Let $F$ be a field and $f(x)$ be an irreducible polynomial in $F[x]$. Prove that $P = (f(x))$ is maximal in $F[x]$.
(Here is what I know: $f(x) \neq 0 \wedge f(x) \not \in U(F[x])$, since $f(x)$ is irreducible.
Suppose, $P = (f(x))$ is not a maximal. Then, let $I = (g(x))$ be an ideal of $F(x)$ such that ${P \subseteq I \subseteq F[x]}$...(not sure there. What to show when something is not a maximal ideal?) Then, $g(x) \mid f(x) \Rightarrow f(x) = g(x)h(x)$...Here, I am trying to contradict the fact that $f(x)$ is irreducible but not sure how to do so. Any help would be much appreciated.
I know I can quote some theorem to prove what I am doing. But, I also need to know how to prove it directly.
First, you need the inclusions to be proper. Likely, you know this but you should also write it, so write: ${P \subsetneq I \subsetneq F(x)}$.
Second, when you say $I = (g(x))$ you use that $F[X]$ is a PID. Otherwise you could not just assume that $I$ is a principal ideal. Again, this is likely fine as you should have seen the result that it is a PID, but you should be aware that you need that and say so.
Then, you do the right thing. You say that you get $f(x)= g(x)h(x)$. Then, since $f(x)$ is irreducible. One of $g(x)$ and $h(x)$ must be a unite/invertible. If it is $g$, then $I = F[X]$ contradiction. If it is $h$, then $(f(x))= (g(x))$, contradiction.
You can also set up the last step differently. Like so: Since $I \neq F[X]$, we have $g(x)$ is not invertible. Since $(f(x)) \neq (g(x))$ we get $h$ is not invertible. A contradiction to $f$ irreducible.