27
$\begingroup$

Let $k$ be a field. I am wondering if there is an easy description of the ring

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

that is the tensor product of the power series ring $k[[x]]$ with itself over the ring of polynomials $k[x]$.

Any hints would be appreciated.

Edit: As suggested in the comments I am asking: is this ring isomorphic to $k[[x]]$?

$\endgroup$
24
  • 1
    $\begingroup$ @user10676, as far as I know, tensor products does not commute with inverse limits. $\endgroup$
    – the L
    Aug 10, 2011 at 13:00
  • 2
    $\begingroup$ It's pretty clear that the "obvious" map $k[[x]]\to k[[x]]\otimes_{k[x]}k[[x]]$ given by $\alpha\mapsto\alpha\otimes 1$ is not an isomorphism. It's trickier to rule out the existence of some other isomorphism. You'd have to find a property of rings that distinguishes the two sides. $\endgroup$ Aug 10, 2011 at 14:36
  • 1
    $\begingroup$ @Jim : I can see that the map is injective, but why is it not surjective ? $\endgroup$
    – user10676
    Aug 10, 2011 at 15:09
  • 3
    $\begingroup$ Your example doesn't work : $1 \otimes (1+x+x^2 + \cdots) = (1-x).(1-x)^{-1} \otimes (1-x)^{-1}$ $=(1-x)^{-1} \otimes (1-x).(1-x)^{-1} = (1+x+x^2 + \cdots) \otimes 1$. I have thought about this kind of argument but I can't find a proof. Let $P \in k[[x]]$, write $P=A+x^kB$, with $A$ polynomial, then $1 \otimes P - P \otimes 1 = x^k (B \otimes 1 - 1 \otimes B)$. So $1 \otimes P - P \otimes 1$ is $x$-divisible. I suspect that $k[[x]] \otimes_{k[x]} k[[x]]$ has no non-zero divisible element (since $k[[x]]$ doesn't). $\endgroup$
    – user10676
    Aug 10, 2011 at 16:03
  • 3
    $\begingroup$ @Jim: This span is $k(x)$. $\endgroup$ Aug 10, 2011 at 17:50

2 Answers 2

22
$\begingroup$

No, the rings $k[[x]] \otimes_{k[x]} k[[x]]$ and $k[[x]]$ are not isomorphic because they have different Krull dimensions:$\; \infty$ and $1$ respectively

I) The ring $k[[x]]$ has Krull dimension one since it is a discrete valuation ring.

II) The ring $k[[x]] \otimes_{k[x]} k[[x]]$ has at least the Krull dimension of its localization $k((x)) \otimes_{k(x)} k((x))$. It is thus sufficient to prove that the latter ring has infinite Krull dimension.
This results from Grothendieck's formula for the Krull dimension of the tensor product of two field extensions $K,L$ of a field $k$ as a function of the transcendence degrees of the extensions: $$ \dim (K \otimes_k L) =\min(\operatorname{trdeg}_k K, \operatorname{trdeg}_k L) $$

Since $\operatorname{trdeg}_{k(x)} k((x))=\infty$, we deduce that, as anounced, $$\dim(k[[x]] \otimes_{k[x]} k[[x]]) \geq \dim(k((x)) \otimes_{k(x)} k((x))) = \infty$$

Addendum: some properties of our tensor products
i) Let me show, as an answer to Pierre-Yves's first question in the comments, that $R=k[[x]] \otimes_{k[x]} k[[x]]$ is not a noetherian ring.
Any ring of fractions of a noetherian ring is noetherian and since, as already mentioned, the ring $T=k((x)) \otimes_{k(x)} k((x))$ is such a ring of fractions , it is enough to show that the latter tensor product $T$ of extensions is not noetherian.
This results from the following theorem of Vamos: given a field extension $F\subset K$, the tensor product $K\otimes_F K$ is noetherian iff the extension is finitely generated (in the field sense). Since in our case the extension $k(x) \subset k((x))$ is not finitely generated , we conclude that $T$ is not noetherian.
By the way, since the discrete valuation ring $k[[x]]$ is noetherian this gives another proof that it $R$ is not isomorphic to $k[[x]]$

ii) Pierre-Yves also asks if the ring $T$ is local. It is not because a theorem of Sweedler states that a tensor product of algebras over a field is local only if one of the factors is algebraic.
Since $k(x) \subset k((x))$ is not algebraic, non-locality of $T$ follows.

iii) The ring $R=k[[x]] \otimes_{k[x]} k[[x]]$ is not a domain (another thing Pierre-Yves has asked about) because if it were, its ring of fractions $T=k((x)) \otimes_{k(x)} k((x))$ would also be a domain ( that reduction again!) and I'm going to show that actually $T$ has zero divisors.

