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Exercises on chapter 3
1. If M is a left R-module and X ⊆ M , the annihilator of X, Ann R (X), denotes the set of all
elements of R that act as zero on all elements of X, i.e.
Ann R (X) = {a ∈ R | ax = 0 for all x ∈ X}.
(i) Check that Ann R (X) is always a left ideal of R, and that if in fact X is a submodule of
M then Ann R (X) is a two-sided ideal of R.
(ii) If M is a cyclic R-module, say M = Rx for x ∈ M , then M ∼= R/AnnR (x) as R-modules.
(iii) Prove that every non-zero cyclic module M = Rx has a maximal submodule (a proper
submodule that is maximal amongst all proper submodules of M ).
(iv) Prove that every finitely generated left R-module M has a maximal submodule. (Hint:
Let x 1 ,... , xn be a minimal spanning set for M and let K = Rx 2 + · · · + Rxn , noting that
x 1 ∈/ K.)
(v) Prove that an R module M is irreducible if and only if M ∼= R/I for a maximal left ideal
I of R.
(i) You can check this. (ii) Define a map R → M, r "→ rx. It is an R-module homomorphism with kernel AnnR (x). Hence by the first isomorphism theorem you get induced an isomorphism of left R-modules R/Ann (^) R (x) ∼= M. (iii) By (ii), this follows from the fact that every left ideal of R can be embedded into a maximal left ideal. We proved almost the same thing in class for two-sided ideals using Zorn’s lemma last term... the same argument works for this. (iv) As in the hint, we just need to show that M/K has a maximal submodule. Then by the lattice isomorphism theorem, its pre-image in M is a maximal submodule of M. But that follows by (iii) since M/K is cyclic, generated by the image of x 1. (v) Note R/I is irreducible if and only if I is a maximal left ideal of R, by the lattice isomorphism theorem. This proves one implication easily. Conversely, suppose that M is irreducible. Take 0 $= x ∈ M. Then Rx is a non-zero R-submodule of M , hence Rx = M by irreducibility. Hence M ∼= R/I for some left ideal of R, necessarily maximal as M is irreducible.
2. Let I be a two-sided ideal of a ring R and let M be a left R-module. Explain a natural way
to make M/IM into a left R/I-module.
Define an action of r + I ∈ R/I on M/IM by
(r + I)(m + IM ) = rm + IM.
We need to check this is well-defined. Say r + I = r ′^ + I and m ′^ + IM = m + IM , i.e. (r − r ′^ ) ∈ I and (m − m ′^ ) ∈ IM. Then (r ′^ + I)(m ′^ + IM ) = r ′^ m ′^ + IM. So we need to see that r ′^ m ′^ − rm ∈ IM. But r ′^ m ′^ − rm = (r ′^ − r)m ′^ + r(m ′^ − m) and the right hand side belongs to IM. It is clear that this makes M/IM into a left R/I-module.
3. The goal of this question is to prove directly that Z 60 = Z 4 ×Z 3 ×Z 5 = Z 4 ⊕Z 3 ⊕Z 5. Of course
we know this already by the Chinese remainder theorem since products and coproducts are
the same thing in the category of abelian groups...
(i) Find submodules M 4 , M 3 and M 5 of Z 60 isomorphic to Z 4 , Z 3 and Z 5 respectively with
Z 60 = M 4 ⊕ M 3 ⊕ M 5.
(ii) For each i = 4, 3 , 5, write down the maps p i : Z 60 → Zi so that Z 60 together with these
maps is the direct product of Z 4 , Z 3 and Z 5.
(i) Let M 4 = Z 15 , M 3 = Z20 and M 5 = Z12. Note M 4 ∼= Z 4 , M 3 ∼= Z 3 and M 5 ∼= Z 5. Note M 3 +M 4 = Z5 (because 5 is the GCD of 15 and 20), hence since 5 and 12 are coprime M 5 ∩ (M 3 + M 4 ) = 0. Similarly M 3 ∩ (M 5 + M 4 ) = M 4 ∩ (M 3 + M 5 ) = 0. Hence the sum M 4 + M 3 + M 5 is direct. It is all of Z 60 because GCD(15, 20 , 12) = 1 so you can write 1 = a15 + b20 + c12 for some a, b, c ∈ Z. (ii) Define pi by n "→ n + Zi, i.e. reduce modulo i, for each i = 4, 3 , 5. Since Z 60 = Z/Z60 it follows by the second isomorphism theorem that for i|60, Z 60 /iZ 60 ∼= Z/Zi. So this really does the job. To see that it is the direct product you need to check the universal property. Take maps q (^) i from an abelian group A to Z 4 , Z 3 and Z 5. We need to construct a map from A to Z 60 ... Well, a ∈ A maps to the unique element of Z 60 that is congruent to q (^) i (a) modulo i for each i = 4, 3 , 5...
4. Let {M i }i∈I be a family of left R-modules and M be another left R-module.
(i) Prove that
Hom R (
i∈I
M i , M ) ∼=
i∈I
Hom R (M i , M ).
(ii) Prove that
Hom R (M,
i∈I
M i ) ∼= Hom R
i∈I
(M, M i ).
(iii) Is it true that
Hom R (M,
i∈I
M i ) ∼=
i∈I
Hom R (M, M i )?
How about if M is finitely generated?
