The isomorphism between two complete ordered fields is unique
$begingroup$
The isomorphism between two complete ordered fields is unique.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle,mathfrak{A}=langle A,prec,oplus,odot,0',1' rangle$ be complete ordered fields where $Bbb R$ is the set of real numbers. Let $mathfrak{B}=langle B,prec,oplus,odot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$ and $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle$.
Lemma 1: $mathfrak{R}$ is isomorphic to $mathfrak{A}$.
Lemma 2: $mathfrak{Q}$ is uniquely isomorphic to $mathfrak{B}$.
By Lemma 1, let $Phi:Bbb R to A,Psi:Bbb R to A$ be isomorphisms between $mathfrak{R}$ and $mathfrak{A}$. By Lemma 2, let $f:Bbb Q to B$ be the unique isomorphism between $mathfrak{Q}$ and $mathfrak{B}$.
Let $X subseteq Bbb Q$ be bounded from above and $sup,sup'$ supremums w.r.t $<,prec$ respectively. We next prove that $Phi(sup X) = sup' f[X]$.
$forall xin X: x le sup X implies forall xin X: Phi(x) preccurlyeq Phi(sup X) implies forall xin X: f(x) preccurlyeq Phi(sup X) implies sup' f[X] preccurlyeq Phi(sup X).$
Assume the contrary that $sup' f[X] prec Phi(sup X)$. Since $B$ is dense in $A$, there exists $bin B$ such that $sup' f[X] prec b prec Phi(sup X)$. Then there exists $pin Bbb Q$ such that $f(p)=b$. Thus $sup' f[X] prec f(p)=Phi(p) prec Phi(sup X).$
We have $Phi(p) prec Phi(sup X) implies p<sup X implies p<p'$ for some $p'in X implies$ $f(p) prec f(p')$ for some $p'in X$ $implies f(p) prec sup' f[X]$. This is a contradiction.
Hence $Phi(sup X)=sup' f[X]$. Similarly, $Psi(sup X)=sup' f[X]$.
Let $X_x={pinBbb Q mid p<x} subseteq Bbb Q$. Since $Bbb Q$ is dense in $Bbb R$, $x=sup X_x$ for all $xinBbb R$. Then $Phi(x)=Phi(sup X_x)=sup' f[X_x]=Psi(sup X_x)=Psi(x)$ for all $xinBbb R$. It follows that $Phi=Psi$.
real-analysis proof-verification real-numbers ordered-fields
asked Jan 16 at 0:42
Le Anh DungLe Anh Dung
1,4461621
$endgroup$
add a comment |
$begingroup$
The isomorphism between two complete ordered fields is unique.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle,mathfrak{A}=langle A,prec,oplus,odot,0',1' rangle$ be complete ordered fields where $Bbb R$ is the set of real numbers. Let $mathfrak{B}=langle B,prec,oplus,odot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$ and $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle$.
Lemma 1: $mathfrak{R}$ is isomorphic to $mathfrak{A}$.
Lemma 2: $mathfrak{Q}$ is uniquely isomorphic to $mathfrak{B}$.
By Lemma 1, let $Phi:Bbb R to A,Psi:Bbb R to A$ be isomorphisms between $mathfrak{R}$ and $mathfrak{A}$. By Lemma 2, let $f:Bbb Q to B$ be the unique isomorphism between $mathfrak{Q}$ and $mathfrak{B}$.
Let $X subseteq Bbb Q$ be bounded from above and $sup,sup'$ supremums w.r.t $<,prec$ respectively. We next prove that $Phi(sup X) = sup' f[X]$.
$forall xin X: x le sup X implies forall xin X: Phi(x) preccurlyeq Phi(sup X) implies forall xin X: f(x) preccurlyeq Phi(sup X) implies sup' f[X] preccurlyeq Phi(sup X).$
Assume the contrary that $sup' f[X] prec Phi(sup X)$. Since $B$ is dense in $A$, there exists $bin B$ such that $sup' f[X] prec b prec Phi(sup X)$. Then there exists $pin Bbb Q$ such that $f(p)=b$. Thus $sup' f[X] prec f(p)=Phi(p) prec Phi(sup X).$
We have $Phi(p) prec Phi(sup X) implies p<sup X implies p<p'$ for some $p'in X implies$ $f(p) prec f(p')$ for some $p'in X$ $implies f(p) prec sup' f[X]$. This is a contradiction.
