Proving maximal ideals when considering an irreducible polynomial For $F$ a field, and $q(x)\in F[x]$
Suppose that $q(x)$ is a irreducible polynomial within the ring. Prove that 
$ \langle q(x) \rangle$  is a maximal ideal of $F[x]$
I've already proved that $F[x]$ is a principal ideal domain, i'm not quite sure about the rest.  
 A: Since $\;\Bbb F[x]\;$ is an Euclidean domain, an element there is irreducible iff it is prime, so then $\;\langle q\rangle\;$ is a prime ideal in $\;\Bbb F[x]\;$, and thus $\;\Bbb F[x]/\langle q\rangle\;$ is an integer domain. Put $\;I:=\langle q\rangle\;$ for simplicity.
Now, if $\;p(x)+I\neq 0\in\Bbb F[x]/I\;$ , then $\gcd (p,q)=1\;$ , so that there exist $\;g,h\in\Bbb F[x]\;$ such that
$$pg+qh=1\implies pg=1-gh\implies p(x)g(x)=1+I$$
so that $\;p(x)+I\;$ is invertible, and thus the integral domain is in fact a field, and this means $\;I\;$ is a maximal ideal.
A: Let $R$ be a principal ideal domain.
If $a$ is an irreducible element, then < a > = $Ra$ is a maximal ideal.
Proof: Firstly, $a$ is irreducible, then $a$ is not an unit. Then Ra $\neq$ R.
Secondly, suppose J an ideal of R and $Ra \subset J \subset R$. Claim: either $J=Ra$ or $J=R$. Because R an principle ideal domain, there is $b\in R$ such that $J=Rb$. And $a\in Ra \subset J = Rb$, then a can be written as $a=rb$ for some $r\in R$. But a irreducible, and then either $r\in R^*$ or $b\in R^*$. Suppose $r\in R^*$, then $b=r^{-1}a \in Ra$, then $J= Rb \subset Ra$, but then $J=Ra$. Suppose $b\in R^*$ then $J=Rb=R$. Done.
