Theorem: The following are equivalent on an R-module P-
(i) is split exact.
(ii) P is a direct summand of a free module
(iii) P is projective.
Proof- Assume that P is projective and let be a short exact sequence. By the exactness we have that the map g is surjective, and for P we have the identity map from P to P. Since P is projective there exists a map such that . By one of the equivalent properties we know that this gives that the sequence splits.
Assume that every short exact sequence splits. Every module is the homomorphic image of a free module, so there exists some free module F and some surjection . Then we can construct a short exact sequence, where the map is the inclusion map. Then by assumption the sequence splits so that .
Suppose that P is a direct summand of a free module F so that for some R module K. Then we have the canonical projection and inclusion maps, and . Let and . Since F is free and any free module is projective, there exists a map such that . Then consider . We have that . Therefore the diagram commutes so that P is projective.
2009 Qual #3: Let R be a ring. Let M be a finitely generated R-module and let N be a projective R -module. Prove that if is a surjective homomorphism, then the is finitely generated.
Proof- First consider the sequence . This is an exact sequence because f is surjective and is the canonical inclusion map which is injective. Since N is projective we know that the sequence splits so that . Since M is finitely generated it must follow then that is also finitely generated.
2010 Qual #7: Suppose that are R-modules. Prove that is projective iff are projective.
Proof- Suppose that is projective. Then there eixsts a free module F and an R module K such that . Then by the same reason we have that are projective.
Suppose that are projective. Then there exists free modules and R-modules such that and . Then we have that . Since the direct sum of two free modules is still free this gives that is projective.
Lemma 4.3.9(general case): A R-module D is a divisible R module iff D is an injective R-module.
Proof- . Let D be an injective R-module. Take and . We want to find some such that . Consider the principle ideal . Then we have is an exact sequence where is the inclusion map. Define by . Then since D is injective there exists a map such that . So we have that . Therefore we choose and we are done.
2005 Qual #4: Decide whether they are injective -modules.
Yes because it is a divisible
No, it is not divisible because , since 2 is not a unit. Thus it is also not injective.
Final #5b (Hungerford 4.4.3): Let be the canonical epimorphism. The induced map is the zero map. Since we have that is not an epimorphism.
(Give an example of a commutative ring R, R-modules M, M’, N and such that f is surjective, but is not surjective.)
Proof- Let . Suppose that . Then . This is a contradiction, so that f must be the zero map and so . Thus must be the zero map. However we have that because the . Thus cannot be an epimorphism.
Final #6 (Hungerford 4.4.11(a)): Let P be a finitely generated projective left module. Prove that P* is a projective right R module.
Proof- Let P be a finitely generated projective left R module. Then there exists some free module left module F and an epimorphism since every module is the homomorphic image of a free R module. Since P is finitely generated, F can be chosen to be finitely generated. Also, there is some left R module K such that , since P is projective. Then we get that . Since F is a free left module which is finitely generated we get that is a free right R module. Thus is a direct summand of a free module and so is projective.
Final #5a (Hungerford 4.5.7): The usual injection map is a monomorphism of abelian groups. Show that is the zero map.
Proof- Consider what the map does to each generator: and . Thus it is indeed the zero map. Note that and by exercise 2b. Thus we have the zero map between two nonzero spaces so that it is not a monomorphism. This also tells us that tensoring two injective maps need not result in an injective map.
Qual 2013 #3a, Final #4: a) Give an example of a divisible -module. Prove that if N is a torsion right R module, M is a divisible left R module, then .
Proof- Consider the element . Since N is torsion there is some such that . M is divisible so there exists some such that . So then . Thus an arbitrary generator is zero so that .
b) Let R be a commutative ring and let S be a multiplicative set. Prove that if the natural homomorphism of R-modules, by is an isomorphism, then M is divisible with respect to S. Conversely if M is divisible with respect to S then the map f is surjective.
Proof- Suppose that f is an isomorphism. Let and . Since f is an isomorphism we have that is an isomorphism. and so that M is divisible with respect to S.
Now suppose that M is divisible with respect to S. Let then there exists some such that . So then . Thus f is surjective.
Hungerford 4.5.3b: .
Proof- Consider the short exact sequence . Since is flat we have that is also exact. Then since is torsion and is a divisible module we have that . Thus . And since we are done.
2002 Qual #4(ii): Prove or disprove, .
Proof- False , but obviously the rationals are not isomorphic to the integers.
2013 Qual #5: R a ring, and . Is it always true that .
Proof- False, consider the maps and . They are both injective however is not injective so that but .
Qual 2011 #3: Let R be a ring and let M, N be left R-modules. Construct a natural homomorphism of abelian groups . If R = K is a field and M, N are finite dimensional K vector spaces prove that the natural homomorphism is in fact an isomorphism of K vector spaces.
Proof- Define the map by sending the element where . Check that this is a homomorphism of abelian groups.
Then, assuming that R = K is a field we show that is an isomorphism. First,
by Corollary 4.5.12
by Thm 4.4.11(ii)
Thus it is sufficient to show that is surjective by previous exercise. To show that is surjective, let be a basis of M and be a basis of N. By theorem 4.4.11, for each we have that given by . Then these maps form a basis for since M is finitely generated. Now, has a basis where with sending to and all other to zero. Then by definition of we have . This tells us that . So every basis element gets hit by through . Therefore is onto so we are done.