Hence $Phi(sup X)=sup' f[X]$. Similarly, $Psi(sup X)=sup' f[X]$.
Let $X_x={pinBbb Q mid p<x} subseteq Bbb Q$. Since $Bbb Q$ is dense in $Bbb R$, $x=sup X_x$ for all $xinBbb R$. Then $Phi(x)=Phi(sup X_x)=sup' f[X_x]=Psi(sup X_x)=Psi(x)$ for all $xinBbb R$. It follows that $Phi=Psi$.
real-analysis proof-verification real-numbers ordered-fields
asked Jan 16 at 0:42
Le Anh DungLe Anh Dung
1,4461621
$endgroup$
$begingroup$
You should use that a complete ordered field has a unique ordering, because an element is nonnegative if and only if it has a square root.
$endgroup$
– egreg
Jan 17 at 0:05
$begingroup$
Hi @egreg, DanielWainfleet has utilized your idea and posted it as an answer below. I am reading his answer. Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 0:10
add a comment |
$begingroup$
The isomorphism between two complete ordered fields is unique.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle,mathfrak{A}=langle A,prec,oplus,odot,0',1' rangle$ be complete ordered fields where $Bbb R$ is the set of real numbers. Let $mathfrak{B}=langle B,prec,oplus,odot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$ and $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle$.
Lemma 1: $mathfrak{R}$ is isomorphic to $mathfrak{A}$.
Lemma 2: $mathfrak{Q}$ is uniquely isomorphic to $mathfrak{B}$.
By Lemma 1, let $Phi:Bbb R to A,Psi:Bbb R to A$ be isomorphisms between $mathfrak{R}$ and $mathfrak{A}$. By Lemma 2, let $f:Bbb Q to B$ be the unique isomorphism between $mathfrak{Q}$ and $mathfrak{B}$.
Let $X subseteq Bbb Q$ be bounded from above and $sup,sup'$ supremums w.r.t $<,prec$ respectively. We next prove that $Phi(sup X) = sup' f[X]$.
$forall xin X: x le sup X implies forall xin X: Phi(x) preccurlyeq Phi(sup X) implies forall xin X: f(x) preccurlyeq Phi(sup X) implies sup' f[X] preccurlyeq Phi(sup X).$
Assume the contrary that $sup' f[X] prec Phi(sup X)$. Since $B$ is dense in $A$, there exists $bin B$ such that $sup' f[X] prec b prec Phi(sup X)$. Then there exists $pin Bbb Q$ such that $f(p)=b$. Thus $sup' f[X] prec f(p)=Phi(p) prec Phi(sup X).$
We have $Phi(p) prec Phi(sup X) implies p<sup X implies p<p'$ for some $p'in X implies$ $f(p) prec f(p')$ for some $p'in X$ $implies f(p) prec sup' f[X]$. This is a contradiction.
Hence $Phi(sup X)=sup' f[X]$. Similarly, $Psi(sup X)=sup' f[X]$.
Let $X_x={pinBbb Q mid p<x} subseteq Bbb Q$. Since $Bbb Q$ is dense in $Bbb R$, $x=sup X_x$ for all $xinBbb R$. Then $Phi(x)=Phi(sup X_x)=sup' f[X_x]=Psi(sup X_x)=Psi(x)$ for all $xinBbb R$. It follows that $Phi=Psi$.
real-analysis proof-verification real-numbers ordered-fields
asked Jan 16 at 0:42
Le Anh DungLe Anh Dung
1,4461621
$endgroup$
The isomorphism between two complete ordered fields is unique.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle,mathfrak{A}=langle A,prec,oplus,odot,0',1' rangle$ be complete ordered fields where $Bbb R$ is the set of real numbers. Let $mathfrak{B}=langle B,prec,oplus,odot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$ and $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle$.
