# The "extending the automorphism" theorem in Galois theory

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!

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.
• 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? Apr 28, 2017 at 21:57
• 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. Apr 28, 2017 at 22:06
• 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). Apr 28, 2017 at 22:18
• 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). Apr 28, 2017 at 22:24