Consider the map $\phi:R\to \R$ defined by
\[\phi(f)=f(1),\]
for every $f(x)\in R$.

Proof.

Let us consider the map $\phi$ from $R$ to the field of real numbers $\R$ defined by
\[\phi(f)=f(1),\]
for each $f(x)\in R$. Namely, the map $\phi$ is the evaluation at $x=1$.

We claim that $\phi:R \to \R$ is a ring homomorphism. In fact we have for any $f(x), g(x)\in R$,
\begin{align*}
\phi(fg)&=(fg)(1)=f(1)g(1)=\phi(f)\phi(g)\\
\phi(f+g)&=(f+g)(1)=f(1)+g(1)=\phi(f)+\phi(g),
\end{align*}
hence $\phi$ is a ring homomorphism.

Next, consider the kernel of $\phi$. We have
\begin{align*}
\ker(\phi)&=\{ f(x)\in R\mid \phi(f)=0\}\\
&=\{f(x) \in R \mid f(1)=0\}=I.
\end{align*}
Since the kernel of a ring homomorphism is an ideal, it follows that $I=\ker(\phi)$ is an ideal of $R$.

Next, we claim that $\phi$ is surjective. To see this, let $r\in \R$ be an arbitrary real number.
Define the constant function $f(x)=r$. Then $f(x)$ is an element in $R$ as it is continuous function on $[0, 2]$.
We have
\begin{align*}
\phi(f)=f(1)=r,
\end{align*}
and this proves that $\phi$ is surjective.

Since $\phi: R\to \R$ is a surjective ring homomorphism, the first isomorphism theorem yields that
\[R/\ker(\phi) \cong \R.\]
Since $\ker(\phi)=I$ as we saw above, we have
\[R/I \cong \R.\]
Thus, the quotient ring $R/I$ is isomorphic to the field $\R$.
It follows from this that $I$ is a maximal ideal of $R$.

(Recall the fact that an ideal $I$ of a commutative ring $R$ is maximal if and only if $R/I$ is a field.)

Related Question.

Problem.
Let $\Z[x]$ be the ring of polynomials with integer coefficients.

Prove that
\[I=\{f(x)\in \Z[x] \mid f(-2)=0\}\]
is a prime ideal of $\Z[x]$. Is $I$ a maximal ideal of $\Z[x]$?

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