ALGEBRA - LECTURE V 1. Bilinear forms Let R be a commutative ring with 1, and M and N two R-modules. A map T : M → N is a homomorphism of R-modules if (i) T (v + u) = T (v) + T (u) for any two u, v in M. (ii) T (rv) = rT (v) for any r in R and v in M. Of course, if R is a field than this is simply a definition of a linear transformation. The set of all such homomorphism is denoted by HomR(M, N). This set is an R-module itself, under the addition (T1 + T2)(v) = T1(v) + T2(v) and scalar multiplication (rT )(v) = r(T (v)). For m n example, if M = R and N = R , then HomR(M, N) is simply the set of all n × m matrices with coefficients in R. Let M, N and L be three R-modules. A bilinear form on M × N with values in L is a map B : M × N → L which is linear in each variable, that is, (i) B(v + u, w) = B(v, w) + B(u, w) for any v, u in M and w in N. (ii) B(v, u + w) = B(v, u) + B(v, w) for any v in M and u, w in N. (iii) B(rv, u) = rB(v, u) = B(v, ru) for any v in M, u in N and r in R. The set of all such bilinear forms on is denoted by BilR(M × N, L). It is an R module with respect to the obvious operations. m n Assume that M = R with a basis e1, . , em and N = R with a basis f1, . , fn. Then we can write any v in M as v = x1e1 + ··· + xmem and any w in N as w = y1f1 + ··· ynfn. Thus, we can identify v with an m × 1 matrix x and w with an n × 1 matrix y. If B is a bilinear form, then the axioms imply that X T B(v, w) = xiyjB(ei, fj) = x Ay i,j where A is an n × m matrix with entries B(ei, fj). Thus we can identify BilR(M × N, L) with Mn×m(L), the set of n × m matrices with coefficients in L. If M = N = Rn then A is a square matrix. In this case B is said to be symmetric if B(v, w) = B(w, v). This is equivalent to A = AT . 2. Tensor products In this section we shall define the tensor product M ⊗R N of two R-modules M and N. This operation closely related to bilinear forms on M × N. The tensor product of M and N is defined as follows. Let F be the free R-module generated by all pairs (v, w) where v is in M and w in N. (In other words, F is a free module with basis given by elements in M × N.) Let I ⊆ F be a submodule of F generated by elements (i) (v + u, w) − (v, w) − (u, w) for any v, u in M and w in N, (ii) (v, u + w) − (v, u) − (v, w) for any v in M and u, w in N, 1 2 ALGEBRA - LECTURE V (iii) (rv, u) − r(v, u) and (v, ru) − r(v, u) for any v in M, u in N and r in R. Then M ⊗R N is defined to be the quotient F/I. The image of a pair (v, w) in M ⊗R N is denoted by v ⊗ w. These elements are called pure tensors and they generate the tensor product. Due to the relations contained in I we have relations between pure tensors such as (v + u) ⊗ w = v ⊗ u + u ⊗ w and (rv) ⊗ v = r(v ⊗ u) = v ⊗ (ru) in M ⊗R N. It is important to notice that the identity element for addition in M ⊗R N is equal to the pure tensor 0 ⊗ 0. In order to see this we need to show that the pair (0M , 0N ) ∈ F is contained in the submodule I. (I am just for a moment using subscripts to differentiate among 0’s of different modules.) This is easy to see. Substituting 0M = 0R · 0M and subtracting 0F = 0R · (0M , 0n) we get (0M , 0N ) = (0R · 0M , 0N ) − 0R · (0M , 0N ) ∈ I. As the first example of a tensor product, we shall compute Z/nZ ⊗Z Z/mZ assuming that m and n are relatively prime. Consider a pure tensor x ⊗ y. Since n is invertible modulo m, we can write y = nz for some z modulo m. Then x ⊗ y = x ⊗ nz = n(x ⊗ z) = nx ⊗ z = 0 ⊗ z = 0 ⊗ 0 where the last identity is derived using 0 = 0 · 0 for the first factor. We have shown that every pure tensor is trivial. It follows that Z/nZ ⊗Z Z/mZ = 0. As this example shows, tensor products are tricky matter. However, the following is clear. If M