Let $k[X]$ be the space of polynomials over a field $k$ (regarded as a vector space over $k$). What is the dual space of this vector space? My guess is that it is somehow generated by the derivations $d/dX$?
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1I don't think is a good practice to change a question after an answer was accepted. – Quimey Jul 17 '13 at 06:38
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edited Jul 12 at 13:49. answered Jul 12 at 13:52. – Jul 17 '13 at 16:46
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It is possible to identify $k[X]^*$ with $k[[X]]$ (power series).
Consider the map $\phi:k[X]^* \to k[[X]]$ such that $\phi(f)=\sum f(x^i)x^i$. It is clear that $\phi$ is a monomorphism. It is also surjective because if $w=\sum a_i x^i\in k[[X]]$ then we can define $f$ in the standard basis of $k[X]^*$ so that $f(x^i)=a_i$ and now $\phi(f)=w$.
It is more or less the same as the identification of the dual of eventually zero sequences with the space of all sequences.
EDIT: The space of derivations of $k[X]$ is isomorphic to $k[X]$ as a vector space. One possible isomorphism is given by $p\in k[X] \mapsto d_p :k[X]\to k[X]$ where $d_p$ is the unique derivation such that $d_p(X)=p$. As $k[X]$ is not isomorphic to $k[[X]]$ as vector spaces the dual cannot be identified with derivations.
Quimey
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