Let $f(x)$ be an irreducible polynomial over $F$ of degree $n$, and let $K$ be a field extension of $F$ with $[K:F]=m$. If $\gcd(n,m)=1$, show that $f$ is irreducible over $K$.

I thought suppose $f$ reducible over $K$, so $$f(x) = \underbrace{p(x)}_{(\deg = r)}\underbrace{q(x)}_{(\deg = s)}$$ where $1<r,s<n$ and both divides $n$. I imagine that $r$ or $s$ divides $m$ too, but I don't know how to prove this.

Any hint? Or is there a better way?


If $f=pq$ where $p$ is irreducible in $K[x]$ with degree $r$ and $0<r<n$.select a root $\alpha$ of $p$ .then $[K[\alpha]:K][K:F]=[K[\alpha]:F[\alpha]][F[\alpha]:F]$,that is $rm=[K[\alpha]:F[\alpha]]n$.Hence $n\mid r$ since n and m are coprime. Contradiction

  • $\begingroup$ ${}$Nice! Thanks! $\endgroup$ – Corrêa May 4 '18 at 6:27

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