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I want to know if the notion of completed tensor product in Stacks Project Tag 0AMU is the one that yields $$k[[x]] \widehat{\otimes} k[[y]]≅k[[x,y]].$$ Here I should be considering the inverse limit topology in the power series rings, and $R=k$ a field (with the trivial topology?). If it is not, then what is the right notion of $\widehat{\otimes}$ ?

I believe this is broadly used when talking of formal groups laws (eg defining its contravariant bialgebra).

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  • $\begingroup$ It might be the same as you say but is not it more immediate to use the variable-adic valuation? (Two series very near means their difference is divisible by a very high power of the variable) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Oct 23, 2020 at 12:08
  • $\begingroup$ I have seen the completed tensor product you mention, but as far as I know you can perform such a tensor product of (let me write $h$ for the variable) $k[[h]]$-modules. However if I see $k[[x]]$ and $k[[y]]$ as $k[[h]]$-modules then $k[[x]] \widehat{\otimes} k[[y]]≅k[[h]],$ so only one variable. I think the completion should be of $\otimes_k$, not of $\otimes_{k[[h]]}$. $\endgroup$
    – Minkowski
    Commented Oct 23, 2020 at 13:41

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$k[[x]]$ and $k[[y]]$ are topological $k$-algebras, and their completed tensor product is indeed isomorphic to $k[[x,y]]$ (with the topology defined by the maximal ideal $(x,y)$). This is because the completed tensor product is the projective limit of the algebras $k[x]/(x^m) \otimes k[y]/(y^n) = k[x,y]/(x^m,y^n)$ over $m,n \geq 1$.

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