A tensor product of power series

Solution 1:

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.

Solution 2:

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.