Picard-Vessiot Extension over a Differential Field?

KarmonEuloid

Given a differential field F and a linear algebraic group G over the constant field C of F, find a Picard-Vessiot extension of E of F with G(E/F)=G:

This isn't homework, just something I saw in a book that I was curious about. The author says that this can be shown but doesn't illustrate how. Can anyone help?

morphism

Caveat: I know nothing about this subject. I checked Wikipedia for the relevant definitions, and I believe this works. First find a Picard-Vessiot extension E/F with G(E/F)=GL_n(C) (such a thing does exist, right??). Next, given a linear algebraic group G, view it as sitting in some GL_n(C)=G(E/F), and then consider the fixed field E^G (this notion makes sense, right??). If the Galois theory of Picard-Vessiot extensions works like normal Galois theory (i.e. if you have an analogue of Artin's theorem), then E/E^G should be Picard-Vessiot and G(E/E^G) should be G.

Note that this proof is identical to the standard proof that every finite group G is the Galois group of some extension. (The role of GL_n above is played by S_n here.)

KarmonEuloid

This seems to work to me. I reposted it on MathOverflow (giving you credit of course) to see if they could verify it (as that is where the question originally came from), although I would also appreciate it if someone here could check this or provide an alternate answer. Thanks.

morphism

I read a bit more about this topic and can now confirm that the above proof is correct, provided the field of constants C is algebraically closed. This assumption is apparently necessary for the analogue of Artin's theorem to hold for Picard-Vessiot extensions. See Chapter 6 of Crespo and Hajto, Algebraic Groups and Differential Galois Theory (AMS 2011), freely available here. [Also note: Exercise 7 shows that, for any n, there is a Picard-Vessiot extension E/F with G(E/F)=GL_n(C).]

I don't know what happens if C is not algebraically closed.