There is a unique isomorphism from $langle Bbb Q,<,+,cdot,0,1 rangle$ to an ordered field












3












$begingroup$



Let $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle,mathfrak{A}=langle A,<,+,cdot,0',1' rangle$ be ordered fields where $Bbb Q$ is the set of rationals, and $f,g$ be isomorphisms from $mathfrak{Q}$ to subfields of $mathfrak{A}$. Then $f=g$.




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



It suffices to prove that $f(p)=g(p)$ for all $pinBbb Q$ and $p>0$.



It is easy to prove that $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$.



For $pinBbb Q$ and $p>0$, $p=dfrac{m}{n}$ for some $m,ninBbb N$.




  • $f(p)=fleft(dfrac{m}{n}right)=dfrac{f(m)}{f(n)}=dfrac{f(underbrace{1+cdots+1}_{mtext{ times}})}{f(underbrace{1+cdots+1}_{ntext{ times}})}=dfrac{underbrace{f(1)+cdots+f(1)}_{mtext{ times}}}{underbrace{f(1)+cdots+f(1)}_{ntext{ times}}}=dfrac{underbrace{1'+cdots+1'}_{mtext{ times}}}{underbrace{1'+cdots+1'}_{ntext{ times}}}$


  • $g(p)=gleft(dfrac{m}{n}right)=dfrac{g(m)}{g(n)}=dfrac{g(underbrace{1+cdots+1}_{mtext{ times}})}{g(underbrace{1+cdots+1}_{ntext{ times}})}=dfrac{underbrace{g(1)+cdots+g(1)}_{mtext{ times}}}{underbrace{g(1)+cdots+g(1)}_{ntext{ times}}}=dfrac{underbrace{1'+cdots+1'}_{mtext{ times}}}{underbrace{1'+cdots+1'}_{ntext{ times}}}$



It follows that $f(p)=g(p)$ and thus $f=g$.






proof-verification ordered-fields







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes. It's ok. Or you can show from $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$ and by induction on $|n|,$ that $f(n)=g(n)$ for all $nin Bbb Z$ and also that $f(n)ne 0' $ for all $0ne nin Bbb Z....$ Then for $m,nin Bbb Z$ with $0ne n,$ we have $f(n)f(m/n)=f(m)=g(m)=g(n)g(m/n)=f(n)g(m/n),...$ so $f(n)f(m/n)=f(n)g(m/n)$ with$ f(n)ne 0'$, so $f(m/n)=g(m/n)$.
    $endgroup$
    – DanielWainfleet
    Jan 9 at 4:35








  • 1




    $begingroup$
    Hi @DanielWainfleet, I actually did what you suggested here :)
    $endgroup$
    – Le Anh Dung
    Jan 9 at 4:37
















3












$begingroup$



Let $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle,mathfrak{A}=langle A,<,+,cdot,0',1' rangle$ be ordered fields where $Bbb Q$ is the set of rationals, and $f,g$ be isomorphisms from $mathfrak{Q}$ to subfields of $mathfrak{A}$. Then $f=g$.




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



It suffices to prove that $f(p)=g(p)$ for all $pinBbb Q$ and $p>0$.



It is easy to prove that $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$.



For $pinBbb Q$ and $p>0$, $p=dfrac{m}{n}$ for some $m,ninBbb N$.




  • $f(p)=fleft(dfrac{m}{n}right)=dfrac{f(m)}{f(n)}=dfrac{f(underbrace{1+cdots+1}_{mtext{ times}})}{f(underbrace{1+cdots+1}_{ntext{ times}})}=dfrac{underbrace{f(1)+cdots+f(1)}_{mtext{ times}}}{underbrace{f(1)+cdots+f(1)}_{ntext{ times}}}=dfrac{underbrace{1'+cdots+1'}_{mtext{ times}}}{underbrace{1'+cdots+1'}_{ntext{ times}}}$


  • $g(p)=gleft(dfrac{m}{n}right)=dfrac{g(m)}{g(n)}=dfrac{g(underbrace{1+cdots+1}_{mtext{ times}})}{g(underbrace{1+cdots+1}_{ntext{ times}})}=dfrac{underbrace{g(1)+cdots+g(1)}_{mtext{ times}}}{underbrace{g(1)+cdots+g(1)}_{ntext{ times}}}=dfrac{underbrace{1'+cdots+1'}_{mtext{ times}}}{underbrace{1'+cdots+1'}_{ntext{ times}}}$



