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Let $K$ be an arbitrary field, and $K(t)$ denote the field of rational functions in $t$, i.e. function field on $K$.

If $K$ is algebraically closed field, then $\mathrm{Gal}(K(t),K)\cong \mathrm{PGL}_2(K)$ (is this correct?)

My question is about the Galois group for non-algebraically closed fields, say for example $\mathbb{R}$, $\mathbb{Q}$. What is the Galois group $\mathrm{Gal}(K(t),K)$ in this case of $K$? I have no idea about this kind of problems and research on it. It simply came to my mind when I saw the above stated result.

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    $\begingroup$ What is your definition for the Galois group of a transcendental extension? Usually, to make sense of the actual Galois theory, the Galois group of an infinite extension is taken as the inverse limit over the Galois groups of its finite subextensions, but this is not useful in your case. It is not clear in what sense $\mathbb C(t)/\mathbb C$ would be a Galois extension. $\endgroup$ – Mathmo123 Jun 28 '16 at 10:01
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I take it that by $\mathrm{Gal}(K(t),K)$ you mean the set of all automorphisms of $K(t)$ preserving $K$ pointwise. This group is more commonly denoted by $\mathrm{Aut}(K(t)/K)$ or $\mathrm{Aut}_K(K(t))$. In any case, it is quite well-known that

Theorem: For any field $K$, $\mathrm{Aut}(K(t)/K)=\mathrm{PGL}_2(K)$.

This theorem can be found in many places, e.g. on Page 647 of Dummit&Foote Abstract algebra, or on Page 8 of the book Introductory Notes on Valuation Rings and Function Fields in One Variable, by Renata Scognamillo, Umberto Zannier (you might be able to read the page on Google Books https://books.google.com/books?id=c-AlBAAAQBAJ).

A related Mathoverflow post is https://mathoverflow.net/questions/131464/relations-between-automorphisms-of-field-of-rational-functions-and-mobius-transf

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  • $\begingroup$ If $L/K$ is a Galois extension, then the group $\operatorname{Gal}(L/K)$ is frequently used to denote the group $\operatorname{Aut}(L/K)$ as a topological group under its corresponding profinite topology (Krull topology) induced from the Galois connection. $\endgroup$ – Geoff Jun 29 '16 at 4:39
  • $\begingroup$ @Geoff: Good point. Sadly, I often see people using $\mathrm{Gal}(L/K)$ instead of $\mathrm{Aut}(L/K)$ even when $L/K$ is not Galois. $\endgroup$ – Peradventure Jun 29 '16 at 5:48

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