Give an example of each of the following. (You don’t need to justify your answers.)
A maximal ideal of \(\mathbf{Z}[x]\).
A ring \(R\) and an ideal \(I\) of \(R\) that is prime but not maximal.
A ring \(R\) and a module \(M\) over \(R\) which is torsion free but isn’t free.
A unique factorization domain which isn’t a PID.
A ring which isn’t Noetherian.
Solution
1. For part (a), take \((2,x)\subseteq\mathbb Z[x]\).
The homomorphism \(\mathbb Z[x]\to\mathbb F_2\) sending a polynomial to its constant coefficient modulo \(2\) is surjective. Its kernel is \((2,x)\): a polynomial has even constant coefficient exactly when it has the form \(2a+xg(x)\) with \(a\in\mathbb Z\) and \(g(x)\in\mathbb Z[x]\). Thus the quotient is the field \(\mathbb F_2\), so the ideal is maximal.
2. For part (b), take \(R=\mathbb Z\) and \(I=(0)\).
This is a prime ideal because a product of integers is zero only when one factor is zero. It is not maximal, since \[ (0)\subsetneq 2\mathbb Z\subsetneq\mathbb Z. \] The first containment is strict because \(2\ne0\), and the second because \(1\) is not even.
3. For part (c), take \(R=\mathbb Z\) and \(M=\mathbb Q\).
If \(n\in\mathbb Z\setminus\{0\}\) and \(nq=0\) in \(\mathbb Q\), then \(q=0\). Thus this module is torsion free.
Suppose it had a \(\mathbb Z\)-basis \(B\). Since \(\mathbb Q\ne0\), choose \(b\in B\). The rational number \(b/2\) belongs to the module and has a finite expansion in \(B\) with integer coefficients. Multiplying that expansion by \(2\) expresses \(b\) with an even coefficient on every basis vector. Its coefficient on \(b\) must instead be \(1\) by uniqueness of basis coordinates, a contradiction. This excludes infinite as well as finite free rank.
4. For part (d), take \(\mathbb Z[x]\).
The integers are a unique factorization domain, and the polynomial ring over a unique factorization domain is again one, by Gauss’s lemma [@DF04]. But the proper ideal \((2,x)\) from step 1 is not principal. If \((2,x)=(d)\), then \(d\) divides the nonzero constant \(2\). Degree additivity in the integral domain \(\mathbb Z[x]\) forces \(d\) to be a constant integer, necessarily \(\pm1\) or \(\pm2\). Since \(d\) must also divide \(x\), it cannot be \(\pm2\), whose polynomial multiples have even coefficients. Hence \(d=\pm1\), which would make \((d)\) the whole ring, contradicting its properness.
5. For part (e), take \(R=\mathbb Q[x_1,x_2,\ldots]\).
Let \(I=(x_1,x_2,\ldots)\). Suppose \(I\) were generated by finitely many polynomials \(f_1,\ldots,f_s\). Each polynomial involves finitely many indeterminates, so choose \(x_j\) occurring in none of them. Every \(f_i\in I\) has zero constant coefficient. The evaluation homomorphism \[ R\longrightarrow\mathbb Q[t],\qquad x_j\longmapsto t,\qquad x_k\longmapsto0\ (k\ne j) \] therefore sends all \(f_i\) to \(0\), but sends \(x_j\in I\) to the nonzero polynomial \(t\). This contradicts \(I=(f_1,\ldots,f_s)\). Thus \(R\) has an ideal that is not finitely generated and is not Noetherian.