Mixing
Dual of $mathbb{Q}$[x] is not isomorphic to $mathbb{Q}$[x]
$begingroup$
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
linear-algebra
asked Jan 8 at 21:40
math studentmath student
2,37411017
$endgroup$
add a comment |
$begingroup$
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
linear-algebra
asked Jan 8 at 21:40
math studentmath student
2,37411017
$endgroup$
3
$begingroup$
Hint: compute the cardinality of the dual
$endgroup$
– Wojowu
Jan 8 at 21:42
1
$begingroup$
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
add a comment |
$begingroup$
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
linear-algebra
asked Jan 8 at 21:40
math studentmath student
2,37411017
$endgroup$
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
linear-algebra
linear-algebra
asked Jan 8 at 21:40
math studentmath student
2,37411017
asked Jan 8 at 21:40
math studentmath student
2,37411017
asked Jan 8 at 21:40
math studentmath student
2,37411017
asked Jan 8 at 21:40
math studentmath student
2,37411017
asked Jan 8 at 21:40
math studentmath student
2,37411017
2,37411017
3
$begingroup$
Hint: compute the cardinality of the dual
$endgroup$
– Wojowu
Jan 8 at 21:42
1
$begingroup$
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
add a comment |
3
$begingroup$
Hint: compute the cardinality of the dual
$endgroup$
– Wojowu
Jan 8 at 21:42
1
$begingroup$
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
3
3
$begingroup$
Hint: compute the cardinality of the dual
$endgroup$
– Wojowu
Jan 8 at 21:42
$begingroup$
Hint: compute the cardinality of the dual
$endgroup$
– Wojowu
Jan 8 at 21:42
1
1
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
$Bbb Q[x] not cong (Bbb Q[x])^ast. tag 8$
answered Jan 9 at 2:30
Robert LewisRobert Lewis
45.6k23065
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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
$Bbb Q[x] not cong (Bbb Q[x])^ast. tag 8$
answered Jan 9 at 2:30
Robert LewisRobert Lewis
45.6k23065
$endgroup$
add a comment |
$begingroup$
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
$Bbb Q[x] not cong (Bbb Q[x])^ast. tag 8$
answered Jan 9 at 2:30
Robert LewisRobert Lewis
45.6k23065
$endgroup$
add a comment |
$begingroup$
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
$Bbb Q[x] not cong (Bbb Q[x])^ast. tag 8$
answered Jan 9 at 2:30
Robert LewisRobert Lewis
45.6k23065
$endgroup$
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
$Bbb Q[x] not cong (Bbb Q[x])^ast. tag 8$
answered Jan 9 at 2:30
Robert LewisRobert Lewis
45.6k23065
answered Jan 9 at 2:30
Robert LewisRobert Lewis
45.6k23065
answered Jan 9 at 2:30
Robert LewisRobert Lewis
45.6k23065
answered Jan 9 at 2:30
Robert LewisRobert Lewis
45.6k23065
45.6k23065
add a comment |
add a comment |
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3
$begingroup$
Hint: compute the cardinality of the dual
$endgroup$
– Wojowu
Jan 8 at 21:42
1
$begingroup$
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