Let $F$ be field. Show that in $F[x]$ a prime ideal is a maximal ideal. Let $F$ be field. Show that in $F[x]$ a prime ideal is a maximal ideal.

can someone help me please to tackle this problem.thanks for your time.
 A: What is important here is that $F[x]$ is a PID.
Let $P$ be a prime ideal in $F[x]$, $P\neq (0)$. Write $P=(p)$.
Suppose that $Q$ is an ideal of $F[x]$ such that $P\subsetneq Q$. Write $Q=(q)$.
Now $p\in Q$, and so $p=aq$ for some $a$. Since $P$ is prime and $p=aq\in P$, we must have either $a\in P$ or $q\in P$; but $q\notin P$, and therefore $a\in P$.
Since $a\in P=(p)$, we can write $a=bp$ for some $b$. Then we have $p=aq=bpq$, or $p(1-bq)=0$. But, since $F[x]$ is an integral domain, and $p\neq 0$, we must have $bq=1$. But, this shows that $q$ is a unit, and so $Q=(q)=F[x]$.
So, if $P$ is a prime ideal and $P\subsetneq Q$, then $Q=F[x]$; hence $P$ is maximal.
A: Another sort of approach:
Let $R=F[x]$ and suppose you are modding out by the nontrivial ideal $I=(f(x))$. Recall that $R/I$ is going to be finite dimensional over $F$, since the quotient reduces things from $R$ to have degree less than $deg(f)$. Consequently, $R/I$ is an Artinian ring.
But $R/I$ is also a domain, if $I$ is prime. However, an Artinian domain is a field, and so $I$ must be maximal.
A: We must assume $P$ is nonzero.
Let $P$ be a prime ideal in $F[x]$. There is a theorem that a polynomial ring over a field is a PID, so $F[x]$ is a PID, and $P$ is generated by a single polynomial $g(x)$. Since $P$ is prime, $g(x)$ must be irreducible. Then the gcd of $g(x)$ and any polynomial that is not a multiple of $g(x)$ is $1$, so using the Bezout relation, you can invert any element of $F[x]/(g(x))$. So it is a field.
This is kind of sketchy. Let me know if you need more detail.
