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Here $k[[t]]$ is the $k$-algebra of formal power series and $A[[t]]$ is a $k[[t]]$-module of formal power series.

Bikram Banerjee
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  • Can you prove that the "obvious" mapping is a homomorphism of $k[[t]]$-modules? By the way, you should not rely on the title of your post to be the question. Check out our guide for new askers. – Jyrki Lahtonen Mar 18 '21 at 07:36
  • I.e trivial...I want to know why has an inverse. By the way, I am sorry for not making the title according to the guidelines – Bikram Banerjee Mar 18 '21 at 07:55
  • All you want to show is that it is a bijection and that is a $k[[t]]$-homomorphism. For the bijective part you can also consider it as a morphism of $k$ vector spaces. Picking a $k$-basis of $A$ would also be helpful. – Severin Schraven Mar 18 '21 at 09:12
  • I would think that the image of the canonical map $A \otimes_k k[[t]] \to A[[t]]$ would only consist of the formal series where the span of the coefficients (as a $k$-vector space) is finite dimensional. So in particular, if $A$ itself is finite dimensional, then the result would hold, but otherwise I don't think it would. – Daniel Schepler Mar 18 '21 at 15:45
  • Thanks a lot Severin. I am quite fine with it. Sorry for delayed reply. – Bikram Banerjee Mar 20 '21 at 06:49

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