0
$\begingroup$

Given a linear order $(E,<)$, let $o(E)$ denote the least ordinal which does not order-embed in $(E,<)$.

It is known that for an ordered field $k$, $o(k)$ is a regular ordinal $\geq \omega_1$.

I am trying to compute $o(k((x^{\Gamma})))$ where $k((x^{\Gamma}))$ is the ordered field of Hahn series over the ordered field $k$ with value ordered group $\Gamma$. The natural valuation is denoted by $v$.

I think that this only depends on $o(k)$ and $o(\Gamma)$. More precisely, I think that if $reg(\alpha)$ for $\alpha \in Ord$ denotes the least regular ordinal above (or equal to) $\alpha$ - so $reg(\alpha) = \alpha$ if $\alpha$ is regular, $reg(\alpha) = \alpha^+$ (successor cardinal) if $\alpha \geq \omega_0$ is singular -, then $o(k((x^{\Gamma}))) = \max(o(k),reg(o(\Gamma)))$. (1)

In fact, I know that $o(k((x^{\Gamma}))) \geq \max(o(k),reg(o(\Gamma)))$ because if $\alpha \in Ord$ embeds in $k$ or $\Gamma$, it embeds in $k((x^{\Gamma}))$, so $o(k((x^{\Gamma}))) \geq o(k),o(\Gamma)$, and by regularity, $o(k((x^{\Gamma}))) \geq reg(o(\Gamma))$.

I had an idea to prove (1), but it was flawed in the beginning so I deleted my answer. I will maybe post another answer here in the days to come because I think there is a way to correct this.

In the meantime, does someone know if (1) is true?

$\endgroup$
4
  • $\begingroup$ In any event, it still sounds dubious to me: unless I'm wrong, it is easy to see that the $o$ is at least $o(k)\cdot o(\Gamma)$ (or the other way around, I don't remember the conventions for ordinal multiplication; just consider monomials). $\endgroup$
    – tomasz
    Oct 20, 2016 at 13:31
  • $\begingroup$ By regular ordinal, I mean ordinal which is equal to its cofinality, so regular ordinals are initial ordinals, infinite if strictly greater that $1$. $\endgroup$
    – nombre
    Oct 20, 2016 at 13:35
  • $\begingroup$ You're right. I neglected the fact that most ordinals are singular. I will delete the meaningless comments. $\endgroup$
    – tomasz
    Oct 20, 2016 at 13:44
  • $\begingroup$ Okay, so if we consider monomials, for each couple $(\alpha,\beta) \in o(k) \times o(\Gamma)$, we get a strictly increasing map $c: \alpha.\beta \rightarrow k((x^{\Gamma}))$ by setting $c(\alpha.\eta + \gamma) = a(\gamma)x^{-b(\eta)}$ where $a: \alpha \rightarrow k_{\geq 0}$, $b: \beta \rightarrow \Gamma$ are strictly increasing. But this only proves that $o(k((x^{\Gamma}))) \geq \underset{\alpha,\beta \in o(\Gamma) \times o(k)} \alpha.\beta$, which is lower than $\max(o(k),reg(o(\Gamma))$. $\endgroup$
    – nombre
    Oct 20, 2016 at 14:15

1 Answer 1

0
$\begingroup$

I have finally found a satisfying way to prove $o(k((x^{\Gamma})) \leq \max(o(k),reg(o(\Gamma)))$.


Let's assume the contrary: let $\kappa$ denote $\max(o(k),reg(o(\Gamma)))$; let $f: \alpha \rightarrow k((x^{\Gamma}))$ be strictly increasing.

The proof requires the following:

$\star$

Lemma:

If $\alpha$ is an ordinal, $(X,<)$ is an ordered set, and $\rho: \alpha \rightarrow X$ is monotone, then either $\rho$ is eventually constant, or there is $\varphi: cof(\alpha) \rightarrow \alpha$ strictly increasing such that $\rho \circ \varphi$ is strictly monotone with same monotony.

[One can arrange that $\varphi$ is cofinal but this won't be needed here.]

$\star$

For any $\alpha < \kappa$, let $u_{\alpha}$ denote the sequence $(v(f(\beta) - f(\alpha)))_{\alpha < \beta < \kappa}$. Since $f$ is strictly increasing, $(f(\beta) - f(\alpha))_{\alpha < \beta < \kappa}$ is strictly positive, strictly increasing, and so $u_{\alpha}$ is decreasing. The order type of $]\alpha;\kappa[$ is $\kappa = cof(\kappa)$, and $\kappa \geq o(\Gamma)$ does not embed in $\Gamma$ so it doesn't "reverse-embed" in it (because $\Gamma$ has a stricty decreasing involution), and $u_{\alpha}$ cannot have a strictly decreasing subsequence. By the previous lemma, $u_{\alpha}$ is eventually constant. Let $g(\alpha)$ denote this constant value.

Let $\alpha < \alpha'$ be ordinals in $\kappa$. There are $\gamma,\gamma' < \kappa$ such that $u_{\alpha},u_{\alpha'}$ are constant from $\lambda,\lambda'$. For $\mu\geq \lambda,\lambda'$, we have:

$f(\alpha) < f(\alpha') < f(\mu)$, so $0 <f(\mu) - f(\alpha') < f(\mu) - f(\alpha)$, and $v(f(\mu) - f(\alpha)) \leq v(f(\mu) - f(\alpha'))$, that is $g(\alpha) \leq g(\alpha')$.

Hence, $g$ is increasing. For the same reason as above, $g$ is eventually constant. Fix $\alpha_0 < \kappa$ from which on $g$ is constant.

We now define a strictly increasing map $h: \kappa \rightarrow [\alpha_0;\kappa[$ such that $\forall \beta < \beta' <\kappa$, $u_{h(\beta)}(h(\beta')) = g(\alpha_0)$. We proceed by induction: set $h(\varnothing) = \alpha_0$. If $h$ is defined on some ordinal $\lambda < \kappa$ and satisfies the conditions on $\lambda$, for each $\beta < \lambda$, let $\beta <\gamma_{\beta} < \kappa$ denote the least ordinal such that $u_{h(\beta)}$ is constant equal to $g(\alpha_0)$ from $\gamma_{\beta}$ on. $\{\gamma_{\beta} \ | \ \beta < \lambda\}$ has a strict upper bound in $\kappa$ by regularity of $\kappa$. Define $h(\lambda)$ as the least of its strict upper bounds. $\forall \beta < \lambda, \beta < \gamma_{\beta} < h(\lambda)$ so $h$ is now strictly increasing on $\lambda+1$. For $\beta < \lambda$, $u_{h(\beta)}(h(\lambda)) = u_{h(\beta)}(\gamma_{\beta}) = g(\alpha_0)$ so we have successfuly prolonged $h$ on $\lambda+1$. By induction, $h$ can be defined on $\kappa$.

$\forall \beta < \beta' < \kappa$, $f(h(\beta)) < f(h(\beta'))$ so $f(h(\beta))(v(f(h(\beta')) - f(h(\beta)))) < f(h(\beta'))(v(f(h(\beta')) - f(h(\beta))))$, that is $f(h(\beta))(g(\alpha_0)) < f(h(\beta'))(g(\alpha_0))$, which is to say that $(f(h(\beta))(g(\alpha_0)))_{\beta < \kappa}$ is strictly increasing in $k$, which is impossible since $\kappa \geq o(k)$.

So there was a contradiction in assuming that such an embedding $f$ existed, and so $o(k((x^{\Gamma}))) = \max(o(k),reg(o(\Gamma)))$.


$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .