Math 121 Homework 2: Notes on Selected Problems

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1 Math 121 Homework 2: Notes on Selected Problems Problem 2. Let M be the ring of 2 2 complex matrices. Let C be the left M-module of 2 1 complex vectors; the module structure is given by matrix multiplication: a b M, c d x C y ax + by C. cx + dy Similarly, let R be the right M-module of 1 2 complex matrices. Compute, with proof, the tensor product R M C. (Remember to give the abelian group the balanced map.) Solution. We prove that the tensor product of R C over M is the M-balanced map R C C given by (r, c) r c where juxtaposition denotes matrix multiplication the resulting 1 1 matrix is identified with C. Let f : R C A be an M-balanced map into some abelian group A. There is at most one group homomorphism C A completing the diagram R C C f A because R C C is surjective. To complete the proof that R C C is the tensor product of R C over M, we must show that there exists a group homomorphism completing the diagram above. We first show that f is constant on the fiber under R C C of any complex number. Consider r in R c in C such that r c =. We will show that f (r, c) =. Additivity of f in each component gives f (r, ) = f (r, ) + f (r, ) f (, c) = f (, c) + f (, c). Thus we are done if either of r or c is zero so we assume otherwise. For convenience identify each of R C with C 2. By extending the nonzero element {c} to a C-module basis {c, r } of C 2 we see that there is a C-linear transformation C 2 C 2 that sends r to r kills c. Letting m be the corresponding 2 2 matrix, we have r m = r mc =. Then since f is M-balanced, f (r, c) = f (r m, c) = f (r, mc) = f (r, ) = by the above. Now assume that r c is some nonzero complex number that r in R c in C are such that r c = r c. Since r r are nonzero 1

2 M acts transitively on the nonzero vectors in C 2, there exists m in M such that r = r m. We then have r (mc c) = r mc r c = r c r c = so f (r, mc c) = by the above. Now since f is additive M- balanced, f (r, c) = f (r, c) + f (r, mc c) = f (r, mc ) = f (r m, c ) = f (r, c ). For any α in C, the preimage under R C C of α is a nonempty subset of R C on which f is constant we define g(α) to be that constant value. We obtain a set map C A that makes the above diagram commute. Let α β be complex numbers. Then ( ) ( ) ( ) 1 ( α = α,, β ) 1 1 = β,, γ = α + β, are preimages of α β α + β, respectively, under R C C. Hence g(α) = f ( α) g(β) = f ( β) g(α + β) = f ( γ) = f ( α) + f ( β) = g(α) + g(β) by additivity of f in the first component. The set map g is a group homomorphism so the proof is complete. 2 Problem 3. Let K be a field V = K n, W = K m as K-modules. Show that every element of V K W can be written as the sum of at most min(n, m) pure tensors v w (v V, w W ); also, show that there exist elements that cannot be written as the sum of any fewer pure tensors. Solution. We show that if v i w i is a sum of r pure tensors in V W either {v 1,..., v r } or {w 1,..., w r } is K-linearly dependent, then this sum can be written with just r 1 pure tensors. This is clear if any v i or w i is zero. Now assume otherwise, so in particular r > 1. Then we may write either v l = i l k li v i or w l = i l k li w i so K- linearity gives vi w i = v l w l + i l v i w i = i l v i k li w l + i l v i w i = i l v i (k li w l + w i )

