Dual of $\mathbb{Q}$[x] is not isomorphic to $\mathbb{Q}$[x] Denote by $\mathbb{Q}$ the set of the rational numbers. Denote by $\mathbb{Q}[x]$ the vector space over $\mathbb{Q}$ of the polynomials with rational coefficients.
Denote by $(\mathbb{Q}[x] )^{\star}$ the dual of $\mathbb{Q}$[x] . I am trying to show that  $(\mathbb{Q}[x] )^{\star}$ and $\mathbb{Q}[x]  $ are not isomorphic (that is: does not exist a linear transformation which is bijective). I really don't know how to start. Someone could help me ?
Thanks in advance
 A: The vector space $\Bbb Q[x]$ may be viewed as consisting of those sequences
$(a_i)_0^\infty, \tag 1$
where each $a_i \in \Bbb Q$ and $a_i = 0$ for all but a finite number of $i$; now consider an arbitrary sequence of the form
$(\lambda_i)_0^\infty \in \Bbb Q^\infty, \tag{2}$
where we allow $\lambda_i \ne 0$ for an infinite number of index values $i$.  Any such sequence $\lambda = (\lambda_i)_0^\infty$ determines a well-defined linear functional on $\Bbb Q[x]$ via the formula
$\lambda(p(x)) = \displaystyle \sum_0^\infty \lambda_i p_i, \tag 3$
where
$p(x) = \displaystyle \sum_0^{\deg p} p_i x^i \in \Bbb Q[x]. \tag 4$
Since only a finite number of the $p_i \ne 0$, the sum in (3) is well-defined and determines a unique element of $(\Bbb Q[x])^\ast$; linearity is easily verified.  
Now  the cardinality $\vert \Bbb Q[x] \vert$ of $\Bbb Q[x]$ is well known to be
$\vert \Bbb Q[x] \vert = \aleph_0, \tag 5$
that is, $\Bbb Q[x]$ is countable; but it is also reasonably well-known that the cardinality of the set of sequences of rationals is $\vert \Bbb R \vert$, the cardinality of $\Bbb R$:
$\vert \{ (\lambda_i)_0^\infty \} \vert = \vert \Bbb R \vert; \tag 6$
since
$\vert \Bbb Q[x] \vert = \aleph_0 \ne \vert \Bbb R \vert = \vert \{ (\lambda_i)_0^\infty \} \vert, \tag 7$
we see that
$\Bbb Q[x] \not \cong (\Bbb Q[x])^\ast. \tag 8$
