Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know that the only non-zero ring homomorphism from $\mathbb{R}$ to $\mathbb{R}$ is identity. From this some questions came in to my mind as follow:

Question $1$: Can we characterize all fields $F$ which only has the identity function as the ring homomorphism of $F$?

If the answer is "no" in general, can we find a partial answer if we restrict ourselves to subfields of $\mathbb{R}$?

Also we can improve our Question 1 to subrings of $\mathbb{R}$, i.e.

Question $2$: Can we find nice subrings of $\mathbb{R}$ (not necessarily subfields) with the aforesaid property?

share|cite|improve this question

I will try to give examples of question 2:

Obviously, $\mathbb{Q}, \mathbb{Z}$ are possible answers. Note that $\mathbb{Z}$ is not a field. So, you have an example of something not necessarily a subfield of $\mathbb{R}$.

In general, take an irreducible polynomial $f \in \mathbb{Q}[x]$ (such as $x^3 - 2$) with only one root $\alpha \in \mathbb{R}$. Since an endomorphism of $\mathbb{Q}(\alpha)$ maps roots of $f$ to other roots of $f$ (note that any such endomorphism fixes $\mathbb{Q}$ because it fixes $\mathbb{Z}$), the only possible field endomorphism of $\mathbb{Q}(\alpha)$ is the identity for it has to map $\alpha$ to itself.

One can further generalize the previous situation, and not require $f$ to be irreducible, but again having only one real root $\alpha$. Then a ring endomorphism of $\mathbb{Q}[\alpha]$ has to be the identity by a similar reasoning.

Similarly, for any non-zero subring $R \subset \mathbb{R}$ such that $End(R)$ is trivial, if you can find a polynomial $f \in R[x]$ with only one root $\alpha$ in $\mathbb{R}$, then $End(R[\alpha])$ is also trivial, or in other words, it just consists of the identity map. (Had an incorrect method of coming up with such polynomials here, which I realized was incorrect just as I was about to fall asleep)

Hence, you get nice examples of both subfields and subrings of $\mathbb{R}$.

I have been looking at fields of characteristic 0, but I think you may be able to generalize this to fields of characteristic $p$ (where $p$ is a positive prime in $\mathbb{Z}$) mainly because $End(\mathbb{F}_p)$ is also trivial.


For Question 1, I don't know a complete characterization, but if you have heard of p-adic numbers, then for a prime $p \in \mathbb{Z}$, the only automorphism of $\mathbb{Q}_p$ is the identity.

Here is a reference for question 1:

share|cite|improve this answer
@OP In general for any field extension $F/\Bbb{Q}$ with $\textrm{Aut}(F/\Bbb{Q})$ trivial we can get an example like Rankeya's above. – user38268 Nov 3 '12 at 8:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.