Let $f(x)$ be an irreducible polynomial over $F$ of degree $n$, and let $E$ be a field
extension of $F$ with $[E:F]=m$. If $\gcd(m, n)=1$, show that $f$ is irreducible over $E$.
I have the following:
We suppose that $f$ is not irreducible over $E$, Then $\exists a \in E$ with $f(a)=0$, $f(x)=g(x) \cdot h(x), g, h \in E[x]$, where $g$ is irreducible.
$$F \leq F(a) \leq E$$
$Irr(a, F)=f(x)$
$[E:F]=[E:F(a)][F)a):F] \Rightarrow m =[E:F(a)] \cdot n \Rightarrow n \mid m$
which is a contradiction, sonce $\gcd(n, m)=1$
At first sight it looks like the integral below
$$\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$$
can be evaluated by using some geometric series. What else can we do? Is there a fast easy way
to finish it?
