Are there as many real-closed fields of a given cardinality as I think there are? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:39:23Z http://mathoverflow.net/feeds/question/12949 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12949/are-there-as-many-real-closed-fields-of-a-given-cardinality-as-i-think-there-are Are there as many real-closed fields of a given cardinality as I think there are? Pete L. Clark 2010-01-25T18:09:15Z 2010-05-05T02:09:38Z <p>Let $\kappa$ be an infinite cardinal. Then there exists at least one <a href="http://en.wikipedia.org/wiki/Real-closed%5Ffield" rel="nofollow">real-closed field</a> of cardinality $\kappa$ (e.g. <a href="http://en.wikipedia.org/wiki/Lowenheim-Skolem" rel="nofollow">Lowenheim-Skolem</a>; or, start with a function field over <code>$\mathbb{Q}$</code> in $\kappa$ indeterminates, choose an ordering and a real-closure). </p> <p>But I think there are many more, namely $2^{\kappa}$ pairwise nonisomorphic real-closed fields of cardinality $\kappa$. This is equal to the number of binary operations on a set of infinite cardinality $\kappa$, so is the largest conceivable number.</p> <p>As for motivation -- what can I tell you, mathematical curiosity is a powerful thing. One application of this which I find interesting is that there would then be $2^{2^{\aleph_0}}$ conjugacy classes of order $2$ subgroups of the automorphism group of the field $\mathbb{C}$. </p> <p><b>Addendum</b>: Bonus points (so to speak) if you can give a general model-theoretic criterion for a theory to have the largest possible number of models which yields this result as a special case.</p> http://mathoverflow.net/questions/12949/are-there-as-many-real-closed-fields-of-a-given-cardinality-as-i-think-there-are/12952#12952 Answer by Joel David Hamkins for Are there as many real-closed fields of a given cardinality as I think there are? Joel David Hamkins 2010-01-25T18:44:15Z 2010-01-25T18:44:15Z <p>In the countable case, the bound of 2<sup>&omega;</sup> is realized, since any countable real-closed field will contain the rational numbers and fill at most countably many cuts in the rationals with LUBs. But we can arrange that any given cut is filled by a real closed subfield of R containing that real. So there must be 2<sup>&omega;</sup> many non-isomorphic countaable real closed fields.</p> <p>In the general case, because the models have an order, you can easily make this order have different cofinalities, by building elementary chains of different lengths. That is, just use your Lowenheim Skolem construction to add another point on top of the previous model, and continue for &delta; steps. This will produce an elementary extension of size &kappa; whose order has cofinality &delta;, for any regular &delta; up to &kappa;. So this gives many more models, but doesn't quite answer your 2<sup>&kappa;</sup> question. I'm inclined to agree with you and expect that it must be the maximal number for all &kappa; on general grounds.</p> http://mathoverflow.net/questions/12949/are-there-as-many-real-closed-fields-of-a-given-cardinality-as-i-think-there-are/12991#12991 Answer by Marty for Are there as many real-closed fields of a given cardinality as I think there are? Marty 2010-01-26T00:20:18Z 2010-01-26T00:20:18Z <p>Hi Pete!</p> <p>There's been a lot of study of this and similar problems. I believe that Shelah's theorem, from his 1971 paper "The number of non-isomorphic models of an unstable first-order theory" (Israel J. of Math) answers your question about real closed fields in the positive.</p> <p>The best big result on such questions that I know of is in the 2000 Annals paper "The uncountable spectra of countable theories." by Hart, Hrushovski, Laskowski.</p> <p>To answer the question on real closed fields specifically (and somewhat cautiously since I'm not a model-theorist):</p> <p>The theory of real closed fields is a complete first order theory, with countable language. It is an unstable (an easy fact, I think, and <a href="http://en.wikipedia.org/wiki/Stable%5Ftheory" rel="nofollow">explained better on wikipedia</a> than I could explain) theory as well. Hence Shelah's result applies, and the bound $2^\kappa$ is realized as you surmised.</p> <p>Bonus points should go to Shelah (and perhaps also to Hart, Hrushovski, Laskowski, whose paper mentions the result of Shelah and proves other things) for proving that this bound is realized (for uncountable cardinals), <strong>except</strong> for theories $T$ which have <strong>all</strong> of the following properties:</p> <ol> <li>$T$ has infinite models.</li> <li>$T$ is superstable.</li> <li>$T$ has prime models over pairs.</li> <li>$T$ does not have the dimensional order property.</li> </ol> <p>I have no clue what the fourth property means. But there are plenty of non-superstable theories to which Shelah's theorem applies, and hence which realize your bound (for uncountable cardinals).</p> <p>For countable cardinality, I think there are still some open problems about how many non-isomorphic models there can be of a given theory, with cardinality $\aleph_0$.</p> http://mathoverflow.net/questions/12949/are-there-as-many-real-closed-fields-of-a-given-cardinality-as-i-think-there-are/23529#23529 Answer by Dave Marker for Are there as many real-closed fields of a given cardinality as I think there are? Dave Marker 2010-05-05T02:01:32Z 2010-05-05T02:09:38Z <p>For real closed fields this is fairly easy.</p> <p>First show that for any infinite cardinal k there are 2^k nonisomorphic linear orders of cardinality k</p> <p>For example if X is a subset of k let A_x be Q+2+Q if x is in k and Q+3+Q if x is not in X. Let L_X be the sum of the A_x for x in k. It is easy to see that L_X is isomorphic to L_Y if and only if X=Y.</p> <p>If F is a real closed and x and y are infinite element of R we say that x and y are comparable if and only if there are natural numbers m and n such that x is less than y^m and y is less than x^n. The ordering of R induces a linear order L_R of the comparability classes, which we call the ladder of R.</p> <p>Suppose L is a linear order. Let F be the real algebraic numbers. Let R_L be the real closure of the transcendental extension of the real algebraic numbers F(x_l:l\in L) ordered such that if i is less than j then x_i^n is less than x_j for all n. It's not hard to show that the ladder of R_L is isomorphic to L.</p> <p>Thus if we start with nonisomorphic orders A and B then the fields R_A and R_B will be nonisomorphic.</p>