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Let $p(x)=x^4+9x+6$. it is irreducible over $\mathbb{Z}[x]$ by Eisenstein criterion(Because $3^2$ does not divide 6). My question is whether $K = \mathbb{Z}[x] / (p(x))$ is a field or not ?

There is a result that says : All maximal ideals of the ring $\mathbb{Z}[x]$ are of the form $(p,f(x))$ where $p$ is a prime number and $f(x)$ is a polynomial whose projection to $\mathbb{Z}/p\mathbb{Z}[x]$ is irreducible.

Can I use the above theorem to deduce that $K = \mathbb{Z}[x] / (p(x))$ is not a field ?

Thanks in advance...

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  • $\begingroup$ Can you prove that in the ideal generated by $x^4+9x+6$ there is no prime number $p$ such that ...? Hint: This is not difficult if you remember what principal ideals in commutative rings look like. In other words, you can use that theorem, if you can prove that your ideal is not of that form. $\endgroup$ Nov 24, 2014 at 9:04
  • $\begingroup$ @JyrkiLahtonen Thanks ! so you agree that $K$ is not a field, right ? $\endgroup$
    – the8thone
    Nov 24, 2014 at 9:07
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    $\begingroup$ $(p(x))$ is a prime ideal (since $p$ is irreducible), but not maximal, and this follows directly from the theorem you stated. $\endgroup$
    – Crostul
    Nov 24, 2014 at 9:12

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Suppose $A = \mathbb{Z}[x]/(x^4 + 9x + 6)$ is a field. Now $2 \in A$ and $2 \neq 0$ in $A$. So there exists a polynomial $p(x) \in \mathbb{Z}[x]$ of degree $\leq 3$ such that $2p(x) - 1 \in (x^4 + 9x + 6),$ which is clearly not possible unless $2p(x) -1 = 0,$ in which case $2$ is invertible in the ring $\mathbb{Z}[x].$

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