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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 $\mathbb{R}$?

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  • $\begingroup$ How do you get from $\phi(1)^2=1$ to $\phi(1)=1$? $\endgroup$ – 5xum Jun 8 '16 at 9:37
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    $\begingroup$ 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 $1.$ $\endgroup$ – Nikolaos Skout Jun 8 '16 at 9:39
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    $\begingroup$ Here is a detailed proof: math.stackexchange.com/a/449449/317129 $\endgroup$ – Maik Pickl Jun 8 '16 at 9:46
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    $\begingroup$ Please do not mix up homeomorphisms with homomorphisms. There are infinitely many homeomorphisms from $\mathbb{R}$ to $\mathbb{R}$ (for example "stretching functions"). $\endgroup$ – lattice Jun 8 '16 at 9:56
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    $\begingroup$ @lattice: True, but to the best of my knowledge there's no such thing as a "ring homeomorphism" anyway. $\endgroup$ – hmakholm left over Monica Jun 8 '16 at 10:05
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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 $\mathbb{R}$.

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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 $f(r)=r$.

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    $\begingroup$ You cannot use continuity, $\phi$ is a ring homomorphism, not a homeomorphism (but the OP made a mistake). $\endgroup$ – Mathematician 42 Jun 8 '16 at 10:09
  • $\begingroup$ @Mathematician 42: You're right, of course. Where did I get the continuity idea from?... $\endgroup$ – ChanaG Jun 8 '16 at 10:48
  • $\begingroup$ Probably from the use of the word homeomorphism, which was misplaced here (and edited by now). $\endgroup$ – Mathematician 42 Jun 8 '16 at 11:31

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