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1. • CommentRowNumber1.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 10th 2011
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexI added a little bit of material to [[ordered field]], namely that a field is orderable iff it is a real field (i.e., $-1$ is not a sum of squares). More importantly, at [[real closed field]], I have addressed an old query of Colin Tan: > Colin: Is it true that real closure is an adjoint construction to the forgetful functor from real closed fields to orderable fields? by writing out a proof (under Properties) that indeed the forgetful functor from category of real closed fields and field homomorphisms to the category of real fields and field homomorphisms has a left adjoint (the real closure). Therefore I am removing this query from that page over to here.

   I added a little bit of material to ordered field, namely that a field is orderable iff it is a real field (i.e., $-1$ is not a sum of squares). More importantly, at real closed field, I have addressed an old query of Colin Tan:

   Colin: Is it true that real closure is an adjoint construction to the forgetful functor from real closed fields to orderable fields?

   by writing out a proof (under Properties) that indeed the forgetful functor from category of real closed fields and field homomorphisms to the category of real fields and field homomorphisms has a left adjoint (the real closure). Therefore I am removing this query from that page over to here.

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeAug 10th 2011
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   Author: Mike Shulman
   Format: MarkdownItexThanks! I am a little confused that the domain of this adjoint is the category of *orderable* fields rather than the category of *ordered* fields. If an orderable field F admits two embeddings into real-closed fields $\bar{F}_1$ and $\bar{F}_2$ which induce different orderings on F, then it seems that these two embeddings would not be isomorphic under F, since any isomorphism of real-closed fields is order-preserving. Does an orderable field only admit one order that can be induced by an embedding into a real-closed field?

   Thanks!

   I am a little confused that the domain of this adjoint is the category of orderable fields rather than the category of ordered fields. If an orderable field F admits two embeddings into real-closed fields $\bar{F}_1$ and $\bar{F}_2$ which induce different orderings on F, then it seems that these two embeddings would not be isomorphic under F, since any isomorphism of real-closed fields is order-preserving. Does an orderable field only admit one order that can be induced by an embedding into a real-closed field?

3. • CommentRowNumber3.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 11th 2011
   • (edited Aug 11th 2011)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexI'll look it over again, to see whether I made an avoidably dumb mistake. It might be less confusing just to call the domain the category of *real* fields (with ordinary field homomorphisms as maps), and the codomain the category of real closed fields (again with field homomorphisms as maps). You're right that those two hypothetical embeddings would not be isomorphic as *ordered fields*, but then I'm not demanding that much in the statement. The place where ordering enters is that we know the extension map $\widebar{F} \to R$ between real closed fields will automatically preserve their canonical orders, and that will fix how the map is supposed to behave on roots/adjoined elements when restricted to finite real extensions of the ground field $F$. **Edit:** I did find a mistake (forgot a bar over one of the $F$'s), and I did a little rewriting which hopefully makes the proof clearer. Let me know if you find anything else wrong. I may sleep on it and come back to it tomorrow.

   I’ll look it over again, to see whether I made an avoidably dumb mistake.

   It might be less confusing just to call the domain the category of real fields (with ordinary field homomorphisms as maps), and the codomain the category of real closed fields (again with field homomorphisms as maps). You’re right that those two hypothetical embeddings would not be isomorphic as ordered fields, but then I’m not demanding that much in the statement. The place where ordering enters is that we know the extension map $\widebar{F} \to R$ between real closed fields will automatically preserve their canonical orders, and that will fix how the map is supposed to behave on roots/adjoined elements when restricted to finite real extensions of the ground field $F$.

   Edit: I did find a mistake (forgot a bar over one of the $F$’s), and I did a little rewriting which hopefully makes the proof clearer. Let me know if you find anything else wrong. I may sleep on it and come back to it tomorrow.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeAug 11th 2011
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   Author: Mike Shulman
   Format: MarkdownItexStill confused, but can't quite articulate a question. I'm not claiming there's anything wrong with your proof, just that I don't really understand real-closure. Maybe if I think about an example it will help. What is an example of an orderable field that admits more than one order?

   Still confused, but can’t quite articulate a question. I’m not claiming there’s anything wrong with your proof, just that I don’t really understand real-closure. Maybe if I think about an example it will help. What is an example of an orderable field that admits more than one order?

