Define a map $\phi: \Z[x] \to \Z$ defined by
\[\phi \left( f(x) \right)=f(-2).\]

We first prove that $\phi$ is a ring homomorphism.
For any $f(x), g(x)\in \Z[x]$, we have
\begin{align*}
\phi(fg)&=(fg)(-2)=f(-2)g(-2)=\phi(f)\phi(g)\\
\phi(f+g)&=(f+g)(-2)=f(-2)+g(-2)=\phi(f)+\phi(g),
\end{align*}
hence $\phi$ is a ring homomorphism.
Then by definition of $\phi$, we see that $\ker(\phi)=I$:
\begin{align*}
\ker(\phi)&=\{f(x)\in \Z[x] \mid \phi(f(x))=0\}\\
&=\{f(x)\in \Z[x] \mid f(-2)=0\}\\
&=I.
\end{align*}

Next, we prove that $\phi: \Z[x] \to \Z$ is surjective.
Let $n$ be an arbitrary integer.
Consider the polynomial $f(x):=x+2+n\in \Z[x]$.
Then we have
\[\phi\left( f(x) \right)=f(-2)=(-2)+2+n=n.\]
Hence $\phi$ is surjective.
(Or we could’ve considered the constant function $f(x):=n$.)

These observations together with the first isomorphism theorem give
\[Z[x]/I\cong \phi(\Z[x])=\Z.\]

It follows that the quotient $\Z[x]/I$ is an integral domain as so is $\Z$.
Hence $I$ is a prime ideal of $\Z[x]$.

On the other hand, since $\Z[x]/I\cong \Z$ is not a field, the ideal $I$ is not a maximal ideal of $\Z[x]$.

Related Question.

Problem.
Let $R$ be the ring of all continuous functions on the interval $[0, 2]$.
Let $I$ be the subset of $R$ defined by
\[I:=\{ f(x) \in R \mid f(1)=0\}.\]
Then prove that $I$ is an ideal of the ring $R$.
Moreover, show that $I$ is maximal and determine $R/I$.

A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring
Let $R$ be the ring of all continuous functions on the interval $[0, 2]$.
Let $I$ be the subset of $R$ defined by
\[I:=\{ f(x) \in R \mid f(1)=0\}.\]
Then prove that $I$ is an ideal of the ring $R$.
Moreover, show that $I$ is maximal and determine […]

Examples of Prime Ideals in Commutative Rings that are Not Maximal Ideals
Give an example of a commutative ring $R$ and a prime ideal $I$ of $R$ that is not a maximal ideal of $R$.
Solution.
We give several examples. The key facts are:
An ideal $I$ of $R$ is prime if and only if $R/I$ is an integral domain.
An ideal $I$ of […]

Irreducible Polynomial Over the Ring of Polynomials Over Integral Domain
Let $R$ be an integral domain and let $S=R[t]$ be the polynomial ring in $t$ over $R$. Let $n$ be a positive integer.
Prove that the polynomial
\[f(x)=x^n-t\]
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Consider the principal ideal $(t)$ generated by $t$ […]

If Every Proper Ideal of a Commutative Ring is a Prime Ideal, then It is a Field.
Let $R$ be a commutative ring with $1$.
Prove that if every proper ideal of $R$ is a prime ideal, then $R$ is a field.
Proof.
As the zero ideal $(0)$ of $R$ is a proper ideal, it is a prime ideal by assumption.
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Characteristic of an Integral Domain is 0 or a Prime Number
Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.
Definition of the characteristic of a ring.
The characteristic of a commutative ring $R$ with $1$ is defined as […]

Every Prime Ideal of a Finite Commutative Ring is Maximal
Let $R$ be a finite commutative ring with identity $1$. Prove that every prime ideal of $R$ is a maximal ideal of $R$.
Proof.
We give two proofs. The first proof uses a result of a previous problem. The second proof is self-contained.
Proof 1.
Let $I$ be a prime ideal […]

Determine the Quotient Ring $\Z[\sqrt{10}]/(2, \sqrt{10})$
Let
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}\]
be an ideal of the ring
\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}.\]
Then determine the quotient ring $\Z[\sqrt{10}]/P$.
Is $P$ a prime ideal? Is $P$ a maximal ideal?
Solution.
We […]

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