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...
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 $\mathbb{Z}[x].$