(i) Suppose we are given an element of
Q
i∈I Hom^ R^ (Mi^ , M^ ).^ This means we are given maps^ fi^ : Mi → M for each i. Define a map f :
L
i∈I Mi^ →^ M^ by^ f^ (
P
i m^ i^ ) :=^
P
i fi^ (m^ i^ ).^ Note this depends on all but finitely manyQ m (^) i ’s being zero to make sense... In this way we get a map from i∈I Hom^ R^ (Mi^ , M^ ) to Hom^ R^ (
L
i∈I Mi^ , M^ ). It is easy to see that it is actually a homomorphism of abelian groups (i.e. better than just a map of sets). Conversely given f ∈ Hom (^) R (
L
i∈I Mi^ , M^ ) define (fi^ )i∈I^ ∈^
Q
i∈I Hom^ R^ (Mi^ , M^ ) by setting^ fi^ (m^ i^ ) := f (m (^) i ) for each i ∈ I. This is a two-sided inverse to the map in the first paragraph. Note here it is NOT true that
Hom (^) R (
M
i∈I
Mi , M ) ∼=
M
i∈I
Hom (^) R (Mi , M )
if I is infinite. Take the case R is a field F , and let M = Mi = F (the one dimensional vector space) and I = N (or any countable set). The left hand side is then the dual vector space F I^ of a countable dimensional vector space, which is of uncountable dimension. The right hand side is just the vector space F I^ again, of countable dimension. So they cannot be isomorphic... (ii) Suppose we are given an element f ∈ Hom (^) R (M,
Q
i∈I Mi^ ). Define (fi^ )i∈I^ ∈^
Q
i∈I Hom^ R^ (M, Mi^ ) by letting fi (m) = π (^) i (f (m)), the ith coordinate of f (m). Conversely given (fi )i∈I ∈
Q
i∈I Hom^ R^ (M, Mi^ ) define f ∈ Hom (^) R (M,
Q
i∈I Mi^ ) by letting^ f^ (m) = (fi^ (m))i∈I^. (iii) This is false in general but I didn’t try to find a counterexample, sorry.... But if M is finitely generated as an R-module then this is true and you can prove it just like in (ii). Indeed the direct sums here sit inside the direct products from (ii) as the tuples with all but finitely many entries being zero. The restriction of the isomorphism from (ii) to these subgroups gives an isomorphism between the direct sums providing M is finitely generated.
(i) False. Take V = Z 2 ⊕ Z 2 and W = {(0, 0), (1, 1)}, X = {(0, 0), (1, 0)} and Y = {(0, 0), (0, 1)}. (ii) TRUE!!!! Note that AnnR (R/I) = I, AnnR (R/J) = J and if R/I and R/J are isomorphic they have the same annihilators in R. (iii) False. For example take the matrix algebra R = M 2 (C). It is a direct sum of two left ideals I 1 ⊕ I 2 where I 1 consists of matrices with zeros in the second column and I 2 consists of matrices with zeros in the first column. Now R/I 1 ∼= I 2 and R/I 2 ∼= I 1. It remains to observe that I 1 and I (^2) are isomorphic as left R-modules (they are both isomorphic to the module of column vectors C 2 ).
7. True or False? (i) Every finitely generated module over a field has finite length.
(ii) Every finitely generated module over a PID has finite length.
(i) Of course: finitely generated modules over fields are finite dimensional vector spaces, and they have finite bases... (ii) False. Take R = Z and consider the regular module (^) R R which is cyclic, so certainly finitely generated. If it has a composition series, you get it by picking a maximal submodule, then a maximal submodule of that, ... and the process stops in finitely many steps. But a maximal submodule of Z is (2), and as a module that is isomorphic to Z again. So you can go on forever...
8. Prove that every short exact sequence of the form
0 −→ K −→ M −→ R R −→ 0
is split. Is it true that every short exact sequence of the form
0 −→ R R −→ M −→ Q −→ 0
is split?
Let π : M → (^) R R be the epimorphism. We just need to show that it is split. Let m ∈ M be any element with π(m) = 1R. Since R is a free R-module on basis 1 R , there is a unique R-module homomorphism τ : (^) R R → M with τ (1R ) = m. Then π ◦ τ = id as required... But the second short exact sequence need not be split! Take for example R = Z and the monomor- phism i : Z → Z given by multiplication by 2. For this to split, 2Z would have to have a complement in Z. But that complement would have to be a submodule of Z isomorphic to Z 2 , and there can’t be such a thing since Z is an integral domain...
9. This exercise is known as the five lemma. Let
0 −→ A 1 −→ A 2 −→ A 3 −→ 0
and
0 −→ B 1 −→ B 2 −→ B 3 −→ 0
be short exact sequences of R-modules. Suppose there are vertical maps γi : Ai → Bi for
each i = 1, 2 , 3 so that the resulting diagram commutes. Prove that:
(i) γ 1 , γ 3 monomorphisms ⇒ γ 2 is a monomorphism;
(ii) γ 1 , γ 3 epimorphisms ⇒ γ 2 is an epimorphism;
(iii) γ 1 , γ 3 isomorphisms ⇒ γ 2 is an isomorphism.
This is an example of “diagram chasing” which is usually not hard but often tedious. I’ll just do (i). Hopefully my notation for the maps is obvious. Take a 2 ∈ A 2 such that γ 2 (a 2 ) = 0. We need to show that a 2 = 0. Well, γ 3 (π (^) A (a 2 )) = π (^) B (γ 2 (a 2 )) =
- As γ 3 is mono, this implies that π (^) A (a 2 ) = 0 already. Hence a 2 ∈ ker π (^) A = im iA. Hence a 2 = iA (a 1 ) for some a 1 ∈ A 1. Now γ 2 (iA (a 1 )) = iB (γ 1 (a 1 )) = 0. As iB and γ 1 are mono, this implies that a 1 = 0 already. Hence a 2 = iA (a 1 ) = 0 too.
10. Quite soon in class we will show that if R is a PID then every submodule of a free module
is free. Actually we’ll only prove this in class for finitely generated modules, but it is true in
general. Bearing this in mind, think about the following True or False?
(i) If R is a commutative ring, then every submodule of a free module is free.
(ii) If R is an integral domain, then every submodule of a free module is free.
(iii) If R is commutative and every submodule of a free R-module is free, then R is a PID.
(i) False. For example take R = Z 6 and consider the submodule 2Z 6. (ii) False. For example take R = F [x, y], F a field. Consider the submodule (x, y). (iii) True. Its obvious that R has to be an integral domain (else you get counterexamples like in (i)). Now suppose (x, y) is an ideal that is not principal. Then (x, y) is not free (like in (ii)).
11. In linear algebra, you often take linearly independent sets and extend them to bases, and
you often take spanning sets and cut them down to a basis.
(i) Use Zorn’s lemma to prove that any set X of linearly independent vectors in an R-module
M is contained in a maximal linearly independent set.
