Take the 2-minute tour ×

Suppose that $k$ is an algebraically closed field. Let $F/k$ be a (possibly non-finitely generated) field extension. Is

$$ k[[x]] \otimes_{k} F $$


If not, is the natural map $k[[x]] \otimes_{k} F \to F[[x]]$ injective?

share|improve this question
The natural map is injective, by a simple argument: Let $T$ be a tensor in $k\left[\left[x\right]\right]\otimes_k F$ which gets mapped to $0$ by this map. Then, we can write $T$ as $\sum\limits_{i\in I} s_i \otimes f_i$ for some finite set $I$, some $s_i\in k\left[\left[x\right]\right]$ and some linearly independent $f_i\in F$. Now, the condition that $T$ gets mapped to $0$ by the natural map rewrites as $\sum\limits_{i\in I} f_is_i=0$. Hence, every $j\in\mathbb N$ satisfies $\sum\limits_{i\in I} f_i\left(s_i\right)_j=0$, where we treat power series in $k\left[\left[x\right]\right]$ ... –  darij grinberg Mar 31 '12 at 3:17
... as sequences of elements of $k$. But due to the linear independence of the $f_i$, this yields that $\left(s_i\right)_j = 0$ for all $i$ and $j$, and thus $T=0$. –  darij grinberg Mar 31 '12 at 3:18
I gave a wrong answer sometime ago, using injectivity. As it seems ok, let me try again:starting with infinitely many power series $\sum_ia_n^{i}x^i$ with $a_n\in\mathbb{C}$ all algebraically independent, the ideal generated by all of these is not generated by finitely many of them, I guess, for an argument of transcendence degree. But I still lack a neat proof, so I do not post it as an answer. –  Filippo Alberto Edoardo Mar 31 '12 at 6:12
Darij's argument show in fact that for any field $k$ and any $k$-vector space $V$, the natural map $k^{\mathbf{N}} \otimes V \to V^{\mathbf{N}}$ is injective (it is bijective if and only if $V$ is finite dimensional). –  François Brunault Mar 31 '12 at 11:33

2 Answers 2

up vote 15 down vote accepted

The answer is no: for example, $k[[x]]\otimes_k k((x))$ is not noetherian.
Indeed, if it were, so would be $ k((x))\otimes_k k((x))$.
But this would contradict the following interesting general theorem of Vámos:

Given an extension of fields $K/F$ the tensor product $K\otimes_F K$ is noetherian if and only if $K$ is finitely generated as a field over $F$.

Full confession
I have only read an abstract of Vámos's article because I have no access to it. Anyway, here is the reference:
P. Vámos, On the minimal prime ideal of a tensor product of two fields, Math. Proc. Cambridge Philos. Soc. 84 (1978), no. 1, p.25-35.

share|improve this answer
Every proper inclusion of subfields $F\subset L_1\subset L_2\subset K$ gives rise to a non-injective surjective ring homomorphism $K\otimes_{F_1} K\to K\otimes_{F_2} K$. So, in Vamos' theorem, the "only if" directly follows, and the "if" condition is equivalent to the fact that every sub-extension of a finitely generated extension of $F$ is itself finitely generated. The latter fact is stated with no reference in Wiki's page on the 14th Hilbert problem en.wikipedia.org/wiki/Hilbert's_fourteenth_problem –  YCor Mar 31 '12 at 13:16
[of course I means $L_i=F_i$]. Note that the immediate implication in my comment is that if $K$ is infinitely generated then $K\otimes_F K$ is not noetherian (this is enough for Georges's example). For the converse, let $L\subset K$ be the field generated by a transcendence basis $x_1,\dots,x_d$, so $B=L\otimes_F L$ is a localization of $F[x_1,\dots,x_d,y_1,\dots,y_d]$ so is noetherian, and $K\otimes_F K$ is a finitely generated $B$-algebra so is noetherian as well. (By my previous remark, as a corollary every subfield of a f.g. field is f.g.) –  YCor Apr 1 '12 at 15:53
Dear @Yves, your comments are quite interesting. Why not upgrade them to a leisurely answer, which will be easier to read than necessarily terse comments? –  Georges Elencwajg Apr 1 '12 at 19:30

I was emailed the following argument:

We prove that $k[[x]]\otimes_{k} k((x))$ is not noetherian by showing directly that $k((x)) \otimes k((x))$ is not noetherian (as suggested by Georges Elencwajg). I will just handle the case $k=\bar{\mathbb{Q}}$ and then make a remark about the general case at the end.

The field $k((x))$ has only countably many finite separable extensions because every such extension is obtained by adjoining a root of $x$. On the other hand, the transcendence degree of $k((x))$ over $k$ must be uncountable because $k((x))$ is uncountable and $k$ is countable. Fix a transcendence basis $( t_i )_{i \in I}$ for $k((x))$ over $k$.

The algebraic extension $k((x))$ of $K := k((t_i))$ is algebraic with infinite separable degree. Indeed, if the separable degree was finite, then $K$ would admit at most countably many finite separable extensions as this is true for $k((x))$. This is absurd because $( t_i )$ is uncountable.

Because the separable degree of $k((x))$ over $k((t_i))$ is infinite, $(k((x)) \otimes_{K} k((x)))_{\text{red}}$ has infinitely many idempotents and so

$$k((x)) \otimes_{K} k((x))$$ and hence $$k((x)) \otimes_{k} k((x))$$

are non-noetherian. This completes the proof.

With work, this proof can be modified to hold when $k$ is a finite field $\mathbb{F}$. In this case, one must argue more carefully to show that $k((x))$ has only countably many finite separable extensions. (The email indicated that one should use local compactness together with Krasner's Lemma.) Finally, one can deduce the case of a more general field $k$ from the case $k=\bar{\mathbb{Q}}$ or $\mathbb{F}$ by using a faithfully flat descent argument.

share|improve this answer

Your Answer


