Show that $I$ is a maximal ideal of $\Bbb Z[x]$.

Question

In $\Bbb Z[x]$ let $I=\{f(x):f(0)\; \text{is even}\}$. Show that $I$ is a maximal ideal of $\Bbb Z[x]$. How many elements does $\Bbb Z[x]/I$ have ?

Attempt:

Let $I\subset J\subset \Bbb Z[x]$ then there exists $g(x)\in J $ such that $g(x)\notin I$. Then $g(0)$ is odd.

To show that $J=\Bbb Z[x]$. Now let $k(x)\in \Bbb Z[x]\implies $ either $k(0) $ is even or odd. If $k(0)$ is even then $k(x)\in I\subset J$.

If $k(0)$ is odd then how to show that $k(x)\in J$. Please help

Answer 1

If $k(0)$ is odd then $k-1\in I$. Then $k-(k-1)=1\in J$.

Let $S$ be the set of polynomials $k$ such that $k(0)$ is odd. If $s_1,s_2\in S$ then $s_1-s_2\in I$. Then $\Bbb Z[x]/I=\{I,S\}\cong\Bbb Z_2$

Answer 2

Hint: Consider the ring homormophism $\mathbb Z[x] \to \mathbb Z/ 2\mathbb Z$ given by $f(x) \mapsto f(0) \bmod 2$.

Equivalently, prove that $f(x) \equiv (f(0) \bmod 2) \bmod I$.