**2011 Qual #8 (Hungerford 3.4.4): Suppose . Prove that is a finite field and identify the field.
Proof-** Recall that is the set of equivalence classes of under the relation that for some . With the sets R and S given we have that all possible elements of are Now we must see which are equivalent under the relation. We have that:

because

because

because

Thus Having only 3 elements we know that so that it is in fact a finite field.

**2013 Qual #1 (Final Exam #8): Let R be a ring and let I, J be ideals. Let be a short exact sequence of R modules.
(a) Prove that
(b) Show that . Give an example of a ring R a short exact sequence of R modules as above, and an ideal I of R such that but .
Proof-** a) Suppose that . Let . Then we have . Therefore . But by being a short exact sequence we have that . Therefore . Since this was arbitrary we have that . Next, is surjective due to exactness so there exists such that . This gives that , but because of the R-mod homomorphism we have , but again since we have that so that . Therefore .

b) Now suppose that . Let elements be the same as in part a and let . Since is onto we have there exists some such that So then by exactness. This gives that there exists such that . Then . Therefore .

For the counterexample take the following short exact sequence:

, where the maps are defined as . Then clearly f and g are homomorphisms, with the property that . Then take to be the ideal of . We have then that but . Note that this counterexample works because the sequence does not split and that the statement would be true for any short exact sequence that splits. This is because if splits we can write And so if an ideal I has the property that , then

**2006 Qual #1 (Hungerford 4.1.9, Takehome midterm #4c): If is an R-module homomorphism such that then .
Proof-** First recall that: Since the kernel and image are submodules of A we have that . To finish we must show the opposite containment. Let . We can write . Then note that by definition and that because of the assumption. Therefore . Thus . Next we must show that . Let . Then and also there exists a such that . But then applying f to both sides of this equality we have . Therefore . This gives that A is the internal direct sum, .

**Qual 2011 #6: Let R be a ring and for be R-modules. Consider the diagram with exact rows of R-mod homomorphisms:
**

**
Suppose that there exist such that . Prove that there exists that makes the diagram commute. Which conditions should and/or satisfy to ensure that is surjective.
Proof-** First we define such a function. Let by where . This can be done because is assumed to be surjective. We must show that this function is well defined. If then there exist which define and . So to show well defined we must show that . We have . So then there exists such that . Then by the commutativity of the first square. Then we have because the bottom row is exact and so always. Therefore . So the map is well defined. It is a composition of R-mod homomorphisms so it is also one. The diagram commutes by construction.

Now, claim that surjective gives that surjective. Let . Since is surjective we have there exists a such that . Since is surjective there exists a such that . Then and hence which gives the map to be surjective.

**Qual 2009 #1 (Final #10): Let R be a ring, M an R-mod, . Define a natural homomorphism where this is the external direct sum. Describe it’s image and kernel and use it to prove the formula where D is a division ring and are vector subspaces.
Proof-** Let be the natural homomorphism defined by . Then clearly , and . Therefore the Kernel is trivial. Now let D be a division ring, then any vector space is free and we have and . Apply to the vector subspaces then by the rank nullity theorem we have Then using the above facts this gives that:

which implies .

**2011 Final #2(i): Is free as a ?
Proof-** Suppose that the statement were true. If it were a free then it would have a basis. Let be a basis for as a . Let , and suppose there exists an element , such that but this is not an integer so contradiction. Thus if the rationals have a basis it cannot have cardinality 1. Suppose then that is a basis. We can write so that the set is not linearly independent. Thus if X will not be linearly independent. Therefore the rationals do not have a basis and so cannot be free as a .

**Qual 2013 #4 (Hungerford 4.2.14, Final #12): Let R be commutative and let E, F be free R-modules of the same finite rank. Prove that if is surjective then it is an isomorphism. Why do we need to assume that R is commutative?
Proof-** Let . Let be a basis for . Then because the kernel is a submodule, we can extend this to a basis for E, . Next, claim that is a basis for . Let . Then because f is an R-module homomorphism we have . This tells us that . Then we can use the basis of the kernel to write: . Then subtracting we get . Because X is a basis for E we get that . Thus is linearly independent. Next, suppose then there exists an element such that . Since X is a basis of E we can write . Then by the homomorphism we have . Then because , we have that the first k elements in the sum are 0. Thus so that b is written as a linear combination of elements in the set . Thus we have that this is a basis of F. However we know . Thus f is one to one. Thus f is an isomorphism.

We need commutativity in order to have the invariant rank property for R, which allows us to have basis’ for E and F with n elements.