This is my solution to problems 1,4,5 and 14 in Atiyah Mcdonald

Problem 1: Let \(S\) be a multiplicatively closed subset of a ring \(A\), and let \(M\) be a finitely generated \(A\)-module. Prove that \(S^{-1}M=0\) if and only if there exists \(s \in S\) such that \(sM = 0\)

Solution: \((\Rightarrow)\) Let \(M\) be a finitely generated \(A\) module. With elements \(b_1 , \ldots, b_n\) as the generators of \(M\) and suppose that \(S^{-1}M = 0\). Then for any \(b_i\), we have that \(\frac{b_i}{1} = \frac{0}{1}\) in \(S^{-1}M\). Hence, there exists an elements \(u_i \in S\) such that \((b_i - 0)u_i=0\). If we take the collection of these such \(u_i\), then for any element \(m \in M\), we can express it as the sum \(m = \sum h_i b_i\), where \(h_i \in M\). Taking the product of the corresponding \(u_i\)'s, call it \(u\). We see that \(0 = um = u(1m - 1\dot 0)\). The necessary form for an element of \(S^{-1}M\). Thus, \(\frac{m}{1} = \frac{0}{1}\) for any \(m \in M\)

In the other direction, suppose that \(sM = 0 \) for some \(s \in S\). Then for any \(\frac{m}{u} \ in S^{-1}M\) we have that \(s(m- u\dot 0)=sm=0\). So, \(\frac{m}{u} = \frac{0}{1}\) in \(S^{-1}M\).

Problem 4: Let \(f : A \rightarrow B\) be a homomorphism of rings and let \(S\) be a multaplicatively closed subset of \(A\). Let \(T = f(S)\). Show that \(S^{-1}B\) and \(T^{-1}B\) are isomorphic as \(S^{-1}A modules\)

Solution: Let \(f: S^{-1}B \rightarrow T^{-1}B\) send \(\frac{b}{s} \rightarrow \frac{b}{f(s)}\) and let \(g: T^{-1}B \rightarrow S^{-1}B\) send \(\frac{b}{t} \rightarrow \frac{b}{f(t)^{-1}}\). Then we see that \(g \circ f (\frac{b}{s}) = g ( \frac{b}{f(s)} ) = \frac{b}{f(f(s))^{-1}} = \frac{b}{s}\) for any \(\frac{b}{s} \in S^{-1}B\). Thus, this is an isomorphism.

Problem 5: Let \(A\) be a ring. Suppose that for each prime ideal \(\mathfrak{p}\), the local ring \(A_{\mathfrak{p}\) has no nilpotent element \(\neq 0\). Show that \(A\) has no nilpotent element \(\neq 0\). If each \(A_\mathfrak{p}\) is an integral domain, is \(A\) necessarily an integral domain?

Solution: From one of the corrolaries in the book, the nilradical, \(N\), of \(A_p\) is exactly \(N_p = \{ \frac{h}{n} | h \in \mathfrak{p}, n \in N \}\). But \(N_p = 0\) for any prime ideal \(\mathfrak{p}\), then we can conclude that \(N =0\) from another previous corrolary. For a counterexample, consider the ring \((\mathbb{Z} / 2 \mathbb{Z}) \times (\mathbb{Z} /2 \mathbb{Z})\). The localization of this ring at its prime ideals is exactly \(\mathbb{Z} / 2 \mathbb{Z}\). However, we also have that \((1,0) * (0,1) = 0\). So this is not an integral domain.

Problem 14: Let \(M\) be an \(A\)-module and \(a\) an ideal of \(A\). Suppose that \(M_m = 0\)for all maximal ideals \(m \supseteq a\). Prove that \(M = \mathfrak{a}M\).

Solution: Since \(M_m = 0\), we can write it as \( 0 = M_m / (aM)_m \cong (M / aM)_m\) by a previous corrolary (3.4). This is then isomorphic to \( A_m \otimes_A (M / aM) \) by corrolary (3.5). If we consider the element \(\frac{x}{p} \otimes m \in A_m \otimes M/aM\), where \(p \in A \setminus m\). This element is equal to \(\frac{1}{p} \otimes xm = 0\). This means that \((A / a)_m \otimes_{A /a} M /aM = 0\). This is because we are using fewer elements in \(A/a\). (3.5) Tells us that we can write \((M/aM)_{m/a} \cong (A/a)_{m/a} \otimes _{A/a} M/aM\). It follows then that \((A / a)_m\) and \((A/a)_{m/a}\) are isomorphic since \((A/a)_m \otimes_{A/a} M/aM = 0 = (A/a)_{m/a} \otimes_{A /a} M/aM\). Since we have a correspondence between maximal ideals \(m \supseteq a\) and maximal ideals \(m' \subseteq A/a\), the localization of \(M/aM\) at a maximal ideal of \(A/a\) is then zero. Now we apply corollary (3.8) to see \(M /aM =0\). Hence, \(M = aM\)

Testing words here