# $K[[X]]$ is not a finitely generated $K[X]$-module.

How can I prove that $K[[X]]$ is not finitely generated over $K[X]$ as a module, where $K$ is a field.

What I tried: if above is not true then $K[[X]]$ is integral extension over $K[X]$. But I failed to draw any contradiction.

Help me. Thanks

• For finite $K$, counting is enough. - Commented Apr 8, 2016 at 18:39
• counting of what ? @ Hagen von Eitzen
– user185640
Commented Apr 8, 2016 at 18:41
• @Panja I'm pretty sure what Hagen meant here "counting sizes", i.e. cardinalities of the sets. To be more specific, for $K$ at most countable, $K[X]$ is countable and thus every finitely generated module over $K[X]$ is. On the other hand, $K[[X]]$ is easily seen to be uncountable. Commented Jul 9, 2016 at 13:02
• Thank you for explaining that counting argument . I understand. @Wojowu
– user185640
Commented Jul 10, 2016 at 10:32
• I don't know if anyone is still interested in it, but your original idea can be made to work, too. The going-up theorem holds for an integral extension, but your extension ring only has two prime ideals while the base rings has infinitely many, so that's a contradiction. Commented Mar 19 at 19:12

The inclusion of rings $j:K[X] \hookrightarrow K[[X]]$ induces a map $j^\ast: \operatorname {Spec}(K[[X]])\to \operatorname {Spec}(K[X])$ between the corresponding sets of prime ideals of these rings.
If $K[[X]]$ were module-finite (or even just integral) over $K[X]$, the map $j^\ast: \operatorname {Spec}(K[[X]])\to \operatorname {Spec}(K[X])$ would be surjective: Atiyah-Macdonald, Theorem 5.10 .
But this is impossible since $K[[X]]$ (like all discrete valuation rings) has only two prime ideals, whereas $K[X]$ has infinitely many (by a variation of Euclid's proof of the infinitude of prime integers).

• You are welcome, dear Panja. And thanks for the kind words. Commented Apr 8, 2016 at 19:02
• I think it's also true that $K[[X]]$ isn't of finite type over $K[X]$, and that the transcendental degree of $\operatorname{Frac}(K[[X]])$ over $K(X)$ is infinite. Commented Apr 8, 2016 at 19:16
• I have no simple proof, but if $K[[X]]$ is of finite type over $K$, then it should be an affine variety over $K$, but $K[[X]]$ is local. Commented Apr 8, 2016 at 19:23
• @FrankScience Why can't an affine variety over $K$ be local? e.g., $k[X]/(X^2)$? Commented Apr 8, 2016 at 23:26
• @FrankScience If the ring of formal power series was finite type over the affine line, then Chevalley implies that the image is constructible. But this would imply that the generic point is constructible (by intersecting the image with the complement of the closed point in the image), which is false. (This mimics a proof of the Nullstellensatz.) Commented May 20, 2016 at 6:21

In $K[[X]]$, the element $f=\sum_{i=0}^\infty X^i$ is an inverse of $1-X$. Since $K[X]$ is integrally closed (even a PID), $(1-X)^{-1}$ can't be integral over $K[X]$. Therefore $K[X][f] \subseteq K[[X]]$ is a not f.g submodule. Since $K[X]$ is Noetherian (by Hilbert Basis Theorem), we can conclude that $K[[X]]$ can't be finitely generated either.

• I really like Georges Elencwajg's geometric solution, I just wanted to add an algebraic one.
– benh
Commented Apr 8, 2016 at 19:44
• a) Fantastic proof, benh: +1. b) You have proved that $(1-X)^{-1}$ is not integral, so $K[[X]]$ is not integral and thus a fortiori not finitely generated. You don't have to use noetherianness nor Hilbert's theorem, which makes your proof more elementary. c) Users are strongly encouraged to give alternative interesting proofs, like benh's, even of accepted answers: nobody should have a monopoly for answers here! Commented Apr 8, 2016 at 20:42
• Oh - right! Nice, thank you! :)
– benh
Commented Apr 8, 2016 at 23:23

Inspired by the counting suggested by Hagen, here is a proof that shows $K[[X]]$ is not even of finite type over $K[X]$.

First, note that this holds when $K$ is at most countable, for in this case $K[X]$ is countable while $K[[X]]$ is uncountable, and any ring of finite type over a countable ring is still countable.

Now the general case. Suppose $K[[X]]=K[X,f_1,\dots,f_n]$. Let $k$ be the field generated by the prime field of $K$ (namely $\mathbb Q$ or $\mathbb F_p$) and the (countably many) coefficients of $f_i$. Then $k$ is countable. By the above argument, $k[[X]]$ is not of finite type over $k[X]$, so there is $g\in k[[X]]$ such that $g\notin k[X,f_1,\dots,f_n]$. This means that, for any $N\in\mathbb N$, the equation

$$g=\sum_{\alpha_i\le N\atop\deg p_\alpha\le N} p_\alpha f_1^{\alpha_1}\cdots f_n^{\alpha_n} \tag{*}$$

has no solution $p_{\alpha}\in k[X]$. This can be thought of an (infinite) system of linear equations in the finitely many (in fact $(N+1)^{n+1})$ coefficients of $p_\alpha$, whose own coefficients lie in $k$. Since it has no nonzero solution in $k$, finitely many of the equations already has no nonzero solution in $k$. Since the existence of nonzero solutions to a system of linear equations does not depend on the ground field, the same finitely many equations has no nonzero solution in the larger field $K$, and hence the system as a whole has no nonzero solution $K$. This implies that (*) is not satisfiable either in $K$. Since this is true for any $N\in\mathbb N$, $g\notin K[X,f_1,\dots,f_n]$.

• Actually this is a system of linear equations in the coefficients of $p_\alpha$, so no need for the Nullstellensatz. Commented Sep 24, 2016 at 17:34

An alternative approach showing that $K[[X]]$ is not a finitely generated $K$-algebra, and hence not a finitely generated $K[X]$-algebra.

If a finitely generated algebra over a field is a local ring then it is artinian.

Suppose that $K[X_1,\dots,X_n]/I$ is a local ring, for some ideal $I$. Then $I$ is contained in a single maximal ideal, say $M$, so $\sqrt I=M$ (since a polynomial ring over a field is a Jacobson ring). In particular, $\dim K[X_1,\dots,X_n]/I=0$, and thus $K[X_1,\dots,X_n]/I$ is an artinian local ring.

In the following context we assume everything is commutative.

Definition: A Jacobson ring is a ring where every prime ideal is an intersection of some maximal ideals.

Then we have the fact (cf. Eisenbud Commutative Algebra with a View towards Algebraic Geometry Theorem 4.19) that:

Fact (Nullstellensatz): Let $A$ be a Jacobson ring, then any finitely generated $A$-algebra is a Jacobson ring.

Now apply this result to the question.

First note that a field $K$ is obviously a Jacobson ring, by Nullstellensatz $K[X]$ is a Jacobson ring. Next note that if a local ring has more than one prime ideal, then it cannot be a Jacobson ring. Now $K[[X]]$ is a local ring with prime ideals $(X)$ and $(0)$, hence it cannot be a Jacobson ring. Hence it cannot be a finitely generated $K[X]$-algebra. (A fortiori not finite $K[X]$-module.)

• Really need Nullstellensatz to show that $K[X]$ is Jacobson? Commented Mar 11, 2018 at 19:03
• @user26857 Perhaps not! But the point is if K[[X]] is finitely generated over K[X] then again by Nullstellensatz it is Jacobson, and this contradiction is somehow not so obvious.
– Y_q
Commented Mar 11, 2018 at 21:06