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PicardVessiot Extension over a Differential Field? 
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#1
Apr812, 07:16 PM

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Given a differential field F and a linear algebraic group G over the constant field C of F, find a PicardVessiot 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? 


#2
Apr812, 07:26 PM

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Caveat: I know nothing about this subject. I checked Wikipedia for the relevant definitions, and I believe this works. First find a PicardVessiot 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 PicardVessiot extensions works like normal Galois theory (i.e. if you have an analogue of Artin's theorem), then E/E^G should be PicardVessiot 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.) 


#3
Apr912, 02:54 PM

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#4
Apr1112, 11:29 AM

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PicardVessiot Extension over a Differential Field?
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 PicardVessiot 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 PicardVessiot extension E/F with G(E/F)=GL_n(C).]
I don't know what happens if C is not algebraically closed. 


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