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 ?

• Hint: compute the cardinality of the dual – Wojowu Jan 8 at 21:42
• the dual will consist of the sequences $a_n = 1, n \in I$ where $I \subset \mathbb{N}$ is finite and zero otherwise? – math student Jan 8 at 21:50
• 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. – 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

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

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

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