To clarify the problem, consider the field ${\Bbb R}$ as a field extension of ${\Bbb Q}$ using some sort of Hamel basis. Does there exist a field $F\subsetneq{\Bbb R}$ such that $F(\sqrt{2})={\Bbb R}$ (or $F(x)={\Bbb R}$ for any other irrational $x$)? If $(z_\alpha)_{\alpha<{\frak c}}$ is a basis for ${\Bbb R}$ as a ${\Bbb Q}$-vector space such that $z_0=\sqrt 2$, then $F={\Bbb Q}(z_1,z_2,\dots,z_\alpha,\dots)$ (where all $z_\alpha$ are adjoined for $0\ne \alpha<{\frak c}$) should be a field not containing $\sqrt2$, even though $F(\sqrt2)={\Bbb R}$. (I am using that $\sqrt{2}$ is algebraic of degree $2$ in this argument to prove that $\sqrt{2}\notin F$, although it is probably not necessary.)

My question relates to the process used to find this field. As you can see, I've used a nonconstructive proof using a Hamel basis (and the axiom of choice, implicitly), and this is not philosophically satisfactory for me. Is there an "explicit" proof of existence of such a field $F$, and do its elements have any nice characterization? Is $F$ uniquely determined by the properties $F\subsetneq{\Bbb R}$ and $F(\sqrt{2})={\Bbb R}$? What are the closure properties of $G:={\Bbb R}\setminus F$?

Edit: it seems the construction above doesn't work, and there are good reasons why the $F$ described above doesn't exist. I would like to patch my construction instead by specifying only that $F\subseteq{\Bbb R}$ and $\sqrt2\notin F$, and $F$ is maximal in the sense that there is no proper field extension $F(\alpha)$ satisfying the same properties. Surely this definition will work by Zorn's lemma, although I doubt that such an $F$ is unique. Am I right?

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    $\begingroup$ You may want to replace Hamel basis by a transcendence basis; but then you end up with $[\Bbb R:F]$ being infinite (for example, $\sqrt[2n]2\notin F$ for all $n$, so we can easily manufacture intermediate extensions). $\endgroup$
    – Asaf Karagila
    Jun 5, 2013 at 7:53
  • $\begingroup$ Even though you excluded $z_0$ from the set of generators, you will still get it as an algebraic combination of the other elements of a Hamel basis. $\endgroup$ Jun 5, 2013 at 8:03
  • $\begingroup$ If you try your second route, then you get the problem that $F(\sqrt2)$ won't be all of $\mathbb{R}$. For example, why would that field contain $\root4\of 2$? $\endgroup$ Jun 5, 2013 at 8:30
  • $\begingroup$ @AsafKaragila I noticed my subtle switch from field extensions to vector space extensions in the description of the Hamel basis, but I was hoping that $\sqrt2^{-1}=\frac12\sqrt2$ would assure that you couldn't make $\sqrt2$ from the other $z_\alpha$. I should have used a transcendence basis as you say (didn't know the name or technique), but as the answers below attest, it seems my construction is doomed from the start. I've edited the OP with a slightly different tack. $\endgroup$ Jun 5, 2013 at 8:32
  • $\begingroup$ I have changed the title, which I found confusing ("field restrictions" is not the proper term here, also real numbers should be mentioned). If the result is not appropriate (for instance "algebraic" might be redundant), please improve it. $\endgroup$ Jun 5, 2013 at 8:33

3 Answers 3


Such a field $F$ does not exist.

Assume contrariwise that $F$ is a field with the properties $\sqrt2\notin F$, $F(\sqrt2)=\mathbb{R}$. In that case there exists a non-trivial $F$-automorphism $\sigma$ of $F(\sqrt2)$ with the property $\sigma(z)=z$ for all $z\in F$ and $\sigma(\sqrt2)=-\sqrt2$.

This contradicts the known fact that the field $\mathbb{R}$ has no non-trivial automorphisms. I outline the steps in the argument in case you have not seen them before. Below $\tau$ is an arbitrary automorphism of $\mathbb{R}$.

