# Is $\mathbf{C}$ the algebraic closure of any field other than $\mathbf{R}$?

It seems to me (intuitively) that there should be no other fields whose algebraic closure is $\mathbf{C}$, even though I have no reason to believe it. The facts I've been using to formulate an argument are $[\mathbf{C}\mathbin{:}\mathbf{R}]=2$ and $\mathbf{R}$ is the only field with the usual analytic properties. I mean, it seems that for the complexes to be even defined we need to reference the analytic properties of the reals. Also, we know that such a field would have to be uncountable, right?

This question might end up being trivial, but any information or references would be appreciated.

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Why did you change $\Bbb{R,C}$ to $\bf R,C$? – Asaf Karagila Feb 18 '13 at 21:31
@AsafKaragila I'm having an internal struggle as to which of \mathbb and \mathbf is more aesthetically pleasing. Being required to do homework in $\LaTeX$ has made me worry about these things more than I should. – Daenerys Naharis Feb 19 '13 at 7:58
@Joseph: I'm not sure if this is what Asaf was getting at, but I have come across a convention in logic that $\mathbb{R}$ is the set, whereas $\mathbf{R}$ is the entire structure (with its arithmetic operations, distinguished zero and unity, etc.). – Isaac Solomon Jan 3 '14 at 23:56

Let $S$ be a transcendence basis for $\mathbf{C}$ over $\mathbf{Q}$. Then $\mathbf{C}$ is the algebraic closure of $\mathbf{Q}(S)$. You can choose $S$ to contain any transcendental number, so take one in $\mathbf{C}\setminus\mathbf{R}$, such as $\pi\sqrt{-1}$, and then $\mathbf{Q}(S)\neq\mathbf{R}$. Also $\mathbf{Q}(S)$ can't contain $\sqrt{-1}$, so $\mathbf{Q}(S)\neq\mathbf{C}$.

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Yes, but one should also argue that $\mathbb{Q}(S)$ is not isomorphic to $\mathbb{R}$. – Martin Brandenburg Feb 15 '13 at 4:01
Dear @Martin, But this can't be because $\mathbf{R}$ is not purely transcendental over $\mathbf{Q}$, right? – Keenan Kidwell Feb 15 '13 at 11:36
Dear Keenan, @Martin is right: there is something to be proved. Your idea is correct: $\mathbb R$ is not purely transcendental over $\mathbb Q$ because $\mathbb Q$ is algebraically closed in any of its purely transcendental extensions, whereas of course there are many numbers in $\mathbb R\setminus \mathbb Q$ algebraic over $\mathbb Q$. – Georges Elencwajg Feb 15 '13 at 12:39
The OP did not say up to isomorphism... Just any other field. – GEdgar Mar 3 '13 at 17:57

Consider a transcendental extension of $\mathbb C$, $\Bbb C(t)$. Since $\sqrt t\notin\Bbb C(t)$ it is not algebraically closed and therefore the fields are different. Clearly $\Bbb C(t)$ is not isomorphic to $\Bbb R$ as well.

The algebraic closure of $\Bbb C(t)$ is of cardinality $2^{\aleph_0}$ and is therefore isomorphic to the complex numbers. This follows from the fact that the theory of algebraically closed fields in a fixed characteristics is categorical for uncountable cardinalities, that is to say once we chose the characteristics of the field there is one model up to isomorphism. So every algebraically closed field of characteristics zero whose cardinality is $2^{\aleph_0}$ must be isomorphic to the complex numbers.

Other examples of non-$\Bbb R$ fields whose algebraic closure is isomorphic to $\Bbb C$ include the $p$-adic numbers, and any other field of characteristic zero and cardinality continuum.

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So does this mean that there are subfields of $\mathbf{C}$ which are isomorphic to $\mathbf{C}(t)$? And if so is there any explicit construction of such objects? – Daenerys Naharis Feb 20 '13 at 6:25
@Joseph: There is no explicit construction to my knowledge, but I'm not too sure about that. – Asaf Karagila Feb 20 '13 at 14:49

Perhaps this is too obvious, but: Itself?

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@YACP: If you do not like the answer, downvote it and move on. – Eric Naslund Feb 15 '13 at 22:29
@YACP The question is: "is $\mathbb C$ the algebraic closure of a field other than $\mathbb R$?" Now answer me this: What is the algebraic closure of $\mathbb C$? – kahen Feb 15 '13 at 23:54

there should be no other fields whose algebraic closure is C,

Artin and Schreier proved, approximately one century ago, that the only situation where an algebraically closed field is a finite degree extension of a subfield is the degree 2 extension of a real-closed field (one in which every sum of squares is nonzero, every odd degree polynomial has a root, and every sum of squares has a square root). In logic terms, up to elementary equivalence, the extension of $R$ to $C$ is unique. Without the finite degree condition, it is very non-unique.

for the complexes to be even defined we need to reference the analytic properties of the reals.

