Assume the field extension $L:K$ is Galois (i.e. finite, normal and separable), with $M$ an intermediate field.

  • Show that $L:M$ and $M:K$ are finite separable field extensions.

Attempt: Both $M:K$ and $L:M$ are algebraic ($M:K$ since $M \subset L$, so every $\alpha \in M$ is algebraic over $K$ and $L:M$ since any $\alpha \in L$ is algebraic over $K$, so also over $M$).

Then $M:K$ is separable again since $M \subset L$ (every element of $L$ so in particular every element of $M$ is separable over $K$).

For $L:M$, take any $\alpha \in L$, then min$_M(\alpha)$ divides min$_K(\alpha)$ in $M[X]$. Since min$_K(\alpha)$ has no multiple zeros in a splitting field, neither does min$_M(\alpha)$, i.e. $\alpha$ is separable over $M$.

I'm not quite about the proof for finite, would it involve the Tower Law?

  • Show that $L:M$ is a normal extension.

I'm a bit stuck here, any help for this one?

  • 1
    $\begingroup$ Your proof will depend on the definition of normality that you’re using. What’s yours? $\endgroup$ – Lubin Jun 2 '18 at 19:37
  • $\begingroup$ @Lubin "A field extension $L:K$ is normal if every irreducible polynomial over $K$ which has at least one zero in $L$ splits in $L$." $\endgroup$ – user12002 Jun 2 '18 at 21:28
  • $\begingroup$ Take an $M$-irreducible polynomial $f$ with a root $\alpha$ in $L$. Look at the minimal (irreducible) $K$-polynomial for $\alpha$, call it $g$. What do you know about $f$ versus $g$? (You might look at some examples.) $\endgroup$ – Lubin Jun 3 '18 at 1:56

Yes, to show that $L/M$ and $M/K$ are finite extensions use the Tower Law, $$ [L:K] = [L:M][M:K]. $$

To show that $L/M$ is a normal extension, let $f \in M[x]$ be an irreducible polynomial which has a root $\alpha$ in $L$. We want to show that $f$ splits over $L$. Let $g \in K[x]$ be the minimal polynomial of $\alpha$ over $K$. Then, $f$ divides $g$. Since $L/K$ is normal, $g$ splits over $L$ and therefore so does $f$.

  • 1
    $\begingroup$ Happy crusading :-) $\endgroup$ – Jyrki Lahtonen Nov 29 '18 at 4:50
  • $\begingroup$ @JyrkiLahtonen Thank you for the encouragement :) $\endgroup$ – Brahadeesh Nov 29 '18 at 7:10

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.