Is the only homomorphism $\phi:\mathbb{R} \rightarrow \mathbb{R}$ onto the ring of reals the identity?

Question

Let $\phi:\mathbb{R} \rightarrow \mathbb{R}$ be a ring homomorphism onto. Prove or disprove: $\phi(r)=r$ for all $r\in \mathbb{R}.$

Attempt: I believe it is true! Since $\phi(1)=\phi(1\cdot 1)=\phi(1)^2$ we get $\phi(1)=1$, since $\phi$ is onto. Then quite easily we get $\phi(m)=m$ for $m\in \mathbb{Z}$ and $\phi(q)=q$ for $q\in \mathbb{Q}$. How do we go on from this point, to conclude for $\mathbb{R}$?

We have $\phi(1)^2=\phi(1)$, so $\phi(1)(1-\phi(1))=0$ on the integral domain of the reals, so $\phi(1)=0$ or $1.$ — Nikolaos Skout, Jun 08 '16 at 09:39
Here is a detailed proof: http://math.stackexchange.com/a/449449/317129 — Maik Pickl, Jun 08 '16 at 09:46
True or false? a homeomorphism onto the reals is one to one. (inspired from the above links, for $x>0,~\phi(x)>0$ so for $a<b,~$ $\phi(b)-\phi(a)=\phi(a-b)>0$ so one to one, in fact, strictly increasing). — Nikolaos Skout, Jun 08 '16 at 09:52
Please do not mix up homeomorphisms with homomorphisms. There are infinitely many homeomorphisms from $\mathbb{R}$ to $\mathbb{R}$ (for example "stretching functions"). — lattice, Jun 08 '16 at 09:56
@lattice: True, but to the best of my knowledge there's no such thing as a "ring homeomorphism" anyway. — hmakholm left over Monica, Jun 08 '16 at 10:05

score 2 · Answer 1 · answered Jun 08 '16 at 09:50

Notice that if $a\leq b$, then $b-a\geq 0$, hence there exists a $c\in \mathbb{R}$ such that $c^2=b-a$, thus $\phi(b-a)=\phi(c^2)=\phi(c)^2\geq 0$, it follows that $\phi(b)\geq \phi(a)$. Hence $\phi$ is order preserving. Now any $r\in \mathbb{R}$ is equal to $\sup_{q\in \mathbb{Q}, q< r}q=r=\inf_{q\in \mathbb{Q}, q>r}q$. Hence $$\sup_{q\in \mathbb{Q}, q< r} \phi(q) \leq r \leq \inf_{q\in \mathbb{Q}, q>r}\phi(q).$$ Using that $\phi$ is order preserving, we find that $\phi(r)=r$ for all $r\in\mathbb{R}$.

Edit: I nowhere used continuity of $\phi$, this is not needed. However I do use the supremum property of $\mathbb{R}$.

score 1 · Answer 2 · answered Jun 08 '16 at 09:50

1

Let $r\in\mathbb{R}$, and suppose by contradiction that $f(r)\neq r $. Then there are disjoint neighbourhoods $V,U$ of $f(r),r$ respectively.

Using the continuity, there is a neighbourhood $W$ of $r$ such that $f(W)\subset V$.

Consider $q\in U\cap W$ such that $q\in\mathbb{Q}$. Then on one hand, $f(q)=q\in U$ and on the other hand, $f(q)\in V$, but $V\cap U=\emptyset$.

Hence $f(r)=r$.

answered Jun 08 '16 at 09:50

ChanaG

325

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You cannot use continuity, $\phi$ is a ring homomorphism, not a homeomorphism (but the OP made a mistake). – Mathematician 42 Jun 08 '16 at 10:09
@Mathematician 42: You're right, of course. Where did I get the continuity idea from?... – ChanaG Jun 08 '16 at 10:48
Probably from the use of the word homeomorphism, which was misplaced here (and edited by now). – Mathematician 42 Jun 08 '16 at 11:31

Is the only homomorphism $\phi:\mathbb{R} \rightarrow \mathbb{R}$ onto the ring of reals the identity?

2 Answers2