The key remark is that if $F\subset F(a)$ is a non trivial simple algebraic extension the tensor product $F(a)\otimes _F F(a)$ is not a domain. Indeed we have an isomorphism $F(a)\otimes _F F(a)=F(a)[T]/(m(T))$ where $m(T)$ is the minimal polynomial of $a$ over $F$. Since $m(T)$ has $a$ as a root, it is no longer irreducible over $F(a)$ and the quotient $F(a)[T]/m(T)$ has zero-divisors.

And now the rest is easy. Just take an element $a\in k((x))\setminus k(x)$ algebraic over $k(x)$ . Since $(k(x))(a)\otimes _{k(x)} (k(x))(a)$ is a subring of $T=k((x)) \otimes_{k(x)} k((x))$ containing zero-divisors, the ring $T$ a fortiori has zero-divisors.

$\endgroup$
11
  • $\begingroup$ Dear Georges: great answer! One nitpick, though: I was curious about the formula you mention, couldn't find it in the section of EGA that I expected, and googled to find this: jlms.oxfordjournals.org/content/s2-19/3/391.extract It seems the formula you mention is due to R. Y. Sharp. $\endgroup$ Aug 10, 2011 at 23:57
  • 10
    $\begingroup$ Dear Akhil,thanks for the kind words. I knew of Sharp's article and I stand by my attribution to Grothendieck of that wonderful formula. Indeed it was proved ten years before Sharp's article in EGA IV, Quatrième Partie, (4.2.1.5), page 349. But I am not surprized you could not find it: it is hidden in the Errata et Addenda of that volume! And you are in good company, because ten years after its publication the formula was so little known that Sharp could publish it in the excellent journal you mention, he and the referee being obviously unaware of Grothendieck's priority! $\endgroup$ Aug 11, 2011 at 1:57
  • $\begingroup$ Great, thanks for the EGA reference. That's a nice result to be buried in an appendix. (By the way, although you addressed me as "Dear Akhil" and not "@Akhil," it seems that I was nonetheless notified; this seems to be a pleasant improvement to the software.) $\endgroup$ Aug 11, 2011 at 2:27
  • $\begingroup$ Dear Georges, could you tell me if $k[[x]]\otimes_{k[x]}k[[x]]$ is noetherian? (And why?) (Or direct me to a reference.) [If you’re kind enough to answer, could you use the @ sign? To follow up on Akhil, my personal experience is that things work out usually without it, but they always do with it.] $\endgroup$ Aug 11, 2011 at 5:15
  • 1
    $\begingroup$ Dear Georges, thank you very much again for everything! I've just posted a somewhat long comment as a community wiki answer. [I didn't get a notification for your last comment. I guess "Dear (at)Pierre-Yves" is better - for the software - than (at)Dear Pierre-Yves. ;)] $\endgroup$ Aug 12, 2011 at 7:16
5
$\begingroup$

The rings $R:=k[[x]]\otimes_{k[x]}k[[x]]$ and $k[[x]]$ are not isomorphic. Here is a mild simplification of Georges Elencwajg's proof.

Assume by contradiction $R\simeq k[[x]]$. Let $K$ be the fraction field of $R$, and $S$ the multiplicative system $k[x]\backslash\{0\}$. As $R$ is a maximal subring of $K$, and $x\otimes1$ is invertible in $$S^{-1}R=k((x))\otimes_{k(x)}k((x)),$$ but not in $R$, we have $$K=S^{-1}R=k((x))\otimes_{k(x)}k((x)).$$ If $a$ is in $k((x))\backslash k(x)$, then $a\otimes1-1\otimes a$ is a nonzero element of $K$ which is mapped to $0$ by the natural morphism to $k((x))$, a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.