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In this post, Georges Elencwajg says that

Erdős-Kaplansky tell us that there does not exist a real vector space whose dual is isomorphic to R[X] !

I would like to know if there is a characterization of those fields $\mathbb{F}$ satifying the following condition: there exist a vector space $V$ such that $V^{\star}$ is isomorphic to $\mathbb{F}[X]$

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No field satisfies this condition. See… . – Qiaochu Yuan Feb 22 '12 at 18:46
up vote 5 down vote accepted

As is mentioned in the post you referenced, the dimension of $V^*$ is the same as $V$ when $V$ is finite dimensional and when $V$ is infinite dimensional $\mathrm{dim}(V^*)=|V^*|=|\mathbb{F}|^{\mathrm{dim}(V)}$. This holds for any field $\mathbb{F}$ (finite or otherwise). [To see a proof of this fact see N. Jacobson's Lectures in Abstract Algebra Volume II Theorem 2 in Chapter IX (page 247 in my copy).]

The dimension of $\mathbb{F}[x]$ is always $\aleph_0$ regardless of the field. [Why? $\{1,x,x^2,\dots\}$ is a basis.]

If we wish to have a dual space isomorphic to a space of polynomials, we would need $|\mathbb{F}|^{\mathrm{dim}(V)}=\aleph_0$. But this cannot happen (ever).

The smallest possible field is $\mathbb{F}_2=\mathbb{Z}_2$ and the smallest possible $\dim(V)$ (among infinite dimensional vector spaces) is $\aleph_0$. So the smallest you can get is $2^{\aleph_0}$ ($ > \aleph_0$).

In other words, no matter what field you're working over, the dimension of a dual space of an infinite dimensional vector space is at least the cardnality of the continuum.

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