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Let $F \subset K$ be a field extension of degree $n$, then $F^n \cong K$ as $F$-vectorspaces. Now $K^\times$ acts on $F^n$, by multiplication on $K$, and so $K^\times$ embeds into $GL_n(F)$, and every Galois element gives an automorphism of $K^\times$.

Question: Under which conditions can it be extended to an automorphism of $GL_n(F)$? How?

Cyclic extension, abelian extension, solvalabe extension, general extension?

I am mostly interested in the case, where $F$ is a local field.

Example $\mathbb{R} \subset \mathbb{C}$: We fix an $\mathbb{R}$-basis $\\{ 1,i \\}$. The multipliaction by $a+ib$ correspond to the matrix $$ \begin{pmatrix} a & -b \newline b & a \end{pmatrix}.$$ The Galois element is complex conjugation and corresponds to transpose on the above matrices. Can it be extended to the group $GL_2(\mathbb{C})$?

Motivation: I am actually hoping for an explanation of the Caley transform introduced here on page 2:

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Small typo - $K^\times$ embeds into $GL_n(F)$. – user29743 Apr 24 '12 at 20:32
I think I am misunderstanding the question. $Gal(K/F)$ respects the $F$-space structure on $K$, and so lives in $GL_n(F)$, where it acts on $K^\times$ via conjugation. Clearly this extends to all of $GL_n(F)$. – user641 Apr 24 '12 at 20:50
Could you include one worked example in the question so it is clearer exactly what you're looking for? Are you assuming the $K/F$ is a Galois extension? You don't say that in the question, but just mention a "Galois element", so it's not clear if you mean an $F$-automorphism of $K$, whether or not $K$ is actually Galois over $F$. – KCd Apr 25 '12 at 0:25
For $n=2$, when $K/F$ is quadratic (and the embedding is obvious), if I understand the question correctly, "conjugation by the longest Weyl element" extends the action of the non-trivial element of the Galois group to $GL_2(F)$. – B R Apr 25 '12 at 5:55
What @SteveD said is correct. Read his comment carefully again. Then work it out in the example of $\mathbb{C}/\mathbb{R}$: transposing $\left(\begin{smallmatrix}a&b\\-b&a\end{smallmatrix}\right)$ is the same as conjugating by $\left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)$, which is precisely how Galois acts with respect to your chosen basis. – Alex B. Apr 25 '12 at 8:47
up vote 3 down vote accepted

Maybe I am wrong, but consider rather the more canonical group $\mathrm{Aut}_F(K)$ of $F$-linear automorphisms on $K$. The Galois group naturally embedds and acts as inner automorphisms. The elements of $K^\times$ correspond to homothecies and the (inner) Galos action on the homothecies correspond to the natural action on $K$.

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