Studying Galois Theory I have finally done the Fundamental Theorem of Algebra, which simply states that $\overline{\mathbb{R}} = \mathbb{C}$.

My question is: do there exist other fields different from $\mathbb{R}, \mathbb{C}$ such that their algebraic closure is $\mathbb{C}$? Moreover, can we find the least (with respect to set inclusion) field with such property?

Any help is appreciated.

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2 Answers2


Yes to the first question, no to the other one. This answer needs some theory of trascendental extensions, though. $\Bbb R$ is algebraic over some purely trascendental extension of $\Bbb Q$ by $\beth_1$-many variables: namely, there is a field $\Bbb Q\subseteq k\subseteq \Bbb R$ such that $\overline k\supseteq \Bbb R$ and such that $k\cong \Bbb Q(X_\alpha\,:\,\alpha\in\beth_1)$, the field of rational functions in uncountably many variables.

Now, $\Bbb R$ cannot be $k$, because there are several ring homomorphisms $f:k\to k$ such that $\left. f\right\rvert_{\Bbb Q}=id$. On the other hand, any homomorphism of rings $\Bbb R\to\Bbb R$ which fixes $\Bbb Q$ must be the identity.

The least (or even a minimal) field $k$ such that $\overline k=\Bbb C$ does not exist: as before, such a field $k$ ought to contain $k'=\Bbb Q(\xi_\alpha\,:\,\alpha\in\beth_1)$ with $\overline{k'}=\Bbb C$ and $\xi_i\in k$ algebraically independent over $\Bbb Q$. By minimality, $k=k'$. Now, $k''=\Bbb Q(\xi^2_\alpha\,:\,\alpha\in\beth_1)$ is strictly contained in $k'$ and it has the same algebraic closure. At a closer inspection, the only fields which are minimal in the family of fields which have their same algebraic closure are $\Bbb Q$ and $\Bbb F_p$ (where $p$ is a prime).

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    This is really nice! The fact, casually used here, that any ring homomorphism of $\mathbb{R}$ that fixes $\mathbb{Q}$ is the identity is not obvious but it is true: the key (for me) was to realize that each such homomorphism must preserve the ordering because the statement $a > b$ is (in $\mathbb{R}$, not in any other equally famous ring!) equivalent to the 'ring-theoretic' statement that $(a - b)$ is a square. – Vincent Dec 20 '17 at 13:00
  • I am having trouble reconciling the following three statements: $k\subseteq\mathbb{R}$, $k\supseteq\mathbb{R}$, and $k\neq\mathbb{R}.$ – ziggurism Dec 20 '17 at 13:12
  • @ziggurism It says $\bar{k} \supseteq \mathbb R$, not $k \supseteq \mathbb R$. – Erick Wong Dec 20 '17 at 13:13
  • Nevermind, I see now that I misread the second statement. It is $\bar k\supseteq\mathbb{R}$, not $k\supseteq\mathbb{R}$ – ziggurism Dec 20 '17 at 13:13
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    @Vincent Thinking about it, there are more direct ways to prove that $\Bbb R$ is not a field of rational functions over $\Bbb Q$. Namely, because the polynomial $T^2-2$ has a root. –  Dec 21 '17 at 07:00

According to the Axiom of Choice, there is a field automorphism $\tau$ of $\mathbb C$ that does not map $\mathbb R$ into itself. Then $F := \tau(\mathbb R)$ has the property that $\overline{F} = \mathbb C$.

It is interesting that such a field $F$ cannot be a Borel set in $\mathbb C$, however. It cannot be constructed (and proved correct) in ZF set theory only.

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  • Well, that answer is actually right according to my question, but as many other students (and teachers) often do, I wrote different but i meant non-isomorphic Anyway it seems an intresting fact, can you provide any reference for it? – JayTuma Dec 29 '17 at 17:54