Lemma 1: $mathfrak{R}$ is isomorphic to $mathfrak{A}$.
Lemma 2: $mathfrak{Q}$ is uniquely isomorphic to $mathfrak{B}$.
By Lemma 1, let $Phi:Bbb R to A,Psi:Bbb R to A$ be isomorphisms between $mathfrak{R}$ and $mathfrak{A}$. By Lemma 2, let $f:Bbb Q to B$ be the unique isomorphism between $mathfrak{Q}$ and $mathfrak{B}$.
Let $X subseteq Bbb Q$ be bounded from above and $sup,sup'$ supremums w.r.t $<,prec$ respectively. We next prove that $Phi(sup X) = sup' f[X]$.
$forall xin X: x le sup X implies forall xin X: Phi(x) preccurlyeq Phi(sup X) implies forall xin X: f(x) preccurlyeq Phi(sup X) implies sup' f[X] preccurlyeq Phi(sup X).$
Assume the contrary that $sup' f[X] prec Phi(sup X)$. Since $B$ is dense in $A$, there exists $bin B$ such that $sup' f[X] prec b prec Phi(sup X)$. Then there exists $pin Bbb Q$ such that $f(p)=b$. Thus $sup' f[X] prec f(p)=Phi(p) prec Phi(sup X).$
We have $Phi(p) prec Phi(sup X) implies p<sup X implies p<p'$ for some $p'in X implies$ $f(p) prec f(p')$ for some $p'in X$ $implies f(p) prec sup' f[X]$. This is a contradiction.
Hence $Phi(sup X)=sup' f[X]$. Similarly, $Psi(sup X)=sup' f[X]$.
Let $X_x={pinBbb Q mid p<x} subseteq Bbb Q$. Since $Bbb Q$ is dense in $Bbb R$, $x=sup X_x$ for all $xinBbb R$. Then $Phi(x)=Phi(sup X_x)=sup' f[X_x]=Psi(sup X_x)=Psi(x)$ for all $xinBbb R$. It follows that $Phi=Psi$.
real-analysis proof-verification real-numbers ordered-fields
real-analysis proof-verification real-numbers ordered-fields
asked Jan 16 at 0:42
Le Anh DungLe Anh Dung
1,4461621
asked Jan 16 at 0:42
Le Anh DungLe Anh Dung
1,4461621
edited Jan 16 at 9:27
Le Anh Dung
asked Jan 16 at 0:42
Le Anh DungLe Anh Dung
1,4461621
asked Jan 16 at 0:42
Le Anh DungLe Anh Dung
1,4461621
asked Jan 16 at 0:42
Le Anh DungLe Anh Dung
1,4461621
1,4461621
$begingroup$
You should use that a complete ordered field has a unique ordering, because an element is nonnegative if and only if it has a square root.
$endgroup$
– egreg
Jan 17 at 0:05
$begingroup$
Hi @egreg, DanielWainfleet has utilized your idea and posted it as an answer below. I am reading his answer. Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 0:10
add a comment |
$begingroup$
You should use that a complete ordered field has a unique ordering, because an element is nonnegative if and only if it has a square root.
$endgroup$
– egreg
Jan 17 at 0:05
$begingroup$
Hi @egreg, DanielWainfleet has utilized your idea and posted it as an answer below. I am reading his answer. Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 0:10
$begingroup$
You should use that a complete ordered field has a unique ordering, because an element is nonnegative if and only if it has a square root.
$endgroup$
– egreg
Jan 17 at 0:05
$begingroup$
You should use that a complete ordered field has a unique ordering, because an element is nonnegative if and only if it has a square root.
$endgroup$
– egreg
Jan 17 at 0:05
$begingroup$
Hi @egreg, DanielWainfleet has utilized your idea and posted it as an answer below. I am reading his answer. Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 0:10
$begingroup$
Hi @egreg, DanielWainfleet has utilized your idea and posted it as an answer below. I am reading his answer. Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 0:10
add a comment |
1 Answer
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oldest
votes
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This is an extended comment on the uniqueness of the isomorphism.
