For each of the following, either give an example or indicate that none exists. You do not need to supply proofs. a. Two non-zero \(\mathbf{Z}\)-modules \(A\) and \(B\) such that \(A\otimes_{\mathbf{Z}} B=0\). b. Two non-zero \(\mathbf{Q}\)-modules \(V\) and \(W\) such that \(V\otimes_{\mathbf{Q}} W=0\). c. A ring \(R\) and a module \(M\) over \(R\) which is torsion free but is not free. d. A ring \(R\) and an ideal \(I\) of \(R\) that is maximal but not prime. e. A ring which is not Noetherian. f. A ring whose center is not an ideal.
Solution
1. For part (a), take\[ A=\mathbb Z/2\mathbb Z, \qquad B=\mathbb Z/3\mathbb Z. \]Then \(A\otimes_{\mathbb Z}B=0\).
Using \((\mathbb Z/m)\otimes_{\mathbb Z}B\cong B/mB\), \[ (\mathbb Z/2)\otimes_{\mathbb Z}(\mathbb Z/3) \cong (\mathbb Z/3)/2(\mathbb Z/3)=0, \] because multiplication by \(2\) is an automorphism of \(\mathbb Z/3\).
2. For part (b), no such nonzero \(\mathbb Q\)-modules exist.
A \(\mathbb Q\)-module is a vector space. If \(0\ne v\in V\) and \(0\ne w\in W\), choose linear functionals \(\lambda:V\to\mathbb Q\) and \(\mu:W\to\mathbb Q\) with \(\lambda(v)=\mu(w)=1\). The induced map \[ \lambda\otimes\mu:V\otimes_{\mathbb Q}W\to\mathbb Q \] sends \(v\otimes w\) to \(1\), so \(V\otimes_{\mathbb Q}W\ne0\).
3. For part (c), take \(R=\mathbb Z\) and \(M=\mathbb Q\).
The \(\mathbb Z\)-module \(\mathbb Q\) is torsion-free. It is not free: \(\mathbb Q\) is divisible, whereas a nonzero free abelian group is not divisible.
4. For part (d), no example exists in a ring with identity; without identity, take\[ R=2\mathbb Z, \qquad I=4\mathbb Z. \]
In a ring with identity, a maximal two-sided ideal \(I\) is prime: if \(a,b\notin I\) and \(aRb\subseteq I\), then \(R=I+RaR\) and \(R=I+RbR\), so \(R=R\cdot R\subseteq I+RaR\,RbR\subseteq I+R(aRb)R\subseteq I\), a contradiction. In \(R=2\mathbb Z\), the ideal \(4\mathbb Z\) has index \(2\), so no ideal lies strictly between it and \(R\), and it is maximal. It is not prime, since \(2\cdot2=4\in I\) while \(2\notin I\).
5. For part (e), take\[ R=k[x_1,x_2,x_3,\dots] \]for any field \(k\).
The ascending chain \[ (x_1)\subsetneq(x_1,x_2)\subsetneq(x_1,x_2,x_3)\subsetneq\cdots \] does not stabilize, so \(R\) is not Noetherian.
6. For part (f), again take \(R=M_2(\mathbb Q)\).
Its center is \[ Z(R)=\{aI_2:a\in\mathbb Q\}. \] This is a proper subset of \(R\) but contains the identity matrix. Any two-sided ideal containing \(1\) is all of \(R\), so \(Z(R)\) is not an ideal.