# Categorical viewpoint of $k[x]^\ast\cong k[[x]]$

An exam question which I have found on the internet asks for a "categorical" explanation of $k[x]^\ast\cong k[[x]]$. Could someone help me here? Maybe we have some functorial isomorphism but I don't see how, after all the dual space functor is contravariant.

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Could you please give the link to the exam question? –  magma Jan 31 '14 at 12:05

The notation is terrible. Essentially, you are being asked to show that $$\mathrm{Hom}_k (k^{\oplus \mathbb{N}}, k) \cong k^{\times \mathbb{N}}$$ and this is just the fact that $\mathrm{Hom}_k (-, k)$ sends coproducts to products.

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this is typical for most exam questions: look strange, trivial at heart. sad but true... –  user88576 Jan 30 '14 at 18:39
One also uses that this functor maps $k$ to $k$. –  Martin Brandenburg Jan 30 '14 at 23:36

Zhen writes "The notation is terrible." In fact, this is because most of the notation is missing. Some forgetful functors are involved here, and because they are not written at all, the isomorphism is just wrong or at least very confusing: "Why should the unit group of the polynomial ring be the power series ring?!" was my first thought.

So let me make clear what is really going on here, using the language of forgetful functors: Let $U : \mathsf{Alg}(k) \to \mathsf{Vect}(k)$ be the forgetful functor from $k$-algebras to $k$-vector spaces. We have a contravariant functor $\mathsf{Vect}(k) \to \mathsf{Vect}(k)$ which takes $V$ to its dual vector space $V^*$. In his answer, Zhen explains why $(k^{\oplus \mathbb{N}})^* \cong k^{\mathbb{N}}$. We have $k^{\mathbb{N}} \cong U(k[ [x] ])$ and $k^{\oplus \mathbb{N}} \cong U(k[x])$ (essentially by definition of these $k$-algebras $k[x]$ and $k[[x]]$). Hence, we have $U(k[x])^* \cong U(k[[x]])$. Please don't write this as $k[x]^* \cong k[[x]]$!

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Since every vector space is isomorphic to its dual, we could also write $U(k[x]) \cong U(k[[x]])$. Correct? –  magma Jan 31 '14 at 14:19
"Since every vector space is isomorphic to its dual" <- This is wrong. –  Martin Brandenburg Jan 31 '14 at 14:44
True, I was thinking at finite dimensional spaces, which is not the case here –  magma Feb 1 '14 at 0:40