n), (m0, 0)) + β((m, 0), (m0, 0)) = βb(0, n0)(m, 0) + βb(0, n)(m0, 0) + β((m, 0), (m0, 0)) ³¡ ¢ ¡ ¢´ = µ (m, 0), βb(0, n) , (m0, 0), βb(0, n0) . The converse follows from Lemma 10.2. ¤ In view of Proposition 10.3 some authors define metabolic spaces as the ones isometric with M(P ) for some bilinear module P thus avoiding the geometric approach we have adopted in these notes. We now suplement the Proposition 10.3 by proving the parallel characterization of hyperbolic spaces. For this we will consider a special type of metabolic spaces M(P ) with the finitely generated projective module P equipped with the zero bilinear form ρ = 0. In that case we shall write H(P ) instead of M(P ). Thus (H(P ), µ) := (P ⊕ P ∗, µ), µ((p, f), (p0, f 0)) = f(p0) + f 0(p) for all p, p0 ∈ P , f, f 0 ∈ P ∗. By Lemma 10.1 the bilinear space (H(P ), µ) is an inner product space. Lemma 10.4. For any finitely generated projective module P the inner product space (H(P ), µ) is hyperbolic. Proof. We verify the condition (c) of Definition 5.5 for H(P ). We shall use (10.1) to determine the images of P and P ∗ under µb. From (10.1) we infer that µb : P ⊕P ∗ → P ∗ ⊕ P ∗∗ acts as follows: µb(P ) = P ∗∗ and µb(P ∗) = P ∗. Moreover, P ∗ is totally isotropic subspace of H(P ) (that is, P ∗⊥ = P ∗) as shown in the proof of Lemma 10.1. Thus identifying P ∗∗ and P we have checked all the requirements of the condition (c) of Definition 5.5. ¤ WITT RING OF A COMMUTATIVE RING 33

Proposition 10.5. An inner product space (S, β) is hyperbolic if and only if there exists a ﬁnitely generated projective module P so that (S, β) is isometric to (H(P ), µ). Proof. Suppose S is hyperbolic. Then it is metabolic, and according to the proof of Proposition 10.3, if S = M ⊕ N and N = N ⊥, then S is isometric to M(M) = M ⊕ M ∗. Since S is hyperbolic we can assume that M and N satisfy condition (d) of the Deﬁnition 5.5, that is, we can assume that β(N,N) = 0 and β(M,M) = 0. Thus µ((m, f), (m0, f 0)) = f(m0) + f 0(m) + β(m, m0) = f(m) + f 0(m0) for all m, m0 ∈ M and f, f 0 ∈ M ∗. It follows that the metabolic space M(M) actually equals H(M). Thus S is isometric to the hyperbolic space H(M), as desired. The converse follows from Lemma 10.4. ¤

11. Matrices of bilinear spaces In this section we introduce the matrix of a bilinear space with respect to a direct sum decomposition of the underlying projective module. This will be used in the next section to prove an important result on metabolic and hyperbolic spaces known as Knebusch’s Lemma. We consider a bilinear space (M, β) and fix a direct sum decomposition for the finitely generated projective module M, M` M = Mi. i=1 We consider the adjoint homomorphism βb : M → M ∗. Since we are going to consider the matrix of βb we need a direct sum decomposition of the dual module M ∗. From Lemma 3.1 we know that M` ∗ ∗ M = Mi . i=1 We define the matrix of the bilinear space (M, β) with respect to these decompositions of M and M ∗ to be the matrix Mat(βb) of the adjoint homomorphism βb with respect to the given direct sum decompositions. Since by definition (M, β) is an inner product space if and only if the adjoint homomorphism βb is an isomorphism, we conclude that a bilinear space (M, β) is an inner product space if and only if its matrix (relative to a direct sum decomposition of M) is invertible.

