Problem 1: Show that \(\mathbb{Z} / m \mathbb{Z} \otimes \mathbb{Z} / n \mathbb{Z} = 0\) if \(m,n\) are coprime.

Solution: Since \(m\) and \(n\) are coprime. There exists integers \(p,q\) such that \(pm +qn = 1\). Let \(x\ in \mathbb{Z} / m \mathbb{Z}\) and \(y \in \mathbb{Z} / n \mathbb{Z}\)/. Then \(x \otimes y = (pm +qn)(x \otimes y) = pm(x \otimes y) + qn(x \otimes y) = p( mx \otimes y) + q(x \otimes ny) = p(0 \otimes y) + q(x \otimes 0) = 0\)

Problem 2: Let \(A\) be ring, \(\mathfrak{a}\) and ideal, \(M\) is an \(A\)-module. Show that \((A / \mathfrak{a}) \otimes_A M\) is isomorphic to \(M / \mathfrak{a}M\)

Solution: If we tensor the short exact sequence \(0 \rightarrow a \rightarrow A \rightarrow A/\mathfrak{a} \rightarrow 0\) with \(M\), notice that \( a \otimes M \xrightarrow{\phi} A \otimes M \xrightarrow{\psi} (A/\mathfrak{a}) \otimes M \rightarrow 0 \otimes M\) is exact sequence. So then \(ker(\psi) = Im(\phi)\), and \((A / \mathfrak{a} ) \otimes M \cong (A \otimes M) / Im(\phi)\). However, since \(A \otimes M \cong M\), this would mean that \(Im(\phi) \xrightarrow{\phi} \mathfrak{a}M \subset M\). Hence, \((A \otimes M) / Im(\phi) \cong M / \mathfrak{a} M\)

Problem 3: Let \(A\) be a local ring, \(\mathfrak{a}\) an ideal, \(M\) and \(N\) finitely generated \(A\)-modules. Prove that if \(M \otimes N = 0\), then \(M=0\) or \(N=0\).

Solution:Let \(m\) be the maximal ideal of \(A\), and \(k = A / m\) the residue field. We have the following isomorphisms \(( k \otimes M) \otimes (k \otimes N) \cong k \otimes (M \otimes N) \cong k \otimes 0 = 0\). Forcing one of \(k \otimes M \) or \(k \otimes N\) to be equal to 0. Without loss of generality, suppose then that \(k \otimes M\) is 0. Then the previous exercise implie that \(k \otimes M = 0 \cong M / mM\). This implies that \(M = mM\). Since \(M\) was finitely generated and \(A\) was assumed to be local, it has only one maximal ideal. making \(m\) the jacobson radical of \(A\). Applying Nakayamas Lemma, we conclude that \(M = 0\).

Problem 6: For any \(A\)-module, let \(M[x]\) denote the set of all polynomials in \(x\) with coefficients in \(M\), that is to say expressions of the form \(m_0 + m_1x + \ldots + m_r x^r , (m_i \in M)\). Defining the product of an element of \(A[x]\) and an element of \(M[x]\) in the obvious way, show that \(M[x]\) is an \(A[x]\) module. Show that \(M[x] \cong A[x] \otimes_A M\)

Solution: Let \(M[x] = \{ \sum_{i=0} ^r m_i x^i | m_i \in M, r \in \mathbb{Z} \} \). Let the product of an element of \(A[x]\) and \(M[x]\) be given by \((\sum a_j x^j)(\sum m_i x^i) = \sum_{k=1} ((\sum_{i+j = k} a_j m_) x^k)\). To see why \(M[x]\) is an \(A[x]\) module, we need to check that it is distributive and associative. The identity is obviously contained in \(A[x]\) and \(1*p(x) = p(x)\) for any \(p(x) \in M[x]\).

Checking associativity: If \(a(x), b(x) \in A[x]\) and \(m(x) \in M[x]\). Note that \((a(x) b(x)) m(x) = (\sum (\sum_{i+j =l} a_i b_j) x^{l}))(\sum m_k x^k) = \sum ( \sum ( \sum a_i b_j) m_k ) x^r = \sum_r (\sum_{i+j+k =r} a_i b_j m_k) x^r \)

In the other direction: \(a(x) (b(x) m(x)) = (\sum a_i x^i) ( \sum_r ( \sum_{j+k =r} b_j m_k) x^r) = \sum (\sum a_i ( \sum b_j m_k) ) x^l = \sum_l (\sum_{i+j+k =l} a_i b_j m_k) x^ l \)

Checking distributivity: \((a(x) +b(x)) m(x) = (\sum (a_i +b_i) x^i) (\sum m_k x^k) = (\sum a_i x^i + \sum b_i x^i)(\sum m_k x^k) = (\sum a_i x^i)(\sum m_k x^k) + (\sum b_i x^i)(\sum m_k x^k) = a(x) m(x) + b(x) m(x) \) similarly in the other direction.

Thus, \(M[x]\) is also an \(A[x]\) module. Further, we can observe that \(M = AM\) from what we just shown. Then we can write \(M[x]\) as \((AM)[x]\).

Now we show that \(M[x] \cong A[x] \otimes M\). To see this, if we consider the homomorphism \(\varphi: A[x] \otimes M \rightarrow (AM)[x] = M[x]\) where \(\varphi(a(x) \otimes m ) \rightarrow a(x)\). We can see the inverse map clearly, \(\varphi(a(x) m)^{-1} \rightarrow a(x) \otimes m\). This gives us an isomorphism between \(A[x] \otimes M\) and \((AM)[x] = M[x]\).