Section 10.3.
Homework #3
Masaya Sato
Let R be a ring with 1 and M be an R-module. 1. Prove that if A and B are sets of the same cardinality, then the free modules F ( F (A) and
F ( F (B ) are isomorphic. Proof. Let A and B both consist of n elements, i.e.
A = {a1 , . . . , an } and
B = {b1 , . . . , bn }.
Then free R-modules generated by A and B are F ( F (A) = {r1 a1 + · · · + rn an |ri ∈ R} and
F ( F (B ) = {r1 b1 + · · · + rn bn |ri ∈ R}.
Now define an R-module homomorphism ϕ : F ( F (A) → F ( F (B ) by ϕ(ai ) = bi
for i = 1, . . . , n. n.
Then for every b ∈ F ( F (B ) there are r1 , . . . , rn such that b = r1 b1 + · · · + rn bn . So for a = r1 a1 + · · · + rn an ∈ F ( F (A) ϕ(a) = ϕ(r1 a1 + · · · + rn an ) = r1 ϕ(a1 ) + · · · + rn ϕ(an ) = r1 b1 + · · · + rn bn =b and thus ϕ is surjective. Moreover for a = r1 a1 + · · · rn an and a = r1a1 + · · · + rn an in F ( F (A), where ri , ri ∈ R for i = 1, . . . , n suppose that
ϕ(a) = ϕ(a ) ⇒ ϕ(r1a1 + · · · + rn an ) = ϕ(r1a1 + · · · + rn an ) ⇒ r1 ϕ(a1 ) + · · · + rn ϕ(an ) = r1 ϕ(a1 ) + · · · + rn ϕ(an ).
This equation reduces to
(r1 − r1 )b1 + · · · + (r (rn − rn )bn = 0. Since F ( F (B) is a free R-module generated by B , ri = ri for i = 1, . . . n. n. Therefore a = a and hence ϕ is injective. Thus ϕ is an isomorphism and F ( F (A) is isomorphic to F ( F (B ).
3. Show that the F [ F [x]-modules in Exercises 18 and 19 of Section 1 are both cyclic. Exercise 18. Let V = R2 be the F [ ]-module, where where F = R, with the linear Proof. Exercise F [x]-module,
transformation T =
0 −1 1
0
.
Abstract Abstract Algebra Algebra by Dummit and Foote 1
Section 10.3.
Homework #3
Masaya Sato
Then let a = (1, (1, 0)T ∈ V . V . So for every nonzero v = (r, s)T =∈ V , V , where r, s ∈ R, (1, 0)T + s(0, (0, 1)T = rI (a) + sT ( v = (r, s)T = r(1, sT (a) = (rI + sT )( sT )(a a) and thus V is generated by the single element a ∈ V . V . 2 Exercise 19. Next let V = R be the F [ F [x]-module, where F = R, with the linear transformation 0 0 T = 0 1
and let a = (1, (1, 1) ∈ V . V . Then for every nonzero v = (r, s)T ∈ V with r, s ∈ R, (0, s − r)T = r(1, (1, 1)T + (s (s − r)(0, )(0, 1)T v = (r, s)T = (r, r)T + (0, = rI (a) + (s ( s − r )T ( T (a) = (rI + (s (s − r)T )( T )(a a). Therefore V is generated by a and thus V is cyclic. 4. An R-module M is called a torsion module if for each m ∈ M there is a nonzero element Tor(M ) in the notation of r ∈ R such that rm = 0, where r may depend on m (i.e. M = Tor(M Exercise Exercise 8 of Section Section 1). Prove Prove that every every finite abelian group is a torsion torsion Z-module. Give an example of an infinite abelian group that is a torsion Z-module.
Proof. Let G be a finite abelian group. Define a ring action Z × G → G by k
g k.g = kg = i=1
for all k ∈ Z and all g ∈ G. Then 1.g = 1g = g and
rs
r
s
(rs) ) = ( ) = = rs .g rs) rs g g g = r (sg) sg) = r.( r.(s.g) s.g) i=1
i=1 j =1
for all r, s ∈ Z. Therefore for each g ∈ G there exists n ∈ Z>0 such that ng = ng = g + · · · + g = 0 since g has finite order. Hence n.g = ng = 0 and g is a torsion element. Thus G is a torsion Z-module. And an example of infinite abelian group that is a torsion Z-module is Q/Z since every element a ∈ Q/Z has finite order. 5. Let R be an integral integral domain. domain. Prove Prove that every finitely finitely generated generated torsion torsion R-module has a nonzero annihilator i.e. there is a nonzero element r ∈ R such that rm = 0 for all m ∈ M
– here r does not depend on m (the annihilator of a module was defined in Exercise 9 of Section 1). Give an example of a torsion R-module whose annihilator is the zero ideal. Abstract Abstract Algebra Algebra by Dummit and Foote 2
Section 10.3.
