Is an automorphism of the field of real numbers $\mathbb{R}$ the identity map? If yes, how can we prove it?

Remark An automorphism of $\mathbb{R}$ may not be continuous.

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    $\begingroup$ Hint: it's fairly easy to see that any order-preserving automorphism is the identity. $\endgroup$ Jul 22, 2013 at 12:05
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    $\begingroup$ I was sure this is a duplicate given that I have a recollection of spelling this out here at least twice. But the best match I could find right away is this. I'm not sure we can call it a duplicate. Sure, the answers of that question also answer this, but... $\endgroup$ Jul 22, 2013 at 13:27
  • $\begingroup$ Also, Jacobson's Basic Algebra I :-) $\endgroup$ Jul 22, 2013 at 13:28
  • $\begingroup$ @JyrkiLahtonen It is an interesting fact that there are infinitely many automorphisms of $\mathbb{C}$, even though $[\mathbb{C}:\mathbb{R}]=2$. Why is this fact not a contradiction to this problem? $\endgroup$
    – user425181
    Sep 4, 2017 at 20:50
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    $\begingroup$ @user425181 It means that $\Bbb{C}$ has infinitely many subfields isomorphic to $\Bbb{R}$. Mind you, this result depends heavily on the axiom of choice. IIRC (ask a set theorist) if we drop out the axiom of choice, it may (or may not?) happen that $\Bbb{C}$ has only finitely many automorphisms. $\endgroup$ Sep 4, 2017 at 21:04

7 Answers 7


Here's a detailed proof based on the hint given by lhf.

Let $\phi$ be an automorphism of the field of real numbers. Let $x \gt 0$ be a positive real number. Then there exists $y$ such that $x = y^2$. Hence $\phi(x) = \phi(y)^2 \gt 0$.

If $a \lt b$, then $b - a \gt 0$. Hence $\phi(b) - \phi(a) = \phi(b - a) \gt 0$ by the above. Hence $\phi(a) \lt \phi(b)$. This means that $\phi$ is strictly increasing.

If $n$ is a natural number, it can be written in the form $1 + \ldots + 1$, so $\phi(n) = n$. Now, any rational number is of the form $r = (a - b)c^{-1}$, for $a, b, c$ natural numbers, so it follows that $\phi(r) = r$ for any rational number.

Let $x$ be a real number. Let $r, s$ be rational numbers such that $r \lt x \lt s$. Then $r \lt \phi(x) \lt s$. Since $s - r$ can be arbitrarily small, $\phi(x) = x$. This completes the proof.

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    $\begingroup$ @MakotoKato It is an interesting fact that there are infinitely many automorphisms of $\mathbb{C}$, even though $[\mathbb{C}:\mathbb{R}]=2$. Why is this fact not a contradiction to this problem? $\endgroup$
    – user425181
    Sep 4, 2017 at 20:49
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    $\begingroup$ @user425181 Because in one case you're looking at automorphisms of $\mathbb{C}$, and in the other you're only looking at those which restrict to the identity on $\mathbb{R}$. The restriction in the second case is strong enough to only leave two options (and this is one of the key realizations in Galois theory). $\endgroup$
    – Ducky
    Mar 8, 2021 at 2:52

Hint: Let $\phi$ be a field automorphism of $\mathbb R$. Then prove:

  • $\phi$ sends positive numbers to positive numbers

  • $\phi$ is increasing

  • $\phi$ is continuous

  • $\phi$ is the identity on $\mathbb Q$

  • $\phi$ is the identity on $\mathbb R$.

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    $\begingroup$ I just feel tempted to write how one's intuition might develop in such cases. $\phi$ is an automorphism $\implies$ it's a function. To know about this function, we must know in every possible way how this function will behave in the real domain. And that's why we want to investigate all properties of $\phi$. A real domain follows the laws of trichotomy, it has positive and real numbers. It's worthy to note how the function will react to such kinds of inputs. Again, a beautiful question :-) $\endgroup$
    – MathMan
    Jul 17, 2014 at 19:55
  • $\begingroup$ How do you prove the continuity? $\endgroup$
    – user422112
    May 5, 2018 at 12:50
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    $\begingroup$ @ThatIs, actually, as Makoto's answer shows, continuity is not needed. But my point was that an increasing function can only have jump discontinuities and then it cannot be surjective. See math.stackexchange.com/questions/1207135/…. $\endgroup$
    – lhf
    May 5, 2018 at 13:28
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    $\begingroup$ My intuition for this proof is that once we know $\phi$ is the identity on the rational numbers, we want to extend $\phi$ by continuity. One way to do that is to show $\phi$ is increasing. But an automorphism is something that only "knows" about algebraic properties of the field, involving the field operations. Why would it "know" about the order relation on $\mathbb R$? We have to find a way to encode the order relation in algebraic terms, which we do by noting an algebraic difference between positive and negative numbers: having square roots. $\endgroup$
    – Jack M
    Jul 22, 2018 at 11:41

I have never liked the proofs of this that use analysis more than necessary. Once you get that $\phi$ is order-preserving and the identity map on the rationals, take an arbitrary real number $a$. If $\phi(a) \neq a$, then there is a rational $q \in \mathbb{Q}$ between $a$ and $\phi(a)$. If $a \leq q \leq \phi(a)$, then $\phi(a) \leq \phi(q) = q$, so $\phi(a) \leq q$ and $q \leq \phi(a)$, so $\phi(a) = q = \phi(q)$, so $a = q$, contradicting the fact that $\phi(a) \neq a$. Similarly if $\phi(a) \leq q \leq a$.


For related but slightly stronger results, see $\S$ 16.7 of these field theory notes.


