This is a follow-up to the question on field reductions.

(EDIT: Originally this question used the notation $\{\mathbb{R}(\setminus a)\}$, but now uses $\mathbb{R}(\setminus a)$ instead.)

There isn't a unique largest subfield of $\mathbb{R}$ not containing $a$, so let $\mathbb{R}(\setminus a)$ denote the set of maximal subfields of $\mathbb{R}$ not containing $a$,

EDIT: The question "What is the cardinality of $\mathbb{R}(\setminus a)$ ?" has now been asked on Mathoverflow.

From the previous question if $F\in \mathbb{R}(\setminus a)$ then $F$ is uncountable, but what is the size of $\mathbb{R}\setminus F$ the complement of $F$ in $\mathbb{R}$

Each $F \in \mathbb{R}(\setminus a)$ should be arrived at by removing elements of $\mathbb{R}$, but if you start with $\mathbb{R}$ and remove $a$ and then remove more elements until you get a field then surely you arrive at a single field, so where do the multiple maximal fields come from ? - It must be because you get different fields depending on the order in which elements are removed from $\mathbb{R}$, in which case what is the most natural order to remove elements to get a definition of a single canonical subfield $\mathbb{R}(\setminus a)$ of $\mathbb{R}$ that doesn't contain $a$, that we could call 'the' field reduction of $\mathbb{R}$ by $a$ ?

EDIT: Given Arturo Magidin's comments and answer, if we abandon the idea of a distinguished element in the abstract/in general, then since the notation $\{\mathbb{R}(\setminus a)\}$ looks like a set containing a single element called $\mathbb{R}(\setminus a)$, it makes more sense to drop the brackets and just write $\mathbb{R}(\setminus a)$ on the understanding that this is a set of field reductions i.e. a set of subfields not a single subfield.

From Pete L. Clark's answer to the previous question, if $F(a)=\mathbb{R}$ then $F=\mathbb{R}$, so let's define the notion of a super-extension of $F$ by $a$, $F(a..)=R$ being the sequence of field extensions to get to $\mathbb{R}$ from $F\in\mathbb{R}(\setminus a)$. More generally, let $@K(\setminus a)$ be an element of $K(\setminus a)$ then $@K(\setminus a)(a..)=K$.

We could also reduce $K$ by a set $A$, with $K(\setminus A)$ being the set of maximal subfields that don't contain any elements from $A$, and again write the super-extension $@K(\setminus A)(A..)=K$.

Let $\mathbb{A}'$ be the set of non-rational algebraic numbers.

What is the cardinality of $\mathbb{R}(\setminus \mathbb{A}')$ ?

  • $\begingroup$ @Msw: Please try to use some formatting. $\endgroup$ Nov 28, 2010 at 12:11
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    $\begingroup$ @Msw: It's not the order of removal that matters, is the choice of what you remove. Think about the Klein 4-group: if you want to obtain subgroups that don't contain $(1,1)$, then you remove $(1,1)$, and then you have a choice on whether to remove $(1,0)$ or remove $(0,1)$ to get a subgroup; different choices lead to different subgroups. To get a canonical choice, you would need at least to well-order $\mathbb{R}$ to know what to remove at each step, but well-ordering $\mathbb{R}$ requires AC: there isn't even an explicit well-ordering, let alone a canonical one. $\endgroup$ Nov 28, 2010 at 12:20
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    $\begingroup$ @Arturo: Perhaps it is more accurate to say that consistently there is not an explicit well-ordering. There are certainly many models where there are explicit well-orderings of ${\mathbb R}$, but their complexity (there are several natural ways of measuring this) varies depending on the specific model. $\endgroup$ Nov 28, 2010 at 23:25
  • $\begingroup$ @Andres Caicedo: fair enough. Of course, we would have to agree on a specific model to use, and a specific well-ordering within that model (if there is one explicit well-ordering, surely there are many (infinitely many?)), and then try to proceed from there. In any case, it's important to note that it is not the "order" so much as the choice of what to "remove" that matters. The fact that the very existence of such maximal fields relies on Zorn's Lemma seems to me to pretty much kill any chance of a consistent, natural choice for something to be called the field reduction. $\endgroup$ Nov 28, 2010 at 23:30
  • $\begingroup$ @Msw: The attempt at defining "super-extension" seems to me to be rather poor; first, the definition is contextual but the nomenclature does not reflect it (you have to specify what the "target" field is); second, you again use the singular definite article ("the sequence of field extensions") when there is more than one such object and no way to choose among the many different ones. Thirdly, just where exactly does it lead anyway? What you know is that if $F\in K(\setminus a)$, then for any $b\in K\setminus F$, you have $a\in F(b)$. That's about it. $\endgroup$ Nov 29, 2010 at 18:11

