For k a field, the localization

et3atissb
2022-04-23
Answered

Is k[[x]] ever a finitely generated k[x](x) module?

For k a field, the localization$k{\left[x\right]}_{\left(x\right)}$ naturally includes into k[[x]]. I can prove that if $k=\mathbb{C},$ , then this inclusion is not surjective, and k[[x]] is not even finitely generated over $k{\left[x\right]}_{\left(x\right)}$ , because $\mathbb{C}{\left[x\right]}_{\left(x\right)}$ corresponds to rational functions holomorphic at 0, and $\mathbb{C}\left[\left[x\right]\right]$ 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?

For k a field, the localization

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haarplukxjf

Answered 2022-04-24
Author has **16** answers

Explanation:

In fact, k[[x]] is never even countably generated as a$k{\left[x\right]}_{\left(x\right)}$ -module.

To prove this, suppose k[[x]] were generated by countably many elements$\left\{{c}_{i}\right\}}_{i\in \mathbb{N}$ as a $k{\left[x\right]}_{\left(x\right)}$ -module, and let ${k}_{0}\subseteq k$ be the subfield generated by all the coefficients of the $c}_{i$ . For any $f\in k\left[\left[x\right]\right]$ , then, there exist $a,{b}_{1},\cdots ,{b}_{n}\in k\left[x\right]$ with $a\ne 0$ such that

$af=\sum _{i=1}^{n}{b}_{i}{c}_{i}$

(here a is the common denominator of the coefficients when you write f as a$k{\left[x\right]}_{\left(x\right)}$ -linear combination of the $c}_{i$ ). If $f\in {k}_{0}\left[\left[x\right]\right]$ , we can consider this equation as a system of linear equations with coefficients in $k}_{0$ (coming from the coefficients of f and the $c}_{i$ ) in the (finitely many) coefficients of a and the $b}_{i$ . 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 $k}_{0}{\left[x\right]}_{\left(x\right)$ -linear combination of the $c}_{i$ .

Thus${k}_{0}\left[\left[x\right]\right]$ is also countably generated as a $k}_{0}{\left[x\right]}_{\left(x\right)$ -module (by the same countably many elements $c}_{i$ ). But $k}_{0$ is countable, so $k}_{0}{\left[x\right]}_{\left(x\right)$ is countable and thus so is any countably generated $k}_{0}{\left[x\right]}_{\left(x\right)$ -module. Since ${k}_{0}\left[\left[x\right]\right]$ is uncountable, this is a contradiction.

In fact, k[[x]] is never even countably generated as a

To prove this, suppose k[[x]] were generated by countably many elements

(here a is the common denominator of the coefficients when you write f as a

Thus

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