24
$\begingroup$

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 $k[[x]]$?

  • 0
    I was wondering if it would be possible to ask more precise questions about this ring. One is tempted to ask you: What do you mean by “description”? What exactly do you want to know about this ring? [Sorry for not being more helpful.]2011-08-10
  • 0
    As I understand it, although $k[x]\otimes_k k[x] \cong k[x,y]$, it seems that it is not in general true that $k[[x]]\otimes_k k[[x]] \cong k[[x,y]]$. I don't know anything about $k[[x]]\otimes_{k[x]} k[[x]]$ though.2011-08-10
  • 0
    @Pierre-Yves Gaillard, I was secretly hoping that it will turn out to be isomorphic to $k[[x]]$...2011-08-10
  • 0
    @John M, I am quite sure this is not true. For this to be true you need to take the complete tensor product.2011-08-10
  • 0
    @John : yes, you are right. I've fogot to complete the tensor product for the first isomorphism.2011-08-10
  • 1
    @user10676, as far as I know, tensor products does not commute with inverse limits.2011-08-10
  • 0
    I'll delete my comments.2011-08-10
  • 0
    I’m sure you know much more than I about the subject, but I believe that if you asked: “Is $k[[x]]\otimes_{k[x]}k[[x]]$ isomorphic to $k[[x]]$?” you’d get more answers (or at least more attention). [That’s just an opinion.]2011-08-10
  • 0
    @Pierre-Yves Gaillard, thank you. I acted as you suggested.2011-08-10
  • 0
    You're welcome. I hope it'll work. We'll see...2011-08-10
  • 2
    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.2011-08-10
  • 0
    A probably silly question: Do you know if $k[[x]]\otimes_{k[x]}k[[x]]$ is complete (in the $x$-adic topology)?2011-08-10
  • 1
    @Jim : I can see that the map is injective, but why is it not surjective ?2011-08-10
  • 0
    @user10676: $1\otimes (1+x+x^2+\cdots)$ is not in the image. You can pass an arbitrarily high number of terms across the tensor product, but you'll never be able to pass them all across.2011-08-10
  • 3
    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).2011-08-10
  • 1
    @user10676: Aha! Excellent point.2011-08-10
  • 0
    @user10676: I wonder if this is equivalent to the problem of whether the subspace of $k[[x]]$ generated by polynomials and inverses of polynomials is actually all of $k[[x]]$.2011-08-10
  • 1
    @Jim: Irrational series exist. (E.g. the exponential series.) (Is $1\otimes e^x$ in the image?)2011-08-10
  • 0
    @user10676: I guess my last comment is impossible in general for cardinality reasons. If $k=\mathbb Q$, then $k[[x]]$ is uncountable but the span of polynomials and their inverses is countable. So my new guess for something not in the image is $1\otimes \alpha$ where $\alpha$ is not in the span of polynomials and their inverses.2011-08-10
  • 3
    @Jim: This span is $k(x)$.2011-08-10
  • 0
    Dear anonymous: You ask: “is this ring isomorphic to $k[[x]]$?” The literal interpretation is: as rings. Other possible interpretations are: as $k[[x]]$-algebras, as $k[x]$-algebras, as $k$-algebras. [You can put 0, 1 or 2 pairs of brackets! (Or work over $\mathbb Z$.)]2011-08-10
  • 0
    Perhaps it’s not too hard to rule out the existence of a $k[[x]]$-algebra (or even a $k[[x]]$-module) isomorphism. --- At any rate, it's crucial to know in which category we're working.2011-08-10
  • 0
    @Pierre-Yves Gaillard, I would be happy to hear about any of the interpretations. For my applications I actually care about ring isomorphisms.2011-08-10
  • 1
    Thanks! For the non-existence of a $k[[x]]$-module isomorphism: $1\otimes1$ and $1\otimes e^ x$ are linearly independent over $k[[x]]$ (because $1$ and $e^x$ are linearly independent over $k[x]$).2011-08-10

2 Answers 2

21

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.

  • 0
    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: http://jlms.oxfordjournals.org/content/s2-19/3/391.extract It seems the formula you mention is due to R. Y. Sharp.2011-08-10
  • 10
    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!2011-08-11
  • 0
    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.)2011-08-11
  • 0
    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.]2011-08-11
  • 0
    Very nice! thank you!2011-08-11
  • 0
    @Pierre-Yves Gaillard: Dear Pierre-Yves, the tensor product you mention is *never* noetherian. I'll post a proof as an addendum to my answer.2011-08-11
  • 0
    Dear Georges, thanks a lot! I’m looking forward to your addendum. I was wondering if (in the addendum) you could also mention some other basic properties of this ring, like: Is it local? Is it a domain? ...2011-08-11
  • 0
    Dear Georges, I don’t know how to thank you! Three details: (1) Let’s call $R$ the ring $k[[x]]\otimes_{k[x]}k[[x]]$. The ring $T$ is **not** local. What about $R$? (2) Do you know if $R$ has zero divisors? (3) It seems to me that you found various **independent** proofs of the non-isomorphism of $R$ and $k[[x]]$. Is that right? [My first name is Pierre-Yves, not Jean-Pierre (as you wrote in ii)).]2011-08-11
  • 0
    @Dear Pierre-Yves, sorry for my absent-mindedness: I have corrected ii) so that your real first name now appears there. You are absolutely right that ii) gives a new proof of non-isomorphism. I had actually intended to mention that fact but somehow forgot (absent-mindedness again!): thanks for reminding me . I have added a third addendum iii) proving that neither $R$ nor $T$ are domains but haven't examined whether $R$ is local.2011-08-11
  • 1
    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. ;)]2011-08-12
4

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.