**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.

2**005 Qual #4: Decide whether they are injective -modules.**

a)

Yes because it is a divisible

b)

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.