Math 593: Problem Set 7

Math 593: Problem Set 7 Feng Zhu, Punya Satpathy, Alex Vargo, Umang Varma, Daniel Irvine, Joe Kraisler, Samantha Pinella, Lara Du, Caleb Springer, Jiahua Gu, Karen Smith 1 Basics properties of tensor product i. We have a bilinear map M N N M given by (m, n) n m by construction of the tensor product, (m, n 1 + n 2 ) n 2 m + n 1 m, (m, rn) r(n m), (m 1 + m 2, n) n m 1 + n m 2 and (rm, n) r(n m). Then, by the universal property of tensor, we have a unique R-module map : M N N M s.t. M N M N N M such that is given by n m m n. In the same way, we can construct an R-linear map φ : N M M N which is inverse to. Hence M N = N M. ii. For any fixed l L, we have a bilinear map f l : M N M (N L) defined by (m, n) m (n l). By the universal property of tensor products, we have a well-defined R-linear map f l : M N M (N L). It is then easy to check that the map (M N) L M (N L) given by (m n, l) f l (m n) = m (n l) is well-defined and bilinear. Then, applying again the universal property of tensor products, we have a R-module map : (M N) L M (N L) s.t. (M N) L (M N) L On simple tensors, we have ((m n) l) = m (n l). M (N L) Similarly, we get an R-module map φ in the other direction, by first fixing an m M and defining a bilinear map N L (M N) L by (n, l) (m n) l. We get an analogous an R-module map φ : M (N L) (M N) L satisfying φ(m (n l)) = m (n l). So φ and are inverses and we conclude that (M N) L = M (N L). iii. We have a bilinear map (M N) L (M L)(N L) given by (mn, l) (m l, n l) by construction of the tensor product and the properties of the direct sum, (m 1 + m 2 n 1 + n 2, l) ((m 1 + m 2 ) l, (n 1 + n 2 ) l) = (m 1, n 1 ) l + (m 2, n 2 ) l, (rm rn, l) (rm l, rn l) = r(m l, n l), (m n, l 1 + l 2 ) (m (l 1 + l 2 ), n (l 1 + l 2 )) = (m l 1, n l 1 ) + (m l 2, n l 2 ), (m n, rl) (m rl, n rl) = r(m l, n l). 1

Then, by the universal property of the tensor product, we have a unique R-module map : M N N M s.t. (M N) L (M N) L (M L) (N L) and we may verify by following the arrows that is given by (m, n) l (m l, n l). We may verify that such a is bijective on the simple tensors it is injective since m l = n l = 0 implies m + n l = 0; it is surjective because we may hit any m l by starting with (m, 0) l, and any n l by starting with (0, n) l and is hence a bijective R-module map between the tensor products. Hence (M N) L = (M L) (N L). iv. Let µ : M M and ν : N N be isomorphisms. We have a bilinear map M N M N given by (m, n) µ(m) ν(n) by construction of the tensor product and since µ and ν are R-module morphisms, (m, n 1 + n 2 ) µ(m) ν(n 1 ) + µ(m) ν(n 2 ), (m, rn) r(µ(m) ν(n)), (m 1 + m 2, n) µ(m 1 ) ν(n) + µ(m 2 ) ν(n) and (rm, n) r(µ(m) ν(n)). Then, by the universal property of the tensor product, we have a unique R-module map : M N M N s.t. M N M N M N and we may verify by following the arrows that is given by m n µ(m) ν(n). Analogously, by working with the maps µ 1 : M M and ν 1 : N N, we construct the inverse map. Hence M N = M N whenever M = M and N = N (as R-modules.) 2 Change of rings Since A B C, every C module is a B-module, and every B-module is an A-module. We abuse notation by writing b for both b B and its image in C, and likewise a will denote a A or its image in B or its image in C. In all cases, the meaning should be clear from context. There is a B-bilinear map C (B A M) C A M given by (c, b m) cb m (where [in the left tensor factor] on the RHS b denotes the image thereof in C by the abuse mentioned above). This map is well-defined by the linearity properties of the tensor product. Now by the universal property of the tensor product, we can find a unique B-module map : C B (B A M) C M s.t. C (B A M) C B (B A M) C A M In fact, it is easy to see that this map is indeed C-linear, since c(c (b m) = c(c b m) = cc b m = (c(c (b m))). There is also an A-bilinear map C M C B (B A M) given by (c, m) c (1 m) We verify this is bilinear by checking that 2