(ii) Suppose that D is a division ring. We know already that all D-modules are semisimple,
hence free. Prove that every maximal linearly independent set of vectors in an D-module M
is a basis.
(iii) In general you need to be careful: if M is a free R-module and X is a maximal linearly
independent set of vectors in M , then X need not be a basis. For example, suppose that R is
an integral domain with the property that every maximal linearly independent set of vectors
in a free R-module M is a basis. Prove that R is a field.
(iv) Let M be a module over a division ring D and let X ⊆ M be a subset of vectors that
span M as a D-module. Prove that X contains a basis of M. (Hint: use the fundamental
lemma about semisimple modules.)
(i) Let F be the set of all sets Y of linearly independent vectors M containing X. Note F is non-empty as X ∈ F. We just need to prove that F has a maximal element. This follows by Zorn’s lemma if we show every chain (Yi )i∈I in F has an upper bound in F. Well just take Y =
S
i∈I Yi^. Its an upper bound, but why is it in F? Well if not, there is a dependency r 1 y (^) i 1 + · · · + rk y (^) ik = 0 for some i 1 ,... , ik ∈ I and r 1 ,... , rk $= 0. But then you can find i ∈ I such that y (^) i 1 ,... , y (^) ik all already lie in Yi. But that contradicts the linear independence of the set Yi. (ii) Let X be a maximal linearly independent set of vectors in M. Take m ∈ M. Since X is maximal linearly independent, X ∪ {m} is linearly dependent. Hence we can write m as a linear combination of elements of X. Hence X spans M , so its a basis. (iii) Take M = (^) R R. Take 0 $= x ∈ R. This is a linearly independent set of vectors in M , since R is an integral domain. I claim its maximal. Well, if not you can find another 0 $= y ∈ R such that x, y is still linearly independent. But that’s a contradiction: (y)x + (−x)y = 0 is a non-trivial linear relation between x and y. Hence by assumption x is a basis for R. So we can write 1 = yx for some y ∈ R. Hence x is a unit, and R is a field. (iv) Note M =
P
0 $ =x∈X Dx. Each^ Dx^ is simple. Hence by the fundamental lemma on semisimple modules, there exists Y ⊆ X such that M =
L
y∈Y Dy. That is,^ y^ is a basis.
12. Let R and S be rings, let M be a left R-module and let B be an R, S-bimodule.
(i) Prove that Hom R (B, M ), the abelian group of all left R-module homomorphisms from B
to M , can be made into a left S-module by defining the action of s ∈ S on f ∈ Hom R (B, M )
by letting sf be the unique element of Hom R (B, M ) with (sf )(b) := f (bs) for all b ∈ B.
(ii) Prove that Hom R (M, B), the abelian group of all left R-module homomorphisms from M
to B, can be made into a right S-module by defining the action of s ∈ S on f ∈ Hom R (M, B)
by letting f s be the unique element of Hom R (M, B) with (f s)(m) = f (m)s for all m ∈ M.
to R R – but the fact that these are equivalent definitions is far from obvious if R is not a
commutative ring!)
(i) Show that if R is SBN then so is every quotient ring R/I.
(ii) Let A = ZN^ , the free abelian group on the set N. Let R = EndZ (A), the ring of all
Z-module endomorphisms of A. Prove that R is SBN.
(i) LetL M be a finitely generated free left (R/I)-module. Pick a basis x 1 ,... , xn for M , so M ∼= n i=1 (R/I)xi^ and each (R/I)xi^ ∼=^ R/I. View^ M^ now as an^ R-module. Let^ P^ =^
L (^) n i=1 Rxi^ be the free R-module on basis x 1 ,... , xn. Define an R-module homomorphism P! M mapping xi to xi for each i = 1,... , n. Note the kernel of this is IP. So, M ∼= P/IP. Now R has IBN, so P ∼= R R. Hence M ∼= P/IP ∼= R/I. Hence R/I has IBN. (ii) Note A is a left R-module by definition. For i ∈ N, let ei be the projection of A onto its ith component. Note ei ∈ R is an idempotent and Rei ∼= A as a left R-module. (The isomorphism here maps r ∈ Rei to r (^1) i where 1 (^) i is the identity in the ith slot in the direct sum A.). So
R =
Y
i∈N
Rei ∼=
Y
i∈N
A
as a left R-module. But this is a direct product of countably many copies of A. And a finite direct sum (aka direct product) of copies of R is still direct product of countably many copies of A. Hence they’re isomorphic.
16. Let A be the abelian group
Zp^ r 11 ⊕ Zp^ r 22 ⊕ · · · ⊕ Zprnn
where p 1 ,... , p n are distinct primes and r 1 ,... , r n ≥ 1. Compute the abelian groups
Hom Z (Z, A) and HomZ (A, Z).
Since Z is a free Z-module of rank 1, evaluation at 1 defines an isomorphism between Hom (^) Z (Z, A) and A. Hence the first abelian group is just A again. For the second one, pr 11 acts as zero on Zp r 11 , and the only element in Z with this property is zero. So the only map here is the zero map, and this is the zero module.
17. In class, we proved (using Fitting’s lemma) that if M is an indecomposable module of finite
length, then EndR (M ) is a local ring, i.e. the sum of two non-units is a non-unit. Instead
prove Schur’s lemma: if M is an irreducible module then EndR (M ) is a division ring, i.e. 0
is the only non-unit.
Let f : M → M be a non-zero R-module endomorphism with M simple. The kernel of f is not all of M , but it is an R-submodule of M , hence it is zero as M is irreducible. Therefore f is injective. The image of f is not zero, but it is an R-submodule of M , hence it is M as M is irreducible. Therefore f is surjective. Therefore f is invertible.
18. Let e, f ∈ R be idempotents, and consider the summands Re and Rf of R R.
(i) Prove that Re = Rf if and only if f = e + (1 − e)ae for some a ∈ R. (Hint: (⇒) If f = ae
and e = bf then ef e = e and 1f 1 = f where 1 = e + (1 − e).)
(ii) Prove that Re ∼= Rf if and only if there exist a, b ∈ R with ab = e and ba = f.
(iii) Deduce that Re ∼= Rf as left R-modules if and only if eR ∼= f R as right R-modules.
