Showing prime and maximal ideal

In $\mathbb{Z}\left [ x \right ]$, let $I = \left \{ f\left ( x \right ) \in \mathbb{Z}\left [ x \right ]: f\left ( 0 \right ) \text{ is an even integer} \right \}$.

Is $I$ a Prime ideal of $\mathbb{Z}\left [ x \right ]$?

Let $f_{1}\left ( x \right ),f_{2}\left ( x \right ) \in \mathbb{Z}\left [ x \right ].$

Let $f_{1}\left ( x \right )f_{2}\left ( x \right ) \in I$.

Clearly, I is a proper ideal of $\mathbb{Z}\left [ x \right ]$ since the elements in $\mathbb{Z}\left [ x \right ]$ evaluated at 0 can produce an odd integer but this cannot be true in I.

Now, $f_{1}\left ( 0 \right )f_{2}\left ( 0 \right )$ is an even integer in I. Hence, either $f_{1}\left ( 0 \right )$ has an even constant integer or $f_{2}\left ( 0 \right )$ has an even constant integer. Thus, I is a prime ideal by definition.

Is this correct?

I would now like to show if I is a maximal ideal.

Again, clearly, I is a proper ideal by the above argument. Let B be an ideal of $\mathbb{Z}\left [ x \right ]$ and I $\subseteq B\subseteq \mathbb{Z}\left [ x \right ]$. How should B be defined so that I would conclude either $B=I$ or $B=\mathbb{Z}[x]$?

• Hint: $I = (2, x)$. What is $\mathbb{Z}[x]/(2,x)$ isomorphic to? – André 3000 Jul 14 '17 at 6:05
• It is isomorphic to the image of Z[x].@Quasicoherent – Mathematicing Jul 14 '17 at 6:10
• @Mathematicing Isn't it isomorphic to $\mathbb{Z}/2\mathbb{Z}$? – RideTheWavelet Jul 14 '17 at 7:07
• "Clearly, I is a proper ideal of $\mathbb{Z}\left [ x \right ]$ since the elements in $\mathbb{Z}\left [ x \right ]$ evaluated at 0 can produce an odd integer but this cannot be true in I." If I were writing this, I would use a specific example, such as $f(x) = 1$, to demonstrate. There are other minor details like this, but I think your proof is correct. – Arthur Jul 14 '17 at 7:29
• @Quasicoherent are you referring to my question asking for hep to prove that I is a maximal ideal? – Mathematicing Jul 14 '17 at 7:36

Clearly if $f(0)$ is not even, then $f(0)-1$ is even. This shows: If $f \notin I$, then $f-1 \in I$, hence $1 \in (I,f)$. Thus $I$ is maximal by the very definition.
This argument also immediately yields $\mathbb Z[X]/I = \{0,1\} \cong \mathbb F_2$.
For maximality, consider the two ring homomorphisms \begin{align} \alpha&\colon \mathbb{Z}[x]\to\mathbb{Z} & p(x)&\mapsto p(0) \\[6px] \beta&\colon \mathbb{Z}\to\mathbb{Z}/2\mathbb{Z} & m&\mapsto m+2\mathbb{Z} \end{align} and look at the kernel of $\beta\circ\alpha$.