(1a). Let $F, G$ be sub-fields of $Bbb R$ such that $psi:Fto G$ is a field-isomorphism. ($psi$ is not assumed to preserve order.) If $forall xin F,(0le ximplies sqrt xin F) $ then $psi=id_F$ and $G=F.$
Proof: $Bbb Qsubset F$ and $psi|_{Bbb Q}=id_{Bbb Q}$ so for any $xin F$ and any $qin Bbb Q$ we have $$xge qiff psi(x)-q=psi(x)-psi(q)=(psi(sqrt {x-q}))^2ge 0iff$$ $$iff psi(x)ge q$$ so $Bbb Q cap (-infty,x]=Bbb Qcap (-infty,psi(x)],$ so $x=psi(x).$
(1b).In particular, letting $F=Bbb R$ in (1a), the only sub-field of $Bbb R$ that is field-isomorphic to $Bbb R$ is $Bbb R$ itself, and the only field-isomorphism of $Bbb R$ to $Bbb R$ is $id_{Bbb R}.$
(2). If $ B $ is a field and $psi_1, psi_2$ are field-isomorphisms from $Bbb R$ to $B$ then by (1b), $psi_2^{-1}psi_1=id_{Bbb R},$ so $psi_1=psi_2.$
answered Jan 16 at 22:51
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
$begingroup$
A proper sub-field of $Bbb R$ can be an ordered field according to an order that is not the usual order $<$ of $Bbb R$. For example ${a+bsqrt 2,:a,bin Bbb Q}.$ For $a,bin Bbb Q$ let $a+bsqrt 2,>^*0iff a-bsqrt 2<0.$
$endgroup$
– DanielWainfleet
Jan 16 at 23:06
$begingroup$
I think you meant $psi$ rather than $f$. Please check if my reasoning is correct: For all $xin F$ and $qinBbb Q$: $psi(x)-q=psi(x)-psi(q)$ [since $psi|_{Bbb Q}=text{id}_{Bbb Q}$] $=psi(x-q)$. Then $x ge q iff x-q ge 0 iff x-q$ $=sqrt {x-q} cdot sqrt {x-q} iff psi(x-q)=psi(sqrt {x-q} cdot sqrt {x-q})=psi(sqrt {x-q})cdot psi(sqrt {x-q})=$ $(psi(sqrt {x-q}))^2 ge 0 iff psi(x)-q=psi(x-q) ge 0$. To sum up, $xge q iff psi(x)-q ge 0$ $iff psi(x)ge q$.[...]
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
$begingroup$
[...] It follows that ${qinBbb Q mid qle x}={qinBbb Q mid qle psi(x)}$. Hence $sup {qinBbb Q mid qle x}=$ $sup {qinBbb Q mid qle psi(x)}$ and thus $x=psi(x)$. As a result, the isomorphism between two fields is unique.
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
$begingroup$
Perfectly correct. And $f$ was a typo for $psi$.
$endgroup$
– DanielWainfleet
Jan 17 at 13:58
$begingroup$
Thank you so much for your verification! Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 14:04
add a comment |
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1 Answer
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$begingroup$
This is an extended comment on the uniqueness of the isomorphism.