3 3 or vi w i = v l w l + i l v i w i = i l k liv l w i + i l v i w i = i l (k liv l + v i ) w i, respectively. This shows every element of V W has a representation vi w i with {v 1,..., v r } {w 1,..., w r } both K-linearly independent. In particular, since r > n vectors in V = K n are dependent r > m vectors in K m are dependent, there is a representation with at most n pure tensors there is a representation with at most m pure tensors. Since min(n, m) is either n or m, one of the two representations has at most min(n, m) pure tensors. Choose linearly independent elements e 1,..., e min(n,m) in V = K n linearly independent elements f 1,..., f min(n,m) in W = K m. We will show that the element x = 1 i min(n,m) e i f i of V W cannot be written as a sum of fewer than min(n, m) pure tensors. To accomplish this, we demonstrate, for each sum of fewer than min(n, m) pure tensors, a functional on V W, that is a linear map V W K, that annihilates that sum of fewer than min(n, m) pure tensors, but does not annihilate x. For r < min(n, m), consider a sum y = 1 i r v i w i of r pure tensors. There exists an index l with 1 l min(n, m) such that e l is not in the span of {v 1,..., v r }. Then there exists a functional λ on V that sends e l to 1 is zero on the span of {v 1,..., v r }. Let µ be a functional on W that sends f l to 1 sends each f i for i l to. The bilinear map (v, w) λ(v)µ(w) gives, by the universal property, a functional from the tensor product satisfying v w λ(v)µ(w). By construction this functional kills y, but does not kill x. In particular x y. Since y was an arbitrary sum of fewer than min(n, m) pure tensors, this proves that x cannot be written as a sum of fewer than min(n, m) pure tensors. Problem 4. Let R be a commutative ring with 1 let f, g : R 2 R 2 be R-module homomorphisms. Your goal is to compute f g. More precisely: f g defines an R-module homomorphism R 2 R R 2 R 2 R R 2. Let θ be the isomorphism R 2 R R 2 R 4 given in class, so that θ : (x, y) (x, y ) (xx, xy, yx, yy ). Therefore, we obtain an R-module homomorphism h θ (f g) θ 1 : R 4 R 4.

4 Compute the matrix of h in terms of the matrices of f g. For fun, not for credit: how does the determinant of h relate to the determinants of f g? Solution. Write the matrices of f g as a11 a 12 b11 b 12, a 21 a 22 b 21 b 22 respectively. We compute h(1,,, ) = θ (f g)((1, ) (1, )) = θ(f (1, ) g(1, )) = θ((a 11, a 21 ) (b 11, b 21 )) = (a 11 b 11, a 11 b 21, a 21 b 11, a 21 b 21 ) h(, 1,, ) = θ (f g)((1, ) (, 1)) = θ(f (1, ) g(, 1)) = θ((a 11, a 21 ) (b 12, b 22 )) = (a 11 b 12, a 11 b 22, a 21 b 12, a 21 b 22 ) h(,, 1, ) = θ (f g)((, 1) (1, )) = θ(f (, 1) g(1, )) = θ((a 12, a 22 ) (b 11, b 21 )) = (a 12 b 11, a 12 b 21, a 22 b 11, a 22 b 21 ) h(,,, 1) = θ (f g)((, 1) (, 1)) = θ(f (, 1) g(, 1)) = θ((a 12, a 22 ) (b 12, b 22 )) = (a 12 b 12, a 12 b 22, a 22 b 12, a 22 b 22 ). Therefore the matrix for h is a 11 b 11 a 11 b 12 a 12 b 11 a 12 b 12 a 11 b 21 a 11 b 22 a 12 b 21 a 12 b 22 a 21 b 11 a 21 b 12 a 22 b 11 a 22 b 12, a 21 b 21 a 21 b 22 a 22 b 21 a 22 b 22 which is the matrix of f after replacing each entry with itself times the matrix for g. 4

5 Using, for example, determinants of block matrices, we argue that det(θ (f 1 R 2) θ 1 ) = (det f ) 2 det(θ (1 R 2 g) θ 1 ) = (det g) 2. To compute the determinant of h, use the identity h = θ [(f 1 R 2) (1 R 2 g)] θ 1 [ = θ (f 1 R 2) θ 1] [ θ (1 R 2 g)] θ 1], the first equality of which follows from the universal property of the tensor product. Then multiplicativity of the determinant gives det h = (det f ) 2 (det g) 2. (The exponent on det f comes from the rank of the space of g the exponent on det g comes from the rank of the space of f.) 5