(i) Every Archimedean ordered field $K$ admits a unique homomorphism of ordered fields $K \hookrightarrow \mathbb{R}$.
(ii) Let $(F,<)$ be an ordered field in which every positive element is a square (e.g. any real-closed field, e.g. $\mathbb{R}$). Then the ordering $<$ is unique, so every homomorphism of fields between two such fields is necessarily a homomorphism of ordered fields. Thus:

The identity map on $\mathbb{R}$ is the unique field homomorphism from $\mathbb{R}$ to $\mathbb{R}$: "$\mathbb{R}$ is strongly rigid".

(In the Lemma that occurs just before the "Main Theorem on Archimedean Ordered Fields" -- currently numbered Lemma 192 and on p. 106, but both of these are subject to change -- where it says "topological rings", I think it should say "Hausdorff topological rings".)


Here is a quick formulation of the proof which bypasses explicitly showing the ordering is preserved. Let $f:\mathbb{R}\to\mathbb{R}$ be an automorphism. We know $f$ fixes all rational numbers. Suppose $f$ is not the identity; say $f(x)\neq x$ for some $x$. We may assume $f(x)>x$ (if $f(x)<x$, then $f(-x)>-x$, so we can replace $x$ by $-x$). Now let $q$ be a rational number such that $x<q<f(x)$ and let $y=\sqrt{q-x}$. Note that $$f(q)=f(x+y^2)=f(x)+f(y)^2\geq f(x).$$ But $f(q)=q$ since $q$ is rational, so this contradicts the fact that $q<f(x)$. Thus $f$ must be the identity.

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    $\begingroup$ How do we know, that f fixes all rational numbers ? $\endgroup$
    – martina
    Mar 13, 2018 at 8:17
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    $\begingroup$ @martina: $f$ fixes $1$ and then additive and multiplicative properties forces it to fix all rationals. $\endgroup$
    – Bumblebee
    May 20, 2018 at 0:08

I think the correct way to use density of rationals in this problem is this:

Let $a$ be a real number and suppose $\phi(a) \neq a$. Then, there exists a rational number $q$ such that $a<q<\phi(a)$. Hence, $q < \phi(a)$ and $q>\phi(a)$. This is a contradiction since $\phi$ is a bijection.

  • $\begingroup$ Of course you first must show that $\phi$ preserves the ordering. $\endgroup$
    – GEdgar
    Jun 15, 2016 at 14:57
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    $\begingroup$ Yes. This is done above. $\endgroup$ Jun 15, 2016 at 15:43
  • $\begingroup$ @GEdgar It is an interesting fact that there are infinitely many automorphisms of $\mathbb{C}$, even though $[\mathbb{C}:\mathbb{R}]=2$. Why is this fact not a contradiction to this problem? $\endgroup$
    – user425181
    Sep 4, 2017 at 20:51
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    $\begingroup$ @user425181 ... only one non-identity automorphism of $\mathbb C$ maps $\mathbb R$ into itself. The others don't. Thus, no contradiction. $\endgroup$
    – GEdgar
    Sep 4, 2017 at 22:24

Since $f$ preserves addition, $f(0+0)=f(0)+f(0)$, and so $f(0)=0$. Since $f$ preserves mutliplication, for all $x\in\mathbb R$ we have $f(x)=f(1\cdot x)=f(1)\cdot f(x)$. Therefore, either $f(1)=1$ or $f$ is identically $0$; the second possibility is ruled out by the fact that $f$ maps $\mathbb R$ onto $\mathbb R$. Let $n\in\mathbb N$. It can be easily proven by induction that if $a_1,\dots,a_n$ is any list of numbers, then $$ f\left(\sum_{i=1}^{n}a_i\right)=\sum_{i=1}^{n}f(a_i) \, , $$ and so in particular $$ f(n)=f\left(\sum_{i=1}^{n}1\right)=\sum_{i=1}^{n}f(1)=\sum_{i=1}^{n}1=n \, . $$ Since $f(n)+f(-n)=f(0)=0$, we have $f(-n)=-f(n)=-n$. This shows that $f$ is the identity map on $\mathbb Z$. Let $a,b\in\mathbb Z$, with $b\neq0$. We have $f(1/b)\cdot f(b)=f(1)$, so $f(1/b)=f(1)/f(b)=1/b$. Hence $f(a/b)=f(a)\cdot f(1/b)=a/b$. This shows that $f$ equals the identity on $\mathbb Q$.

For the final stage of the proof, we utilise three facts that rest upon $\mathbb R$ being a complete ordered field:

  • Every positive real has one positive square root.
  • The square of a nonzero real number is positive.
  • $\mathbb Q$ is dense in $\mathbb R$: between any two distinct real numbers there is a rational number in between them.

Let $x>0$. Then $f(x)=f(\sqrt x)f(\sqrt x)>0$. This means that if $x>y$, then as $x-y>0$, $$ f(x-y)=f(x)-f(y)>0\implies f(x)>f(y) \, . $$ Therefore, $f$ preserves order. Now, let $z$ be irrational. If it were the case that $f(z)<z$, then there would an $r\in\mathbb Q$ such that $f(z)<r<z$. But as $f(r)=r$, this contradicts the fact that $f$ is order-preserving. The case $f(z)>z$ similarly leads to a contradiction. Hence $f(z)=z$ and so $f$ is the identity on $\mathbb R\setminus\mathbb Q$, completing the proof.

  • $\begingroup$ I failed to see what this answer brings new. $\endgroup$
    – user26857
    Mar 15 at 16:49
  • $\begingroup$ @user26857: It doesn't present a different method to the other answers, but it does flesh things out a bit more, which might be helpful for readers who find it difficult to fill in the details of the other answers. $\endgroup$
    – Joe
    Mar 15 at 16:54

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