(rc 1 + c 2, m) (rc 1 + c 2 ) (1 m) = r(c 1 (1 m)) + c 2 (1 m) (c, rm 1 + m 2 ) c (1 (rm 1 + m 2 )) = r(c (1 m 1 )) + c (1 m 2 ) Now by the universal property of the tensor product, we can find a unique A-module map ψ : C A M C B (B A M) s.t. C M C A M C B (B A M) Claim. and ψ are inverses, so in particular they are bijective. To see this, note that (ψ )(c (b m)) = ψ(cb m) = cb (1 m) = c (b m) and ( ψ)(cb m) = cb (1 m) = cb m. Hence C B (B A M) = C A M. ψ 3 Tensor products in the category of A-algebras (a) First we show that R A S has a ring structure. R A S, being an A-module, is already an abelian group under +. Now define multiplication on R A S by (r s) (r s ) = rr ss extended linearly. This is well-defined by the following argument: the universal property of the tensor product gives us well-defined A-linear maps R A R R and S A S S, from R R R A R S S S A S (r,r ) rr (s,s ) ss R S and then, again by the universal property of the tensor product, we have an A-linear map : (R A S) A (R A S) R A S s.t. (R A R) (S A S) (R A R) (S A S) = (R A S) (R A S) (r r,s s ) rr ss R A S and now we may compose the natural map (R A S) (R A S) (R A S) (R A S) with to obtain the above-described map (R A S) (R A S) R A S. We note that multiplication hence defined is associative, by the associativity of multiplication in R and S, and has 1 1 as an identity element. Moreover multiplication distributes across addition by bilinearity. Now define an A-algebra structure on R A S by a a(1 1). This is a ring homomorphism because (a + b) (a + b)(1 1) = a(1 1) + b(1 1), ab ab(1 1) = a(b(1 1)), 1 (1 1), and 0 0(1 1) = 0. Moreover the image of A is clearly in the center of B (with multiplication as given above.) (b) Define a map R R A S by r r 1 and a map S R A S by s 1 s. These are ring maps by the way we have defined ring operations on R A S. To see that they are also A-algebra maps, we need a commutating diagram 3

R R A S A but this is easy since A R R A S sends a a a 1 = a(1 1). Similarly with S R S sending s 1 s is an A-algebra map. Now suppose we have an commutative A-algebra T s.t. we have A-algebra maps R T and S T. We wish to show that we can find a unique A-algebra map : R A S T s.t. T R A S R so that R A S is indeed a coproduct (of R and S) in the category of commutative A-algebras. Since R S T sending (r, s) rs is A-bilinear, the universal property of tensor products gives the map via: S R S R S We may verify that is a ring map: (1 1) = 1 and (0 1) = 0, and respects ring addition by definition, and ((r s)(r s )) = rr ss = (rs)(r s ). Moreover is an A-algebra map, since (a(1 1)) = a T. (c) Commutative rings are Z-algebras, so the result of the previous part applied to A = Z shows that coproducts exist in the category of commutative rings: in particular, given two rings R and S, the tensor product R Z S is a coproduct of R and S in the category of commutative rings. T 4 A-algebra isomorphisms a. Define the map : B A A[x, y] B[x, y] by (b f) bf. This is an isomorphism, from the chain of isomorphisms B A A[x, y] = B A A(x i y j ) i,j Z 0 = (B A A)(x i y j ) i,j = B(x i y j ) = B[x, y] b. We start with the exact sequence 0 (f 1,..., f t ) R[x 1,..., x d ] R[x 1,..., x d ]/(f 1,..., f t ) 0 i,j 4

