Math 323. Practice problems on modules.
(1) Let R be an integral domain, and let F be a free module over R of finite rank, and let R2 be the free module of rank 2 over R. Prove that
HomR (F, R2 ) ≃ F ⊕ F .
Solution. This problem is basically the same as the earlier homework
problem about HomR (F, R) ≃ F .
First, let us establish a bijection of sets between HomR (F, R2 ) and
F ⊕ F . Let v1 , . . . vn be a basis of F . Then by the universal property, a homomorphism φ : F → R2 is uniquely determined by n elements (a1 , b1 ), . . . , (an , bn ) of R2 - indeed, we set φ(vi ) = (ai , bi ) for i =
1, . . . , n, and that uniquely determines φ. Thus we get a bijection between
HomR (F, R2 ) and {(ai , bi ) | ai , bi ∈ R, i = 1, . . . n}. Now note that the
latter set has a natural R-module structure, and as an R-module it is isomorphic to F ⊕ F (make sure you know why). Now it just remains to check
that our bijection of sets constructed above is actually a module homomorphism; and that is again easy, because it behaves well with respect to
addition and the action of R on the basis v1 , . . . , vn , and then by the universal property, this works on all of F (see the mentioned above homework
problem for details).
(2) Let R be an integral domain, and let M be a torsion module over R, and F
– a free module over R. Prove that HomR (M, F ) = {0}. Is this statement
still true if R is not an integral domain?
Solution. Suppose φ : M → F is a module homomorphism. Since M is
a torsion module, for every m ∈ M , there exists r ∈ R such that rm = 0.
Then rφ(m) = 0. Since we know that a free module over an integral domain
is torsion-free (make sure you know why this is so), this forces φ(m) = 0,
so φ is a zero homomorphism.
If R is not an integral domain, then it is, of course, false: take, for
example, R = Z/4Z, F = R, and M = Z/2Z. Then F is free, M is torsion.
Let φ : M → R be defined by φ(m) = 2m (i.e, φ(0) = 0, φ(1) = 2). It is
clearly a non-zero homomorphism.
(3) Give an example of three modules A, B and C over a PID such that A/B ≃
C, but A is not isomorphic to B ⊕ C.
Solution. For example, take R = Z, A = Z, B = nZ, C = Z/nZ.
(4) Let p and q be primes, and let m be an arbitrary integer. Compute
HomZ (Z/pZ × Z/qZ, Z/mZ).
Solution. Note that by the Chinese Remainder Theorem, Z/pZ ×
Z/qZ ≃ Z/(pq)Z, and this reduces the problem to the special case of
Problem 6 in Section 10.2 (see homework 8). The answer is Z/dZ, where
d = gcd(pq, m). (The same answer can, of course, be obtained by looking
at Hom(Z/pZ, Z/mZ) and Hom(Z/qZ, Z/mZ) separately).
(5) Let I and J be two ideals in Z[i]. Then we can think of them as Z[i]modules. Find HomZ[i] (I, J).
Solution. This problem was meant to emphasize that as a module, any
principal ideal in an integral domain R is just a free rank 1 module (the
1
2
R-module isomorphism from R to I = (a) is determined by 1 7→ a, where
a is a generator of I). Since Z[i] is Euclidean and therefore, a PID, all
ideals are principal. Let R = Z[i]. Then as R-modules, I ≃ J ≃ R. So,
HomR (I, J) ≃ HomR (R, R) ≃ R.
(6) Consider the submodule N of Z2 generated by the vectors h2, 4i and h4, 10i.
Describe the quotient module Z2 /N .
Solution. We do row and column operations on the relations matrix
for the generators of N to get it to diagonal form. We have:
4 ] → [2 4] → [2 0],
[ 24 10
0 2
02
where the first operation was subtracting row 1 multiplied by 2 from Row
2, and the second was a similar operation on columns. Thus, Z2 /N ≃
Z/2Z ⊕ Z/2Z.
(7) Prove that any module over Z[i] is a direct sum of √
a free module and a
torsion module. Is the same true for modules over Z[ −5]?
Solution.Since Z[i] is a PID, by the Theorem on structure of modules
over PIDs, any module is isomorphic to
Z[i]r ⊕ sum of cyclic modules of the form Z[i]/(a).
The sum of modules of the form Z[i]/(a) is a torsion module.
Over a non-principal ideal domain
not true. For example, √
take a
√ this is √
non-principal ideal (e.g. I = (2, 1+ 5) in Z[ −5]). Then I, as an Z[ −5]module, is torsion-free, but not free. Since it is torsion-free, it does not have
any non-zero torsion submodule, so there cannot be a torsion summand;
on the other hand, it is not free either. (Note: you need to know why the
non-principal ideal is not a free√module; this appeared in homework 7; also,
need to know why I = (2, 1 + 5) is not a principal ideal).
(8) Consider the linear operator T : R3 → R3 which acts as the orthogonal
projection onto the x-axis (in the standard basis), that is, in the standard
coordinates, if v̄ = hx, y, zi, then T v̄ = hx, 0, 0i. Consider R3 as an R[x]module via T .
(a) Decompose it into a direct sum of cyclic modules (using any method
you like, including guessing).
(b) Find the invariant factors of this module.
(c) Write the rational canonical form for T , and Jordan canonical form.
