Prime & Maximal Ideal in $\mathbb{Z}[x]$

Question

Question:

Is $<x^2+x+1>$ a prime and\or maximal ideal of $\mathbb{Z}[x]$

Background:

• Relatively new to abstract algebra, hence, would appreciate beginner level answers.

My Solution:

For Prime:

$$ R = \frac{\mathbb{Z}[x]}{<x^2+x+1>} $$
Elements of R are of the class:
$[0]$, $[a]$, $[x]$, $[a+x]$
where $[<>]$ represents $<> + <x^2+x+1>$

$[x]\cdot[a+x]$ = $[a+x]$
This was the only possibility for a zero divisor to exist, thus R is an integral domain. Hence, I is prime

For Maximal:

$\mathbb{Z}$ is not a field, implying R is not a field
Hence, $<x^2+x+1>$ is not maximal

Question:

This doesn't seem like an acceptable solution, more of a rough work, so how can i prove it "correctly"

Answer 1

Your characterization of elements of $R$ is not exactly correct: the coefficient of $x$ need not be $1$, eg $[2x + 1]$ is an element of $R$ that is distinct from the elements you have written down. But every element of $R$ is represented by some linear polynomial in $\mathbb{Z}[x]$, since you can perform long division by the monic polynomial $x^2 + x + 1$.

Now suppose $\bar{p}, \bar{q} \in R$ are such that $\bar{p} \bar{q} = 0$ and let $p, q \in \mathbb{Z}[x]$ be of $\deg \le 1$ such that $\bar{p} = [p]$ and $\bar{q} = [q]$. Now $\bar{p}\bar{q} = [pq]$, so $[pq] = [0]$. This means that $pq \in \langle x^2 + x + 1 \rangle$, ie is divisible by $x^2 + x + 1$. If you know that $\mathbb{Z}[x]$ has unique factorization then you can conclude from here that one of $\bar{p}$ and $\bar{q}$ is $0$, but let's avoid using that as a black box.

If $pq = 0$ then we are done since one of $p$ and $q$ has to be $0 \in \mathbb{Z}[x]$, so one of $\bar p$ and $\bar q$ has to be $0 \in R$. Otherwise, $\deg pq \ge 2$, so comparing degrees, each of $p$ and $q$ is degree $1$ and $pq$ is degree $2$ hence must be $n(x^2 + x + 1)$ for some $n \in \mathbb{Z}$. But $n(x^2 + x + 1)$ has no rational roots, while each of $p$ and $q$ does, so this is impossible.

So $R$ is an integral domain, or equivalently $\langle x^2 + x + 1\rangle$ is a prime ideal in $\mathbb{Z}[x]$.

To check maximality, you could check if $R$ is a field. Using arguments similar to above you can verify that the only elements with inverses are $[\pm 1]$, $[\pm x]$ and $[\pm x^2] = [\mp (x + 1)]$. But let's show that $\langle x^2 + x + 1 \rangle$ is not maximal by finding a $p \in \mathbb{Z}[x]$ such that $p \notin \langle x^2 + x + 1 \rangle$ and $1 \notin \langle p, x^2 + x + 1\rangle$. To make the calculations easier, let's pick $p \in \mathbb{Z}$. The exact details of the argument will depend on which $p$ you pick, but anything other than $0, \pm 1$ will work.

We will take $p = 3$, and we already know that $3 \notin \langle x^2 + x + 1\rangle$. Note that $x^2 + x + 1 = (x - 1)(x + 2) + 3$, so $\langle{3, x^2 + x + 1\rangle} \subset \langle{3, x - 1 \rangle}$ and therefore it's enough to show that $1 \notin \langle 3, x - 1\rangle$. So suppose $1 = 3 f(x) + (x - 1) g(x)$. Then plugging in $x = 1$, $1 = 3 f(1)$, which is impossible.

This argument actually generalizes quite easily, if you are trying to show that $\langle F(x) \rangle$ is not maximal for a non-constant $F(x) \in \mathbb{Z}[x]$, find $a \in \mathbb{Z}$ such that $p = F(a) \ne 0, \pm 1$ and show that $\langle F(x) \rangle \subsetneq \langle p, x - a \rangle \subsetneq \mathbb{Z}[x]$.

Answer 2

$x^{2}+x+1$ is irreducible over $\mathbb{Z}$ and $\mathbb{Z}[x]$ being an UFD, $x^{2}+x+1$ is prime. So, $<x^{2}+x+1>$ is a prime ideal of $\mathbb{Z}[x]$. Now consider [3] in $R$, suppose $[p(x)]$ is the inverse of $[3]$ in $R$. Then $3p(x) -1 = (x^{2} + x +1)q(x)$ for some $q(x)$ in $\mathbb{Z}[x]$ . Put $x=1$ in both sides, we obtain $3p(1) -1 = 3q(1)$, which is absurd. Hence, $R$ is not a field i.e., $<x^{2}+x+1>$ is not a maximal ideal of $R$.