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$1$ has to map to $1$, right? So $n$ has to map to $n$ (for $n \in \mathbb{Z}$), and $\frac{1}{n}$ maps to $\frac{1}{n}$, so $\frac{n}{m}$ maps to itself, so the only possible automorphism is the identity. Is this true or am I deceiving myself? Because I feel like there should definitely be more automorphisms of $\mathbb{Q}$.

Also, if you have some extension of $\mathbb{Q}$ (call it $E$), does every automorphism of $E$ automatically fix $\mathbb{Q}$?

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Here's a much less intuitive fact: There is only one field automorphism of $\mathbb{R}$. (Although certainly there are subfields of $\mathbb{R}$ that have nontrivial automorphisms.) – Charles Staats Apr 21 '11 at 3:15
up vote 20 down vote accepted

When one says "automorphisms", it is important to specify automorphisms of what. There are a lot of automorphisms of $\mathbb{Q}$ as an abelian group (equivalently, as a $\mathbb{Q}$-vector space).

However, there is one and only one field automorphism (equivalently, one and only one ring automorphism). Indeed: If you are doing a ring automorphism, then $1$ must map to an idempotent (an element equal to its square); there are only two idempotents in $\mathbb{Q}$, $1$ and $0$; but if you map $1$ to $0$, then you map everything to $0$ and the map is not an automorphism. So $1$ must map to $1$ (you can skip this step if your definition of "homomorphism of rings" requires you to map $1$ to $1$).

Since $1$ maps to $1$, by induction you can show that for every natural number $n$, $n$ maps to $n$. Therefore, $-n$ must map to $-n$ (since the map sends additive inverses to additive inverses), and must sent $\frac{1}{n}$ to $\frac{1}{n}$ (because it maps $1 = n(\frac{1}{n})$ to $n$ times the image of $\frac{1}{n}$, and the only solution to $nx = 1$ in $\mathbb{Q}$ is $x=\frac{1}{n}$. And from here you get that any field automorphism of $\mathbb{Q}$ must be the identity.

As to your second question, yes: if $E$ is an extension of $\mathbb{Q}$, then any field automorphism of $E$ restricts to the identity automorphism of $\mathbb{Q}$. More generally, if $E$ is any field, then any automorphism of $E$ restricts to the identity of its prime field (which is $\mathbb{Q}$ in the case of characteristic 0, and $\mathbb{F}_p$ in the case of characteristic $p$).

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Arturo: I am not sure I understand how ${\mathbb Q}$ is not rigid as a ${\mathbb Q}$-vector space. Perhaps you mean something different from what I have in mind... – Andrés E. Caicedo Apr 21 '11 at 3:25
@Andres: The map $1\to 2$ defines a nonidentity linear automorphism from $\mathbb{Q}$ to itself (as a $\mathbb{Q}$-vector space). Remember that if you specify where to send a basis, there is always an automorphism that does that; here, 1 is a basis for $\mathbb{Q}$. In general, a 1-dimensional vector space over $\mathbf{F}$ has automorphism group isomorphic to $\mathbb{F}^{\times}$. The map $x\mapsto \alpha x$ is linear ($\alpha(x+y) = \alpha x + \alpha y$; $\alpha(qx) = q(\alpha x)$). – Arturo Magidin Apr 21 '11 at 3:28
@Andres: Correction: "there is always an automorphism" should be "there is always a linear map" – Arturo Magidin Apr 21 '11 at 3:33
Arturo: Aha! Thanks. – Andrés E. Caicedo Apr 21 '11 at 4:46

It's true for both questions. $\mathbb Q$ is a prime field: everything is fixed for those.

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