Solution. For (a), note that this module is not cyclic, since for any vector
v, the space generated by v, T v is at most 2-dimensional, and T k v = T v
for all k ≥ 1. So, we need three vectors w1 , w2 , w3 such that T w1 = w2 ,
and T w3 = 0. Then w1 , w2 will be the basis vector of a 2-dimensional
vector space which is a cyclic submodule, and w3 will be the basis vector
of a 1-dimensional subspace that also gives a cyclic submodule. Take w3 =
h0, 0, 1i; w1 = h1, 1, 0i, w2 = T w1 = h1, 0, 0i.
For (b), write the matrix A for this operator in the standard basis (it
has one 1 in the top left corner and 0s elsewhere); then do row and column
operations on the matrix xI − A to get it to canonical form. The invariant
factors come out to be 1, x, and x2 − x.
3
For (c), rational canonical form consist of the companion matrix for x
(which is a 1 × 1-matrix with its only entry equal to 0, and the companion
matrix for x2 −x (recall that the invariant factor 1 gives the quotient module
F [x]/F [x], which is zero, so it is invisible in the decomposition of V as a
direct sum of cyclic modules). The answer for rational canonical form is:
h0 0 0i
000 .
011
To get Jordan form, decompose the invariant factors into irreducible
factors; we get x, x, and x − 1. So, we have two zero 1 × 1-blocks, and a
1 × 1-block for 1, i.e.
h0 0 0i
000 .
001
(9) (This requires Jordan canonical form). Classify all nilpotent linear operators in a 4-dimensional vector space over C, up to conjugacy.
Solution. We proved that two linear operators are conjugate if and only
if they have the same Jordan canonical form. Thus, we just have to find all
possible Jordan forms for nilpotent liner operators in a 4-dimensional space.
First, note that if λ is an eigenvalue of a nilpotent operator, then λ = 0
(if λ 6= 0, then for the corresponding eigenvector v, we’ll have T n v = λn v,
and so T n v 6= 0 for all n, which contradicts the assumption that T is
nilpotent). Thus, all Jordan blocks have 0s on the diagonal, and so the
only remaining matter is the sizes of the blocks. The sizes of the blocks
have to add up to 4. In fact, we just proved that in general, conjugacy
classes of nilpotent linear operators in an n-dimensional vector space over
C are in bijection with partitions of n. Specifically, for n = 4, the Jordan
forms of the nilpotent operators are: the zero operator (which has 4 blocks
of size 1, so corresponds to the partition 4 = 1 + 1 + 1 + 1), and:
0 1 0 0
0 1 0 0
0 1 0 0
0 1 0 0
0
0
0
0
0
0
0
0
0
0
1
0
T2+1+1 = 0 0 0 0 , T2+2 = 0 0 0 1 , T3+1 = 0 0 0 0 , T4 = 00 00 10 01 .
0000
0000
000 0
00 00
(10) (a) Make an example of a linear operator on some vector space V over R,
such that the corresponding R[x]-module has invariant factors x − 2,
(x − 2)2 , (x − 2)3 (x − 3).
(b) For this linear operator, write down its characteristic and minimal
polynomials.
(c) Write the Jordan canonical form for the matrix from (a).
Solution. For (a), we just have to write a companion matrix for each
of the polynomials, and put these blocks together (it’ll give us an operator
in a 1 + 2 + 4 = 7-dimensional space).
(this is
The first two blocks are: 1 × 1-block with 2, and the block 10 −4
4
the companion matrix of (x− 1)2 ); the companion matrix for (x− 2)2 (x− 3)
is an exercise for the reader :)
For part (b): the minimal polynomial is the last invariant factor (x −
2)3 (x − 3). The characteristic polynomial is the product of them all:
(x − 2)(x − 2)2 (x − 2)3 (x − 3) = (x − 2)6 (x − 3).
For (c), each factor of the form (x − a)k that appears in any invariant
factor, gives a size k Jordan block. So, we get: the size 1 Jordan block with
λ = 2; then a size 2 Jordan block with λ = 2 from the factor (x − 2)2 in
4
the second invariant factor; then a size 3 Jordan block with λ = 2 from the
third invariant factor; and a size 1 block with λ = 3. Putting it together,
we get:


2000000
0210000
0 0 2 0 0 0 0
 00 00 00 20 12 01 00  .
0000020
0000003
(11) (a) Prove that if two 3 × 3-matrices have the same characteristic polynomial and the same minimal polynomial, then they are similar.
(b) Make an example of two 4 × 4-matrices that have the same characteristic and minimal polynomials, but are not similar.
Solution. For a 3 × 3 matrix the characteristic polynomial p has degree
3. Since the invariant factors a1 (x), a2 (x), a3 (x) satisfy a1 |a2 |a3 , we have:
case 1: a1 = a2 = a3 = x − λ for some λ. So, if we see that the minimal
polynomial a3 (x) is of degree 1, we can recover the remaining invariant
factors, which determines Jordan form (and therefore, the similarity class
of A) uniquely.
Case 2: a1 = 1, and deg(a2 ) = 1, deg(a3 ) = 2. If we see that a3 has
degree 2, then we can conclude that a1 = 1, and can recover a2 from a3 and
the product p = a1 a2 a3 . Finally, if a3 has degree 3, it means a1 = a2 = 1,
and we again have recovered the invariant factors.
For 4 × 4, consider, for example, the matrices T2+1+1 and T2+2 from
problem 9. Both have characteristic polynomial x4 , and minimal polynomial x2 , but they are not similar (since their Jordan forms are not the
same).