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 $\Z[x]$.

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Let $R$ be the ring of all continuous functions on the interval $[0, 2]$.
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