Prove that if $p(x)$ is irreducible, then $\langle p(x) \rangle$ is a maximal ideal of $F[x]$

Question

Problem:

Let $F$ be a field. Prove that if $p(x)$ is irreducible, then $\langle p(x) \rangle$ is a maximal ideal of $F[x]$.

Attempt:

Let $d(x),a(x),p(x) \in F[x]$ and suppose $\text{gcd}[a(x), p(x)] = d(x)$. Then $d(x) \mid p(x)$. So $p(x) = d(x)c(x)$ for some $c(x) \in F[x]$. Because $p(x)$ is irreducible, either $d(x)$ or $c(x)$ is a constant.

If $d(x)$ is a nonzero constant in $F$, then $\langle p(x) \rangle = F[x]$ by [previous proof]* and $\langle p(x) \rangle$ is not maximal because it generates all of $F[x]$.

If $c(x)$ is a nonzero constant, then: \begin{align*} p(x) &= d(x)c(x) \\ p(x) &= d(x)c && \text{$c(x) = c$ is a constant}\\ p(x)c^{-1} &= d(x)cc^{-1} \\ p(x)c^{-1} &= d(x) \\ d(x) &= p(x)c^{-1} \end{align*}

Since $d(x) \mid a(x)$, we have that $a(x) = d(x)e(x)$ for some $e(x) \in F[x]$. Then: \begin{align*} a(x) &= d(x)e(x) \\ a(x) &= p(x)c^{-1}e(x) && \text{substitution}\\ a(x) &= p(x)e(x)c^{-1} \end{align*}

This implies that $a(x)$ is a multiple of $p(x)$.

Questions:

My understanding is that if $d(x)$ is a nonzero constant, then $\langle p(x) \rangle$ cannot be maximal because it generates the entire ring. But it also seems to me that $c(x)$ being a constant presents a contradiction.

Does $a(x)$ being a multiple of $p(x)$ present a contradiction, because $a(x)$ and $p(x)$ are relatively prime?

If not, how can I use that $c(x)$ being constant shows that $\langle p(x) \rangle$ is maximal?

Second, guided attempt:

Let $J$ be an ideal containing $\langle p(x) \rangle$ that isn't equal to $\langle p(x) \rangle$. Then there must exist a polynomial $a(x)$ that is in $J$ but not in $\langle p(x) \rangle$. We'll show that $J = F[x]$ using that $p(x)$ and $a(x)$ are relatively prime.

Let $d(x),a(x),p(x) \in F[x]$ and suppose $\text{gcd}[a(x), p(x)] = d(x)$. Then $d(x) \mid p(x)$. So $p(x) = d(x)c(x)$ for some $c(x) \in F[x]$. Because $p(x)$ is irreducible, either $d(x)$ or $c(x)$ is a constant.

By [previous proof]*, if $d(x)$ is a nonzero constant in $F$, then $J=F[x]$ and we are done.

If, on the other hand, $c(x)$ is a nonzero constant, then: \begin{align*} p(x) &= d(x)c(x) \\ p(x) &= d(x)c && \text{$c(x) = c$ is a constant}\\ p(x)c^{-1} &= d(x)cc^{-1} \\ p(x)c^{-1} &= d(x) \\ d(x) &= p(x)c^{-1} \end{align*}

Since $d(x) \mid a(x)$, we have that $a(x) = d(x)e(x)$ for some $e(x) \in F[x]$. Then: \begin{align*} a(x) &= d(x)e(x) \\ a(x) &= p(x)c^{-1}e(x) && \text{substitution}\\ a(x) &= p(x)e(x)c^{-1} \end{align*}

This implies that $a(x)$ is a multiple of $p(x)$, which is a contradiction, because $a(x)$ and $p(x)$ are relatively prime. Thus, $\langle p(x) \rangle$ is a maximal ideal of $F[x]$.

* Previous proof: If $J$ is an ideal of $A$ and $J$ contains an invertible element $a$ of $A$, then $J = A$.

Answer 1

Here is the proof, maybe that will settle your questions?

If $(p(x))$ is not maximal, then there is an ideal $I$ containing it. We'll show $I=F[x]$ or $I=(p(x)).

Since $F[x]$ is a principal ideal domain, we know $I=(f(x))$ for some polynomial $f(x)$. (This is probably where your argument involving $a(x)$, $d(x)$, and $c(x)$ comes in --- your $a(x)$ is my $f(x)$.)

Now the containment of ideals $(p(x))\subseteq (f(x))$ implies that $f(x)$ divides $p(x)$. But $p(x)$ is irreducible --- so this is only possible if $f(x)$ is a scalar multiple of $p(x)$, or if $f(x)$ is constant. These cases correspondingly imply $(f(x))=(p(x))$, or $(f(x))=F[x]$.

Answer 2

Every ideal of $F[x]$ is principal. If $\langle p(x)\rangle$ is not maximal, then $p(x)$ must have a non-constant factor with a lower degree, which is impossible.

Answer 3

Assume the contrary. Suppose that $<p(x)>=I$ (say) is not a maximal ideal. Let $<q(x)>=J$ (say) be another ideal property contained in $<p(x)>$, that is, $<p>\subset <q(x)>$. Then $q(x)=r(x)p(x)$ for some $r(x)\in F[x]$. Now consider the necessary cases.