2

since V ∗ is in bijection with set maps B → F . We now derive another expression for the cardinality of V ∗ assuming that V is infinite dimensional, so the same is true of V ∗ . Write B for a basis of V ∗ . Since V ∗ is infinite dimensional, the cardinality of V ∗ can also be written as #B × #F . (This is true because an F -vector space is in bijection with the disjoint union indexed by nonnegative integers N of the set of pairs consisting of an N element subset of a fixed basis and a mapping from that set to F \ {0}, a set of cardinality equal to that of a basis for the vector space times #F when the vector space is infinite dimensional.) Recall that (since at least one of B or #F is infinite) #B × #F = max{#B, #F }. Pick a linearly independent subset of V indexed by N: say v1 , v2 , . . . . Define a map F \ {0} → V ∗ by sending the nonzero element a of F to a functional satisfying vi , ai . By the formula for the determinant of a Vandermonde matrix, the image is a linearly independent subset of V ∗ and so #F ≤ #B and thus the cardinality of V ∗ can be written as #B = max{#B, #F }. Equating the two expressions for the cardinality of V ∗ gives #B = (#F )#B ≥ 2#B > #B since #F ≥ 2 and so the cardinality of B is strictly greater than that of B, that is dim V ∗ > dim V . (g) True. If B is a basis of V , then the matrix of A with respect to B, B is the transpose of the matrix of A∗ with respect to B ∗ , B ∗ , so the traces, sums of diagonal entries, agree. To see this without coordinates, consider a rank one element A of Hom(V , V ), which we write as w , α0 (w)v0 , so A corresponds to α ⊗ v in V ∗ ⊗ V and thus has trace α0 (v0 ). Then by definition A∗ takes a functional β in V ∗ to the functional w , β(α0 (w)v0 ) = α0 (w)β(v0 ) = α0 (w)v0 (β), so A∗ corresponds to v0 ⊗ α0 : V ∗∗ ⊗ V ∗ → F , where we have identified V with the subspace V > V ∗∗ , and hence has trace v0 (α0 ) = α0 (v0 ). This proves the result for A in a generating set, which is sufficient by linearity of A , A∗ .

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2. An R-module M is called simple if it has no submodules except 0 and M. Describe, with complete proof, all simple modules for R = Z/6Z. Solution. The image of a module homomorphism is a submodule of the codomain so a homomorphism into a simple module is either zero or surjective. In particular if R is a ring and M is an R-module with some element m0 , then the R-module homomorphism R → M given by r , r m0 is either zero or surjective. If M is a simple module (convention: simple modules are nonzero), then for m0 in M nonzero the R-module homomorphism r , r m0 is a surjective R-module homomorphism R → M and hence by the first isomorphism theorem induces an isomorphism R/I → M where I is the annihilator of m0 . Moreover, by the correspondence theorem, R/I is simple if and only if I is a maximal proper ideal of R. Therefore each simple R-module is isomorphic to R/I for some maximal proper ideal I of R. The ideals of R = Z/6Z are all principal (since the same is true of Z) and fit into the diagram of inclusions 0(Z/6Z) ⊂

⊂

2(Z/6Z) = 4(Z/6Z)

3(Z/6Z) ⊂

⊂

1(Z/6Z) = 5(Z/6Z). The maximal proper ideals are 2(Z/6Z) and 3(Z/6Z) so the simple modules over R = Z/6Z are Z/6Z Z/2Z 2(Z/6Z)

and

by the third isomorphism theorem.

Z/6Z Z/3Z 3(Z/6Z)

3. Let M be an R-module and let K ⊆ M ⊗R M be the submodule generated by all elements m ⊗ n + n ⊗ m (m, n ∈ M). (a) Prove that the map f : M × M → (M ⊗R M)/K is a bilinear map of R-modules and satisfies f (m, n) = −f (n, m). (b) Formulate a universal property that characterizes the pair (f , (M ⊗R M)/K) uniquely. (You do not need to justify it, but you should formulate the property precisely.)

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Solution. (a) The canonical map M × M → M ⊗R M given by (m, n) , m ⊗ n is R-bilinear and the canonical map M ⊗R M → (M ⊗R M)/K is R-linear so the composition f : M × M → (M ⊗R M)/K is an R-bilinear map given by (m, n) , m ⊗ n + K. For any m and n in M, m ⊗ n + n ⊗ m is in K so f (m, n) = m ⊗ n + K = −n ⊗ m + (m ⊗ n + n ⊗ m) + K = −n ⊗ m + K = −f (n, m). (b) Say a map g from M × M to an abelian group is alternating if g(m, n) = −g(n, m) for all m and n in M. The pair (f , (M ⊗R M)/K) consisting of f : M × M → (M ⊗R M)/K is universal among R-balanced alternating maps to abelian groups, that is, if g : M × M → A is an alternating R-balanced map into the abelian group A, then there exists a unique homomorphism of abelian groups completing the diagram f

/ (M ⊗R M)/K M × MO OOO OOO O g OOO OO' A

(1)

The standard universal property argument shows that if for i = 1, 2, gi : M × M → Ai also satisfies the above property, then there exists a unique pair of inverse isomorphisms of abelian groups completing the diagram M ×M HH

g1

HH HH H g2 HH $

/ A1 O

A2 .