Example 11.1. Let M be a vector space over the ﬁeld K and let {v1, . . . , v`} be a basis for the space. So we have M` M` ∗ ∗ M = Kvi and M = (Kvi) . i=1 i=1 b For a bilinear form β on M the matrix Mat(β) = [aij] consists of maps b β ∗ ∗ aij : Kvj → M −→ M → (Kvi) and these act on basis vectors as follows: b b aij : vj 7→ vj 7→ β(vj) 7→ β(vj)|Kvi . 34 KAZIMIERZ SZYMICZEK

b In the last step the action of the functional β(vj) on the subspace Kvi is completely determined by the image of the basis element vi, and this equals b β(vj)(vi) = β(vj, vi). Thus up to the identification of scalar endomorphisms with the scalars, the matrix [aij] is the matrix of the bilinear vector space (M, β) with respect to the basis {v1, . . . , v`} as defined in the classical case of bilinear spaces over fields. We want to find a necessary and sufficient condition for the isometry of two bilinear spaces in terms of the matrices of the spaces. For this we need the concept of the dual homomorphism. So given a module homomorphism ϕ : M → N we define the dual morphism ϕ∗ by setting ϕ∗ : N ∗ → M ∗, ϕ∗(g) = g ◦ ϕ. It is easy to see that ϕ∗ is a module homomorphism. Moreover, ϕ∗ is an isomorphism if and only if ϕ is an isomorphism. For, if ϕ is isomorphism, then for each f ∈ M ∗ we have ϕ∗(f ◦ ϕ−1) = f so that ϕ∗ is surjective. As to injectivity, if ϕ∗(g) = 0 ∈ M ∗, then g ◦ ϕ = 0 and since ϕ is invertible it follows that g = 0. The converse statement is proved similarly. Another property we shall need says that for module homomorphisms ϕ : M → N and ψ : N → L the dual homomorphisms ϕ∗ : N ∗ → M ∗ and ψ∗ : L∗ → N ∗ satisfy the relation ϕ∗ ◦ ψ∗ = (ψ ◦ ϕ)∗. This follows from the computation (ϕ∗ ◦ ψ∗)(λ) = ϕ∗(ψ∗(λ)) = ϕ∗(λψ) = λψ ◦ ϕ = λ ◦ ψϕ = (ψϕ)∗(λ) for all λ ∈ L∗. Proposition 11.2. Let (M, β) and (N, γ) be bilinear spaces and let ϕ : M → N be a module isomorphism. A necessary and sufficient condition for ϕ to be an isometry is that the following equation holds: βb = ϕ∗ ◦ γb ◦ ϕ. Proof. Suppose ϕ is an isometry. Then for all m, m0 ∈ M we have β(m, m0) = γ(ϕ(m), ϕ(m0)) and hence βb(m)(m0) = β(m, m0) = γ(ϕ(m), ϕ(m0)) = γb(ϕ(m))(ϕ(m0) ¡ ¢ = γb(ϕ(m)) ◦ ϕ (m0) = ϕ∗(γb(ϕ(m))(m0) ¡ ¢ = ϕ∗ ◦ γb ◦ ϕ (m)(m0).

This proves that βb = ϕ∗ ◦ γb ◦ ϕ. The argument reverses and so we are done. ¤ WITT RING OF A COMMUTATIVE RING 35

In order to interpret the condition βb = ϕ∗ ◦ γb ◦ ϕ in terms of matrices we must recall the relation between the matrices of ϕ and ϕ∗. Given the canonical injections and projections ρj πj Mj −→ M −→ Mj we take the duals and get the sequence π∗ ρ∗ ∗ j ∗ j ∗ Mj −→ M −→ Mj . ∗ ∗ It is routine to check that now πj is the canonical injection and ρj is the canonical projection. Hence, if [aij] is the matrix of ϕ, then from the formula (9.1) X 0 ϕ = ρiaijπj i,j we get X ∗ ∗ ∗ 0 ∗ ϕ = πj aijρi i,j and so according to the pattern of (9.1) we read oﬀ the (ji)−entry bji of the matrix of ϕ∗ as follows: ∗ bji = aij . ∗ Thus the matrix Mat(ϕ ) = [bji] is obtained from the matrix Mat(ϕ) = [aij ] by taking the duals of all entries and then transposing the matrix. Put together with Proposition 11.2 this gives the following result. Proposition 11.3. Let (M, β) and (N, γ) be bilinear spaces and let A and B be matrices of the spaces with respect to some decompositions of the spaces into direct sums of the same number of direct summands. The spaces M and N are isometric if and only if there exists an invertible matrix C so that A = (C∗)tBC. We shall need one more fact on the matrices of bilinear spaces. Proposition 11.4. Let (M, β) and (N, γ) be bilinear spaces and let A and B be matrices of the spaces with respect to some decompositions of the spaces into direct sums of the same number of direct summands. Then · ¸ A 0 Mat((M, β) ⊥ (N, γ)) = . 0 B Proof. This is left as an exercise. ¤

