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Let $\mathbb Q(x)$ denote, as usual, the field of rational functions with rational coefficients. Any element of $\mathbb Q(x)$ can be written in the form $$f=\frac{a_n x^n + \cdots +a_0}{b_m x^m + \cdots +b_0}$$ with $a_n, b_m \ne 0$ and where all coefficients are integers. Let $P \subset F$ be defined by the condition $f\in P \iff \frac{a_n}{b_m}> 0$, and define $f \prec g \iff g-f \in P$. Then $(F, \prec)$ is an ordered field.

As is discussed in this answer, under this order $\mathbb Q(x)$ is non-Archimedean. In particular, $x$ is a transfinite element (in the sense that $r \prec x$ for all $r \in \mathbb Q$) and $\frac 1x$ is infinitesimal (in the sense that $0 \prec \frac 1x \prec r$ for all positive $r \in \mathbb Q$). The fact that $\mathbb Q(x)$ is non-Archimedean means it is not complete, which is easily witnessed by considering the set $\mathbb Q \subseteq \mathbb Q(x)$; this subset is bounded (for example, by $x$) but has no least upper bound.

What happens if we now perform the Dedekind cut construction on $\mathbb Q(x)$? What do we get?

Any complete ordered field is isomorphic to $\mathbb R$, and certainly the Dedekind completion of $\mathbb Q(x)$ is larger than $\mathbb R$, so this must not be a complete ordered field. But why not? Is the result no longer a field? Or do we end up with a field for which the axioms for an ordered field are somehow broken?

Related question here, but that question starts with the hyperreals, which (I think?) is not the same question as mine. (It's possible that the answer to my question is "You get the completion of the hyperreals"; if so, I'd love to understand that better!)

mweiss
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    If $\alpha=\inf_{q\in\mathbb Q^+} qx$ and $\beta=\sup\mathbb Q$ in your dedekind completion, I think you get $\alpha<\beta.$ There is essentially a 'gap' between the infinite sets, which, on an intuitive level only, is the gap where you might put $x^{1/2}$ and more generally $x^{a}$ for $a$ real numbers. In particular, if you got a field, then $(\alpha+\beta)/2$ would be strictly between $\alpha$ and $\beta.$ So the definition of multiplication will fail. – Thomas Andrews Oct 24 '22 at 18:09
  • @ThomasAndrews thanks, that's very helpful! If you have time to expand this into an answer I'd appreciate it very much. – mweiss Oct 24 '22 at 18:15
  • Well, it was just a start of an intuition/idea for an answer. I don't have it worked out just yet how multiplication fails... – Thomas Andrews Oct 24 '22 at 18:18
  • It's definitely addition which fails. – Thomas Andrews Oct 24 '22 at 18:55
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    Check the answers here. Not quite the same question, since the OP asks about an ordered field of cardinality larger than $|\mathbb{R}|$. But the answers are quite general and explain why the Dedekind completion of any non-Archimedean ordered field fails to be a field (and how to fix this, at the expense of non getting full completeness). If this answers your question, I'll close it as a duplicate. – Alex Kruckman Oct 24 '22 at 19:25
  • The set $\mathbb N = {1,2,3,4,\dots}$ is bounded above. Let $k$ be its supremum in the completion. Then $k=1+k$. So our result is definitely not a field (or even a ring). – GEdgar Oct 24 '22 at 21:01

1 Answers1

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If we define $\deg f=n-m.$ Then if $f,g\in P,$ and $\deg f<\deg g,$ then $f\prec g.$ Define $P_d=\{f\in P\mid \deg f=d\}.$

Define $\beta=\inf P_1.$ This would be the Dedekind cut $\{f\in P\mid \deg f<1.\}$

Now, given any $f\in P,$ $f\in P_1$ iff $f+1\in P_1.$

So, if the result of the Dedekind process is a field, we'd get $\beta+1=\beta,$ or $1=0.$

So there must be a problem with the definition of addition - that it no longer has the property that if $\alpha\neq 0$ then $\alpha+\beta\neq \beta.$

I'm still thinking about what specific property of $\mathbb Q$ lets us get this property, while this field does not. But I think it is related to the fact that, for rationals $q,r>0$ then $q+q+q+q+\cdots + q>r$ for some number finite sum of values $q.$ This is essentially the Archimedean property on the rational numbers.

The rational numbers have the property that in a nonzero cut, $\alpha,$ there is a sequence $(m_k,M_k)$ with $m_k\in\alpha, M_k\notin\alpha,$ and $\inf_k (M_k-m_k)=0.$

This is usually constructed by starting with any pair $(m_0,M_0)$ and then recursively computing $m=(m_k+M_k)/2,$ and replacing either $m_k$ or $M_k$ with $m,$ depending on whether $m\in\alpha$ or not.

Given a positive $p\in\beta,$ then, there is some $k$ such that $p>M_k-m_k$ and thus $p+m_k\in\alpha+\beta,$ but $p+m_k\notin \alpha.$

But in your totally ordered abelian group, we cannot deduce that $\inf (M_k-m_k)=0.$ Indeed, $\deg (M_k-m_k)=\deg(M_0-m_0)=d,$ and thus $x^{d-1}\in P$ is a positive lower bound for $\{M_k-m_k\}.$

And, indeed, if $m_0\in P_0,M_0\in P_1,$ we'd get all $m_k=m_0$ and all $M_k\in P_1.$ So you can deduce that $M_{k}-m_0>m_0$ for all $k.$

Thomas Andrews
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  • And if you try to prove that the Dedekind completion has additive inverses, you'll find that what you need exactly is that if $(L,R)$ is a Dedekind cut then there are elements of $L$ and $R$ that are arbitrarily close together (i.e., basically the property you observed at the end). This tells you exactly that $(-R,-L)$ is an additive inverse to $(L,R)$. – Eric Wofsey Oct 24 '22 at 21:53
  • @EricWofsey Yeah, I was originally expecting multiplication to be the problem, so I was assuming the "positives-only" Dedekind cuts, where you define $\mathbb R^{\geq 0}$ from $\mathbb Q^{\geq 0},$ since the definition there of multiplication is easier. But addition of cuts is easy over the whole additive group, so I could have restarted with that approach. – Thomas Andrews Oct 24 '22 at 22:03
  • This is exactly what I wanted, thanks! – mweiss Oct 25 '22 at 00:07