1 Answer 1


I'll place something as an answer in the interest of not having questions with no answers.

I don't know what the cardinality of $\{\mathbb{R}(\setminus a)\}$ is, but I would be surprised if it is not uncountable. I'll see if I can think about it some in the next couple of days.

As for your final paragraph, the different fields come not from the order in which you would remove elements, but rather on the choices you make as to what to remove at each (of infinitely many) steps. As I noted in the comments, think of the similar process whereby we look for subgroups of a given group that are maximal among those that do not contain a given element. If we start with the Klein $4$-group, $C_2\times C_2$, and remove the element $a=(1,1)$, then we get two different choices of what other element to remove in order to get a group: we can remove either $(1,0)$ or $(0,1)$; each of the choices will lead to a different subgroup.

Similarly, consider for instance the case where $a=\sqrt{2}$; then as noted in your previous question, a maximal subfield of $\mathbb{R}$ that does not contain $a$ may contain either $\sqrt{3}$ or $\sqrt{6}$, but not both, and you get one field that is maximal and does contain $\sqrt{3}$ (simple application of Zorn's Lemma), and one that is maxiimal and does contain $\sqrt{6}$ (ditto); so with your process you can either remove $\sqrt{3}$ at some point, or remove $\sqrt{6}$ at some point, and these choices lead to different maximal subfields.

The very word "choices" suggests how the Axiom of Choice is playing a role in this "top-down" construction of elements of $\{\mathbb{R}(\setminus a)\}$ (in the "bottoms-up" approach, it plays a role via Zorn's Lemma). In order to be able to obtain something one could call the "field reduction", we would need some procedure or algorithm that determines which choice we make at any stage. One way would be by endowing $\mathbb{R}$ with a well-ordering, and then at each step we can consider the set of all elements we can remove; if it is empty, we are done; if it is not empty, we remove the least element of the set. But, of course, well-ordering $\mathbb{R}$ is a bit of a problem in and of itself: there is no consistently agreed-upon well-ordering of $\mathbb{R}$ (and in some cases, not even an explicit one). We could try to pick a particular model in which there are explicit well-orderings (as mentioned by Adres Caicedo in the comments, such models exist), and a particular well-ordering in there, and then define "the" field reduction, but that seems to be as much of a problem in and of itself: which model, and which order, and why? Perhaps there are other mechanisms whereby one could try to establish a distinguished element in $\{\mathbb{R}(\setminus a)\}$ (or perhaps, a distinguished element for specific instances of $a$), but it seems to me to be a hopeless task in the abstract/in general.

Added and edited: Recall that any field extension $K/k$ can be broken up as $K/F/k$, where $K/F$ is algebraic and $F/k$ is a purely transcendental extension. In particular, if $T$ is a transcendence basis for $K$ over $k$, then $k(T)$ will be a maximal subfield of $K$ that contains $k$ and does not contain any elements of $K$ that are algebraic over $k$ but not in $k$. So for any field extension $K/k$, if we let \begin{equation*} \mathbb{A}_{K/k} =\bigl\{ a \in K\ \bigm|\ \mbox{$a$ is algebraic over $k$, $a\notin k$}\bigr\}, \end{equation*} then $$K(\setminus \mathbb{A}_{K/k}) = \{ k(T)\ |\ \mbox{$T$ is a transcendence basis for $K$ over $k$}\}.$$ If $K$ is algebraic over $k$, then you just get $k$, of course; if $K$ is purely transcendental over $k$, then $\mathbb{A}_{K/k}$ is empty, so you get $K$.

