Prove : If $I = (p(x))$ is a prime ideal in $F[x]$ then $p(x)$ is irreducible. I have to show :
If $I = (p(x))$ is a prime ideal in $F[x]$, where F is a field, then $p(x)$ is irreducible.
In the book I use, there is the proof of the converse which uses Euclid's Lemma. 
I tried to suppose $I=(p(x))$ is an ideal, then every polynomial in I is divisible by $p(x)$. Then for $a(x)b(x) \in F[x]$ we have  $a(x)b(x)=p(x)q(x)$ for $q(x) \in F[x]$ with $\delta a<\delta p$ and $\delta b<\delta p$.
From there, I don't see how to deduce $p(x)$ is irreducible ie there is no factorisation over F $p(x)=c(x)d(x) \in F[x]$ with $\delta c < \delta p$ and $\delta d < \delta p$
We could express $p(x)$ in function of $a(x),b(x)$ and $g(x)$ but it does not help.
Any hints ? :)
Thanks
 A: Suppose $p(x)=c(x)d(x)$, i.e. $p(x)$ is reducible.  Since $c(x)d(x)=p(x)\in(p(x))$, either $c(x)\in(p(x))$ or $d(x)\in(p(x))$.  Without loss the former, so $p(x)|c(x)$, and also $c(x)|p(x)$.  Hence $c,p$ are associates, so $d$ is a unit, which contradicts the reducibility assumption.
A: Another way could be to observe the following facts:


*

*$I$ is prime iff $F[x]/I$ is a domain.

*If $p(x)$ is reducible it could be only a product of two coprime polynomials or the power of a single irreducible polynomials.


Then, using the CRT and some others results about quotients, you can find a contradiction looking at properties of the ring $F[x]/I$.
A: First:
In integral domains primes are irreducible.
Proof. let $D$ be and integral domain and $p\in D$ a prime element. Suppose that $$p=ab,$$ for some $a,b\in D$. Certainly, $p\mid ab$, so since $p$ is prime $p\mid a$ or $p\mid b$. Without lost of generality, assume $p \mid a$. Then $a=pc$, for some $c\in D$. So
\begin{align*} p&= pcb\\ 1&= cb,\end{align*}
The last step being justified because in integral domains cancellation is valid. Therefore, $b$ is a unit and $a$ it's an associate to $p$. we have proved that the only possible divisors of $p$ are units or associates to $p$, this says that $p$ is irreducible. 

Back to your problem, $F$ is a field so $F[x]$ it's an integral domain, so it's enough to show that $p(x)$ is prime. 
To this end, suppose $p(x)\mid f(x)g(x)$. This means that $f(x)g(x)\in I$ and since $I$ it's a prime ideal, this says that $f(x)\in I$ or $g(x)\in I$. In other words: $p(x)\mid f(x)$ or $p(x)\mid g(x)$. Therefore, $p(x)$ is prime.