The universal property for (g1 , A1 ) gives a unique map h12 : A1 → A2 fitting into the above diagram. Similarly the universal property for (g2 , A2 ) gives a unique map h21 : A2 → A1 fitting into the above diagram. The compositions h21 ◦ h12 and h12 ◦ h12 complete the diagrams g1

/ A1 GG GG g1 GGG #

M ×M GG

A1

and

M ×M HH

g2

HH HH H g2 HH $

/ A2

A2 ,

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respectively, but the same is true for the identities of A1 and A2 , respectively, so h21 ◦ h12 is the identity of A1 and h12 ◦ h21 is the identity of A2 , as desired. It remains to show that (f , (M ⊗R M)/K) satisfies the property. As above, let g : M × M → A be an alternating R-balanced map into the abelian group A. By the universal property of the tensor product, there exists a unique map completing the diagram / M ⊗R M MMM MMM M MMM g M&

M × MM (2)

A

Since g is alternating, the kernel of the above map contains K so there exists a unique map completing the diagram / M ⊗R M / (M ⊗R M)/K M × ML LLL n LLL nn n g LLL L& wn n n A.

(3)

This gives the existence of a map completing (1), but only a slightly weaker uniqueness property: we have proved uniqueness of maps completing (3), but we need uniqueness of maps completing (1). Precomposition with the canonical projection M ⊗R M → (M ⊗R M)/K is an injective map from arrows completing (1) to arrows completing (2) so uniqueness for arrows completing (2) completes the proof.

4. Let V be a finite-dimensional vector space over the field F , and A ∈ Hom(V , V ). (a) Prove that X , X ◦ A is a linear map from Hom(V , V ) to itself. (b) Compute its trace in terms of the trace of A. Solution. (a) For X in Hom(V , V ) and a scalar c in F , [(cX) ◦ A](v) = (cX)(A(v)) = c(X(A(v)) = c(X ◦ A(v)) = [c(X ◦ A)](v) for any v in V by definition of scalar multiplication in the vector space Hom(V , V ), and so (cX) ◦ A = c(X ◦ A). (Alternatively,

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this follows from associativity of composition, viewing multiplication by the scalar c as application of the corresponding scalar endomorphism of Hom(V , V ).) For X1 and X2 in Hom(V , V ), [(X1 + X2 ) ◦ A](v) = (X1 + X2 )(A(v)) = X1 (A(v)) + X2 (A(v)) = X1 ◦ A(v) + X2 ◦ A(v) = (X1 ◦ A + X2 ◦ A)(v) be definition of addition in the vector space Hom(V , V ), and so (X1 + X2 ) ◦ A = X1 ◦ A + X2 ◦ A. (b) Consider an element A of Hom(V , V ) given by v , α0 (v)w0 for α in V ∗ and w0 in V . This corresponds to the simple tensor α0 ⊗ w0 in V ∗ ⊗ V so has trace α0 (w0 ). Choose a basis v1 , . . . , vn for V so n = dim V . For any X in Hom(V , V ) and index i, let λi (X) be the ith component of X(v0 ) in the basis v1 , . . . , vn , that is, X(v0 ) = λ1 (X)v1 + · · · + λn (X)vn . Then each λi is an element of Hom(V , V )∗ . Now the linear transformation of Hom(V , V ) to itself given by precomposition with A, X , X ◦ A, may be written as X , (v , X(α0 (v)v0 )) , (v , α0 (v)X(v0 )) , (v , α0 (v)λ1 (X)v1 + · · · + α0 (v)λn (X)vn ) , which has image λ1 ⊗ (v , α0 (v)v1 ) + · · · + λn ⊗ (v , α0 (v)vn ) in Hom(V , V )∗ ⊗ Hom(V , V ). Note that for each i, λi (v , α0 (v)vi ) = α(v0 ) so the trace of X , X ◦ A is nα(v0 ), which is n = dim V times the trace of A. Since Hom(V , V ) is generated by such rank one maps A = (v , α0 (v)v0 ), it follows that for any A in Hom(V , V ), the trace of X , X ◦ A is dim V times the trace of A. Note. Solving part (b) as above without the “sum of the diagonal entries” interpretation is a good exercise to check understanding of the

