The Surreal numbers satisfy all the field axioms except that its elements constitute a proper class. Is it safe to call it a field? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T11:55:02Z http://mathoverflow.net/feeds/question/115749 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115749/the-surreal-numbers-satisfy-all-the-field-axioms-except-that-its-elements-constit The Surreal numbers satisfy all the field axioms except that its elements constitute a proper class. Is it safe to call it a field? Taliberius 4 2012-12-07T21:49:08Z 2012-12-08T01:57:18Z <p>Do all the field theorems apply to surreal numbers? If fields were redefined so that their elements were allowed to come from an arbitrary class, would the theory look different to an algebraist?</p> http://mathoverflow.net/questions/115749/the-surreal-numbers-satisfy-all-the-field-axioms-except-that-its-elements-constit/115757#115757 Answer by Daniel Briggs for The Surreal numbers satisfy all the field axioms except that its elements constitute a proper class. Is it safe to call it a field? Daniel Briggs 2012-12-07T23:56:23Z 2012-12-07T23:56:23Z <p>In ZFC+I, where I asserts the existence of a strongly inaccessible cardinal, we can create a Grothendieck universe out of that inaccessible cardinal and have a set that satisfies all the properties of the surreal numbers, where quantification is taken over elements of that universe. From the perspective of the outside universe, this would be a field. Since no one has found a contradiction in ZFC+I, it's safe to say at present that the field theorems apply to surreal numbers and that an algebraist cannot detect any essential difference, only relative differences that you probably know such as "This Field contains all totally ordered fields."</p> http://mathoverflow.net/questions/115749/the-surreal-numbers-satisfy-all-the-field-axioms-except-that-its-elements-constit/115761#115761 Answer by Joel David Hamkins for The Surreal numbers satisfy all the field axioms except that its elements constitute a proper class. Is it safe to call it a field? Joel David Hamkins 2012-12-08T00:29:52Z 2012-12-08T01:57:18Z <p>First, let me say that the set/class issue is not a problem to deal with properly, and so one shouldn't be very worried about it. It is true as you say that the surreal numbers No are a proper class, and they do not form a set. So in a purely technical sense, they are not a field. But nevertheless, they do satisfy all the field axioms and have all the usual kinds of structure that one would want in a field, and so one can correctly describe them as a proper class field, or Field as Conway writes it, in much the same way that the class of all ordinals is regarded as a proper class well-order. Although this set/class issue may seem mysterious or irritating, in practice it is a routine matter to handle correctly for those familiar with the set/class distinction.</p> <p>But since you seem particularly interested in what might go wrong, let me suggest on the negative side, one issue that could make a difference is that when dealing with the surreal field No, one will want to strengthen the background set theory from ZFC to GBC, which includes the global axiom of choice, the assertion that there is a proper class well-ordering of the universe. The reason is that it is consistent with ZFC that the surreal numbers do not admit any proper class well-ordering, and actually, the assertion that they have a definable such well-ordering is equivalent to the set-theoretic axiom known as V=HOD, as I proved in my answer to David Feldman's question on a <a href="http://mathoverflow.net/questions/93468/definable-map-from-all-the-ordinals-to-the-surreal-numbers-with-a-dense-image/93485#93485" rel="nofollow">Definable map from all the ordinals to the surreal numbers</a>. So if one wants to undertake algebraic constructions requiring one to have a well-ordering of the field itself, such as finding a proper class maximal ideal inside a particular subring of No, then there could be difficulties undertaking such a construction in ZFC as opposed to GBC. But nevertheless, the theory GBC is conservative over ZFC and one may thereby freely assume the global axiom of choice. (This is used in the various arguments showing that No is universal for class-sized objects, such as the assertion that every class order embeds to a suborder of No.) In particular, in GBC one has a well-ordering of the entire universe, including the surreals, and this situation would address such issues. Much of the theory undertaken by Ehrlich on the surreals, for example, works in GBC as the background theory. Beyond this issue, even in GBC one does not have any sense of a well-ordering of the (meta-class) collection of all class-sized subrings of No, if this were desired for any algebraic construction, and this is the kind of issue that would arise with the set/class issue. </p> <p>But meanwhile, there is also a positive answer. The situation is that if one wants set versions of the surreal numbers, they are abundantly available in increasingly powerful and accurate approximations, which are well-understood and studied. Specifically, we have numerous set-sized approximations to the surreal numbers, simply by considering the set of surreal numbers born before a given ordinal birthday. For any ordinal $\lambda$, let $\text{No}(\lambda)$ be the set of surreal numbers born before $\lambda$. One should regard $\text{No}(\lambda)$ as the version of the surreal numbers as constructed inside the set-theoretic universe $V_\lambda$, which can satisfy increasing fragments of our set theory, as $\lambda$ is chosen to exhibit increasingly strong closure properties.</p> <p>Philip Ehrlich mentioned in his recent talk at the CUNY Logic Workshop that he and Lou van den Dries prove in their article <a href="http://www.ohio.edu/people/ehrlich/EhrlichvandenDries.pdf" rel="nofollow">Fields of surreal numbers with exponentiation</a> the following facts:</p> <ul> <li><p>$\text{No}(\lambda)$ is an additive subgroup of No if and only if $\lambda=\omega^\alpha$ for some ordinal $\alpha$; that is, if and only if $\lambda$ is additively indecomposable.</p></li> <li><p>$\text{No}(\lambda)$ is an additive subring of No if and only if $\lambda=\omega^{\omega^\alpha}$ for some ordinal $\alpha$; that is, if and only if $\lambda$ is multiplicatively indecomposable.</p></li> <li><p>$\text{No}(\lambda)$ is a subfield of No if and only if $\lambda$ is an <a href="http://cantorsattic.info/Epsilon_naught" rel="nofollow">$\epsilon$-number</a>, that is, if and only if $\lambda=\epsilon_\alpha$ for some $\alpha$.</p></li> </ul> <p>These facts are proved by giving a careful analysis of exactly how long it takes to add the inverse of a given surreal number, based on its birthday, and so when the ordinal $\lambda$ is closed under those waiting times, then the resulting $\text{No}(\lambda)$ contains the requisite inverses. </p> <p>Ultimately, we obtain set-sized approximations $\text{No}(\lambda)$ to the surreals by truncating at sufficiently powerful ordinals. Indeed, for any particular natural number $n$, there will be a closed unbounded proper class of ordinals $\lambda$ such that $\text{No}(\lambda)$ has all the same $\Sigma_n$-expressible properties as the full class of surreal numbers No. This can be proved as an immediate consequence of the reflection theorem. So in fact, No is the union of a proper class chain of increasingly elementary subfields $\text{No}(\lambda)$. </p> <p>One can think of the situation as corresponding to the small/large distinction that one finds in category theory with the use of Grothendieck universes, as in Daniel's answer. But in fact one doesn't need a whole Grothendieck universe just to have a subfield, since a mere epsilon number suffices in comparison with an inaccessible cardinal (which are all epsilon numbers). </p>