Why is $I[x]$ not maximal $\mathbb{Z}[x]$?

Question

We have that $I=(2)$ is maximal in $\mathbb{Z}$ because $(2)\subseteq (4)\subseteq \dots \subseteq (2^k)$, right?

Why is $I[x]$ not maximal $\mathbb{Z}[x]$ ?

Answer 1

It's true that $I = (2)$ is maximal in $\mathbb{Z}$. It is not true that it is because $(2) \subseteq (4) \subseteq \dots$ because

1. $(2)$ is not a subset of $(4)$: $2$ belongs to $(2)$ but not to $(4)$;
2. If $(2)$ were included in the nontrivial ideal $(4)$, then it wouldn't be a maximal ideal, by definition...

The ideal $(2)$ is maximal e.g. because the quotient ring $\mathbb{Z}/2\mathbb{Z}$ is a field. Other ways of proving this fact include, for example, Bézout's theorem.

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Now, $I[x] = \{ P \in \mathbb{Z}[x] \mid \text{the coefficients of } P \text{ are even}\} = (2) \subset \mathbb{Z}[x]$ is not maximal because it is for example contained in the ideal $(2,x)$ generated by $2$ and $x$, a nontrivial ideal. Another way of seeing this is that the quotient $\mathbb{Z}[x] / I[x] \cong (\mathbb{Z}/2\mathbb{Z})[x]$ is the ring of polynomials with coefficients in $\mathbb{Z}/2\mathbb{Z}$; this is not a field as e.g. $x$ is not invertible.

Answer 2

The quotient $\mathbb{Z}[x]/I\cong (\mathbb{Z}/2)[x]$ is not a field, hence $I$ is not maximal.