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Let $k$ be a field of characteristic zero and $A$ be a $k$-algebra. In so many literatures on classical algebraic deformation theory it is stated that $A \otimes _{k} k[[t]] \cong A[[t]]$ as a $k[[t]]$-module (Here $A[[t]]$ and $k[[t]]$ are formal power series over $A$ and $k$ respectively).

Now if $A$ is finite dimensional then it is easy to show that the canonical map is an isomorphism. Now if $A$ is not finite dimensional then I think $A[[t]] \cong A \mathbin{\hat{\otimes}} k[[t]]$, i.e. $A[[t]]$ is isomorphic to the completed tensor product of $A$ with $k[[t]]$ as a $k[[t]]$-module. Am I correct or not? If not then what is the argument?

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    $\begingroup$ Indeed. If $k=\mathbb{Q}$ and $A=\bar{\mathbb{Q}}$ is an algebraic closure of $\mathbb{Q}$, then $\bar{\mathbb{Q}}\otimes_{\mathbb{Q}} \mathbb{Q}[[t]]$ is isomorphic to the subring of $\bar{\mathbb{Q}}[[t]]$ consisting of power series whose coefficients all belong to a finite extension of $\mathbb{Q}$. $\endgroup$
    – Chris
    Jul 1, 2021 at 8:47
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    $\begingroup$ TeX note: Putting a \hat on \otimes cancels out its \mathbinness, which you have manually to re-instate. Compare $A \hat\otimes k[[t]]$ A \hat{\otimes} k[[t]] to $A \mathbin{\hat\otimes} k[[t]]$ A \mathbin{\hat\otimes} k[[t]]. I have edited accordingly. $\endgroup$
    – LSpice
    Jul 1, 2021 at 12:55
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    $\begingroup$ 1.Algebraic deformation theory: W.Stephen Piper (J. Diff Geometry) 1967, Page 135 $\endgroup$ Jul 2, 2021 at 8:01
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    $\begingroup$ An introduction to algebraic deformation theory, Thomas F. Fox, page 19, Journal of pure and applied algebra, 84, 1993 $\endgroup$ Jul 2, 2021 at 8:03
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    $\begingroup$ Deformation theory (Lecture notes) M. Doubek, M. Markl and P. Zima, page 26 $\endgroup$ Jul 2, 2021 at 8:04

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Yes, you are correct. $A [[t]] \simeq A \mathbin{\hat\otimes} \Bbbk [[t]]$ and $A \otimes \Bbbk [[t]]$ without $\hat{}$ consists of the formal power series in $t$ whose coefficients span a finite-dimensional subspace of $A$.

Long answer. It seems like you are looking at deformations of associative algebras, where $A$ is the $\Bbbk$-algebra to be deformed and $B$ is your base of deformation — in your example $B = \Bbbk [[t]]$. (If not, let me just illustrate the discussion for this case.)

If either $A$ or $B$ are finite-dimensional, then $A \mathbin{\hat\otimes} B = A \otimes B$, but if both $A$ and $B$ are infinite-dimensional one should work with $\hat\otimes$.

Since classical deformation functors are often defined on (commutative) local Artinian $\Bbbk$-algebras (which are finite-dimensional), it is enough to look at $A \otimes B$ when $B$ is local Artinian. (These are sometimes called infitesimal deformations.)

However, more generally a formal deformation over $B$ should be given by an associative algebra structure on $A \mathbin{\hat\otimes} B$, where $B$ is a complete local Noetherian $\Bbbk$-algebra such that modulo the maximal ideal $\mathfrak m$ of $B$ you recover the original algebra structure on $A$. Since not every complete local Noetherian $\Bbbk$-algebra is finite-dimensional, one should really write $\hat\otimes$ when $A$ is not necessarily finite-dimensional, but the extra $\hat{}$ is easy to forget.

A nice source for reading about this is Chapter 1 in:

  • Markl, Martin, Deformation theory of algebras and their diagrams, CBMS Regional Conference Series in Mathematics 116. Providence, RI: American Mathematical Society (AMS); Washington, DC: Conference Board of the Mathematical Sciences (CBMS) (ISBN 978-0-8218-8979-4/pbk). ix, 129 p. (2012). ZBL1267.16029.
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