# Polynomial Ring with Integer Coefficients and the Prime Ideal $I=\{f(x) \in \Z[x] \mid f(-2)=0\}$

## Problem 573

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 $\Z[x]$?

Contents

## Proof.

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 $R/I$.

The proof of this problem is given in the post ↴
A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring

### 1 Response

1. 09/27/2017

[…] For a proof, see the post ↴ Polynomial Ring with Integer Coefficients and the Prime Ideal $I={f(x) in Z[x] mid f(-2)=0}$ […]

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