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeAug 11th 2011
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   Author: Mike Shulman
   Format: MarkdownItexAh, I have a question! I don't understand how you know that f has n roots in R. You say that "R, being real closed, has no nontrivial real algebraic extensions", but how do you know that the algebraic extension in which f splits would be real? And to answer my question, here's an example (I think) of an orderable field that admits more than one order: $\mathbb{Q}[\sqrt{2}] = \mathbb{Q}[x]/(x^2-2)$. You can take $x\gt 0$ or $x\lt 0$. So we have two field embeddings of $\mathbb{Q}[\sqrt{2}]$ into $\overline{\mathbb{Q}}$, but no automorphism of $\overline{\mathbb{Q}}$ which commutes with them. I think.

   Ah, I have a question! I don’t understand how you know that f has n roots in R. You say that “R, being real closed, has no nontrivial real algebraic extensions”, but how do you know that the algebraic extension in which f splits would be real?

   And to answer my question, here’s an example (I think) of an orderable field that admits more than one order: $\mathbb{Q}[\sqrt{2}] = \mathbb{Q}[x]/(x^2-2)$. You can take $x\gt 0$ or $x\lt 0$. So we have two field embeddings of $\mathbb{Q}[\sqrt{2}]$ into $\overline{\mathbb{Q}}$, but no automorphism of $\overline{\mathbb{Q}}$ which commutes with them. I think.

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 11th 2011
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexOkay, the first question in 5. does point to a problem with the proof. But not an insurmountable one. I'll fix it up today. Your answer to your own question in 5 is exactly right.

   Okay, the first question in 5. does point to a problem with the proof. But not an insurmountable one. I’ll fix it up today.

   Your answer to your own question in 5 is exactly right.

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 11th 2011
   • (edited Aug 11th 2011)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexOh, oh, oh. You're absolutely right, this proof is seriously effed. I'm rolling back. Man, I am seriously annoyed with myself. **Edit:** Well, hopefully I salvaged it, but maybe you (or anyone for that matter) should take a look. I'm not proud of what I've written, but let's see whether it holds up.

   Oh, oh, oh. You’re absolutely right, this proof is seriously effed. I’m rolling back.

   Man, I am seriously annoyed with myself.

   Edit: Well, hopefully I salvaged it, but maybe you (or anyone for that matter) should take a look. I’m not proud of what I’ve written, but let’s see whether it holds up.

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeAug 11th 2011
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   Author: Mike Shulman
   Format: MarkdownItexThanks! I think I believe this version. This is still an intriguing theorem, to me -- it means that real-closure is somehow very different from algebraic closure. I don't quite understand at an intuitive level why $\sqrt{-1}$ should make so much of a difference.

   Thanks! I think I believe this version. This is still an intriguing theorem, to me – it means that real-closure is somehow very different from algebraic closure. I don’t quite understand at an intuitive level why $\sqrt{-1}$ should make so much of a difference.

9. • CommentRowNumber9.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 12th 2011
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   Author: Todd_Trimble
   Format: MarkdownItexI imagine that the model theorists would have a lot to say about this phenomenon. I know they have studied both real closures and algebraic closures _a lot_, and I think they use them to illustrate aspects of some very general theory (classification and stability theory). There was a really nice Book Review article by Gregory Cherlin in the BAMS, sometime in the 90's, that had a very nice overview of classification theory. I will see if I can dig it up and reproduce some of it here. But one general point I vaguely remember is that if the theory introduces a definable asymmetry (such as a total order in the theory of real closed fields, where the definable relation $a \lt b$ is asymmetric), then one can generally expect that the classification of models of the theory will not be very tractable. We may be getting a whiff of that here: a real field may have many nonisomorphic real closures (non-isomorphic as extension fields), because the ordering imposes a certain rigidity which obstructs the existence of an isomorphism. Whereas the theory of algebraically closed fields doesn't introduce a definable asymmetry, and any isomorphism between fields $$F \to F'$$ extends to an isomorphism between chosen algebraic closures in which these fields are embedded $$\widebar{F} \to \widebar{F'}$$ I guess this might have something to do with amalgamations that model theorists are always talking about.

   I imagine that the model theorists would have a lot to say about this phenomenon. I know they have studied both real closures and algebraic closures a lot, and I think they use them to illustrate aspects of some very general theory (classification and stability theory).

