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Consider a variety of algebras $\mathbb{V} = \mathbb{V}(\sigma,\tau)$ which consists of the set of algebras defined over a fixed signature $\sigma$ and satisfying a set of identities $\tau$. We may then define its ordered version $\mathbb{V}_{or}(\sigma,\tau)$ cas the set of algebras admitting a total order $\leq$ compatible with $\sigma$. By this I mean that:

(*) for each $f : X^n \rightarrow X$ in $\sigma$, for each tuples $t,t' \in X^n$, if $t_i \leq t'_i$ for every $i \in [n]$ then $f(t) \leq f(t')$.

Given such an ordered algebra $A$, we may define its order type $ot(A)$ as the maximum order type of a well-founded order $\leq' \subseteq \leq$ without accumulation points. We may then define the spectrum of the variety $\mathbb{V}$ as the set of ordinals obtainable in that way.

Is this notion known in universal algebra, and if so, which varieties have a known spectra?

An easy case seems to be the variety of ordered fields: for instance, the field of reals $\mathbb{R}$ would have order type $\omega$, and the field of rational fractions $\mathbb{R}(x)$ would have order type $\omega^{\omega}$. So is the following true:

Conjecture: the spectrum of the ordered fields consists of the ordinals closed under addition and multiplication?

(NOTE: I added the requirement that $\leq'$ has no accumulation points, as otherwise any countable ordinal would be embeddable in $\mathbb{R}$. The precise meaning is relative to $\leq$: if $S$ is the set of elements of $\leq'$, we ask that for every $x \in S$, there is an interval $I$ of $\leq$ properly containing $x$ s.t. $I \cap S = \{x\}$.)

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    $\begingroup$ Your questions may have a higher chance of being answered at Mathoverflow, where they would be seen by a broader spectrum of mathematicians. $\endgroup$ – Sasho Nikolov Apr 7 '14 at 23:08
  • $\begingroup$ Thanks for the advice, but I guess that when you've lost yourself, it doesn't really matter where you end... $\endgroup$ – NisaiVloot Apr 8 '14 at 5:37
  • $\begingroup$ @HeloLobo: I agree with Sasho Nikolov. (As a side note, you might want to call you notion something other than "order type" as that term already has a specific meaning: en.wikipedia.org/wiki/Order_type). $\endgroup$ – Joshua Grochow Apr 8 '14 at 13:22
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    $\begingroup$ Regarding the conjecture, Dana Scott seems to have asked a similar question: cs.nyu.edu/pipermail/fom/2003-August/007150.html $\endgroup$ – András Salamon Apr 8 '14 at 16:37
  • $\begingroup$ @Andras: I do not think it is consistent. I would think the consistent usage would be that the order type of $A$ is the isomorphism class of the total ordering $(A, \leq)$, considered up to isomorphism of total orders (note that this notion makes sense for arbitrary total orderings, not just well-orderings). Instead, HeloLobo is considering the order type of the "largest" well-subordering of $(A, \leq)$. $\endgroup$ – Joshua Grochow Apr 11 '14 at 17:59

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