12. Knebusch’s lemma Now we are ready to prove the following sophisticated result. Theorem 12.1 (Knebusch’s Lemma). Let (S, β) be a metabolic space and let S = M ⊕ N with N = N ⊥. Then there is an isometry (S, β) ⊥ (S, −β) =∼ H(N) ⊥ (S, −β). Proof. By Proposition 10.3 it is sufficient to establish the following isometry (M(M), µ) ⊥ (M(M), −µ) =∼ H(N) ⊥ (M(M), −µ), and by Proposition 11.3 this can be done by establishing a suitable matrix identity. So we must write down explicitly the matrices of the involved bilinear spaces. The 36 KAZIMIERZ SZYMICZEK matrix of the metabolic space (M(M), µ) has been determined in the proof of Lemma 10.2. In our present notation · ¸ βb 1 ∗ Mat(M(M), µ) = Mat(βb) = M . 1M 0 Hence also · ¸ −βb −1 ∗ Mat(M(M), −µ) = Mat(−βb) = M . −1M 0 On the other hand · ¸ 0 1 ∗ Mat(H(N)) = Mat(b0) = N . 1N 0 In order to make the computations easier we simplify notation and write 1 for any identity map, and also we write β instead of βb. With these simplifications the space (M(M), µ) ⊥ (M(M), −µ) has the matrix   β 1 0 0 1 0 0 0  B =   , 0 0 −β −1 0 0 −1 0 and the space H(N) ⊥ (M(M), −µ) has the matrix   0 1 0 0 1 0 0 0  A =   . 0 0 −β −1 0 0 −1 0 Now we construct the invertible matrix   0 1 0 0 1 0 β 1 C =   0 1 1 0 0 0 0 1 and by a direct computation check that A = (C∗)tBC. According to the Proposition 11.3 this finishes the proof. ¤ While the metabolic space (S, β) is not in general hyperbolic, it is, according to the Knebusch’s Lemma, stably isometric to the hyperbolic space H(N). This has some interesting consequences in the Witt ring theory. Recall that one of the (equivalent) approaches to the Witt ring constructs the Witt ring W (R) of a ring R as the factor ring G(R)/M(R) of the Grothendieck ring G(R) modulo the ideal M(R) generated by the set of all classes [S], where S is a metabolic space. Here [S] denotes the class of inner product spaces stably isometric to S. Now the Knebusch’s Lemma says that each metabolic space is stably isometric to a hyperbolic space. Hence the classes of stable isometry of metabolic and hyperbolic spaces coincide. In other words, if we write H(R) for the ideal of the Grothendieck ring G(R) generated by the set of all classes [H], where H is a hyperbolic space, then M(R) = H(R) and so W (R) = G(R)/M(R) = G(R)/H(R). WITT RING OF A COMMUTATIVE RING 37

Thus when we construct the Witt ring via Grothendieck ring the metabolic spaces can be replaced with hyperbolic spaces with no difference at all to the final product, the Witt ring. On the other hand, when we view the Witt ring as the ring of similarity classes of inner product spaces with operations induced by the orthogonal sum and tensor product of spaces, then the role of metabolic spaces is irreplacable. The class of hyperbolic spaces is usually narrower than the class of metabolic spaces and so replacing in the definition of the similarity relation metabolic spaces by hyperbolic spaces we get another splitting into equivalence classes. While this seems to be nothing wrong since we still have the possibility of defining addition and multiplication of the classes in the usual way, there is a difficulty we cannot overcome. The point is that there is no way to get the additive inverses for the classes of hyperbolic-similar spaces. The natural candidate hM, −βi for the negative of hM, βi, which works all right in the metabolic-similarity context, here fails to work since, as we know, (M, β) ⊥ (M, −β) is metabolic but not necessarily hyperbolic space.

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Instytut Matematyki, Uniwersytet Sla¸ski,´ Katowice, Poland E-mail address: [email protected]