(1a). Let $F, G$ be sub-fields of $Bbb R$ such that $psi:Fto G$ is a field-isomorphism. ($psi$ is not assumed to preserve order.) If $forall xin F,(0le ximplies sqrt xin F) $ then $psi=id_F$ and $G=F.$
Proof: $Bbb Qsubset F$ and $psi|_{Bbb Q}=id_{Bbb Q}$ so for any $xin F$ and any $qin Bbb Q$ we have $$xge qiff psi(x)-q=psi(x)-psi(q)=(psi(sqrt {x-q}))^2ge 0iff$$ $$iff psi(x)ge q$$ so $Bbb Q cap (-infty,x]=Bbb Qcap (-infty,psi(x)],$ so $x=psi(x).$
(1b).In particular, letting $F=Bbb R$ in (1a), the only sub-field of $Bbb R$ that is field-isomorphic to $Bbb R$ is $Bbb R$ itself, and the only field-isomorphism of $Bbb R$ to $Bbb R$ is $id_{Bbb R}.$
(2). If $ B $ is a field and $psi_1, psi_2$ are field-isomorphisms from $Bbb R$ to $B$ then by (1b), $psi_2^{-1}psi_1=id_{Bbb R},$ so $psi_1=psi_2.$
answered Jan 16 at 22:51
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
$begingroup$
A proper sub-field of $Bbb R$ can be an ordered field according to an order that is not the usual order $<$ of $Bbb R$. For example ${a+bsqrt 2,:a,bin Bbb Q}.$ For $a,bin Bbb Q$ let $a+bsqrt 2,>^*0iff a-bsqrt 2<0.$
$endgroup$
– DanielWainfleet
Jan 16 at 23:06
$begingroup$
I think you meant $psi$ rather than $f$. Please check if my reasoning is correct: For all $xin F$ and $qinBbb Q$: $psi(x)-q=psi(x)-psi(q)$ [since $psi|_{Bbb Q}=text{id}_{Bbb Q}$] $=psi(x-q)$. Then $x ge q iff x-q ge 0 iff x-q$ $=sqrt {x-q} cdot sqrt {x-q} iff psi(x-q)=psi(sqrt {x-q} cdot sqrt {x-q})=psi(sqrt {x-q})cdot psi(sqrt {x-q})=$ $(psi(sqrt {x-q}))^2 ge 0 iff psi(x)-q=psi(x-q) ge 0$. To sum up, $xge q iff psi(x)-q ge 0$ $iff psi(x)ge q$.[...]
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
$begingroup$
[...] It follows that ${qinBbb Q mid qle x}={qinBbb Q mid qle psi(x)}$. Hence $sup {qinBbb Q mid qle x}=$ $sup {qinBbb Q mid qle psi(x)}$ and thus $x=psi(x)$. As a result, the isomorphism between two fields is unique.
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
$begingroup$
Perfectly correct. And $f$ was a typo for $psi$.
$endgroup$
– DanielWainfleet
Jan 17 at 13:58
$begingroup$
Thank you so much for your verification! Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 14:04
add a comment |
$begingroup$
This is an extended comment on the uniqueness of the isomorphism.
(1a). Let $F, G$ be sub-fields of $Bbb R$ such that $psi:Fto G$ is a field-isomorphism. ($psi$ is not assumed to preserve order.) If $forall xin F,(0le ximplies sqrt xin F) $ then $psi=id_F$ and $G=F.$
Proof: $Bbb Qsubset F$ and $psi|_{Bbb Q}=id_{Bbb Q}$ so for any $xin F$ and any $qin Bbb Q$ we have $$xge qiff psi(x)-q=psi(x)-psi(q)=(psi(sqrt {x-q}))^2ge 0iff$$ $$iff psi(x)ge q$$ so $Bbb Q cap (-infty,x]=Bbb Qcap (-infty,psi(x)],$ so $x=psi(x).$
(1b).In particular, letting $F=Bbb R$ in (1a), the only sub-field of $Bbb R$ that is field-isomorphic to $Bbb R$ is $Bbb R$ itself, and the only field-isomorphism of $Bbb R$ to $Bbb R$ is $id_{Bbb R}.$
(2). If $ B $ is a field and $psi_1, psi_2$ are field-isomorphisms from $Bbb R$ to $B$ then by (1b), $psi_2^{-1}psi_1=id_{Bbb R},$ so $psi_1=psi_2.$
answered Jan 16 at 22:51
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
$begingroup$
A proper sub-field of $Bbb R$ can be an ordered field according to an order that is not the usual order $<$ of $Bbb R$. For example ${a+bsqrt 2,:a,bin Bbb Q}.$ For $a,bin Bbb Q$ let $a+bsqrt 2,>^*0iff a-bsqrt 2<0.$
$endgroup$
– DanielWainfleet
Jan 16 at 23:06
$begingroup$
I think you meant $psi$ rather than $f$. Please check if my reasoning is correct: For all $xin F$ and $qinBbb Q$: $psi(x)-q=psi(x)-psi(q)$ [since $psi|_{Bbb Q}=text{id}_{Bbb Q}$] $=psi(x-q)$. Then $x ge q iff x-q ge 0 iff x-q$ $=sqrt {x-q} cdot sqrt {x-q} iff psi(x-q)=psi(sqrt {x-q} cdot sqrt {x-q})=psi(sqrt {x-q})cdot psi(sqrt {x-q})=$ $(psi(sqrt {x-q}))^2 ge 0 iff psi(x)-q=psi(x-q) ge 0$. To sum up, $xge q iff psi(x)-q ge 0$ $iff psi(x)ge q$.[...]
