Ordered ring and mutiplicative ordinal

Question

If $(E,<)$ is a linear order, let $s(E)$ denote the supremum of the set of ordinals which (order-)embed in $(E,<)$. $s(E)$ is also the set of ordinals which embed in $E$ with a non cofinal range.

If $(G,+,0,<)$ is an ordered group, then $s(G)$ is an additive ordinal because given $\alpha < s(G)$ and $f_{\alpha}: \alpha \rightarrow G$ a positive order embeding with strict upper bound $a$, $f_{\alpha.2}:= \{(\alpha.t + \beta,f_{\alpha}(\beta) + t.a) \ | \ (t,\beta) \in 2 \times \alpha\}$ is an order embedding $\alpha.2 \rightarrow G$ with $2.a$ as a strict upper bound.

If $(R,+, \ ,0,1,<)$ is a discrete ordered ring, then $s(A)$ is a multiplicative ordinal because given $\alpha < s(G)$ and $f_{\alpha}: \alpha \rightarrow R$ a positive order embeding with strict upper bound $a$, $f_{{\alpha}^2}:= \{(\alpha.\beta + \delta,af_{\alpha}(\beta) + f_{\alpha}(\delta)) \ | \ \beta,\delta \in \alpha\}$ is an order embedding ${\alpha}^2 \rightarrow R$ with $a^2 + a$ as a strict upper bound.

If $R$ is an ordered ring but $R$ is not discrete, $f_{{\alpha}^2}$ need not be an embedding because two distinct points in the range of $f_{\alpha}$ may be very close to each other.

Is $s(R)$ multiplicative nonetheless whenever $R$ is a ordered ring?

Answer 1

The question actually has an easy positive answer:

Let $\alpha$ be an element of $s(R)$. $s(R)$ being additive, either $s(R) = \omega_0$, in which case it is also multiplicative, or $\forall \gamma \in s(R), \gamma + \omega_0 \in s(R)$, so one can assume that $\alpha$ is limit. Let $f_{\alpha}: \alpha \rightarrow R$ be a positive order emdedding with strict upper bound $a \in R$. Let $\Delta(f_{\alpha}): \alpha \rightarrow R$ denote $\gamma \longmapsto f_{\alpha}(\gamma+1) - f_{\alpha}(\gamma)$.

We consider two cases:

1)$\Delta(f_{\alpha})$ takes some value $\varepsilon$ such that $a\varepsilon < 1$. In this case, define $f_{\alpha^2}: \alpha^2 \rightarrow R$ by $\forall \beta,\gamma < \alpha, f_{\alpha^2}(\alpha.\beta + \gamma) := f_{\alpha}(\beta) + \varepsilon\Delta(f_{\alpha})(\beta)f_{\alpha}(\gamma)$. Then $f_{\alpha^2}$ is strictly increasing and has $2.a$ as a strict upper bound, so ${\alpha}^2 \in s(R)$.

2)Otherwise, define $f_{\alpha^2}(\alpha.\beta + \gamma) := a^2f_{\alpha}(\beta) + f_{\alpha}(\gamma)$. $f_{\alpha^2}$ is an order embedding, with $a^3 + a$ as a strict upper bound, and ${\alpha}^2 \in s(R)$.

So $s(R)$ is multiplicative.