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Math 121 Practice Midterm
In this exam, R denotes a commutative ring with 1, and F denotes
a field. Z denotes the integers, and ◦ denotes the composition of two
functions. Give complete proofs, except where indicated.
1. True or false? No justification needed.
(a) Any module over the ring of real numbers is free.
(b) If M is finitely generated and N ⊆ M is any submodule, then
M/N is finitely generated.
(c) If every ideal of R is principal, then every R-module is free.
(d) For A a finite abelian group, the sizes of A ⊗Z A and A are the
same.
(e) If M is an R-module and f , g : M → M are R-module homomorphisms, then f ◦ g = 0 if and only if g ◦ f = 0.
(f) dim(V ) = dim(V ∗ ) for every vector space V over F .
(g) If V is a vector space and 0 6= v ∈ V , there is ` ∈ V ∗ so that
`(v) = 1.
(h) If A : V → V is a linear map, then the trace of A and A∗ coincide.
Solution.
(a) True. The ring of real numbers is a field, and any module over
a field, that is any vector space, has a basis (on which it is free).
(b) True. If a finite set S ⊆ M generates M, then the image of S in
M/N is finite and generates M/N.
(c) False. Every ideal of the ring R = Z of integers is a principal,
but the Z-module Z/2Z is not free because, in particular, it is
not torsion free.
(d) False. For example if A = (Z/2Z)⊕2 , the direct sum of two copies
of Z/2Z, then
A ⊗Z A › (Z/2Z ⊗Z Z/2Z)⊕4 › (Z/2Z)⊕4
since tensor products commute with direct sums. Thus in this
case A ⊗Z A, an abelian group of order 24 = 16, is not the same
size as A, an abelian group of order 4.
(e) False. Let M be the free R-module R ⊕ R with basis e1 = (1, 0)
and e2 = (0, 1). Let f : M → M be the map given by e1 , e2
and e2 , e2 and let g : M → M be the map given by e1 , e2 and
e2 , 0. Then f ◦ g is nonzero as it does not kill e1 , but g ◦ f is
zero as it kills both e1 and e2 .
(f) False. The following proof of the fact that dim V ∗ > dim V
when V is infinite dimensional is not important to know for
now. Let B be a basis for V . The cardinality of V ∗ is (#F )#B
1
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 .