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I recently heard that the following statement is a theorem in Galois theory.

Theorem. Let $F$ be a field and $L/K/F$ be a tower of field extensions of $F$. Suppose that $L/F$ and $K/F$ are both Galois extensions. Then for any $\sigma \in \text{Gal(K/F)}$, there exists $\tilde{\sigma} \in \text{Gal}(L/F)$ such that \begin{equation*} \tilde{\sigma}|_{K} = \sigma \end{equation*}

I have a couple of questions:

  1. Is this theorem (as I have stated it) true?

  2. Is the $\tilde{\sigma}$ unique?

  3. Do you know of an online reference (lecture notes, wikipedia, etc.) that states this result? I would like to read the proof, see the context, etc.

Thanks so much!

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  1. Yes, the theorem is true as stated.
  2. No, in fact, it will never be unique if the inclusion $ K \subset L $ is proper, since you have a surjective (by this result) group homomorphism $ \textrm{Gal}(L/F) \to \textrm{Gal}(K/F) $ given by restriction to $ K $. By order comparisons, it follows that it has nontrivial kernel iff $ K \subset L $ is a proper inclusion, thus the fibers have more than one element.
  3. This result follows from the fact that if $ \phi : K_1 \to K_2 $ is an isomorphism of fields and $ f \in K_1[X] $ is a polynomial, then the isomorphism extends to an isomorphism $ \bar \phi : L_1 \to L_2 $ of splitting fields of $ f $ and $ \phi(f) $ respectively. Since a finite Galois extension of $ K $ is always the splitting field of some polynomial with coefficients in $ K $, we may take $ K_1 = K_2 = K $ and get an automorphism $ \bar \phi : L \to L $ extending $ \phi $. I don't have any specific references, but this result is proved in pretty much any book on Galois theory.
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  • $\begingroup$ Thanks for this answer. Two questions though. It seems that the proof you gave relies on two facts being true: (1) Let $f(x)$ be the separable polynomial in $K[x]$ whose splitting field is $L$. Applying $\phi \in \text{Gal(K/F)}$ to the coefficients of $f(x)$ gives a polynomial in $K[x]$ that is still separable. Call this polynomial $f'(x)$. (2) $L$ is also the splitting field of $f'(x)$. Why are (1) and (2) true? $\endgroup$
    – Sam Y.
    Commented Apr 28, 2017 at 21:57
  • $\begingroup$ I don't see where I used (1), and (2) follows in this case since $ \phi $ fixes the coefficients of a polynomial in $ F[X] $, and $ L $ is the splitting field over $ F $, and thus over $ K $, of such a polynomial. $\endgroup$
    – Ege Erdil
    Commented Apr 28, 2017 at 22:06
  • $\begingroup$ It seems to me that the argument is the following: (a) Let $\phi: K \to K$ be an element of $\text{Gal}(K/F)$. (b) Since $L/K$ is Galois, $L$ is the splitting field of some separable $f(x) \in K[x]$. (c) Apply $\phi$ to the coefficients of $f(x)$ to get a polynomial $f'(x) \in K[x]$. (d) Note that $f'(x)$ remains separable in $K$. (e) Note that $L$ is also the splitting field of $f'(x)$. (f) It follows that $\phi$ extends to a map $\tilde{\phi}: L \to L$. So I thought you used (1) in (d). $\endgroup$
    – Sam Y.
    Commented Apr 28, 2017 at 22:18
  • $\begingroup$ I did not mention "separability" anywhere in my answer. Indeed, it is not necessary at all for the truth of this result, you can replace all mentions of "Galois" with "normal" and both the theorem and the proof remain exactly the same. (Of course, you also replace Galois groups with automorphism groups in that case.) $\endgroup$
    – Ege Erdil
    Commented Apr 28, 2017 at 22:21
  • $\begingroup$ I thought you mentioned separability in the line: "Since a finite Galois extension of $K$ is always the splitting field of some polynomial with coefficients in $K$". I don't think this sentence is true if the polynomial is not separable (though I could be wrong). $\endgroup$
    – Sam Y.
    Commented Apr 28, 2017 at 22:24

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