Question 6

(a)

Consider the composition \(\varphi: \mathbb{Z}[x] \to \mathbb{Z}/2\mathbb{Z}\) \[ \begin{array}{mmmm} \mathbb{Z}[x] &\to & \mathbb{Z}&\to &\mathbb{Z}/2\mathbb{Z}\\ f(x) &\mapsto & f(0) &\mapsto&\overline{f(0)} = f(0) + 2\mathbb{Z} \end{array} \] Now \(\varphi(f(x)) = \overline{0}\) iff \(\overline{f(0)} = \overline{0}\) iff \(f(0) \in 2\mathbb{Z}\), which means \(f(x)\) is of the form \(g(x)\cdot x + 2k\), so \(f(x) \in (2,x)\), therefore \(\ker(\varphi) = (2,x)\). As \(\mathbb{Z}/2\mathbb{Z}= \mathbb{Z}[x]/(2,x)\) is a field it follows that \((2,x)\) is a maximal ideal.

Suppose for a contradiction that \((2,x)\) is a principal ideal, so let \(g(x)\in\mathbb{Z}[x]\) such that \((2,x) = (g(x))\). Let \(2 = g(x)a(x)\) and \(x = g(x)b(x)\), then comparing degrees we have \(\deg(g) = \deg(a) = 0\) and \(\deg(b) = 1\). As \(2\) is prime in \(\mathbb{Z}\) solving we get \(g(x) = 2\), \(a(x) = 1\), and \(x = 2b(x)\) has no solution in \(\mathbb{Z}[x]\), contradiction.

(b)

As \(1 = 2\frac{1}{2} \in (2,x)\) it follows that \((2,x) = (1) = S\). Not maximal as \((2,x)\) is exactly the whole ring, but principal \((1)\).

Question 7

(a)

Let \(P\subseteq S\) be prime ideal and consider \(\varphi^{-1}(P)\). Suppose we have \(ab \in \varphi^{-1}(P)\), then \(\varphi(ab) = \varphi(a)\varphi(b) \in P\) and as \(P\) is a prime ideal we have \(\varphi(a)\in P\) or \(\varphi(b)\in P\), and \(a\in \varphi^{-1}(P)\) or \(b\in \varphi^{-1}(P)\) as desired.

When \(\varphi\) is onto, by basic set theory \(R = \varphi^{-1}(S) = \varphi^{-1}(P \sqcup (S\setminus P)) = \varphi^{-1}(P) \sqcup \varphi^{-1}(S\setminus P)\) and the second term in the disjoint union is nonempty.

When \(1_R\in R\) and \(\varphi(1_R) = 1_S\). Suppose for contradiction \(\varphi^{-1}(P) = R\), then \(1_R \in \varphi^{-1}(P)\) and \(\varphi(1_R) = 1_S \in P\) which means \(P = S\).

(b)

Consider the composition \[ \begin{array}{mmmm} R &\to &S &\to &S/M \\ r &\mapsto &\varphi(r) &\mapsto &\overline{\varphi(r)} \end{array} \] which is a surjective composition of surjective maps. Observe that its kernel is exactly \(\varphi^{-1}(M)\) and by first isomorphism theorem we have \(R/\varphi^{-1}(M) \cong S/M\). As \(M\) is a maximal ideal \(S/M\) is a field, and so is \(R/\varphi^{-1}(M)\), so \(\varphi^{-1}(M)\) is a maximal ideal too.