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
$begingroup$
[...] It follows that ${qinBbb Q mid qle x}={qinBbb Q mid qle psi(x)}$. Hence $sup {qinBbb Q mid qle x}=$ $sup {qinBbb Q mid qle psi(x)}$ and thus $x=psi(x)$. As a result, the isomorphism between two fields is unique.
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
$begingroup$
Perfectly correct. And $f$ was a typo for $psi$.
$endgroup$
– DanielWainfleet
Jan 17 at 13:58
$begingroup$
Thank you so much for your verification! Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 14:04
add a comment |
$begingroup$
This is an extended comment on the uniqueness of the isomorphism.
(1a). Let $F, G$ be sub-fields of $Bbb R$ such that $psi:Fto G$ is a field-isomorphism. ($psi$ is not assumed to preserve order.) If $forall xin F,(0le ximplies sqrt xin F) $ then $psi=id_F$ and $G=F.$
Proof: $Bbb Qsubset F$ and $psi|_{Bbb Q}=id_{Bbb Q}$ so for any $xin F$ and any $qin Bbb Q$ we have $$xge qiff psi(x)-q=psi(x)-psi(q)=(psi(sqrt {x-q}))^2ge 0iff$$ $$iff psi(x)ge q$$ so $Bbb Q cap (-infty,x]=Bbb Qcap (-infty,psi(x)],$ so $x=psi(x).$
(1b).In particular, letting $F=Bbb R$ in (1a), the only sub-field of $Bbb R$ that is field-isomorphic to $Bbb R$ is $Bbb R$ itself, and the only field-isomorphism of $Bbb R$ to $Bbb R$ is $id_{Bbb R}.$
(2). If $ B $ is a field and $psi_1, psi_2$ are field-isomorphisms from $Bbb R$ to $B$ then by (1b), $psi_2^{-1}psi_1=id_{Bbb R},$ so $psi_1=psi_2.$
answered Jan 16 at 22:51
DanielWainfleetDanielWainfleet
35.7k31648
$endgroup$
This is an extended comment on the uniqueness of the isomorphism.
