This is the exercsies from Chapter 2 of Atiyah-Mcdonald

Exercise 9: Given an exact sequence of A-modules, \(0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0\) where A is a ring. If \(M', M''\) are finitely generated. Then so is \(M\).

Solution: Let's call the first map \(f, f:M' \rightarrow M\) and the second map \(g, g: M \rightarrow M''\). Consider the set, \(\{x_n\}\) of generators of \(M''\) and \(\{y_n\}\) of \(M'\). If we consider the submodule, \(N\), of \(M\) generated by \(\{g^{-1}(x_n)\}\) and \(\{f(y_n)\}\), the image of \(N\) under \(g\) is exactly \(M''\) and it has kernel \(f(M')\). But this is exactly \(M\). So we are done.

Exericse 10: Let \(A\) be a ring, \(r\) an ideal contained in the Jacobson radical of \(A\). Let \(M\) be an \(A\)-module and \(N\) a finitely generated \(A\)-module, and let \(u: M \rightarrow N\) be a homomorphism. If the induced homomorphism \(M/rM \rightarrow N/rN\) is surjective. Then \(u\) is surjective.

Solution: We get the induced homomorphism \(M \rightarrow M / rM \rightarrow N /rN\). Which implies the fact that \(u(M) + rN = N\). Note that \(N\) is finitely generated and \(r \subseteq J(A)\). From our corrolary to Nakayama's Lemma. We can conclude that \(u(M) = N\). Hence, \(u\) is surjective.

Exercise 11: To be done at a later date...