I had originally written that the set would always be singleton, but this is not the case. If $T$ is any transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$, you can modify an element of $T$ by multiplying it by a non-rational algebraic number and you get a slightly different subfield. For example, fix $t\in T$ and replace $t$ with $\sqrt{2}t$ to obtain a new set $T'$. This is still a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$; but whereas $\sqrt{2}t\in \mathbb{Q}(T')$, it does not lie in $\mathbb{Q}(T)$. If it did, then we would immediately get $\sqrt{2}\in\mathbb{Q}(T)$, contradicting the fact that $\mathbb{Q}(T)$ is purely transcendental over $\mathbb{Q}$. You can do the same thing taking arbitrary subsets of $T$ and multiplying them through by a nonrational algebraic number. Since $|T|=\mathfrak{c}$, this gives at least $2^{\mathfrak{c}}$ distinct elements in $\mathbb{R}(\setminus\mathbb{A}')$, and since that is the number of subsets of $\mathbb{R}$, it follows that the cardinality of $\mathbb{R}(\setminus\mathbb{A}')$ is $2^\mathfrak{c}$.

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    $\begingroup$ This answer accords with my thoughts on the question. Especially, I also think that the set of "field reductions" $\mathbb{R}(\setminus \sqrt{2})$ is uncountably infinite (presumably of continuum cardinality), but I haven't seriously thought about it. Note that the argument I gave in the comments and you mention above easily extends to show that it is at least infinite. The question feels "ultrafiltery" to me, so I'm leaving it for a little while to those who know a lot about such things. $\endgroup$ Nov 28, 2010 at 23:58
  • $\begingroup$ Let $\mathbb{Q}\# = \mathbb{Q} \setminus\{0,1\}$, and $\mathbb{A}\# = \mathbb{A} \setminus\{0,1\}$ be the sets of rational numbers and the algebraic numbers minus the additive and multiplicative identities. Then consider the field reductions TQ=$\mathbb{R}(\setminus\mathbb{Q}\#)$ and TA=$\mathbb{R}(\setminus\mathbb{A}\#)$. What are the cardinalities of TQ and TA ? Maybe this is easier to calculate than the cardinality of $\mathbb{R}(\setminus\sqrt 2)$. $\endgroup$
    – Msw
    Nov 29, 2010 at 8:37
  • $\begingroup$ @Msw: Putting a long sequence of comments such as these is not very conducive to dialogue. I know you cannot edit old comments, but you can always remove old comments and coalesce in a new one. $\endgroup$ Nov 29, 2010 at 15:29
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    $\begingroup$ @Msw: You keep talking about the "field reductions" as if there was a distinguished one. Aren't we clear by now that in general there isn't one? So either you have to talk about "a" field reduction, and accept that they may be very different depending on which one you take, or else establish that for the particular case you are looking at there is a single "field reduction". As for $\mathbb{R}(\setminus\mathbb{Q}\#)$, note that any subfield of $\mathbb{R}$ must contain $\mathbb{Q}$, so there is no subfield of $\mathbb{R}$ that does not contain $\mathbb{Q}\#$. $\endgroup$ Nov 29, 2010 at 15:31
  • $\begingroup$ OK, in future I'll merge comments like that. In the above comments when I said "the field reductions" I meant the sets of field reductions. About $\mathbb{Q}\#$ - of course, silly me, 1+1=2, 1 div 2 =1/2 etc, And so we can't remove all of $\mathbb{A}\#$. $\endgroup$
    – Msw
    Nov 29, 2010 at 15:44

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