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This is an exercise from Chapter 9. The Sets of Real Numbers from textbook Introduction to Set Theory by Hrbacek and Jech. The textbook does not provide solution and I would like to verify my attempt.

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


We first introduce some definitions.

  • A structure $\mathfrak{A}=\langle A,<,+,\cdot,0,1 \rangle$ where $<$ is a linear ordering, $+$ and $\cdot$ are binary operations, and $0,1$ are constants such that all properties 1-12 are satisfied is called an ordered field.

  • For all $a,b,c\in A$:

    1. $a+b=b+a$.

    2. $(a+b)+c=a+(b+c)$.

    3. $a+0=a$.

    4. $\exists a'\in A:a+a'=0$. We denote $a'=-a$, the opposite of $a$.

    5. $a<b\implies a+c<b+c$.

    6. $a\cdot (b+c)=a\cdot b + a\cdot c$.

    7. $a\cdot b = b\cdot a$.

    8. $(a\cdot b)\cdot c = a\cdot (b\cdot c)$

    9. $a\cdot 1=a$

    10. $a\neq 0 \implies \exists a'\in A:a\cdot a'=1$. We denote $a'=a^{-1}$, the reciprocal of $a$.

    11. $a<b$ and $0<c$ $\implies a\cdot c < b\cdot c$.

    12. $0\neq 1$

  • We then define subtraction $(-)$ and division $(\div)$ as follows:

    1. $\forall a,b\in A: a-b=a+(-b)$.

    2. $\forall (a,b\in A, b\neq 0): a\div b=a\cdot b^{-1}$.


My attempt:

For convenience, we write $ab$ instead of $a\cdot b$, and $a/b$ instead of $a\div b$.

Let $\mathfrak{A}=\langle A,<,+,\cdot,0',1' \rangle$ be an ordered field. $\langle C,<,+,\cdot,0',1' \rangle$ is called a subfield of $\mathfrak{A}$ if $C\subseteq A$ and $\langle C,<,+,\cdot,0',1' \rangle$ is an ordered field. It follows that - if $\langle C,<,+,\cdot,0',1' \rangle$ is a subfield of $\mathfrak{A}$, then $0',1'\in C$ and thus $C\neq\emptyset$.

Let $\mathfrak{F}=\{C \mid \langle C,<,+,\cdot,0',1' \rangle \text{ is a subfield of }\mathfrak{A}\}$. Then $A \in \mathfrak{F}$ and thus $\mathfrak{F} \neq \emptyset$. Moreover, $0',1'\in C$ for any $C\in \mathfrak{F}$. Let $\overline C =\bigcap \mathfrak{F}$. Then $0',1'\in \overline C$ and thus $\overline C \neq \emptyset$. It is tedious to verify that $\langle \overline C,<,+,\cdot,0',1' \rangle$ is a subfield of $\mathfrak{A}$.

We next prove that $\langle \overline C,<,+,\cdot,0',1' \rangle$ is isomorphic to $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$.

Define $f:\Bbb N \to \overline C$ recursively by $f(0)=0'$ and $f(n+1)=f(n)+1'$ for all $n\in\Bbb N$.

  • $f$ is an order embedding. Since $<$ is a linear ordering, it suffices to prove that $m<n\implies f(m)<f(n)$. The inequality is trivially true for $n=0$. Assume it is true for $n$ and $m<n+1$. We have $m<n+1 \implies$ $m<n$ or $m=n$. If $m=n$, then $f(m)=f(n)<f(n)+1'=f(n+1)$. If $m<n$, then by inductive hypothesis $f(m)<f(n)<f(n)+1'<f(n+1)$.

  • $f(m+n)=f(m)+f(n)$. The identity is trivially true for $n=0$. Assume it is true for $n$. $f(m+(n+1))=f((m+1)+n)=f(m+1)+f(n)=(f(m)+1')+f(n)=f(m)+(f(n)+1')=f(m)+f(n+1).$

  • $f(mn)=f(m)f(n)$. The identity is trivially true for $n=0$. Assume it is true for $n$. $f(m(n+1))=f(mn+m)=f(mn)+f(m)=f(m)f(n)+f(m)=f(m)f(n)+f(m)1'=f(m)(f(n)+1')=f(m)f(n+1).$

Define $g:\Bbb Z \to \overline C$ by $g(n)=f(n)$ for all $n\in\Bbb N$ and $g(n)=-f(-n)$ for all $n\in\Bbb Z \setminus \Bbb N$. Similarly, we can prove that $g$ is an order embedding, $g(m+n)=g(m)+g(n)$, and $g(mn)=g(m)g(n)$ for all $m,n\in\Bbb Z$.

Define $h:\Bbb Q \to \overline C$ by $h(p)=g(m)/g(n)$ if $p=m/n$ for some $m,n\in\Bbb Z$. Similarly, we can prove that $h$ is an order embedding, $h(p+q)=h(p)+h(q)$, and $h(pq)=h(p)h(q)$ for all $p,q\in\Bbb Q$. It follows that $h \restriction \Bbb Q$ is an isomorphism between $\Bbb Q$ and $h[\Bbb Q]$. Then $\langle h[\Bbb Q],<,+,\cdot,0',1' \rangle$ is a subfield of $\mathfrak{A}$ and thus $\overline C \subseteq h[\Bbb Q]$. Moreover, $h[\Bbb Q] \subseteq \overline C$. Hence $h[\Bbb Q]=\overline C$ and thus $\langle \overline C,<,+,\cdot,0',1' \rangle$ is isomorphic to $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$.

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    $\begingroup$ I would not call it tedious to confirm that $<\bar C,...>$ is a field, but it's not very exciting. The common intersection of a non-empty family of sub-fields of a field $ F $ is a sub-field of $F.$ And if $F$ is an ordered field ordered by $<,$ and $G$ is a subfield of $F$ then $G$ is also an ordered field ordered by $<$..... BTW. Some ordered fields can have more then one order. E.g. let $F=\{a+b\sqrt 2\,:a,b\in \Bbb Q\}.$ Consider the order $<^* $ where $ 0<^*1$ and $\sqrt 2\,<^*0$. $\endgroup$ – DanielWainfleet Jan 9 '19 at 5:53
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I think this is essentially correct, but you are working too hard.

Every field contains a prime subfield: its smallest subfield. That's the intersection of all the subfields. They all contain the multiplicative identity - the prime subfield is essentially the one generated by the multiplicative identity.

The prime subfield is either (isomorphic to) the rational numbers (characteristic $0$) or to the field $\mathbb{Z}_p$ of integers modulo $p$ (characteristic $p$).

For an ordered field the characteristic must be $0$ so the prime subfield is (isomorphic to) the rational numbers.

Perhaps the existence of the prime subfield was what you were supposed to prove. That's essentially what you did. Even so, you wrote too much. You need not have copied the definitions of a field and then an ordered field: they are standard.

You can search for prime subfield to see standard arguments.

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  • $\begingroup$ Thank you so much @Ethan! I got your points. $\endgroup$ – Navier_Stokes Jan 8 '19 at 14:44

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