(1a). Let $F, G$ be sub-fields of $Bbb R$ such that $psi:Fto G$ is a field-isomorphism. ($psi$ is not assumed to preserve order.) If $forall xin F,(0le ximplies sqrt xin F) $ then $psi=id_F$ and $G=F.$
Proof: $Bbb Qsubset F$ and $psi|_{Bbb Q}=id_{Bbb Q}$ so for any $xin F$ and any $qin Bbb Q$ we have $$xge qiff psi(x)-q=psi(x)-psi(q)=(psi(sqrt {x-q}))^2ge 0iff$$ $$iff psi(x)ge q$$ so $Bbb Q cap (-infty,x]=Bbb Qcap (-infty,psi(x)],$ so $x=psi(x).$
(1b).In particular, letting $F=Bbb R$ in (1a), the only sub-field of $Bbb R$ that is field-isomorphic to $Bbb R$ is $Bbb R$ itself, and the only field-isomorphism of $Bbb R$ to $Bbb R$ is $id_{Bbb R}.$
(2). If $ B $ is a field and $psi_1, psi_2$ are field-isomorphisms from $Bbb R$ to $B$ then by (1b), $psi_2^{-1}psi_1=id_{Bbb R},$ so $psi_1=psi_2.$
answered Jan 16 at 22:51
DanielWainfleetDanielWainfleet
35.7k31648
edited Jan 17 at 14:00
answered Jan 16 at 22:51
DanielWainfleetDanielWainfleet
35.7k31648
answered Jan 16 at 22:51
DanielWainfleetDanielWainfleet
35.7k31648
answered Jan 16 at 22:51
DanielWainfleetDanielWainfleet
35.7k31648
35.7k31648
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A proper sub-field of $Bbb R$ can be an ordered field according to an order that is not the usual order $<$ of $Bbb R$. For example ${a+bsqrt 2,:a,bin Bbb Q}.$ For $a,bin Bbb Q$ let $a+bsqrt 2,>^*0iff a-bsqrt 2<0.$
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– DanielWainfleet
Jan 16 at 23:06
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I think you meant $psi$ rather than $f$. Please check if my reasoning is correct: For all $xin F$ and $qinBbb Q$: $psi(x)-q=psi(x)-psi(q)$ [since $psi|_{Bbb Q}=text{id}_{Bbb Q}$] $=psi(x-q)$. Then $x ge q iff x-q ge 0 iff x-q$ $=sqrt {x-q} cdot sqrt {x-q} iff psi(x-q)=psi(sqrt {x-q} cdot sqrt {x-q})=psi(sqrt {x-q})cdot psi(sqrt {x-q})=$ $(psi(sqrt {x-q}))^2 ge 0 iff psi(x)-q=psi(x-q) ge 0$. To sum up, $xge q iff psi(x)-q ge 0$ $iff psi(x)ge q$.[...]
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– Le Anh Dung
Jan 17 at 1:54
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[...] It follows that ${qinBbb Q mid qle x}={qinBbb Q mid qle psi(x)}$. Hence $sup {qinBbb Q mid qle x}=$ $sup {qinBbb Q mid qle psi(x)}$ and thus $x=psi(x)$. As a result, the isomorphism between two fields is unique.
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– Le Anh Dung
Jan 17 at 1:54
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Perfectly correct. And $f$ was a typo for $psi$.
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– DanielWainfleet
Jan 17 at 13:58
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Thank you so much for your verification! Have you seen any error in my proof?
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– Le Anh Dung
Jan 17 at 14:04
add a comment |
$begingroup$
A proper sub-field of $Bbb R$ can be an ordered field according to an order that is not the usual order $<$ of $Bbb R$. For example ${a+bsqrt 2,:a,bin Bbb Q}.$ For $a,bin Bbb Q$ let $a+bsqrt 2,>^*0iff a-bsqrt 2<0.$
$endgroup$
– DanielWainfleet
Jan 16 at 23:06
$begingroup$
I think you meant $psi$ rather than $f$. Please check if my reasoning is correct: For all $xin F$ and $qinBbb Q$: $psi(x)-q=psi(x)-psi(q)$ [since $psi|_{Bbb Q}=text{id}_{Bbb Q}$] $=psi(x-q)$. Then $x ge q iff x-q ge 0 iff x-q$ $=sqrt {x-q} cdot sqrt {x-q} iff psi(x-q)=psi(sqrt {x-q} cdot sqrt {x-q})=psi(sqrt {x-q})cdot psi(sqrt {x-q})=$ $(psi(sqrt {x-q}))^2 ge 0 iff psi(x)-q=psi(x-q) ge 0$. To sum up, $xge q iff psi(x)-q ge 0$ $iff psi(x)ge q$.[...]
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
$begingroup$
[...] It follows that ${qinBbb Q mid qle x}={qinBbb Q mid qle psi(x)}$. Hence $sup {qinBbb Q mid qle x}=$ $sup {qinBbb Q mid qle psi(x)}$ and thus $x=psi(x)$. As a result, the isomorphism between two fields is unique.
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
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Perfectly correct. And $f$ was a typo for $psi$.
