Give an example of each of the following. Justification is not necessary.
A principal ideal domain \(D\) that is not a field.
A unique factorization domain \(D\) that is not a principal ideal domain.
A commutative ring \(R\), and a torsion-free \(R\)-module \(M\), such that \(M\) is not a free \(R\)-module.
A commutative ring \(R\), and two non-trivial \(R\)-modules \(M\) and \(N\) such that \(M \otimes_R N \cong \{0\}\).
Solution
1. For part (a), take \(D=\mathbb Z\).
The integers form an integral domain. A nonzero ideal \(I\) contains a positive integer; let \(d\) be its least positive element. For \(a\in I\), Euclidean division writes \(a=qd+r\) with \(0\leq r<d\). Since \(r=a-qd\in I\), minimality implies \(r=0\). Hence \(I=(d)\). The zero ideal is also principal, so \(\mathbb Z\) is a PID. It is not a field, because \(2\) has no multiplicative inverse in \(\mathbb Z\).
2. For part (b), take \(D=\mathbb Z[x]\).
The integers are a UFD, and the polynomial ring over a UFD is a UFD by Gauss’s lemma [@DF04]. Thus \(D\) is a UFD. The ideal \(I=(2,x)\) is proper: the homomorphism \(\mathbb Z[x]\to\mathbb F_2\), \(f\mapsto f(0)\bmod2\), annihilates both generators but does not annihilate \(1\).
Suppose \(I=(h)\). Since \(h\) divides the nonzero constant \(2\), degree additivity in the domain \(\mathbb Z[x]\) forces \(h\) to be a nonzero integer. Since it also divides \(x\), comparison of the coefficient of \(x\) in \(x=hu(x)\) shows that \(h\) divides \(1\). Thus \(h\) is a unit, which would imply \(I=D\), contradicting properness. Therefore \(D\) is not a PID.
3. For part (c), take \(R=\mathbb Z\) and \(M=\mathbb Q\).
For \(0\ne n\in\mathbb Z\), the equation \(nq=0\) in \(\mathbb Q\) implies \(q=0\), so \(M\) is torsion-free. But \(2M=M\), since \(q=2(q/2)\) for every rational \(q\). In a nonzero free \(\mathbb Z\)-module, a basis vector cannot lie in twice the module: every coefficient in the basis expansion of \(2v\) is even, whereas that basis vector has coefficient one at itself. Since \(\mathbb Q\) is nonzero and equals twice itself, it cannot be free.
4. For part (d), take\(R=\mathbb Z\), \(M=\mathbb Z/2\mathbb Z\), and\(N=\mathbb Z/3\mathbb Z\).
Both modules are nonzero. For any elementary tensor \(t=a\otimes b\), the tensor relations give \(2t=(2a)\otimes b=0\) and \(3t=a\otimes(3b)=0\). Therefore \(t=(3-2)t=0\). Every element of the tensor product is a finite sum of elementary tensors, so the whole tensor product is zero, as required.