(i) (⇒). If Re = Rf then f = ae and e = bf for some a, b ∈ R. Hence ef = bf f = bf = e and f (1 − e) = ae(1 − e) = 0. So f = ((1 − e) + e)f ((1 − e) + e) = ef e + (1 − e)f e = e + (1 − e)f e. (⇐). If f = e + (1 − e)ae then f ∈ Re so Rf ⊆ Re. And ef = e^2 = e so e ∈ Rf so Re ⊆ Rf. (ii) Note that every element of Hom (^) R (Re, Rf ) is gotten just by right multiplication by some element of eRf. Indeed, right multiplication by such elements certainly gives R-module maps from Re to Rf. Conversely, given an R-module map from Re to Rf , compose with the projection e from R to Re at the beginning and the inclusion of Rf into R at the end and you get an R-module homomorphism from R to R, i.e. an element of End (^) R (R)op^ = R which actually lies in eRf.
Hence an isomorphism Re → Rf is given by right multiplication by some a ∈ eRf and the inverse is given by right multiplication by some b ∈ f Re. Then right multiplication by ab gives the identity map from Re to Re, i.e. ab = e, and similarly ba = f. (iii) Clear as the statement in (ii) is left-right symmetric.
19. Suppose that R = M n (F ) for F a field and n > 1.
(i) Find a decomposition 1R = e 1 +· · ·+em of 1 as a sum of primitive orthogonal idempotents,
and describe the left R-modules Rei and the right R-modules ei R.
(ii) Find another such decomposition.
(iii) Prove that R is a simple ring, i.e. it has no non-trivial two-sided ideals.
(i) Let ei be the matrix with 1F in its ii entry and all other entries zero. These are orthogonal idempotents summing to 1. To see that they are primitive, just note that Rei is the left R-module of column vectors, which is irreducible as you can get any non-zero column vector to any other by multiplying by some matrix. Hence Rei is indecomposable. Hence ei is a primitive idempotent by result from class. (ii) Take any unit in R ×^ , i.e. an invertible matrix u. Then ue (^) i u −^1 also give such things. (iii) Let I be a non-zero two-sided ideal. Take a non-zero matrix m ∈ I. Say the ij-entry of m is not zero. Then ei mej is the matrix equal to this entry in the ij-place and all other entries are zero. Hence I contains the ij-matrix unit ei,j. Now left multiplying you get all other matrix units in the jth column, and right multiplying you get all other matrix units in the ith row. Hence we’ve got all matrix units. Adding we get any matrix. So I = R.
20. Let A = (ai,j ) be an n × n matrix with integer entries and determinant d - = 0.
(i) Let X be the free abelian group on basis x 1 ,... , xn and let Y by the subgroup generated
by y 1 ,... , y n where
y j =
∑^ n
i=
ai,j xi.
Prove that X/Y is a finite abelian group of order |d|. How would you compute the isomor-
phism type of this group explicitly?
(ii) Compute the isomorphism type of the group X/Y explicitly for the following matrices:
[
]
What about for the n × n matrix following the same pattern?
(iii) Do the same for the matrix
(i) Say A ≡ diag(d 1 ,... , dn ) are the invariant factors of A over Z, and assume all di ≥ 0 by rescaling by −1 if necessary. Note d 1 · · · dn = ±d $= 0, since the determinant of an invertible matrix over Z is ±1 (the only units in Z). So all di are non-zero. Then we can pick a basis v 1 ,... , v (^) n for X so that d 1 v 1 ,... , dn v (^) n is a basis for Y. But then its obvious that X/Y ∼= Zd 1 ⊕ · · · ⊕ Zd (^) n. In particular its order is d 1 · · · dn = |d|. (ii) The determinant of the n × n matrix following this pattern is (n + 1), and its invariant factors are (1, · · · , 1 , n + 1). This is an exercise in determinants... Hence the group X/Y is Zn+. (iii) Using the J (^) i trick, you see that it has invariant factors (1, 1 , 2 , 2) and X/Y ∼= Z 2 ⊕ Z 2.
(Thanks to J.B. for this solution.) Suppose that W is a summand of V , say V = W ⊕ U. Take x ∈ R. We need to show that W ∩ xV ⊆ xW. Well, say we have some v ∈ V so that xv ∈ W. Write v = w + u for w ∈ W, u ∈ U. Then xv = xw + xu ∈ W , hence xu = xv − xw ∈ W ∩ U = { 0 }. This shows that xv = xw so it lies in xW. Conversely, suppose that W is a pure submodule of V. The quotient V /W is finitely generated so by structure theorem it is a direct sum of cyclic modules. Say V /W = R(v 1 + W ) ⊕ · · · ⊕ R(v (^) n + W ) for v 1 ,... , v (^) n ∈ V. Suppose that R(v (^) i + W ) is cyclic of order di. If di = 0 set u (^) i = v (^) i. If di $= 0, then di v (^) i ∈ W hence by purity di v (^) i = di wi for some wi ∈ W. Set u (^) i = v (^) i − wi in this case. Note in both cases that v (^) i + W = u (^) i + W , hence V /W = R(u 1 + W ) ⊕ · · · ⊕ R(u (^) n + W ) and each R(u (^) i + W ) is also cyclic of order di. The advantage now is that we actually have di u (^) i = 0 (whereas we only had di v (^) i ∈ W ...). Finally set U = Ru 1 + · · · + Ru (^) n. We claim that V = W ⊕ U , hence W is a summand of V as required. Clearly V = W + U because (U + W )/W = V /W. So we just need to check that W ∩ U = { 0 }. Take r 1 u 1 + · · · + rn u (^) n ∈ W. Then r 1 (u 1 + W ) + · · · + rn (u (^) n + W ) = 0 in V /W. Hence as V /W = R(u 1 + W ) ⊕ · · · ⊕ R(u (^) n + W ), we have that ri (u (^) i + W ) = 0 in V /W for each i. But this means that di |ri , i.e. ri = di si for some si , hence ri u (^) i = si di u (^) i = 0. Hence r 1 u 1 + · · · + rn u (^) n = 0 which is what we wanted...
