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I have this problem

Let $L/K$ be a finite field extension and $|L:K|=m$. Let $f(x)$ be an irreducible polynomial of degree $n$ in $K[x]$ and $\gcd(m,n)=1$. Prove that $f$ is irreducible over $L$.

I'm just a beginner so I'm really confused with the $\gcd(m,n)=1$, I don't know how to use it. So help me with this problem. Thank you

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  • $\begingroup$ $(m, n) = 1$ means $\gcd(m, n) = 1$. This notation is usually used in number theory's text. IMO, $(a, b)$ is one of the most overloaded notation. It can means coordinate, row vector, open interval, gcd, ... $\endgroup$ – Alex Vong Mar 6 '17 at 15:49
  • $\begingroup$ Thank you, I know what it is, just don't know how to use it in this situation $\endgroup$ – chí trung châu Mar 6 '17 at 15:51
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$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$Let $g$ be an irreducible factor of $f$ in $L[x]$, of degree $q \le n$. Let $\alpha$ be a root of $g$ (and thus of $f$) in some extension of $K$, so that $\Size{K[\alpha] : K} = n$, and $\Size{L[\alpha] : L} = q$

We have $$\Size{L[\alpha] : K} = \Size{L[\alpha] : L} \cdot \Size{L : K} = q m.$$

Now $K[\alpha] \subseteq L[\alpha]$, so $$ q m = \Size{L[\alpha] : K} = \Size{L[\alpha] : K[\alpha]} \cdot \Size{K[\alpha] : K} $$ implies $\Size{K[\alpha] : K} = n \mid q m$.

Since $(n, m) = 1$, we have that $n \mid q$, so $n = q$ and $f$ is irreducible in $L[x]$.

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