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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?

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Morphism on Physicsforums provided the following answer to this question (I reposted it there). Can someone check this for validity: "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.)"

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