  1. We have $\tau(q)=q$ for all the rational numbers $q$.
  2. The automorphism $\tau$ maps any square in the field $\mathbb{R}$ to (possibly another) a square.
  3. A real number is a square, iff it is non-negative, so by Step 2 $\tau$ maps any positive real number to a positive real number.
  4. In the field $\mathbb{R}$ we have $x\le y\Leftrightarrow y-x\ge0$. Therefore Step 3 implies that the automorphism $\tau$ is strictly increasing as a real function.
  5. Step 4 implies that $\tau$ is continuous on all of $\mathbb{R}$. Therefore its fixed points form a closed set.
  6. The (topological) closure of $\mathbb{Q}$ is all of $\mathbb{R}$, so combining Steps 1 and 5 shows that $\tau(x)=x$ for all real numbers $x$.
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    $\begingroup$ It should be noted that the axiom of choice is not used in this argument. $\endgroup$
    – Asaf Karagila
    Jun 5, 2013 at 7:51
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    $\begingroup$ I like this answer as being the simplest, conceptually. I find it curious that an algebraic construction is denied by an essentially topological argument, but I suppose that's what makes ${\Bbb R}$ special. I've edited the problem to patch up my construction. $\endgroup$ Jun 5, 2013 at 8:35
  • $\begingroup$ For clarity: The question was edited in response to this answer. Therefore this is not an answer to the latest version, where the requirement $F(\sqrt2)=\Bbb{R}$ was dropped as impossible. I am fine with the edition. Admittedly the present situation may be a bit confusing when there is both an accepted answer and an unanswered question :-) $\endgroup$ Nov 2, 2020 at 7:07

There is a theorem that if $[\bar{F} : F]$ is finite (where $\bar{F}$ is the algebraic closure of $F$), then $\bar{F} = F(\mathbf{i})$. e.g. this means $[\bar{F} : F] \in \{1, 2, \infty\}$.

If there were a field $F$ such that $F(\sqrt{2}) = \mathbb{R}$, then

$$ [\bar{F} : F] = [\mathbf{C} : F] = [\mathbf{C}:\mathbf{R}][\mathbf{R}:F] = 2 [\mathbf{R}:F] $$

All of the above information implies that $[\mathbf{R}:F]=1$, and so $F = \mathbb{R}$.

  • $\begingroup$ Thanks for the reference. Just two questions: (1) is there a specific reason to link to an old version of the WP page (maybe just to ensure the relevant statement does not evaporate over time)? and (2) this is just a statement (number 4) in a long list of equivalent statements, with (I think) no indication of where it comes from, or why this statement even implies the field can be ordered. Does this specific theorem have a name? $\endgroup$ Jun 5, 2013 at 8:21
  • $\begingroup$ @Marc: Yes, the old version was to ensure it doesn't evaporate. And I am invoking the equivalence of #4 and #5. Eyeballing it, #5 -> #3 looks like it would be straightforward, and the ordering is given by "squares are positive". I'm not sure if I've ever actually seen the proof that #4 is equivalent to the other statements, though. (although it might not be too difficult to get #4 -> "squares are positive gives an ordering"... I'm not sure) $\endgroup$
    – user14972
    Jun 5, 2013 at 8:36
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    $\begingroup$ @Marc: Artin and Schreier studied these questions. If you can locate a copy of Michael Rosen's recent book with selected expositions and papers by Artin, I really recommend that. This result is in there, and it is very much accessible to anyone familiar with basic Galois theory. $\endgroup$ Jun 5, 2013 at 8:39

Not only is your construction not philosophically satisfactory, I'm sorry to say it is also just wrong. Though $z_0=\sqrt 2$ is $\Bbb Q$-linearly independent of the other $z_i$, it will actually be in the field $F$, in other words it is a quotient of polynomial expressions in the other $z_i$.

In fact there are no fields such that $\Bbb R$ is a finite degree (other than$~1$) algebraic extension of it. As Jyrki Lahtonen explained the existence of such a field would imply that $\Bbb R$ as a field has non-trivial automorphisms, but it hasn't. This is quite easy to prove directly: since the real numbers that are a square (of a real number) are precisely the non-negative ones, and field automorphism must stabilise the set of positive numbers, hence be an automorphism of ordered fields, and since the rational numbers are (point-wise) fixed under any automorphism, the identity automorphism of $\Bbb R$ is the only possibility.

I think there is an even more sweeping statement (but since I have no idea about its proof I hope that somebody will either correct me or provide a reference) which says:

If any algebraically closed field is a finite degree extension of any other field, then that degree is either $1$ or $2$.

Or equivalently (I guess), the automorphism group of an algebraically closed field can have elements of finite order $1$ or $2$ only. By the way this does not say that complex conjugation is the only automorphism of $\Bbb C$ of order$~2$, there are many of them (giving rise to "strange copies" of $\Bbb R$ inside $\Bbb C$). However any two such elements generate an infinite subgroup of $\operatorname{Aut}(\Bbb C)$.

Since the degree of $\Bbb C$ over your hypothetic field $F$ would be twice the degree of $\Bbb R$ over $F$, it should be clear that the above statement rules out the existence of such$~F\subsetneq\Bbb R$.

  • $\begingroup$ I see that Hurkyl provided the reference I asked for even before my answer was posted! $\endgroup$ Jun 5, 2013 at 8:07

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