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

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    $\begingroup$ Hint: compute the cardinality of the dual $\endgroup$ – Wojowu Jan 8 at 21:42
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    $\begingroup$ the dual will consist of the sequences $a_n = 1, n \in I$ where $I \subset \mathbb{N}$ is finite and zero otherwise? $\endgroup$ – math student Jan 8 at 21:50
  • $\begingroup$ Good guess, but not right. Consider the $\Bbb Q$-linear mapping from $\Bbb Q[x]$ that associates to a polynomial $f$ the sum of the coefficients of $f$. There are finitely many of these, but the above lin.tf. can not be represented by such a finitary construction as appears in your guess. $\endgroup$ – Lubin Jan 8 at 22:44

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

$\lambda(p(x)) = \displaystyle \sum_0^\infty \lambda_i p_i, \tag 3$


$p(x) = \displaystyle \sum_0^{\deg p} p_i x^i \in \Bbb Q[x]. \tag 4$

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 $\Bbb R$:

$\vert \{ (\lambda_i)_0^\infty \} \vert = \vert \Bbb R \vert; \tag 6$


$\vert \Bbb Q[x] \vert = \aleph_0 \ne \vert \Bbb R \vert = \vert \{ (\lambda_i)_0^\infty \} \vert, \tag 7$

we see that

$\Bbb Q[x] \not \cong (\Bbb Q[x])^\ast. \tag 8$


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