and now we apply the functor R/I R to our sequence We know from [a similar argument to] the above, with A = R and B = R/I, that R/I R R[x 1,..., x d ] = R/I[x 1,..., x d ]. Similarly, R/I R (f 1,..., f t ) = ( f 1,..., f t ). By right-exactness of (and implicitly appealing to the above isomorphisms where required), we obtain the exact sequence im(r/i R (f 1,..., f t )) = ( f 1,..., f t ) R/I[x 1,..., x d ] R/I R R[x 1,..., x d ]/(f 1,..., f t ) 0 whence we may conclude R/I R R[x 1,..., x d ]/(f 1,..., f t ) = R/I[x 1,..., x d ]/( f 1,..., f t ). c. This follows from the result of the previous part with R = Z, I = (5), d = 2, t = 1 and f 1 (x, y) = x 5 1 + 5xy + y 5, where we verify that (x + y) 5 = x 5 + 5xy + y 5 when we reduce all coefficients modulo I = (5). 5 True or false True. We may verify that the map R R Z Q given by r (r 1) is an isomorphism (of Z-modules, i.e. of abelian groups): it is certainly injective, and it is surjective since any s a b = as 1 b = bas b 1 b = as b 1. False. The LHS (as a R-vector space) is generated by the basis {1 1, 1 i, i 1, i i}, and so has dimension 4, but the LHS has dimension 2 (being generated by 1 and i.) False, the LHS (as a R-vector space) has basis {i 1, 1 1} (and, in particular, is two-dimensional), but the RHS is a one-dimensional R-vector space. True, z w = wz 1, so C C C = C (as C-modules) by the C-vector space isomorphism z w wz. True. k[x] k k[y] is a k-vector space with basis (x a y b ) a,b Z 0. k[x, y] is a k-vector space with basis (x a y b ) a,b Z 0. Now we can put the bases in bijection with other by sending x a y b x a y b ; hence the two are isomorphic (as k-modules / k-vector spaces.) 6 First note Z[x, y]/(x 3 ) Z[x,y] Z[x, y]/(y 2 ) = Z[x, y]/(x 3, y 2 ); then observe that Z 10 Z Z[x, y]/(x 3, y 2 ) = Z[x, y]/(10, x 3, y 2 ). Now the ideals of Z[x, y]/(10, x 3, y 2 ) are in one-one correspondence with the ideals of Z[x, y] containing (10, x 3, y 2 ); under this correspondence, primes correspond to primes (by the third isomorphism theorem and the fact that P is prime if and only if R/P is omain). But a prime P containing y 2 = y y must contain y; likewise if x 3 = x x 2 must contain x. Also, since 10 P, either 2 or 5 is in P. So the only options for P are (2, x, y) and (5, x, y). Thus our ring has exactly 2 prime ideals. 5

7 (a) Let R denote our PID. We have M = R r1 R/(f 1 ) R/(f t ) N = R r2 R/(g 1 ) R/(g s ) and now M R N = (R r1 R/(f 1 ) R/(f t )) R (R r2 R/(g 1 ) R/(g s )) = R r1r2 R r1 R/(g 1 ) R r1 R/(g s ) R/(f 1 ) R r2 R/(f t ) R r2 (R/(f i ) R R/(g j )) 1 i t,1 j s = R r1r2 R/(g j ) r1 R/(f i ) r2 R/(f i, g j ) 1 j s 1 i t 1 i t,1 j s In particular, the rank is r 1 r 2. The last invariant factor is the annihilator of the torsion part of the module, which is displayed as irect sum of cyclic modules. Recall from a previous problem set that the annihilator of irect sum is the intersection of their annihilators. So we need to find the intersection of all the ideals (f i ), (g j ) (if r 2, respectively r 1 is not zero) and (f i, g j ) = gcd(f i, g j ). There are many inclusions among these ideals, since (f 1 ) (f 2 ) (f t ) and (g 1 ) (g 2 ) (g t ). Indeed, we only need to consider the ideals (f t ), (g s ) and (f t, g s ). Analyzing this, If r 1 r 2 > 0, then the top invariant factor is lcm(f t, g s ), since this generates (f t ) (g s ) (f t, g s ) = (f t ) (g s ). If r 1 0 but r 2 = 0, then the top invariant factor is g s, since (g s ) (f t, g s ) = (g s ). If r 2 0 but r 1 = 0, then the top invariant factor is f t, since (f t ) (f t, g s ) = (f t ). If r 1 = r 2 = 0, then the top invariant factor is gcd(f t, g s ), (provided it is not a unit!) since this generates (f t, g s ), which annihilates all R/(f i, g j ). If f t and g s have no common factors, then all the R/(f i, g j ) = 0, and there are no invariant factors (the tensor product is free). The first invariant factor will be moved to another problem set, with substantial hints. Part b will also appear again. 8 A QR problem Let Mat 2 2 (C) be the space of 2 2 matrices with complex entries. Fix an element A Mat[ 2 2 (C). ] 1 0 Show that the self-map L sending X to AX XA is a C-vector space endomorphism. When A =, 0 0 find its Jordan form. Let L: Mat 2 2 (C) Mat 2 2 (C) be as above. First we show that L is a C linear transformation. This is just a restatement of the fact that matrix multiplication distributes over matrix addition and that scalar multiplication (by λ C) commutes with matrix multiplication. Symbolically (A + B)C = AC + BC, and C (A + B) = C A + C B, and A(λB) = λab. 6

