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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?

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1 Answer 1

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: http://mathoverflow.net/questions/22897/fields-with-trivial-automorphism-group

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@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

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