It follows that $f(p)=g(p)$ and thus $f=g$.






proof-verification ordered-fields







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes. It's ok. Or you can show from $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$ and by induction on $|n|,$ that $f(n)=g(n)$ for all $nin Bbb Z$ and also that $f(n)ne 0' $ for all $0ne nin Bbb Z....$ Then for $m,nin Bbb Z$ with $0ne n,$ we have $f(n)f(m/n)=f(m)=g(m)=g(n)g(m/n)=f(n)g(m/n),...$ so $f(n)f(m/n)=f(n)g(m/n)$ with$ f(n)ne 0'$, so $f(m/n)=g(m/n)$.
    $endgroup$
    – DanielWainfleet
    Jan 9 at 4:35








  • 1




    $begingroup$
    Hi @DanielWainfleet, I actually did what you suggested here :)
    $endgroup$
    – Le Anh Dung
    Jan 9 at 4:37














3












3








3





$begingroup$



Let $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle,mathfrak{A}=langle A,<,+,cdot,0',1' rangle$ be ordered fields where $Bbb Q$ is the set of rationals, and $f,g$ be isomorphisms from $mathfrak{Q}$ to subfields of $mathfrak{A}$. Then $f=g$.




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



It suffices to prove that $f(p)=g(p)$ for all $pinBbb Q$ and $p>0$.



It is easy to prove that $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$.



For $pinBbb Q$ and $p>0$, $p=dfrac{m}{n}$ for some $m,ninBbb N$.




  • $f(p)=fleft(dfrac{m}{n}right)=dfrac{f(m)}{f(n)}=dfrac{f(underbrace{1+cdots+1}_{mtext{ times}})}{f(underbrace{1+cdots+1}_{ntext{ times}})}=dfrac{underbrace{f(1)+cdots+f(1)}_{mtext{ times}}}{underbrace{f(1)+cdots+f(1)}_{ntext{ times}}}=dfrac{underbrace{1'+cdots+1'}_{mtext{ times}}}{underbrace{1'+cdots+1'}_{ntext{ times}}}$


  • $g(p)=gleft(dfrac{m}{n}right)=dfrac{g(m)}{g(n)}=dfrac{g(underbrace{1+cdots+1}_{mtext{ times}})}{g(underbrace{1+cdots+1}_{ntext{ times}})}=dfrac{underbrace{g(1)+cdots+g(1)}_{mtext{ times}}}{underbrace{g(1)+cdots+g(1)}_{ntext{ times}}}=dfrac{underbrace{1'+cdots+1'}_{mtext{ times}}}{underbrace{1'+cdots+1'}_{ntext{ times}}}$



It follows that $f(p)=g(p)$ and thus $f=g$.






proof-verification ordered-fields







share|cite|improve this question









$endgroup$





Let $mathfrak{Q}=langle Bbb Q,<,+,cdot,0,1 rangle,mathfrak{A}=langle A,<,+,cdot,0',1' rangle$ be ordered fields where $Bbb Q$ is the set of rationals, and $f,g$ be isomorphisms from $mathfrak{Q}$ to subfields of $mathfrak{A}$. Then $f=g$.




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



It suffices to prove that $f(p)=g(p)$ for all $pinBbb Q$ and $p>0$.



It is easy to prove that $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$.



For $pinBbb Q$ and $p>0$, $p=dfrac{m}{n}$ for some $m,ninBbb N$.