Homework #3
Masaya Sato
Proof. Let M be a finitely generated torsion R-module, i.e.
M = {r1 m1 + · · · + rn mn |ri ∈ R ∀i = 1, . . . , n}, where {m1 , . . . , mn } is the generating generating set. Since each each mi is an element in M for i = 1, . . . , n, n, there exists nonzero si ∈ R such that si mi = 0. Therefore for every m ∈ M has the form m = r1 m1 + · · · + rn mn for some ri ∈ R. So r = s1 · · · sn is a nonzero element in R, and )(r1 m1 + · · · + rn mn ) rm = (s1 · · · sn )m = (s1 · · · sn )(r = ((s ((s1 · · · sn )r1 )m1 + · · · + ((s ((s1 · · · sn )rn )mn = (s2 · · · sn r1 )(s )(s1 m1 ) + · · · + (s (s1 · · · sn 1 rn )(s )(sn mn ) = (s2 · · · sn r1 )0 + · · · + (s (s1 · · · sn 1 rn )0 =0 −
−
since R is an integral domain. Hence r is a nonzero annihilator. For an example of a torsion R-module, where annihilator is the zero ideal, consider a subset continuou uouss map with compact compact support. support. Then Then M is a module M of the set C (R, R) of all contin over itself, where the ring action is given by the pointwise multiplication f g(x) = f ( f (x)g(x) for every x ∈ R. Then for every every continuo continuous us map f ∈ M there exists a nonzero continuous map g ∈ M so that gf = 0, where g has a distinct compact support from f . f . However there is no nonzero g ∈ M such that gf = 0 identically for every f ∈ M . 6. Prove that if M is a finitely generated R-module that is generated by n elements then
every quotient of M may be generated by n (or fewer elements elements). ). Deduce Deduce that quotients quotients of cyclic modules are cyclic. Then en not notee Proof. Let an R-module M be generated by a spanning set A = {a1 , · · · , an }. Th that every submodule N is generated by a subset {ai , . . . , aik } of A, where {i1 , . . . , ik } ⊂ {1, . . . , n}. So M/N has a spanning set that consists of elements 1
a1 + N , . . . , an + N ∈ M/N . However {ai , . . . , aik } generates N and thus aij + N = N for j = 1, . . . k. Thereforee every every k. Therefor m + N ∈ M/N is of the form 1
(rn an + N ) = r1 a1 + · · · + rn an + N m + N = (r1a1 + N ) + · · · + (r Abstract Abstract Algebra Algebra by Dummit and Foote 3
Section 10.3.
Homework #3
Masaya Sato
and hence M/N can be generated by n or fewer elements since aij + N = N . Now suppose that an R-module M is cyclic i.e. there exists an element a ∈ M such that every m ∈ M is of the form m = ra for some r ∈ R. In other other words, words, M is generated by a single element a. This This deduces deduces that every nontrivial quotient module M/N is genera generated ted by one elemen element. t. Theref Therefore ore M/N is cyclic. 7. Let N be a submodule of M . Prove that if both M/N and N are finitely generated then
so is M . of M , suppose that M/N and N are finitely Proof. For an R-module M and its submodule N of M generated. Then there exist spanning sets {a1 + N , . . . , am + N } and {b1 , . . . , bn } for M/N and N , respectively. So for every m ∈ M m + N can N can be expressed as an R-linear combination of the spanning set, i.e. (r1 am + N ) = r1 a1 + · · · + rm am + N , m + N = (r1 a1 + N ) N ) + · · · + (r N , where r1 , . . . , rm ∈ R and m − (r1 a1 + · · · + rmam ) ∈ N . Therefore there are r1 , . . . , rn ∈ R such that
m − (r1 a1 + · · · + rm am ) = r1 b1 + · · · + rn bn and hence
m = r1 a1 + · · · + rm am + r1 b1 + · · · + rn bn . Thus M is finitely generated since every m ∈ M can be expressed as an R-linear combination of the spanning set {a1, . . . , am } ∪ {b1 , . . . , bn }. 8. Let S be the collection of sequences (a ( a1 , a2 , a3 , . . . ) of integers a1 , a2 , a3, . . . where all but finitely many of the ai are 0 (called the direct sum of infinitely many copies of Z). Recal Recalll
that S is a ring under componentwise addition and multiplication and S does not have a multiplicative identity. Prove that S is not finitely generated as a module over itself. Proof. Suppose by contradiction that S is finitely generated by n elements of S , i.e.