24. Let R be a commutative ring. Is it true that the left regular R-module is indecomposable?
What if R is a PID?
No. Take for example the ring Z 6. By CRT, it is isomorphic to Z 2 ⊕ Z 3. So it is decomposable. But if R is an integral domain, then the left regular R-module is indecomposable. Proof: Suppose R = A ⊕ B is decomposable as a left R-module. Take 0 $= a ∈ A and 0 $= b ∈ B. Consider ab = ba. It lies in A and in B. Hence its zero. This contradicts R being an integral domain.
25. (i) How many abelian groups are there of order 5^6 75 up to isomorphism?
(ii) How many isomorphism classes of abelian groups are there of order 108 having exactly 4
subgroups of order 6?
(i) It is the number of partitions of 6 times the number of partitions of 5. For 6 there are 11: (6), (5, 1), (4, 2), (4, 1 , 1), (3, 3), (3, 2 , 1), (3, 1 , 1 , 1), (2, 2 , 2), (2, 2 , 1 , 1), (2, 1 , 1 , 1 , 1), (1, 1 , 1 , 1 , 1 , 1). For 5 there are 7: (5), (4, 1), (3, 2), (3, 1 , 1), (2, 2 , 1), (2, 1 , 1 , 1), (1, 1 , 1 , 1 , 1). So there are 77. I hope. (ii) The abelian groups of order 108 are as follows: Z 27 ⊕ Z 4 , Z 27 ⊕ Z 2 ⊕ Z 2 , Z 9 ⊕ Z 3 ⊕ Z 4 , Z 9 ⊕ Z 3 ⊕ Z 2 ⊕ Z 2 , Z 3 ⊕ Z 3 ⊕ Z 3 ⊕ Z 4 , Z 3 ⊕ Z 3 ⊕ Z 3 ⊕ Z 2 ⊕ Z 2. A subgroup of order 6 is Z 3 ⊕ Z 2 , and Z 3 , Z 2 must be in the unique Sylow subgroups of order 27 or 2. Now Z 27 has one subgroup of order 3, while Z 9 ⊕ Z 3 has four and Z 3 ⊕ Z 3 ⊕ Z 3 has 13. Similarly, Z 4 has one subgroup of order 2 and Z 2 ⊕ Z 2 has three. So the only possibility is Z 9 ⊕ Z 3 ⊕ Z 4.
26. True or False?
(i) An n × n nilpotent matrix with entries in a field has characteristic polynomial xn^.
(ii) There are exactly 3 similarity classes of 4 × 4 matrices A over the field F 2 satisfying
A^2 = 1.
(iii) The number of similarity classes of n × n nilpotent matrices over a field F is independent
of the particular field F chosen.
(iv) The matrices
and
are similar over^ Q.
(i),(iii) Both true. The minimal polynomial of a nilpotent matrix divides xn^ so all roots of it are zero, so all its eigenvalues are zero, so up to similarity it is just a bunch of Jordan block J (^) d 1 (0),... , J (^) d (^) k (0) down the diagonal with d 1 ≥ · · · ≥ dk > 0 and d 1 + · · · + dk = n. Obviously the characteristic
polynomial is xn^. And the number of similarity classes is equal to the number of partitions of n, independent of the field F. (ii) True. The minimal polynomial divides x^2 − 1 = (x − 1)^2. So it is just J 1 (1) and J 2 (1)’s down the diagonal in JNF. Now 4 = 1 + 1 + 1 + 1 = 2 + 1 + 1 = 2 + 2 (how to put the Jordan blocks down the diagonal). So there are 3 similarity classes. (iv) Well this was typo, I meant to have 4 by 4 matrices here! But as it is its a very easy question, the answer is false I guess: it doesn’t even make sense to say that non-square matrices are similar at all.
27. How many similarity classes are there of 2 × 2 matrices over F 2?
Possible characteristic polynomials: x^2 , x^2 +1 = (x+1)^2 , x^2 +x = x(x+1), x^2 +x+1. The first three give classes diag(0, 0), J 2 (0), diag(1, 1), J 2 (1), diag(0, 1). The final one is irreducible polynomial, with rational normal form (^) » 0 1 1 1
So I make it 6. Note I often use Jordan normal form even if the field is not algebraically closed. PROVIDING all the roots of the characteristic polynomial lie in the field F , it makes sense to work with JNF. Its just that there are some characteristic polynomials which don’t have roots lying in F , and for such ones you have to resort to the rational normal form.
28. Let V be a 7 dimensional vector space over Q.
(i) How many similarity classes of linear transformations on V are there with characteristic
polynomial (x − 1)^4 (x − 2)^3?
(ii) Of the similarity classes you listed in (i), how many have minimal polynomial (x−1)^2 (x−
2)^2?
(iii) Let φ be a linear transformation from V to V having characteristic polynomial (x −
1)^4 (x − 2)^3 and minimal polynomial (x − 1)^2 (x − 2)^2. Find dim ker (φ − 2 id).
(i) Partitions of 4: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1, 5 in total. Partitions of 3: 3, 2 + 1, 1 + 1 + 1, 3 in total. So there are 15 similarity classes of such linear transformations. (ii) Need biggest Jordan blocks of size 2. So just partitions 2 + 2 or 2 + 1 + 1 for eigenvalue 1 and 2 + 1 for eigenvalue 2, giving just 2 classes in total. (iii) We need to compute the dimension of the 2-eigenspace. There are exactly two Jordan blocks of eigenvalue 2. So this dimension is 2.
29. Let V be a finite dimensional vector space over Q and θ : V → V be a linear transformation
having characteristic polynomial (x − 2)^4.
(i) Describe the possible Jordan normal forms for θ and for each one give the minimal
polynomial of θ.
(ii) For each of the possibilities in (i) compute the dimension of the 2-eigenspace of θ.
(iii) Assume that θ leaves stable only finitely many subspaces of V. What can be said about
the Jordan normal form of θ?