   There was a really nice Book Review article by Gregory Cherlin in the BAMS, sometime in the 90’s, that had a very nice overview of classification theory. I will see if I can dig it up and reproduce some of it here. But one general point I vaguely remember is that if the theory introduces a definable asymmetry (such as a total order in the theory of real closed fields, where the definable relation $a \lt b$ is asymmetric), then one can generally expect that the classification of models of the theory will not be very tractable. We may be getting a whiff of that here: a real field may have many nonisomorphic real closures (non-isomorphic as extension fields), because the ordering imposes a certain rigidity which obstructs the existence of an isomorphism. Whereas the theory of algebraically closed fields doesn’t introduce a definable asymmetry, and any isomorphism between fields

   $$F \to F'$$

   extends to an isomorphism between chosen algebraic closures in which these fields are embedded

   $$\widebar{F} \to \widebar{F'}$$

   I guess this might have something to do with amalgamations that model theorists are always talking about.

10. • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 12th 2011
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    Author: Todd_Trimble
    Format: MarkdownItexHere is a <a href="http://www.ams.org/journals/bull/1989-20-02/S0273-0979-1989-15757-1/S0273-0979-1989-15757-1.pdf">link</a> to the article by Cherlin. It's great. The stuff I was vaguely trying to recollect is explained around page 187, and has to do with the model-theoretic notion of independence.

    Here is a link to the article by Cherlin. It’s great. The stuff I was vaguely trying to recollect is explained around page 187, and has to do with the model-theoretic notion of independence.

11. • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeAug 20th 2011
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    Author: Mike Shulman
    Format: MarkdownItexThanks for the link! This is fascinating and totally different from anything I think about all day; I wish I had time to learn all about it. What I was thinking about seems a little different from this, though: what struck me is the fact that real-closures (of ordered fields) have a universal property, whereas algebraic closures do not. As a category-theorist, this seems to make real-closures seem *better* behaved. Whereas it sounds like the model theoretic point of view is that the algebraically closed fields are the "better" behaved ones, because their lack of a definable asymmetry makes them "classifiable"?

    Thanks for the link! This is fascinating and totally different from anything I think about all day; I wish I had time to learn all about it. What I was thinking about seems a little different from this, though: what struck me is the fact that real-closures (of ordered fields) have a universal property, whereas algebraic closures do not. As a category-theorist, this seems to make real-closures seem better behaved. Whereas it sounds like the model theoretic point of view is that the algebraically closed fields are the “better” behaved ones, because their lack of a definable asymmetry makes them “classifiable”?

12. • CommentRowNumber12.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 20th 2011
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexRight. I guess the classification theory aims to delineate those theories whose models can be classified up to isomorphism by a reasonably small number of invariants. In the archetypal case of algebraically closed fields of uncountable cardinality, this would be just the characteristic and transcendence degree. Whereas real closed fields exist in great profusion, and I think the moral is supposed to be that the minimal number of invariants needed to classify real closed fields of a given cardinality up to isomorphism is "worst-case scenario" or something. I guess this sort of makes sense when you consider how easy it is to build real closed valuation fields with a given value group, like [[Hahn series]] whose value group is a given ordered divisible group. Of course, a category theorist might perceive and want to redress an imbalance here: the focus on "up to isomorphism" is throwing away just what sorts of automorphisms we are talking about. In other words, sure you can easily classify uncountable algebraically closed fields, but their automorphisms exist in great profusion, whereas the exact opposite is true of real closed fields! Is there some sort of general reciprocity principle at work here? This is not to diss the model theorists though, who have done tons of extremely impressive work, obviously. I do wish there were closer connections between model theorists and category theorists, which would reach into the core concerns of stability theory and classification theory. I am guessing guys like Makkai know this stuff actually pretty well, but there ought to be a book "Model Theory for Category Theorists", which goes in a different direction than say the work on accessible categories. :-)

    Right. I guess the classification theory aims to delineate those theories whose models can be classified up to isomorphism by a reasonably small number of invariants. In the archetypal case of algebraically closed fields of uncountable cardinality, this would be just the characteristic and transcendence degree. Whereas real closed fields exist in great profusion, and I think the moral is supposed to be that the minimal number of invariants needed to classify real closed fields of a given cardinality up to isomorphism is “worst-case scenario” or something. I guess this sort of makes sense when you consider how easy it is to build real closed valuation fields with a given value group, like Hahn series whose value group is a given ordered divisible group.

    Of course, a category theorist might perceive and want to redress an imbalance here: the focus on “up to isomorphism” is throwing away just what sorts of automorphisms we are talking about. In other words, sure you can easily classify uncountable algebraically closed fields, but their automorphisms exist in great profusion, whereas the exact opposite is true of real closed fields! Is there some sort of general reciprocity principle at work here?