Now, fix A = [ ] 1 0 and fix a basis for Mat 0 0 2 2 (C) as follows e 1 = [ ] 1 0, e 0 0 2 = [ ] 0 1, e 0 0 3 = [ ] 0 0, e 1 0 4 = [ ] 0 0. 0 1 It should be clear that these matrices are linearly independent over C and that they span all of Mat 2 2 (C). We find how L acts on each of these basis vectors. L(e 1 ) = Ae 1 e 1 A = e 2 1 e 2 1 = 0, since e 1 = A. [ ] [ ] [ ] [ ] [ ] 1 0 0 1 0 1 1 0 0 1 L(e 2 ) = = = e 0 0 0 0 0 0 0 0 0 0 2 [ ] [ ] [ ] [ ] [ ] 1 0 0 0 0 0 1 0 0 0 L(e 3 ) = = e 0 0 1 0 1 0 0 0 1 0 3 [ ] [ ] [ ] [ ] 1 0 0 0 0 0 1 0 L(e 4 ) = = 0 0 0 0 1 0 1 0 0 This means that in this ordered basis B = {e 1, e 2, e 3, e 4 } the transformation L is presented by the matrix 0 [L] B = 1 1 0 which is already in Jordan form. 9 Simple tensors (a) Given V and W of dimensions d and e, we fix bases so that they are identified with k d and k e, respectively. We interpret V as the column space k d and W as the row space k e. There is a bilinear map k d k e Mat de (k) given by matrix multiplication:., [ ] b 1 b 2... b e. [ ] b1 b 2... b e. This induces a linear map V k W M de (k) which sends the basis elements v i w j to the basis elements E ij of Mat de (k), the d e matrix with in the de entry and zeros everywhere else. Because this linear map is a bijection on basis elements, it is an isomorphism To understand the simple tensor v w in this set-up, note that v is interpreted as a row and w as a column. Their product is rank one; indeed, the j-column is b j times the column.., so all columns are scalar multiples of each other. Conversely, a rank one matrix corresponds to a 7

(non-zero) simple tensor: take any non-zero column, say the j-th column.. Every other column is a scalar multiple of this one, say column i is b i times column j (note: b j = 1). So the [ ] rank one matrix is (a i b j ), which means that it is the product b1 b 2... b e. This. means it corresponds to the simple tensor v w where v = a i v i and w = b j w j. To count the number of rank one matrices, we use the fact that (k d \ 0) (k e \ 0) maps onto the set of rank one matrices. This is not a bijection (kudos to Bob Lutz for being possibly the only 593 to realize this). Indeed, for any non-zero scalar λ, the pairs (v, w) and (λv, λ 1 w) have the same image (after multiplying column v times row w). But this is the only way two matrices can multiply to the same matrix, so each rank one matrix has exactly q 1 pre images. This means that the set of rank one matrices has cardinality (qd 1)(q e 1) (q 1), and so (adding in the zero tensor) there are exactly (qd+e q d q e +q) (q 1) simple tensors. So the fraction of simple tensors is (qd+e q d q e +q) (q de )(q 1). This is clearly a small fraction, roughly qd+e q de, unless d or e is one, in which case all tensors are simple. In particular, it goes to zero as q gets large. In R d R e = R de, the set of simple tensors corresponds the pre image of zero under the map Mat de (R) R (d 2)( e 2) sending each d e matrix to the list of determinants of all its 2 2 subminors (note that a matrix has rank one or less if and only if all 2 2 minors are zero). So it is a proper closed set (if e and d are greater than one) and by the inverse function theorem (removing 0) can be given the structure of a manifold of strictly smaller dimension. So is has Lebesque measure zero in Mat de (R). Put differently, if we randomly chose an element in V W (with d, e 2), the probability that it is a simple tensor is zero. 8