2
$\begingroup$

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 $\gcd(m,n)=1$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.