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canonical homomorphism V ∗ ⊗ V → Hom(V , V ). The essential step was finding the preimage under the isomorphism Hom(V , V )∗ ⊗ Hom(V , V ) → Hom(Hom(V , V ), Hom(V , V )) of an element of the codomain by expressing it as as a linear combination of images of pure tensors. Alternatively, one may use a basis for Hom(V , V ) as follows. Let v1 , . . . , vn be a basis for V . Define Eij to be the element of Hom(V , V ) killing vk distinct from vj and taking vj to vi . Then {Eij }ij is a basis for Hom(V , V ). Take A to be Er s . We have Eij ◦ Er s = Eis δjr ∗ where δjr is the indicator function of j = r . Thus writing {Eij } for the dual basis of {Eij } we have ∗ ∗ Eij (Eij ◦ Er s ) = δjr Eij Eis = δjr δjs .

Summing over all i and j gives that the trace is nδr s , which is n = dim V times the trace of Er s . As before, linearity of trace gives that in general the trace of X , X ◦ A is dim V times the trace of A. 5. Let V be a finite dimensional vector space over F and U ⊆ V , W ⊆ V ∗ subspaces so that w ∗ (u) = 0 for all w ∗ ∈ W , u ∈ U. Prove that dim(U ) + dim(W ) ≤ dim(V ). Solution. There is a natural injective F -linear map W > (V /U)∗ . Taking dimensions gives dim(W ) ≤ dim((V /U)∗ ) = dim(V /U) = dim(V ) − dim(U)

from which the desired inequality follows.

Note. In detail, the injective F -linear map W > (V /U)∗ is defined as follows. Precomposition with V → V /U gives an F -linear map Hom(V /U, F ) → Hom(V , F ) that has image the subspace of all F -linear maps V → F with kernel containing U . Since W consists of F -linear maps V → F with kernel containing U , there exists a section over W , that is a map completing the diagram / Hom(V , F ) Hom(V /U,hQF ) O Q Q Q Q Q Q

W,

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which is necessarily injective, as is the case with all sections, because the postcomposition with the map Hom(V /U, F ) → Hom(V , F ) is injective. 6. Let X, Y be finite-dimensional vector spaces over F , x, x 0 ∈ X, y, y 0 ∈ Y . Prove that x ⊗ y = x 0 ⊗ y 0 (equality inside X ⊗ Y ) if and only if there exists a scalar α so αx = x 0 , y = αy 0 . Solution. If there exists a scalar α so that αx = x 0 , y = αy 0 , then x ⊗ y = x ⊗ (αy 0 ) = (αx) ⊗ y 0 = x 0 ⊗ y 0 . We now prove the converse. Assume first that there does not exist a scalar α so that αx = x 0 . Then either x and x 0 are independent, or there is a nontrivial dependence relation cx + c 0 x 0 = 0, necessarily with c 0 = 0, in which case x = 0. In either case there exists a functional `1 in X ∗ so that `1 (x) = 0 and `1 (x 0 ) 6= 0. Choose a functional `2 in Y ∗ so that `2 (y 0 ) 6= 0. Then `1 ⊗ `2 : X ⊗ Y → F ⊗ F = F kills x ⊗ y, but does not kill x 0 ⊗ y 0 so the two elements of X ⊗ Y are distinct. Similarly, if there does not exist a scalar α so that y = αy 0 , then x ⊗ y and x ⊗ y 0 are distinct elements of X ⊗ Y . Finally assume that there exist scalars β and γ so that βx = x 0 and y = γy 0 , but there does not exist a single scalar α so that αx = x 0 and y = αy 0 . In particular β cannot be replaced by γ so x is nonzero and γ cannot be replaced by β so y 0 is nonzero. Since β and γ are not equal, at least one is nonzero and hence at least one of x 0 = βx or y = γy 0 is nonzero. Then one of x ⊗y or x 0 ⊗y 0 is nonzero. Therefore one of β(x ⊗ y − x 0 ⊗ y 0 ) = x 0 ⊗ y − βx 0 ⊗ y 0 = γx 0 ⊗ y 0 − βx 0 ⊗ y 0 = (γ − β)(x 0 ⊗ y 0 ) or γ(x ⊗ y − x 0 ⊗ y 0 ) = γx ⊗ y − x 0 ⊗ y = γx ⊗ y − βx ⊗ y = (γ − β)(x ⊗ y) is nonzero, and in either case we conclude that x ⊗ y − x 0 ⊗ y 0 is nonzero, that is x ⊗ y 6= x 0 ⊗ y 0 .