    This is not to diss the model theorists though, who have done tons of extremely impressive work, obviously. I do wish there were closer connections between model theorists and category theorists, which would reach into the core concerns of stability theory and classification theory. I am guessing guys like Makkai know this stuff actually pretty well, but there ought to be a book “Model Theory for Category Theorists”, which goes in a different direction than say the work on accessible categories. :-)

13. • CommentRowNumber13.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 20th 2011
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    Author: Todd_Trimble
    Format: MarkdownItexThe above remark prompted me to add a section on "Infinites and infinitesimals" to [[real closed field]].

    The above remark prompted me to add a section on “Infinites and infinitesimals” to real closed field.

14. • CommentRowNumber14.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 21st 2011
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    Author: Todd_Trimble
    Format: MarkdownItexToby, why did you change $\widebar{\mathbb{Q}}_{real}$ to $\widebar{\mathbb{Q}}$? I think that's just wrong; $\widebar{\mathbb{Q}}$ is typically used to denote the algebraic closure, not the real closure.

    Toby, why did you change $\widebar{\mathbb{Q}}_{real}$ to $\widebar{\mathbb{Q}}$? I think that’s just wrong; $\widebar{\mathbb{Q}}$ is typically used to denote the algebraic closure, not the real closure.

15. • CommentRowNumber15.
    • CommentAuthorTobyBartels
    • CommentTimeAug 21st 2011
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    Author: TobyBartels
    Format: MarkdownItexIn the main part of the article, $\overline{F}$ is introduced as notation for the real closure of an ordered field, so I followed that to be consistent. It seems to me that the overline denotes a variety of completions and closures, so we should be able to let it mean whatever is most appropriate. Another possibility would be to use $\overline{F}_{real}$ throughout the article (although I would suggest something smaller, such as $\overline{F}_r$).

    In the main part of the article, $\overline{F}$ is introduced as notation for the real closure of an ordered field, so I followed that to be consistent. It seems to me that the overline denotes a variety of completions and closures, so we should be able to let it mean whatever is most appropriate.

    Another possibility would be to use $\overline{F}_{real}$ throughout the article (although I would suggest something smaller, such as $\overline{F}_r$).

16. • CommentRowNumber16.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 21st 2011
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    Author: Todd_Trimble
    Format: MarkdownItexWell, I have to admit you're right -- that *was* the earlier-introduced notation. I'll add a little note of warning.

    Well, I have to admit you’re right – that was the earlier-introduced notation. I’ll add a little note of warning.

17. • CommentRowNumber17.
    • CommentAuthorMike Shulman
    • CommentTimeAug 21st 2011
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    Author: Mike Shulman
    Format: MarkdownItexIs there anything interesting that can be said about the value Group of the surreals?

    Is there anything interesting that can be said about the value Group of the surreals?

18. • CommentRowNumber18.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 21st 2011
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    Author: Todd_Trimble
    Format: MarkdownItexMike, I believe the answer is: it's the additive group of the surreals. See theorem 21, page 33, of On Numbers and Games, where each surreal has a unique normal form $\sum_{\beta \lt \alpha} r_\beta \omega^{y_\beta}$ where $\alpha$ is an ordinal, the $r_\beta$ are real numbers, and the $y_\beta$ form a decreasing sequence of surreals. In fact, this normal form theorem is used to prove that the surreals form a real closed field.

    Mike, I believe the answer is: it’s the additive group of the surreals. See theorem 21, page 33, of On Numbers and Games, where each surreal has a unique normal form $\sum_{\beta \lt \alpha} r_\beta \omega^{y_\beta}$ where $\alpha$ is an ordinal, the $r_\beta$ are real numbers, and the $y_\beta$ form a decreasing sequence of surreals. In fact, this normal form theorem is used to prove that the surreals form a real closed field.

19. • CommentRowNumber19.
    • CommentAuthorMike Shulman
    • CommentTimeAug 23rd 2011
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexThat's cute, and makes sense; thanks!

    That’s cute, and makes sense; thanks!

20. • CommentRowNumber20.
    • CommentAuthorSamuel Adrian Antz
    • CommentTimeFeb 17th 2024
    • PermaLink
    Author: Samuel Adrian Antz
    Format: MarkdownItexLinked new page for [[Field]], the [[category]] of [[fields]]. <a href="https://ncatlab.org/nlab/revision/diff/real+closed+field/21">diff</a>, <a href="https://ncatlab.org/nlab/revision/real+closed+field/21">v21</a>, <a href="https://ncatlab.org/nlab/show/real+closed+field">current</a>

    Linked new page for Field, the category of fields.

    diff, v21, current