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Can we decompose the field of formal Laurent series as $$\mathbb C((t))\cong \mathbb C[t^{-1}]\times\mathbb C[[t]]$$ as vector spaces over the field of complex numbers? The map

$$\sum_{i\in\mathbb Z}\lambda_it^i\mapsto (\sum_{i<0}\lambda_it^i,\sum_{i\geq0}\lambda_it^i)$$

seems to satisfy the above decomposition.

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    $\begingroup$ The RHS is not a field, so no such isomorphism exists $\endgroup$ Commented Jul 7, 2023 at 12:18
  • $\begingroup$ Right, I agree with you on that. I made a sloppy mistake. I should have asked if the decomposition exists at the level of complex vector spaces, not algebras. I will edit the question. $\endgroup$ Commented Jul 7, 2023 at 12:21
  • $\begingroup$ I think this map is not well-defined. What is the image of $\sum_{k\in \mathbb{Z}} t^k$ ? I believe you should replace $\mathbb{C}[t^{-1}]$ by $\mathbb{C}[[t^{-1}]]$. $\endgroup$
    – groupoid
    Commented Jul 7, 2023 at 12:25
  • $\begingroup$ @groupoid formal Laurent series are typically defined to only contain series with finitely many terms of negative exponent. $\endgroup$ Commented Jul 7, 2023 at 12:28
  • $\begingroup$ An element in $\mathbb C((t))$ has only finitely many terms $\lambda_k t^{k}$ with $k<0$, not infinitely many. So, the formal sum $\sum_{k\in\mathbb Z}t^k$ is not an element in the formal Laurent field, it seems. $\endgroup$ Commented Jul 7, 2023 at 12:33

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Yes, the map that you defined is an isomorphism of vector spaces. (Obviously not an isomorphism of algebras, but you knew that already.)

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    $\begingroup$ The isomorphism remains true if we replace the field $\mathbb C$ by a commutative ring $R$. Then, we simply have an isomorphism of $R$-modules, don't we? $\endgroup$ Commented Jul 7, 2023 at 12:39

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