Is k[[x]] ever a finitely generated k[x](x) module? For k a field, the localization k[x]_{(x)

et3atissb 2022-04-23 Answered
Is k[[x]] ever a finitely generated k[x](x) module?
For k a field, the localization k[x](x) naturally includes into k[[x]]. I can prove that if k=C,, then this inclusion is not surjective, and k[[x]] is not even finitely generated over k[x](x), because C[x](x) corresponds to rational functions holomorphic at 0, and C[[x]] corresponds to germs of all functions holomorphic at 0, and so with some complex analysis you can see the finite generation is impossible. But over an arbitrary field, is this claim still true?
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Answers (1)

haarplukxjf
Answered 2022-04-24 Author has 16 answers
Explanation:
In fact, k[[x]] is never even countably generated as a k[x](x)-module.
To prove this, suppose k[[x]] were generated by countably many elements {ci}iN as a k[x](x)-module, and let k0k be the subfield generated by all the coefficients of the ci. For any fk[[x]], then, there exist a,b1,,bnk[x] with a0 such that
af=i=1nbici
(here a is the common denominator of the coefficients when you write f as a k[x](x)-linear combination of the ci). If fk0[[x]], we can consider this equation as a system of linear equations with coefficients in k0 (coming from the coefficients of f and the ci) in the (finitely many) coefficients of a and the bi. A basis for the solutions of such a system can be computed by Gaussian elimination, and this process is unchanged by extending the base field. In particular, since there exists a solution to this system over k where one of the coefficients of a is nonzero, such a solution also exists over k0. This just means that f is also a k0[x](x)-linear combination of the ci.
Thus k0[[x]] is also countably generated as a k0[x](x)-module (by the same countably many elements ci). But k0 is countable, so k0[x](x) is countable and thus so is any countably generated k0[x](x)-module. Since k0[[x]] is uncountable, this is a contradiction.
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