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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

math student
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    Hint: compute the cardinality of the dual – Wojowu Jan 08 '19 at 21:42
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    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 08 '19 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 08 '19 at 22:44

1 Answers1

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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

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

Robert Lewis
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    Please, can you clarify this points? You are determining the linear functional as a sum of the coefficients of the polynomial $p(x) \in Q[x]$. Am I right? How do you know the cardinality of $Q[x]$? Shouldn't the cardinality of $Q[x]$ be infinity as there is no limiting degree for the polynomials from $Q[x]$? Why is the cardinality of the set of sequences of rationals is $|R|$? – user13 Jun 16 '20 at 13:17
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    @user13: In general, a functional $\lambda \in (\Bbb Q[x])^\ast$ is not given by the simple sum of the coefficients of a $p(x) \in \Bbb Q[x]$ to which it is applied, but rather by an expression of the form (3); if $\lambda = (1)_0^\infty$ then $\lambda(p(x)) = \sum_0^{\deg p} p_i$ for $p(x) = \sum_0^n p_ix^i \in \Bbb Q[x]$, $\lambda$ is but one example of such a functional; there are many others. – Robert Lewis Jun 17 '20 at 23:19
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    @user13: Yes, the cardinality of $\Bbb Q[x]$ is infinite, indeed it is $\aleph_0$; $\Bbb Q[x]$ is infinite since $x^k \in \Bbb Q[x]$ for $k \in \Bbb N$ and the $x^k$ are distinct. – Robert Lewis Jun 18 '20 at 00:59
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    @user13: We further have that $\vert \Bbb Q[x] \vert = \aleph_0$, that is, that the set $\Bbb Q[x]$ of polynomials with rational coefficients is countable. We argue as follows: the set of monomials $\Bbb Q x^k = {rx^k: r \in \Bbb Q }$ is manifestly countable, since it may easily be put into one-to-one correspondence with $\Bbb Q$ itself; then the Cartesian product of a finite collection of the $\Bbb Q x^k$ is countable, being the Cartesian product of a finite number of countable sets; *to be continued . . .* – Robert Lewis Jun 18 '20 at 09:01
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    @user13: *continuation of previous comment:* We now map the polynomial $\sum_0^n p_j x^j \in \Bbb Q[x]$ to the ordered $n + 1$-tuple $(p_0, p_1 x, p_2 x^2, \ldots, p_n x^n) \in \Bbb Q \times \Bbb Q x \times \Bbb Qx^2 \times \ldots \times \Bbb Q x^n$; that this mapping is injective is easily seen; and, also by what we have seen, this direct product of $\Bbb Q x^k$ is countable hence the set of polynomials in $\Bbb Q[x]$ with degree $n$ or less is also of cardinality $\aleph_0$. *to be continued . . .* – Robert Lewis Jun 18 '20 at 09:19
  • @user13: *continuation of previous comment:* Setting $M_n = \Bbb Q \times \Bbb Q x \times \Bbb Q x^2 \times \ldots \times \Bbb Q x^n$, the previous result translates $(p_0, p_1 x, p_2 x^2, \ldots, p_n x^n) \in M_n$ and thus every polynomial in $\Bbb Q[x]$ is mapped into $\cup_{k \ge 0} M_k$; but this is is a countable union of countable sets and hence is itself countable, that is $\vert \cup_{k \ge 0} M_k \vert = \aleph_0$; it is now seen that every polynomial in $\Bbb Q[x]$ is mapped injectively into the countable set $\cup_{k \ge 0} M_k$; thus $\vert \Bbb Q[x] \vert = \aleph_0$. – Robert Lewis Jun 18 '20 at 20:19
  • @user13: *continuation of previous comment:* for more on countable sets, see https://en.wikipedia.org/wiki/Countable_set. As for $\vert {(\lambda_i)_0^\infty} \vert = \vert R \vert$, (2) implies ${\lambda_i)_0^\infty } = \Bbb Q^\infty$, the set of all rational sequences, the cardinality of whiich is $\vert \Bbb R \Vert$; see https://math.stackexchange.com/questions/1645617/how-many-sequences-of-rational-numbers-in-0-1-exist. – Robert Lewis Jun 18 '20 at 22:02