(i) Note we’re looking for a 4 by 4 matrix with all eigenvalues equal to 2: diag(2, 2 , 2 , 2), diag(J 2 (2), 2 , 2), diag(J 2 (2), J 2 (2)), diag(J 3 (2), 2) and J 4 (2). Their minimal polynomials are (x − 2), (x − 2)^2 , (x − 2)^2 , (x − 2)^3 and (x − 2)^4 respectively. (ii) This is the number of Jordan blocks, so 4, 3 , 2 , 2 , 1 respectively. (iii) If the 2-eigenspace is more than 1 dimensional, there are infinitely many 1-subspaces of the 2-eigenspace, all of which are θ-stable. So we must have just one Jordan block of eigenvalue 2, i.e. θ has JNF J 4 (2).
Define a map from V ∗^ × W ∗^ to (V ⊗ W )∗^ mapping (f, g) to the linear map f ⊗ g : V ⊗ W → F that sends v ⊗ w to f (v)g(w) (constructed in class). This is bilinear so we get induced a map V ∗^ ⊗ W ∗^ → (V ⊗ W )∗^. I claim that this map is injective. Take something x in the kernel. We can write it as x = f 1 ⊗ g 1 + · · · + fn ⊗ g (^) n for linearly independent functions f 1 ,... , fn ∈ V ∗^ and some other (not necessarily independent) functions g 1 ,... , g (^) n ∈ W ∗^. It is in the kernel of our map, which means for every v ∈ V and w ∈ W that (f 1 ⊗ g 1 + · · · + fn ⊗ g (^) n )(v ⊗ w) = f 1 (v)g 1 (w) + · · · + fn (v)g (^) n (w) = 0. Since f 1 ,... , fn are linearly independent, we can find v 1 ,... , v (^) n so that fi (v (^) j ) = δi,j. Then for all i and w ∈ W f 1 (v (^) i )g 1 (w) + · · · + fn (v (^) i )g (^) n (w) = g (^) i (w) = 0. Hence g (^) i = 0 for all i. Hence x = 0. If both are finite dimensional then dim V ∗^ ⊗W ∗^ = (dim V )(dim W ) = dim(V ⊗W )∗^. So the injective map we’ve constructed must actually be an isomorphism. If they’re both infinite dimensional, it definitely won’t be an isomorphism.
34. Recall a bilinear form on a vector space V is a bilinear map from V × V to the ground field
F. Given any two such bilinear forms b, b′^ , you can add them together to get a new form
b + b ′^ with (b + b ′^ )(v, w) = b(v, w) + b ′^ (v, w) and you can multiply by scalars too. This makes
the space of bilinear forms on V into a vector space in its own right.
(i) Prove that the vector space of bilinear forms on V is naturally isomorphic to the vector
space (V ⊗ V )∗^.
(ii) Assuming that V is finite dimensional, prove that (V ⊗ V )∗^ is naturally isomorphic to
Hom F (V, V ∗^ ). (Hence giving a bilinear form on V is equivalent to giving a linear map from
V to V ∗^ .)
(i) Take a bilinear form b. This means a bilinear map from V × V to F. By the universal property of tensor product, this induces a unique linear map from V ⊗ V to F , i.e. an element of (V ⊗ V )∗^. Conversely given an element f ∈ (V ⊗ V )∗^ you get a bilinear form (v, w) = f (v ⊗ w)... (ii) If V is finite dimensional then by (i) and the previous question the space of bilinear forms is isomorphic to V ∗^ ⊗ V ∗^. But from class, V ∗^ ⊗ W ∼= Hom (^) F (V, W ) for any finite dimensional V. Hence V ∗^ ⊗ V ∗^ ∼= Hom (^) F (V, V ∗^ ).
35. Let A and B be abelian groups.
(i) Prove that A ⊗Z Zm ∼= A/mA.
(ii) Prove that Zm ⊗Z Zn ∼= Zk where k = GCD(m, n).
(iii) Assuming that A and B are finitely generated, describe A ⊗ Z B in general.
(i) Define a map A×Zm → A/mA by (a, z) "→ az +mA. Note this is well-defined: if z ≡ z ′^ (mod m) then az ≡ az ′^ (mod mA). Also it is balanced. Hence it induces a map A ⊗ (^) Z Zm → A/mA such that a ⊗ 1 "→ a + mA. It is clearly onto. Conversely, define a map A → A ⊗ (^) Z Zm mapping a to a ⊗ 1. Under this map, ma "→ ma ⊗ 1 = a ⊗ m = 0, so mA lies in the kernel. Hence our map factors through the quotient A/mA to induce a well-defined map A/mA → A ⊗ (^) Z Zm under which a + mA "→ a ⊗ 1. This is a two-sided inverse to the map in the preceeding paragraph. (ii) Define a map Zm × Zn → Zk mapping ([a] (^) m , [b] (^) n ) to [a] (^) k [b] (^) k. This is balanced, so induces a map Zm ⊗ (^) Z Zn → Zk such that [a] (^) m ⊗ [b] (^) n "→ [a] (^) k [b] (^) k. Note it is onto since 1 ⊗ 1 maps to 1 which generates Zk. On the other hand, Zm ⊗ (^) Z Zn is cyclic generated by the vector [1] ⊗ [1] (any pure tensor looks like [a] ⊗ [b] = a[1] ⊗ b[1] = ab[1] ⊗ [1]; any tensor is a sum of pure tensors hence a multiple of [1] ⊗ [1] too). And now if you write k = am + bn you have that k[1] ⊗ [1] = am[1] ⊗ [1] + bn[1] ⊗ [1] = a[m] ⊗ [1] + b[1] ⊗ [n] = 0. Hence Zm ⊗ (^) Z Zn is of order at most k. So the epimorphism from the previous paragraph must be onto. OR: you can use (i) which shows that Zm ⊗ (^) Z Zn ∼= Zm /(nZm ), which is (Z/mZ)/((nZ + mZ)/mZ) ∼= Z/(m, n) ∼= Z/(k). (iii) Just write A and B in terms of the primary decompostion and use the fact that ⊗ commutes with ⊕...
36. True or False?
(i) If V is a torsion Z-module then V ⊗Z Q = 0.
(ii) Q ⊗Z Q ∼= Q.