is generated by e1, . , em and N by f1, . , fn, as R-modules, then M ⊗R N is generated by pure tensors ei ⊗ fj. In the case of free modules this can be sharpened as follows. Proposition 2.1. Rm ⊗ Rn =∼ Rmn m n m n Proof. Let ei and fj be the standard basis vectors in R and R , respectively. Then R ⊗R is generated by pure tensors ei ⊗ fj. In order to prove the proposition we have to show that ei ⊗ fj are linearly independent. This is accomplished as follows. Let Mm,n(R) be the R- module of m×n-matrices with coefficients in R. Consider the map S : F → Mm,n(R) defined on basis elements (u, v) in F by S(u, v) = uvT . Here uvT is the usual product of an m × 1 matrix u and an 1 × n matrix vT . By the distibution property of matrix multiplication, the submodule I is in the kernel of S. Thus m n the map descends to a map S from R ⊗R R to Mm,n(R). Under this map the pure tensor ei ⊗fj goes to the matrix with 1 at the (i, j)position and 0 elsewhere. This shows that ei ⊗fj are linearly independent. m n If R = k is a field, then the last proposition shows that the tensor product k ⊗k k can be identified with the set of m × n matrices. Under this identification pure tensors correspond to rank 1 matrices. Thus a general element of a tensor product can not be written as a pure tensor, which is a common mistake for people with little experience with tensor products. We finish this section by showing a tautological relationship between tensor products and bilinear forms. More precisely, an R-homomorphism from M ⊗R N = F/I to an R-module L can be composed with the natural projection from F onto F/I to obtain a homomorphism from F to L. Clearly, a homomorphism from F to L corresponds to a homomorphism from ALGEBRA - LECTURE V 3 F/I to L if and only if it is trivial on I. Since F is freely generated by pairs (v, u), in order to define an element B in HomR(F, L) one picks, freely, a value B(v, u) in L for any pair (v, u) in M × N. Next, note that B is equal to 0 on the submodule I if and only if B is a bilinear form on M × N. Thus we have shown that ∼ HomR(M ⊗R N, L) = BilR(M × N, L). This relationship is, in essence, what we used to show that Rm ⊗ Rn =∼ Rmn. Indeed, we m n picked L = Mm,n(R) and defined, using matrix multiplication, a bilinear map from R × R to Mm,n(R). In general, this is likely the only way to show that an element in M ⊗R N is non-trivial. That is, one needs to find a module L and a bilinear form with values in L such that the corresponding homomorphism is non-zero when evaluated on the element in question. ∼ As an example, we shall show that Q ⊗Z Q = Q. First, using the usual properties of tensor products, a pure tensor in Q ⊗Z Q can be reduced to a c ad c ac ⊗ = ⊗ = ⊗ 1. b d bd d bd a c Here b and d are usual fractions, and we used the property of tensor products to move the integers d and c across ⊗. Next, if r1, . , rn are rational numbers, then n n X X ri ⊗ 1 = ( ri) ⊗ 1. i=1 i=1 This shows that any element in Q ⊗Z Q is represented by a pure tensor of the form r ⊗ 1. Finally, we use a Z-bilinear form B : Q × Q → Q defined by (r, s) 7→ r · s. Then B defines a homomorphism (of Z-modules) from Q ⊗Z Q to Q where r ⊗ 1 maps to r. In particular, we have constructed a natural isomorphism between Q ⊗Z Q and Q. 3. Quadratic forms Assume that R = k is a field of characteristic 6= 2 and V a vector space of dimension n. Let B be a symmetric bilinear form on V with values in k. Then the function Q(x) = B(x, x) is a quadratic form attached to B. For example, if V = kn and B corresponds to the identity matrix, then 2 2 Q((x1, . , xn)) = x1 + ··· + xn. The bilinear form B can be in turn recovered from the quadratic form Q by 1 B(v, w) = [Q(v + w) − Q(v − w)].