a problem on prime and maximal ideal from gallian contemporary algebra. In $\mathbb{Z}[x]$, let $I =$ $\{$ $f(x) \in  \mathbb{Z}[x] | f (0)$ is an even integer$\}$. Prove that   $I = \langle x, 2\rangle$.
Is I a prime ideal of $\mathbb{Z}[x]$?
Is I a maximal ideal?
How many elements does $\mathbb{Z}[x]/I$ have?

how can I able to solve the problem?can somebody help.I am completely stuck on it.thanks for your time.
 A: You may want to note that if you consider the surjective morphim
$$
\begin{align}
\varphi :\ &\Bbb{Z}[x] \to \Bbb{Z}_{2} \\&f(x) \mapsto [f(0)],
\end{align}$$
(whereas I denote by $[a]$ the class of $a \in \Bbb{Z}$ in $\Bbb{Z}_{2}$), then $I = \ker(g)$, so that $\Bbb{Z}[x] / I \cong \Bbb{Z}_{2}$. Since $\Bbb{Z}_{2}$ is a field, you get that $I$ is maximal, thus prime.
A: Obviously, $x\in I$ and $2\in I$.
And if $f\in I$, then write $f(x)=f(0)+x g(x)$.  If $f\in I$, we see that $f(0)=2k$ for some $k$, hence $f=k\cdot 2+x\cdot g\in\langle 2,x\rangle$. This shows $I=\langle 2,x\rangle$.
Note that for $h\in\mathbb Z[x]$ we either have $h\in I$ (if $h(0)$ is even) or $h\in 1+ I$ (if $h(0)$ is odd), hence the quotient ring $\mathbb Z[x]/I$ has only two elements, is the field of two elements, hence $I$ is maximal (and of course prime).
A: Hint $\ $ Compose natural maps $\rm\ \begin{eqnarray}\rm f(x) &\mapsto&\rm f(0) &\mapsto&\rm f(0)\ mod\ 2\\ \rm \Bbb Z[x] &\to& \ \ \Bbb Z &\to& \Bbb Z/2\end{eqnarray}\ $ then apply Isomorphism$_{1}\!$ Theorem.
