This is a problem in James Milne's text on Galois Theory:

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 a proof which I think contains a mistake since it appears to work under the weaker condition that $n\nmid m$, but I am unable to locate the error:

Suppose $f$ has a root $\alpha$ in $E$. Since $f$ is irreducible in $F$, we have that $[F(\alpha) : F] = \deg f = n$. Since $E$ contains both $\alpha$ and $F$, $F(\alpha)$ is a subfield of $E$. Now, $$m = [E:F] = [E:F(\alpha)][F(\alpha):F]= n[E:F(\alpha)],$$ contradicting that $\gcd(m,n)=1$.

  • $\begingroup$ What about the case when $f$ has no root in $E$? You need to consider that case also. Take $F = \mathbb Q, f(x) = x^2 - 3, E = \mathbb Q(\sqrt[3]2).$ In this case $f$ has no root in $E.$ $\endgroup$ – Krish Jan 19 '15 at 16:38
  • $\begingroup$ @Krish Oh, you're right, thank you. I had been equating irreducibility with the non-existence of a root. $\endgroup$ – Milind Jan 19 '15 at 16:41
  • $\begingroup$ Related. $\endgroup$ – Git Gud Jan 19 '15 at 16:53

Expanding on the comments above.

One also needs to consider the case when $f$ has no root in $E$, because irreducibility of $f$ over $E$ is not implied by non-existence of any roots of $f$ in $E$.

Consider a tower of extensions $F \subseteq E \subseteq L$ where $L$ is algebraically closed. Let $\alpha$ be a root of $f$ in $L$. Since $[F(\alpha):F] = n$, $[E:F] = m$, and $E(\alpha) = E(F(\alpha))$ with $\gcd(m,n) = 1$, we can conclude (exercise) that $[E(\alpha):F] = mn$. Let $p$ be the minimal polynomial of $\alpha$ over $E$, so that $p \mid f$ in $E[x]$. So, if $d = \deg(p)$, we have $$ mn=[E(\alpha):F] = [E(\alpha):E][E:F] = dm \implies n = d. $$ Hence, $p = f$, which means that $f$ is irreducible over $E$ as well.

  • 1
    $\begingroup$ For the exercise part: suppose $[E(\alpha) : F] = x$. We can extend $F$ to $E(\alpha)$ in two ways, $F \subset F(\alpha) \subset E(\alpha)$ or $F \subset E \subset E(\alpha)$. The first way gives $n|x$ and 2nd way gives $m|x$. Hence $x \geq mn$. We also have $[E (\alpha) : E] \leq n$ because the $minpoly_E (\alpha)$ divides $minpoly_F(\alpha)$, so that $[E(\alpha) : F] \leq mn$. $\endgroup$ – eatfood Sep 12 at 4:31
  • $\begingroup$ @eatfood That's right :) Thanks for completing the proof! $\endgroup$ – Brahadeesh Sep 12 at 15:46

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