(iii) If R is a commutative ring and M, N are R-modules, then m ⊗ n = m′^ ⊗ n ′^ in M ⊗R N
implies that m = m ′^ and n = n ′^.
(iv) If R is a commutative ring and M, N are free R-modules, then M ⊗R N ∼= M ⊗Z N as
abelian groups.
(i) TRUE. It suffices to show every pure tensor is zero. Take v ⊗ ab. Since v is torsion, mv = 0 for some m $= 0. But then v ⊗ ab = mv ⊗ (^) mba = 0. (ii) TRUE. Define a map Q×Q → Q, ( ab , (^) dc ) "→ abbd. It is balanced, hence it induces a map f : Q⊗ (^) Z Q → Q. It is an isomorphism, because I can write down a two-sided inverse map: g : ab "→ 1 ⊗ ab. Clearly f ◦ g is the identity, but its not quite clear that g ◦ f is.... Consider pure tensor ab ⊗ cd = b ab ⊗ (^) bdc = a ⊗ (^) bdc = 1 ⊗ acbd. It follows that any element of Q ⊗ (^) Z Q (not just pure tensors) can be written as 1 ⊗ ab. Now on such an element it is clear that g ◦ f is the identity. (iii) FALSE. I mean rm ⊗ n = m ⊗ rn and you neededn’t have rm = m... (iv) FALSE. For instance take R = Z × Z. Then R ⊗ (^) R R ∼= R ∼= Z × Z = Z⊕^2 as abelian group. But (Z ⊕ Z) ⊗ (^) Z (Z ⊕ Z) ∼= Z⊕^4 as abelian group.
37. Let R and S be rings. We have the tensor functor ?⊗R? from mod-R × R-mod to abelian
groups and the tensor functor ?⊗S? from mod-S × S-mod to abelian groups. Composing
them in two different ways gives two different functors F = (?⊗R ?)⊗S? and G =? ⊗R (?⊗S ?)
from mod-R × R-mod-S × S-mod to abelian groups.
(i) Explain what it means to say that the functors F and G are isomorphic functors.
(ii) Now use the universal property of tensor product to prove that F and G are indeed
isomorphic. (This is “associativity of tensor product”).
(i) It means for any triple M, N, P of modules there is an isomorphism η (^) M,N,P : (M ⊗ N ) ⊗ P → M ⊗ (N ⊗ P ) such that for any morphisms f : M → M ′^ , g : N → N ′^ and h : P → P ′^ , we have that η (^) M ′ (^) ,N ′ (^) ,P ′ ◦ ((f ⊗ g) ⊗ h) = (f ⊗ (g ⊗ h)) ◦ η (^) M,N,P. (ii) Of course we just need a hommorphism (M ⊗ N ) ⊗ P → M ⊗ (N ⊗ P ) such that the pure tensor (m ⊗ n) ⊗ p maps to m ⊗ (n ⊗ p). The diagram in (i) will then certainly commute. But we need to prove that there exists such a map using the universal property. For each p ∈ P , define a map fp : M × N → M ⊗ (N ⊗ P ) sending (m, n) to m ⊗ (n ⊗ p). Its balanced so induces a unique map f¯p : M ⊗ N → M ⊗ (N ⊗ P ) sending m ⊗ n to m ⊗ (n ⊗ p). Now given this define a balanced map (M ⊗ N ) × P → M ⊗ (N ⊗ P ) mapping (x, p) to f¯p (x). Its balanced so induces unique map (M ⊗ N ) ⊗ P → M ⊗ (N ⊗ P ) such that (m ⊗ n) ⊗ p "→ f¯p (m ⊗ n) = m ⊗ (n ⊗ p). Phew.
38. Let n > 1. Prove that Zn is both a projective and an injective Zn -module.
Clearly its projective as its free. For injectivity, let’s apply the criterion for injectivity. We just need to show that if I is an ideal of Zn and α : I → Zn is a module homomorphism then α extends to a map Zn → Zn. We know I = dZn for some d|n. Let k be such that α([d]) = [k]. Since (n/d) times [d] equals zero in Zn , we must have that (n/d) times [k] equals zero in Zn , i.e. n divides nk/d, i.e. d divides k. Let e = k/d. Now define β : Zn → Zn by β([x]) = [xe]. Since β([d]) = [k] = α([d]) we get that β extends α. Done.
39. Let R be commutative. True or false?
(a) Every submodule of a projective R-module is projective.
(b) Every submodule of an injective R-module is injective.
(b) Every submodule of an injective Z-module is injective.
(c) Every quotient of a projective Z-module is projective.
(d) Every quotient of an injective Z-module is injective.
(a) True. This follows from the hint because every projective Z-module is free. To prove the latter, let P be projective, take a short exact sequence 0 → K → F → P → 0 with F free. Then it splits as P is projective, hence F ∼= P ⊕ K. But this shows P is isomorphic to a submodule of a free module, hence it is free by the hint. (b) False. Take Z inside Q like last week. (c) False. Take Z mapping onto Z 2 , the latter is not projective. (d) True. Quotients of divisible abelian groups are divisible.
45. A right R-module M is called flat if the functor M ⊗R? is exact.
(i) Prove that a direct sum of modules is flat if and only if each of the modules is flat.
(ii) Prove that the regular module R R is flat.
(iii) Prove that any projective module is flat.
(i) Let f : A → B be a monomorphism. Applying the functor (M ⊕ N )⊗ (^) R? and using the fact that tensor commutes with direct sum (which remember is actually some natural isomorphism of functors...), we get the following commutative diagram:
(M ⊕ N ) ⊗ R A
idM ⊕N ⊗f −−−−−−−→ (M ⊕ N ) ⊗ (^) R B ?? y
y
M ⊗ (^) R A ⊕ N ⊗ (^) R A −−−−−−−−−−−→ (idM ⊗f,id (^) N ⊗f )
M ⊗ R B ⊕ N ⊗ R B
where the vertical maps are isomorphisms. Now the top map is injective if and only if the bottom map is injective, which is if and only if both idM ⊗ f and idN ⊗ f are injective. This is exactly what we needed...M ⊕ N is flat if and only if both M and N are flat. (ii) The functor R⊗ (^) R? is isomorphic to the identity functor. The latter is exact, hence so is R⊗ (^) R ?. (iii) Let P be projective. Its a summand of a free module. Hence by (i) we are done if we can prove that any free module is flat. But free modules are direct sums of the regular module. So by (i) again (which holds even for infinite direct sums ... though I only wrote down the proof for two) we are done if we can prove that the regular module is flat. But that is true by (ii)
46. Let M be a left R-module. I explained in class by general nonsense (properties of adjoint
functors) that the functor Hom R (M, ?) is left exact. Give a direct proof of this statement,
then give an example to show that it is need not be exact.