S = {r1 s1 + · · · + rn sn |ri ∈ Z ∀i = 1, . . . , n}. Consider the map S si → ki ∈ Z that assigns to the number of nonzero entry of the sequence si for i = 1, . . . , n. n. So the number of a nonzero entry s ∈ S is at most k1 + · · · + kn , which which is still finite. finite. Howeve Howeverr this contradict contradictss that S is finitely generated since (2, (2 , 2, 2, . . . ) can not be expresses as any linear combination of the generating set {s1 , . . . , sn }. Therefore S is not finitely generated. Abstract Abstract Algebra Algebra by Dummit and Foote 4
Section 10.3.
Homework #3
Masaya Sato
9. An R-module M is called irreducible if M = 0 and 0 and M are the only submodules
of M . Sh Shoow that that M is irreducible if and only if M = 0 and M is cyclic module with any nonzero nonzero element element as generator. generator. Determine Determine all the irreducible irreducible Z-modules. Proof. (⇒) Suppose first that an R-module M is irreducible. Since M = 0, there is a nonzero element a ∈ M . Consider a submodule N of M generated by the single element a, i.e.
N = {ra|r ∈ R}. However N = M since M is irreducible. Thus M is cyclic because of the single generator a. (⇐) Suppose conversely that M is cyclic. So M has a generator a ∈ M and M = {ra|r ∈ R}. Note that 0 and M are both submodules of M . So suppose by contradiction that there is a proper submodule N of M of M . Since N is a submodule of M of M , N is also generated by an element b ∈ N ≤ M and b = sa for some s ∈ R. However b = sa ∈ M and rb ∈ M implies that rb = r (sa) sa) ∈ N , N , where r ∈ R. Therefore Therefore N = M and this contradicts that N is a proper submodule of M . Hence M is irreducible. commutative. e. Show that an R-module M is irreducible if and only if M 10. Assume R is commutativ is isomorphic (as a R-module) to R/I where I is a maximal ideal of R. [By the the previo previous us exercise, if M is irreducible there is a natural map R → M defined by r → rm, where m is rm, any fixed nonzero element of M .] .] Proof. (⇒) Suppose first that M is irreducible. So there exists a generator m ∈ M so that
M = {rm|r ∈ R}. Define a map ϕ : R → M by ϕ(r) = rm. rm. Then ϕ is an R-module homomorphism since for all r,s,t ∈ R, where R is a module over itself, (ts))m = rm + t(sm) ϕ(r + ts) ts) = (r + ts) ts)m = rm + (ts sm) = ϕ(r) + tϕ( tϕ(s). So the submodule ϕ(R) is 0 or M because M is irreducible. But ϕ(R) = 0 because 1m = m = ϕ(1) = 1m 0. Thus ϕ(R) = M and ϕ(R) is surjective. Now let I = ker ϕ. Then I is a maximal ideal or M and therefore R/ ker ϕ ∼ = M ⇒ R/I ∼ = M . (⇐) Conversely suppose that M is isomorphic to R/I , where I is a maximal ideal of R, via an R-module isomorphism ψ : R/I → M . Recall that R/I is an irreducible quotient module. Therefore M = ψ (R/I ) is also irreducible. Abstract Abstract Algebra Algebra by Dummit and Foote 5
Section 10.3.
Homework #3
Masaya Sato
11. Show that M 1 and M 2 are irreducible R-modules, then any nonzero R-module homomor-
phism from M 1 to M 2 is an isomorphism. Deduce that if M is irreducible then End R (M )is )is a division ring (this result is called Schur’s Lemma ). ). [Consider the kernel and the image.] isomorphism. hism. Since ϕ is nonzero, im ϕ Proof. Let ϕ : M 1 → M 2 be a nonzero R-module isomorp is a nonzero submodule of M 2 . But im im ϕ = M 2 because M 2 is irreducible and thus does not contai contain n any proper submodu submodules les.. So ϕ is surjectiv surjective. e. Moreov Moreover er ker ker ϕ is a submodule of M 1 . This This implies implies that ker ker ϕ = M 1 or ker ϕ = {0} since M 1 is irreduc irreducibl ible. e. So suppose suppose that ker ϕ = M 1 . Then ϕ(M 1 ) = {0}, and this is contradiction because ϕ is a nonzero homomorphism by assumption. Therefore ker ϕ = {0} and hence ϕ is injective. Then for every R-module homomorphism ψ : M → M , where where M is an irreducible Rmodule, ψ is an isomorphism by the argument above. I.e. every ψ has the inverse R-module homomorphism ψ 1 such that −
−1
ϕ◦ϕ
−1
=ϕ
◦ ϕ = idM ,
where idM is the identity R-module homomorphism. homomorphism. Therefor Thereforee EndR (M ) is a division ring with the identity homomorphism idM that is distinct from the zero homomorphism.
Abstract Abstract Algebra Algebra by Dummit and Foote 6
