Mixing
Is it possible for $langle Bbb Q,<,+,cdot,0,1 rangle$ to be isomorphic to a proper subfield of itself?
$begingroup$
It is well-known that every ordered field contains a subfield that is isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$. Let $langle A,<,+,cdot,0',1' rangle$ be an ordered field. I would like to ask two questions.
Is it possible for $langle A,<,+,cdot,0',1' rangle$ to contain two different subfields that are both isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$?
Is it possible for $langle Bbb Q,<,+,cdot,0,1 rangle$ to be isomorphic to a proper subfield of itself? (proper means the underlying set of the subfield is a proper subset of $Bbb Q$)
Thank you for your help!
real-numbers ordered-fields
asked Jan 9 at 0:33
Le Anh DungLe Anh Dung
1,0721521
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add a comment |
$begingroup$
It is well-known that every ordered field contains a subfield that is isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$. Let $langle A,<,+,cdot,0',1' rangle$ be an ordered field. I would like to ask two questions.
Is it possible for $langle A,<,+,cdot,0',1' rangle$ to contain two different subfields that are both isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$?
Is it possible for $langle Bbb Q,<,+,cdot,0,1 rangle$ to be isomorphic to a proper subfield of itself? (proper means the underlying set of the subfield is a proper subset of $Bbb Q$)
Thank you for your help!
real-numbers ordered-fields
asked Jan 9 at 0:33
Le Anh DungLe Anh Dung
1,0721521
$endgroup$
add a comment |
$begingroup$
It is well-known that every ordered field contains a subfield that is isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$. Let $langle A,<,+,cdot,0',1' rangle$ be an ordered field. I would like to ask two questions.
Is it possible for $langle A,<,+,cdot,0',1' rangle$ to contain two different subfields that are both isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$?
Is it possible for $langle Bbb Q,<,+,cdot,0,1 rangle$ to be isomorphic to a proper subfield of itself? (proper means the underlying set of the subfield is a proper subset of $Bbb Q$)
Thank you for your help!
real-numbers ordered-fields
asked Jan 9 at 0:33
Le Anh DungLe Anh Dung
1,0721521
$endgroup$
It is well-known that every ordered field contains a subfield that is isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$. Let $langle A,<,+,cdot,0',1' rangle$ be an ordered field. I would like to ask two questions.
Is it possible for $langle A,<,+,cdot,0',1' rangle$ to contain two different subfields that are both isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$?
Is it possible for $langle Bbb Q,<,+,cdot,0,1 rangle$ to be isomorphic to a proper subfield of itself? (proper means the underlying set of the subfield is a proper subset of $Bbb Q$)
Thank you for your help!
real-numbers ordered-fields
real-numbers ordered-fields
asked Jan 9 at 0:33
Le Anh DungLe Anh Dung
1,0721521
asked Jan 9 at 0:33
Le Anh DungLe Anh Dung
1,0721521
edited Jan 9 at 0:40
Le Anh Dung
asked Jan 9 at 0:33
Le Anh DungLe Anh Dung
1,0721521
asked Jan 9 at 0:33
Le Anh DungLe Anh Dung
1,0721521
asked Jan 9 at 0:33
Le Anh DungLe Anh Dung
1,0721521
1,0721521
add a comment |
add a comment |
1 Answer
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$begingroup$
No. Each element of $mathbb Q$ satisfies an equation of the form
$$ x = frac{1+1+cdots+1+1}{1+1+cdots+1+1} qquadtext{or}qquad
x = -frac{1+1+cdots+1+1}{1+1+cdots+1+1} $$
Since the right-hand side is an equation in the language of field theory (which can be evaluated in every field of characteristic 0), for each rational number there is at most one element of every field that it can map to by isomorphism.
For your (1) this means that there is at most one isomorphism $mathbb Qto A$, so if such an isomorphism exists its range is unique.
The impossiblity of (2) follows from (1) by setting $A=mathbb Q$.
answered Jan 9 at 0:42
Henning MakholmHenning Makholm
239k17304541
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
No. Each element of $mathbb Q$ satisfies an equation of the form
$$ x = frac{1+1+cdots+1+1}{1+1+cdots+1+1} qquadtext{or}qquad
x = -frac{1+1+cdots+1+1}{1+1+cdots+1+1} $$
Since the right-hand side is an equation in the language of field theory (which can be evaluated in every field of characteristic 0), for each rational number there is at most one element of every field that it can map to by isomorphism.
For your (1) this means that there is at most one isomorphism $mathbb Qto A$, so if such an isomorphism exists its range is unique.
The impossiblity of (2) follows from (1) by setting $A=mathbb Q$.
answered Jan 9 at 0:42
Henning MakholmHenning Makholm
239k17304541
$endgroup$
add a comment |
$begingroup$
No. Each element of $mathbb Q$ satisfies an equation of the form
$$ x = frac{1+1+cdots+1+1}{1+1+cdots+1+1} qquadtext{or}qquad
x = -frac{1+1+cdots+1+1}{1+1+cdots+1+1} $$
Since the right-hand side is an equation in the language of field theory (which can be evaluated in every field of characteristic 0), for each rational number there is at most one element of every field that it can map to by isomorphism.
For your (1) this means that there is at most one isomorphism $mathbb Qto A$, so if such an isomorphism exists its range is unique.
The impossiblity of (2) follows from (1) by setting $A=mathbb Q$.
answered Jan 9 at 0:42
Henning MakholmHenning Makholm
239k17304541
$endgroup$
add a comment |
$begingroup$
No. Each element of $mathbb Q$ satisfies an equation of the form
$$ x = frac{1+1+cdots+1+1}{1+1+cdots+1+1} qquadtext{or}qquad
x = -frac{1+1+cdots+1+1}{1+1+cdots+1+1} $$
Since the right-hand side is an equation in the language of field theory (which can be evaluated in every field of characteristic 0), for each rational number there is at most one element of every field that it can map to by isomorphism.
For your (1) this means that there is at most one isomorphism $mathbb Qto A$, so if such an isomorphism exists its range is unique.
The impossiblity of (2) follows from (1) by setting $A=mathbb Q$.
answered Jan 9 at 0:42
Henning MakholmHenning Makholm
239k17304541
$endgroup$
No. Each element of $mathbb Q$ satisfies an equation of the form
$$ x = frac{1+1+cdots+1+1}{1+1+cdots+1+1} qquadtext{or}qquad
x = -frac{1+1+cdots+1+1}{1+1+cdots+1+1} $$
Since the right-hand side is an equation in the language of field theory (which can be evaluated in every field of characteristic 0), for each rational number there is at most one element of every field that it can map to by isomorphism.
For your (1) this means that there is at most one isomorphism $mathbb Qto A$, so if such an isomorphism exists its range is unique.
The impossiblity of (2) follows from (1) by setting $A=mathbb Q$.
answered Jan 9 at 0:42
Henning MakholmHenning Makholm
239k17304541
answered Jan 9 at 0:42
Henning MakholmHenning Makholm
239k17304541
answered Jan 9 at 0:42
Henning MakholmHenning Makholm
239k17304541
answered Jan 9 at 0:42
Henning MakholmHenning Makholm
239k17304541
239k17304541
add a comment |
add a comment |
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