For each of the following, either give an example or explain why no such example exists. One or two sentence answers will suffice; you don’t need to give formal proofs.
A commutative ring \(R\) and an ideal \(I\) of \(R\) that is prime but not maximal.
A commutative ring \(R\) and a finitely generated, torsion-free \(R\)-module \(M\) that is not free.
A PID \(R\) and an \(R\)-module that is torsion-free but not free.
A Euclidean domain which is not a UFD.
A UFD which is not a PID.
An integral domain which is not Noetherian.
(Conventions: All rings are rings with 1, and an integral domain is commutative.)
Solution
1. Two nonzero elements \(a,b\) of an ideal of an integral domain \(R\) satisfy the nontrivial relation \(b\cdot a-a\cdot b=0\); hence an ideal of \(R\) is a free \(R\)-module only if it is \(0\) or principal.
The coefficients \(b\) and \(-a\) are nonzero, so a basis of the ideal has at most one element; an ideal with basis \(\{f\}\) equals \((f)\).
2. In part (a), take \(R=\ZZ[x]\) and \(I=(x)\).
\(R/I\cong\ZZ\) is an integral domain but not a field, so \(I\) is prime but not maximal; indeed \((x)\subsetneq(2,x)\subsetneq R\).
3. In part (b), take \(R=k[x,y]\) for a field \(k\) and \(M=(x,y)\).
\(M\) is generated by \(x,y\), and it is torsion-free as a nonzero ideal of an integral domain. If \((x,y)=(f)\), then \(f\) divides \(x\) and \(y\), so \(f\) is a nonzero constant and \((x,y)=R\), contradicting that every element of \((x,y)\) vanishes at \((0,0)\). By step 1, \(M\) is not free.
4. In part (c), take \(R=\ZZ\) and \(M=\QQ\).
For \(n\ne0\) and \(q\ne0\), \(nq\ne0\), so \(\QQ\) is torsion-free. Any two nonzero rationals \(p/q\) and \(r/s\) satisfy \((qr)\cdot\frac pq-(ps)\cdot\frac rs=0\) with nonzero integer coefficients, so a basis has at most one element; but \(\QQ\) is not cyclic, since \(\frac{1}{2}\cdot\frac pq\notin\ZZ\frac pq\) for \(p/q\ne0\). Hence \(\QQ\) is not free.
5. In part (d), no example exists.
A Euclidean domain is a principal ideal domain, since a nonzero element of minimal Euclidean norm in a nonzero ideal generates it, and a principal ideal domain is a unique factorization domain [@DF04].
6. In part (e), take \(R=\ZZ[x]\).
By Gauss’s lemma, \(\ZZ[x]\) is a UFD because \(\ZZ\) is [@DF04]. If \((2,x)=(f)\), then \(f\mid2\) forces \(f\in\{\pm1,\pm2\}\); \(f=\pm1\) is impossible because every element of \((2,x)\) has even constant term, and \(f=\pm2\) is impossible because \(x\notin(2)\). So \(\ZZ[x]\) is not a PID.
7. In part (f), take \(R=k[x_1,x_2,x_3,\ldots]\) for a field \(k\).
\(R\) is an integral domain, and the chain \((x_1)\subsetneq(x_1,x_2)\subsetneq(x_1,x_2,x_3)\subsetneq\cdots\) does not stabilize: \(x_{n+1}\notin(x_1,\ldots,x_n)\), since setting \(x_1=\cdots=x_n=0\) sends that ideal to \(0\) and \(x_{n+1}\) to \(x_{n+1}\ne0\).
8. Q.E.D.
Steps 2 through 7 answer parts (a) through (f).