Take an exact sequence 0 → A
f → B
g → C. Apply the functor Hom (^) R (M, ?) to get
0 → Hom (^) R (M, A) f¯ → Hom (^) R (M, B) →¯g Hom (^) R (M, C).
The map f¯ sends θ to f ◦ θ, the map ¯g sends φ to g ◦ φ. Note that g ◦ f = 0, hence applying the functor Hom (^) R (M, ?) we get at once that ¯g ◦ f¯ = 0. This shows for free that im f¯ ⊆ ker ¯g. Now we show that f¯ is injective. Say θ : M → A has f¯ (θ) = 0. This means that f (θ(m)) = 0 for all m ∈ M. But f is injective, hence in fact θ(m) = 0 for all m ∈ M , i.e. θ = 0. Finally we show that im f¯ ⊇ ker ¯g. Take φ : M → B such that ¯g(φ) = 0. This means that g(φ(m)) = 0 for all m ∈ M. We need to show that there exists some θ : M → A such that ( f¯ )(θ) = φ, i.e. f (θ(m)) = φ(m) for all m ∈ M. Take m ∈ M. Since g(φ(m)) = 0 we have that φ(m) ∈ ker g = im f , hence there exists a unique element θ(m) ∈ A such that f (θ(m)) = φ(m). This defines a map θ : M → A. It is an R-module homomorphism because f (θ(rm)) = φ(rm) = rφ(m) = rf (θ(m)) = f (rθ(m)), hence θ(rm) = rθ(m).
47. Let GROU P S be the category of all groups, and AB be the category of all abelian groups.
There is an obvious “forgetful functor” F : AB → GROU P S. We also have the functor
α : GROU P S → AB (“abelianization”) defined on objects by G *→ G/[G, G] (and on
morphisms?). The functor α is either left adjoint or right adjoint to F , I always forget which
until I work it out. Well work it out, then prove that its the case.
Take an abelian group A and any old group K. Then,
Hom (^) GROU P S (K, F A) ∼= Hom (^) AB (K/[K, K], A) ∼= Hom (^) AB (αK, A).
The first isomorphism is because any map from K to an abelian group sends the commutator subgroup [K, K] to zero, so by the universal property of quotients must factor through to give a unique map from K/[K, K] to A. Hence, (α, F ) is an adjoint pair of functors, and α is left adjoint to F.
48. Let S be a unital subring of a ring R. There is a functor ResRS (“restriction”) from left R-
modules to left S-modules defined by sending an R-module M to the same thing but viewed
just as an S-module. Prove that R⊗S? is left adjoint to ResRS.
I claim that the functor ResRS is isomorphic to the functor Hom (^) R (R R (^) S , ?). Yes that’s pretty obvious. Once you’ve noticed this it is exactly adjointness of tensor and hom: (^) R R (^) S ⊗ (^) S? is left adjoint to Hom (^) R (R R (^) S , ?). Incidentally, ResRS also has a right adjoint. Indeed, you can also view it as the functor (^) S RR ⊗ (^) R ?. Then by adjointness of tensor and hom, the functor Hom (^) S (S RR , ?) is right adjoint to it.
49. Prove that every module over the matrix ring M n (F ) (F a field) is semisimple.
Well one proof is to note that Mn (F ) is Morita equivalent to F , and every F -module is semisimple. But you can prove this directly quite easily, as follows. The left regular module is semisimple (it is a direct sum of n copies of the “natural” irreducible module of column vectors – each column gives a copy of this module). Hence any free module is semisimple. Hence every module is semisimple (as a quotient of a free module). Sorry this was a bit of a lame question.
50. (Pontrjagin duality) If G is an abelian group, its Pontrjagin dual is the group G ∗^ = HomZ (G, Q/Z).
In fact Hom Z (?, Q/Z) is a contravariant functor from abelian groups to abelian groups.
(i) Prove that Q/Z is an injective abelian group.
(ii) Prove that if 0 → A → B → C → 0 is a short exact sequence of abelian groups, then so
is 0 → C ∗^ → B ∗^ → A∗^ → 0.
(iii) Prove that (Zn )∗^ ∼= Zn.
(iv) If G is a finite abelian group prove that G ∗^ ∼= G.
(i) It is divisible. (ii) That the functor Hom (^) Z (?, Q/Z) is left exact is easy enough to prove (e.g. by direct argument like question 46 above). This means that we certainly have an exact sequence 0 → C ∗^ → B ∗^ → A∗^. The tricky thing is to show that the injective map f : A → B induces a surjective map f ∗^ : B ∗^ → A∗^ , i.e. that Hom (^) Z (?, Q/Z) sends injections to surjections. This follows because Q/Z is an injective module by (i). Indeed, we need to prove that for every map α : A → Q/Z there is a map β : B → Q/Z such that α = β ◦ f. We get that by the definition of injective module... (iii) Take the ses 0 → Z → Z → Zn → 0 where the first map f : Z → Z is multiplication by n. Apply the functor Hom (^) Z (?, Q/Z) using (ii) to get a ses 0 → (Zn )∗^ → Hom (^) Z (Z, Q/Z) → Hom (^) Z (Z, Q/Z) → 0. The map Hom (^) Z (Z, Q/Z) → Hom (^) Z (Z, Q/Z) here sends θ to θ ◦ f. But Hom (^) Z (Z, Q/Z) ∼= Q/Z, via the isomorphism given by evaluation at 1. Making this identification, noting that θ◦f (1) = θ(n) = nθ(1), you get a short exact sequence
0 → (Zn )∗^ → Q/Z → Q/Z → 0.