  • $f(p)=fleft(dfrac{m}{n}right)=dfrac{f(m)}{f(n)}=dfrac{f(underbrace{1+cdots+1}_{mtext{ times}})}{f(underbrace{1+cdots+1}_{ntext{ times}})}=dfrac{underbrace{f(1)+cdots+f(1)}_{mtext{ times}}}{underbrace{f(1)+cdots+f(1)}_{ntext{ times}}}=dfrac{underbrace{1'+cdots+1'}_{mtext{ times}}}{underbrace{1'+cdots+1'}_{ntext{ times}}}$


  • $g(p)=gleft(dfrac{m}{n}right)=dfrac{g(m)}{g(n)}=dfrac{g(underbrace{1+cdots+1}_{mtext{ times}})}{g(underbrace{1+cdots+1}_{ntext{ times}})}=dfrac{underbrace{g(1)+cdots+g(1)}_{mtext{ times}}}{underbrace{g(1)+cdots+g(1)}_{ntext{ times}}}=dfrac{underbrace{1'+cdots+1'}_{mtext{ times}}}{underbrace{1'+cdots+1'}_{ntext{ times}}}$



It follows that $f(p)=g(p)$ and thus $f=g$.






proof-verification ordered-fields




proof-verification ordered-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 2:57









Le Anh DungLe Anh Dung

1,2221621




1,2221621








  • 1




    $begingroup$
    Yes. It's ok. Or you can show from $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$ and by induction on $|n|,$ that $f(n)=g(n)$ for all $nin Bbb Z$ and also that $f(n)ne 0' $ for all $0ne nin Bbb Z....$ Then for $m,nin Bbb Z$ with $0ne n,$ we have $f(n)f(m/n)=f(m)=g(m)=g(n)g(m/n)=f(n)g(m/n),...$ so $f(n)f(m/n)=f(n)g(m/n)$ with$ f(n)ne 0'$, so $f(m/n)=g(m/n)$.
    $endgroup$
    – DanielWainfleet
    Jan 9 at 4:35








  • 1




    $begingroup$
    Hi @DanielWainfleet, I actually did what you suggested here :)
    $endgroup$
    – Le Anh Dung
    Jan 9 at 4:37














  • 1




    $begingroup$
    Yes. It's ok. Or you can show from $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$ and by induction on $|n|,$ that $f(n)=g(n)$ for all $nin Bbb Z$ and also that $f(n)ne 0' $ for all $0ne nin Bbb Z....$ Then for $m,nin Bbb Z$ with $0ne n,$ we have $f(n)f(m/n)=f(m)=g(m)=g(n)g(m/n)=f(n)g(m/n),...$ so $f(n)f(m/n)=f(n)g(m/n)$ with$ f(n)ne 0'$, so $f(m/n)=g(m/n)$.
    $endgroup$
    – DanielWainfleet
    Jan 9 at 4:35








  • 1




    $begingroup$
    Hi @DanielWainfleet, I actually did what you suggested here :)
    $endgroup$
    – Le Anh Dung
    Jan 9 at 4:37








1




1




$begingroup$
Yes. It's ok. Or you can show from $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$ and by induction on $|n|,$ that $f(n)=g(n)$ for all $nin Bbb Z$ and also that $f(n)ne 0' $ for all $0ne nin Bbb Z....$ Then for $m,nin Bbb Z$ with $0ne n,$ we have $f(n)f(m/n)=f(m)=g(m)=g(n)g(m/n)=f(n)g(m/n),...$ so $f(n)f(m/n)=f(n)g(m/n)$ with$ f(n)ne 0'$, so $f(m/n)=g(m/n)$.
$endgroup$
– DanielWainfleet
Jan 9 at 4:35






$begingroup$
Yes. It's ok. Or you can show from $f(0)=g(0)=0'$ and $f(1)=g(1)=1'$ and by induction on $|n|,$ that $f(n)=g(n)$ for all $nin Bbb Z$ and also that $f(n)ne 0' $ for all $0ne nin Bbb Z....$ Then for $m,nin Bbb Z$ with $0ne n,$ we have $f(n)f(m/n)=f(m)=g(m)=g(n)g(m/n)=f(n)g(m/n),...$ so $f(n)f(m/n)=f(n)g(m/n)$ with$ f(n)ne 0'$, so $f(m/n)=g(m/n)$.
$endgroup$
– DanielWainfleet
Jan 9 at 4:35






1




1




$begingroup$
Hi @DanielWainfleet, I actually did what you suggested here :)
$endgroup$
– Le Anh Dung
Jan 9 at 4:37




$begingroup$
Hi @DanielWainfleet, I actually did what you suggested here :)
$endgroup$
– Le Anh Dung
Jan 9 at 4:37










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