$\mathbb{Q}[x]$ is not isomorphic to $(\mathbb{Q}[x])^{*}$

Question

I've proved that $\mathbb{R}[x]$ is not isomorphic to $(\mathbb{R}[x])^{*}$, where $(\mathbb{R}[x])^{*}$ is the dual space of $\mathbb{R}[x]$.

If you define for each $\alpha \in \mathbb{R}$, $f_\alpha \in (\mathbb{R}[x])^{*}$, $f_{\alpha}(p(x)) = p(\alpha)$, it is easy to prove that $\{f_{\alpha}\}_{\alpha \in \mathbb{R}}$ is linearment independent.

Since $\{f_{\alpha}\}_{\alpha \in \mathbb{R}}$ is not a countable set, there is a basis $B$ of $(\mathbb{R}[x])^{*}$ s.t $\{f_{\alpha}\}_{\alpha \in \mathbb{R}} \subset B$, so $B$ is not countable. But $B' = \{1,x,x^2,...\}$ is a countable basis of $\mathbb{R}[x]$.

Therefore $\mathbb{R}[x]$ could not be isomorphic to $(\mathbb{R}[x])^{*}$, because it bases has not the same cardinality.

But the same argument doesn't work from $\mathbb{Q}[x]$ and $(\mathbb{Q}[x])^{*}$, because $\{f_\alpha\}_{\alpha \in \mathbb{Q}}$ is countable.

May you give me a tip?

Answer 1

Consider $a=(a_n\,:\,n\in\Bbb N)$ a sequence of rational numbers and define $$\phi_a(p)=\sum_{j=0}^{\deg p} p_ja_j$$

This is an element of $(\Bbb Q[x])^*$, and the map $a\mapsto \phi_a$ defines an injection (actually, a bijection) $\Bbb Q^{\Bbb N}\to (\Bbb Q[x])^*$.