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First some notation. Let $F$ be a field, $E$ an algebraic extension of $F$ and $\overline{F}$ the algebraic closure of $F$. Let $\{E:F\}$ represents the number of non-zero homomorphisms from $E$ to $\overline{F}$ which leave the field $F$ fixed.

Suppose we have a tower of fields:

$F \subset E \subset K$

How can it be shown that $\{K:F\} = \{K:E\}*\{E:F\}$?

I know of a similar equality that deals with degrees of field extensions, but here we're talking about functions that go from $E$ to a much larger field $\overline{F}$ which makes it seem hard to visualize. What can I do to prove this?

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  • $\begingroup$ "The closure"?? The algebraic, the separable, the normal...what closure? $\endgroup$ – DonAntonio May 18 '16 at 17:02
  • $\begingroup$ The field in which every polynomial with coeff's in F splits, the algebraic closure. $\endgroup$ – Ninosław Brzostowiecki May 18 '16 at 17:05
  • $\begingroup$ So let me see if I understood, and maybe you could explain this in your question: we have F\subset E\;$ (not necessarily an algebraic extension, as far as I can read), and $\;\overline F\;$ is an (or the, as you wish) algebraic closure of $\;F\;$. Would the extension in fact be algebraic in order the equality you want to prove is true? $\endgroup$ – DonAntonio May 18 '16 at 17:07
  • $\begingroup$ Yes, I will add those details. I didn't think to initially because we haven't really worked with non algebraic extensions. $\endgroup$ – Ninosław Brzostowiecki May 18 '16 at 17:16
  • $\begingroup$ This is proved in Lang's Algebra. By the way, a homomorphism of fields is automatically non-zero because it maps $1$ to $1$. $\endgroup$ – Georges Elencwajg May 18 '16 at 18:32
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As I recall, it depends on a strategy like this, modulo suitable care to the hypotheses:
If $k\subset L$ are fields and $\psi\colon L\to\Omega$, where $\psi$ is a field morphism and $\Omega$ is an algebraically closed field, then $\{L:k\}=\{\psi(L):\psi(k)\}$. This needs a proof, but it isn’t hard.

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