Yes, but it is possible for very different fields to be elementarily equivalent, which is enough for many algebraic purposes.

Also, we know that such a field would have to be uncountable, right?

Yes, algebraic closure does not increase the cardinality of an infinite field.

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Let me be clear from the outset that I am assuming the Axiom of Choice.

There are $2^{c} = 2^{2^{\aleph_0}}$ isomorphism classes of subfields $R$ of $\mathbb{C}$ with $[\mathbb{C}:R] = 2$. It follows that $\operatorname{Aut} \mathbb{C}$ has $2^c$ orbits on the set of index $2$ subfields of $\mathbb{C}$. [Note: this contradicts the last line of JSchlather's answer and David Speyer's comment on it.]

It follows from the answers to this old MO question of mine that there are $2^c$ isomorphism classes of real-closed fields of continuum cardinality. By Artin-Schreier, each of these fields $R$ is a degree $2$ subfield of its algebraic closure $C$. Since $C$ is an algebraically closed field of characteristic $0$ and continuum cardinality, it is isomorphic to the complex field $\mathbb{C}$. Composing $R \hookrightarrow C \cong \mathbb{C}$ realizes $R$ as an index $2$ subfield of $\mathbb{C}$.

Although the argument that there are the largest conceivable number of real-closed fields of continuum cardinality is rather technical, it is easier to see that there must be one other than $\mathbb{R}$ and thus that $\mathbb{R}$ is not unique up to isomorphism among index $2$ subfields of $\mathbb{C}$. Namely, we can take an ordering on $\mathbb{R}(t)$ which extends the usual ordering on $\mathbb{R}$ and makes $t$ larger than any real number. This is a non-Archimedean ordered field of continuum cardinality; its real-closure is thus a non-Archimedean ordered field of continuum cardinality. Since two real-closed fields are isomorphic as ordered fields iff they are abstractly isomorphic, this gives a second real-closed field of continuum cardinality.

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Another complication is that it's not obvious whether any index 2 subfield except R (whether isomorphic to it or not) can be proved to exist without the Axiom of Choice. – zyx Mar 3 '13 at 9:12
@zyx: It is consistent that the only automorphisms of $\Bbb C$ are the conjugation and identity, I assume that if there was another subfield of index $2$ we could have found another automorphism of $\Bbb C$. – Asaf Karagila Mar 3 '13 at 9:17
@Asaf: "I assume": sure, this is immediate from Galois theory. – Pete L. Clark Mar 3 '13 at 9:21
@Pete: When a distinguished professor asks me if the collection of topologies on $X$ is a set without the axiom of choice, I take that as a permission to be careful around parts of mathematics I only have basic knowledge in, e.g. no set theory. :-) – Asaf Karagila Mar 3 '13 at 9:23
@zyx: I'm not sure I understand what you're saying. Surely the fact that if $L/K$ is a separable quadratic field extension then $L = K(\sqrt{d})$ and $x+y\sqrt{d} \mapsto x-y\sqrt{d}$ is a nontrivial automorphism with fixed field $K$ doesn't use AC? – Pete L. Clark Mar 3 '13 at 9:43

Let $L/K$ be a field extension. We call $T \subset L$ a transcendence basis for $L$ over $K$ if

• Each $\alpha \in T$ is transcendental over $K$. That is no element of $T$ satisfies a polynomial in $K[x]$.
• The set $T$ is algebraically independent.
• The extension $K(T) \subset L$ is algebraic.

You can prove that a transcendence basis exists using Zorn's lemma without too much work. Now if $k$ is an algebraically closed field and $F$ is its prime subfield we can take a transcendence basis $T$ for $k$ over $F$. Then it follows that the algebraic closure of $F(T)$ is $k$ and in particular any intermediate extension $F(T) \subset L \subset k$ also has algebraic closure $k$.

As a further note any algebraically closed field is determined up to isomorphism by the cardinality of its transcendence basis and its characteristic.

In some sense though you are correct. In particular $\mathbb R$ is the only subfield of $\mathbb C$ with finite index. This follows from the Artin Schreier theorem. Edit: See Pete's answer.

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More carefully stated, every subfield of $\mathbb{C}$ of finite index can be mapped to $\mathbb{R}$ by an automorphism of $\mathbb{C}$. For example, let $\sigma$ be an automorphism of $\mathbb{C}$ that takes $\sqrt[3]{2}$ to $e^{2 \pi i/3} \sqrt[3]{2}$. Then $\sigma(\mathbb{R}) \neq \mathbb{R}$, and $\sigma(\mathbb{R})$ is also an index two subfield. – David Speyer Feb 18 '13 at 17:34
@David, $@$JSchlather: please see my answer below. – Pete L. Clark Mar 3 '13 at 17:30