if $(f(x))$ is a principal prime ideal in a field. Why is $f(x)$ irreducible? I want to do it by contradiction. Suppose it is reducible, then this principal ideal cannot be prime...
suppose $f(x)=a(x)b(x)$ then $(f(x))$ is the set of $c(x)(a(x)b(x))$ for all $c(x)$.
and we know if $p(x)q(x)$ is in $(f(x))$ then either $p(x)$ or $q(x)$ has to be in $(f(x))$ that is one of $p(x)$ and $q(x)$ has to be the form of $c(x)(a(x)b(x))$
but I cannot see any contradiction here.
We show the following: Let $R$ be an integral domain and $(a) \neq \{0\}$ be a principal ideal which is prime. Then $a$ is irreducible.
Let $a = bc$ for some $b, c \in R$. Then since $bc = a \in (a)$ and $(a)$ is prime, wlog $b \in (a)$. But then $b = r a$ for some $r \in R$ and thus $a = r a c$, which is equivalent to $a (1 - rc) = 0$. Since $R$ is an integral domain and $a \neq 0$, we have $1 - rc = 0$ and thus $rc = 1$, which shows that $c$ is a unit. Thus $a$ is irreducible.
In fact, $a$ is even a prime element, which can be seen as follows: Assume $a \mid bc$, i.e. there is some $r \in R$ with $ra = bc$. Then $bc \in (a)$ and since $(a)$ is prime, wlog $b \in (a)$, i.e. $b = r' a$ and hence $a \mid b$.