$endgroup$
– DanielWainfleet
Jan 17 at 13:58
$begingroup$
Thank you so much for your verification! Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 14:04
$begingroup$
A proper sub-field of $Bbb R$ can be an ordered field according to an order that is not the usual order $<$ of $Bbb R$. For example ${a+bsqrt 2,:a,bin Bbb Q}.$ For $a,bin Bbb Q$ let $a+bsqrt 2,>^*0iff a-bsqrt 2<0.$
$endgroup$
– DanielWainfleet
Jan 16 at 23:06
$begingroup$
A proper sub-field of $Bbb R$ can be an ordered field according to an order that is not the usual order $<$ of $Bbb R$. For example ${a+bsqrt 2,:a,bin Bbb Q}.$ For $a,bin Bbb Q$ let $a+bsqrt 2,>^*0iff a-bsqrt 2<0.$
$endgroup$
– DanielWainfleet
Jan 16 at 23:06
$begingroup$
I think you meant $psi$ rather than $f$. Please check if my reasoning is correct: For all $xin F$ and $qinBbb Q$: $psi(x)-q=psi(x)-psi(q)$ [since $psi|_{Bbb Q}=text{id}_{Bbb Q}$] $=psi(x-q)$. Then $x ge q iff x-q ge 0 iff x-q$ $=sqrt {x-q} cdot sqrt {x-q} iff psi(x-q)=psi(sqrt {x-q} cdot sqrt {x-q})=psi(sqrt {x-q})cdot psi(sqrt {x-q})=$ $(psi(sqrt {x-q}))^2 ge 0 iff psi(x)-q=psi(x-q) ge 0$. To sum up, $xge q iff psi(x)-q ge 0$ $iff psi(x)ge q$.[...]
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
$begingroup$
I think you meant $psi$ rather than $f$. Please check if my reasoning is correct: For all $xin F$ and $qinBbb Q$: $psi(x)-q=psi(x)-psi(q)$ [since $psi|_{Bbb Q}=text{id}_{Bbb Q}$] $=psi(x-q)$. Then $x ge q iff x-q ge 0 iff x-q$ $=sqrt {x-q} cdot sqrt {x-q} iff psi(x-q)=psi(sqrt {x-q} cdot sqrt {x-q})=psi(sqrt {x-q})cdot psi(sqrt {x-q})=$ $(psi(sqrt {x-q}))^2 ge 0 iff psi(x)-q=psi(x-q) ge 0$. To sum up, $xge q iff psi(x)-q ge 0$ $iff psi(x)ge q$.[...]
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
$begingroup$
[...] It follows that ${qinBbb Q mid qle x}={qinBbb Q mid qle psi(x)}$. Hence $sup {qinBbb Q mid qle x}=$ $sup {qinBbb Q mid qle psi(x)}$ and thus $x=psi(x)$. As a result, the isomorphism between two fields is unique.
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
$begingroup$
[...] It follows that ${qinBbb Q mid qle x}={qinBbb Q mid qle psi(x)}$. Hence $sup {qinBbb Q mid qle x}=$ $sup {qinBbb Q mid qle psi(x)}$ and thus $x=psi(x)$. As a result, the isomorphism between two fields is unique.
$endgroup$
– Le Anh Dung
Jan 17 at 1:54
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Perfectly correct. And $f$ was a typo for $psi$.
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– DanielWainfleet
Jan 17 at 13:58
$begingroup$
Perfectly correct. And $f$ was a typo for $psi$.
$endgroup$
– DanielWainfleet
Jan 17 at 13:58
$begingroup$
Thank you so much for your verification! Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 14:04
$begingroup$
Thank you so much for your verification! Have you seen any error in my proof?
$endgroup$
– Le Anh Dung
Jan 17 at 14:04
add a comment |
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You should use that a complete ordered field has a unique ordering, because an element is nonnegative if and only if it has a square root.
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– egreg
Jan 17 at 0:05
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Hi @egreg, DanielWainfleet has utilized your idea and posted it as an answer below. I am reading his answer. Have you seen any error in my proof?
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– Le Anh Dung
Jan 17 at 0:10