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let $\mathbb{R}$ be the field of real numbers. I found stated in this pretty work On Groups that Are Isomorphic to a Proper Subgroup, that there is no proper subfield $K$ of $\mathbb{R}$ which is isomorphic to $\mathbb{R}$ itself. Does someone have a proof of this fact?

Thank you very much for your help in advance.

NOTE1. Contrast this situation with the case of the field $\mathbb{C}$ of complex numbers, for which there exist proper subfields isomorphic to $\mathbb{C}$ itself: see e.g. Automorphisms of the Complex Numbers, Concluding Remark 2.

NOTE2. This issue arouse in my post Proper Subgroup of O_2(R) Isomorphic to O_2(R) about whether the orthogonal group $O_2(\mathbb{R})$ is co-Hopfian or not.

Maurizio Barbato
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3 Answers3

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Let $K$ be a subfield of $\Bbb R$ which is isomorphic to $\Bbb R$. Then $K$ has a canonical ordering: $x\ge y$ iff $x=y+t^2$ for some $t\in K$. As $K$ is order-isomorphic to $\Bbb R$, then $K$ is a complete ordered field, where we take the completeness axiom as "a nonempty bounded-above subset has a supremum".

The embedding of $K$ in $\Bbb R$ must be order preserving. Suppose $\xi\in\Bbb R$. The set $S=\{a\in \Bbb Q: a<\xi\}$ has a supremum in $K$ and in $\Bbb R$, but these must be the same, namely $\xi$. So $\xi\in K$ for all $\xi\in\Bbb R$.

Angina Seng
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Say $K$ is a proper subfield of $\mathbb{R}$ and let $\phi:\mathbb{R}\rightarrow\mathbb{R}$ be the isomorphism mapping $\mathbb{R}$ to $K$. Then, $\phi$ is an injective ring map.

As $\phi$ is a ring map, $\phi(1)=1$ so $\phi(n) =n$ for $n\in\mathbb{Z}$ and also $\phi(q)=q$ for $q\in\mathbb{Q}$.

First, I want to show that $\phi$ preserves the order : If $x>0$, then there exists $y>0$ such that $y^2=x$. Then, $\phi(x)=\phi(y^2)=\phi(y)^2>0$. Since $\phi$ preserves positivity, it preserves the order, i.e. if $a<b$ then $\phi(a)<\phi(b)$.

Now, I want to show that $\phi$ is continuous : Say $a_i$ is a convergent sequence with limit $a\in\mathbb{R}$. Then, for each $\epsilon\in\mathbb{Q}_{>0}$, there exists $N\in\mathbb{N}$ such that $|a-a_i|<\epsilon$ for $i>N$. But then, $|\phi(a)-\phi(a_i)|=|\phi(a-a_i)|<\phi(\epsilon)=\epsilon$. Hence, $\phi(a_i)$ is also convergent and converges to $\phi(a).$

Each real number $x\in\mathbb{R}$ can be written as a limit of a rational sequence $(q_i)$. But now, $\phi(x) = \phi(\lim_i q_i) = \lim_i \phi(q_i)=\lim_i q_i=x$. Hence, $\phi$ is the identity map.

Levent
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  • This is almost the same proof that identity is the only automorpshism on $\mathbb{R} $. +1 Does this apply to other fields like $\mathbb{Q} (\sqrt [3]{2})$ which has identity as the only automorpshism? – Paramanand Singh Aug 08 '20 at 15:29
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Hint: This has to do with the fact that the ordering in $\mathbb R$ can be expressed purely as a logical statement expressed in the language of rings/fields: $x\le y\equiv (\exists z)y=x+z^2$. Now, let $\phi:\mathbb R\to\mathbb R$ be a field monomorphism. Using the previous, you can prove that $\phi$ preserves ordering (i.e. is monotonically increasing). As you can (trivially) prove that $\phi$ maps rational numbers to themselves, i.e.$\phi(q)=q, q\in\mathbb Q$, then (due to preserving the ordering, and the fact that irrational numbers can be seen as Dedekind cuts in $\mathbb Q$), you can conclude that $\phi$ also maps every real number to itself