The Ideal $(x)$ is Prime in the Polynomial Ring $R[x]$ if and only if the Ring $R$ is an Integral Domain

Problem 198

Let $R$ be a commutative ring with $1$. Prove that the principal ideal $(x)$ generated by the element $x$ in the polynomial ring $R[x]$ is a prime ideal if and only if $R$ is an integral domain.

Prove also that the ideal $(x)$ is a maximal ideal if and only if $R$ is a field.

We claim that we have a ring isomorphism
\[ R[x]/(x) \cong R.\]
Let us for the moment assume that this claim is true and prove the problem.
If the ideal $(x)$ is a prime ideal of $R[x]$, then $R[x]/(x)$ is an integral domain.
Hence, by the claim $R\cong R[x]/(x)$ is also an integral domain.

On the other hand, if $R$ is an integral domain, then $R[x]/(x)$ is also an integral domain.
This yields that the ideal $(x)$ is a prime ideal.

Similarly, we see that the ideal $(x)$ is a maximal ideal if and only if $R\cong R[x]/(x)$ is a field.

Thus it remains to prove the claimed isomorphism of rings.

Proof of the claim

Define
\[\Psi: R[x] \to R\]
by mapping $f(x) \in R[x]$ to $f(0)\in R$. (Evaluating the polynomial $f(x)$ at $x=0$.)
Then the map $\Psi$ is a ring homomorphism since we have for $f, g\in R[x]$ and $r\in R$
\begin{align*}
\Psi(f+g)&=(f+g)(0)=f(0)+g(0)=\Psi(f)+\Psi(g)\\
\Psi(rf)&=(rf)(0)=rf(0)=r\Psi(f).
\end{align*}

The homomorphism $\Psi$ is surjective since for any $r\in R$, letting $f(x)=r$ we see that $\Psi(f)=r$.
Therefore by the isomorphism theorem for rings, we have
\[R[x]/\ker(\Psi) \cong R. \tag{*}\]
It remains to show that
\[\ker(\Psi)= (x).\]

$(\implies)$ Let $f\in \ker(\Psi)$. Then we have $\Psi(f)=f(0)=0$.
Let us write
\[f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1 x +a_0.\]
Then since $f(0)=0$, we have $a_0=0$ and thus we have
\[f(x)=x(a_nx^{n-1}+a_{n-1}x^{n-2}+\cdots +a_1) \in (x).\]
Thus $\ker(\Psi) \subset R$.

$(\impliedby)$ Let $f\in (x)$. Then we can write $f$ as $f=xg$ for some $g\in R[x]$.
It follows that we have
\[\Psi(f)=f(0)=0\cdot g(0)=0\]
and $f\in \ker(\Psi)$.
Hence $R \subset \ker(\Psi)$.

Thus putting the two inclusions together gives $\ker(\Psi)= (x)$, and combining this with the isomorphism (*) we obtain
\[R[x]/(x)\cong R\]
as claimed.

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