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We Know that the only non zero ring Homomorphism $\phi$:$\mathbb R\to \mathbb R$,is the identity map. by this property we can show that the only nozero ring homomorphism of $\mathbb R$ is onto(Surjective). Can we have an example of a field $F$ and a nonzero homomorphism $\Psi$:$F\to F$ , that is not onto?

Note that the definition of ring homomorphism is as follows:

for all $x,y \in F, \Psi(x+y)=\Psi(x)+\Psi(y)$ and $\Psi(x.y)=\Psi(x).\Psi(y)$

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    $\begingroup$ $\overline{\mathbb{C}(t)}$ is abstractly isomorphic to $\mathbb{C}$ by standard properties of algebraically closed fields, so the obvious embedding $\mathbb{C} \to \overline{\mathbb{C}(t)}$ works. $\endgroup$
    – Qiaochu Yuan
    May 2, 2012 at 5:15
  • $\begingroup$ Hi dear Qiaochu Yuan . thank you for your example. but I don't Know anything about $\overline{\mathbb{C}(t)}$. can you define $\overline{\mathbb{C}(t)}$ for me. $\endgroup$
    – Ali Reza
    May 2, 2012 at 5:20
  • $\begingroup$ For a field $K$ we denote by $\overline{K}$ its algebraic closure. $\mathbb{C}(t)$ is the fraction field of $\mathbb{C}[t]$. $\endgroup$
    – Qiaochu Yuan
    May 2, 2012 at 5:35
  • $\begingroup$ Do you mean the algebraic closure of the quotint field of the polynomial ring $\mathbb{C}[x]$? .thats oky for me. but can we find a subfield of $\mathbb{R}$ with the mentioned property? $\endgroup$
    – Ali Reza
    May 2, 2012 at 5:36
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    $\begingroup$ How about the map $\mathbb{Q}(\pi)\to \mathbb{Q}(\pi)$ sending $\pi$ to $\pi^2$? $\endgroup$
    – Daniel Litt
    May 2, 2012 at 5:39

2 Answers 2

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Consider the field of rational functions in the infinitely many variables $x_0, x_1, x_2, ...$ (each particular rational function using only finitely many of these), and the endomorphism which shifts variables' indices up by 1.

(Originally, I had asked here "Also, who says the only nonzero ring homomorphism from $\mathbb{R}$ to $\mathbb{R}$ is the identity? I find that hard to believe, at least given Choice", but now, all I can say is "D'oh!")

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  • $\begingroup$ It's true. The key is to observe that we have $a \ge 0$ if and only if there exists $b$ such that $a = b^2$, so any ring endomorphism of $\mathbb{R}$ preserves order, and from that and the fact that it preserves $\mathbb{Q}$ it follows that it must be the identity. $\endgroup$
    – Qiaochu Yuan
    May 2, 2012 at 5:34
  • $\begingroup$ Ah, good point! I should have thought of that. I kept implicitly thinking $\mathbb{R}$ was algebraically closed... $\endgroup$
    – Sridhar Ramesh
    May 2, 2012 at 5:41
  • $\begingroup$ thats perfect dear Daniel! for what property of the field $\mathbb{Q}(\pi)$ we can find a nontrivial homomorphism on it.? which property of $\mathbb{R}$ doesn't work for your example? $\endgroup$
    – Ali Reza
    May 2, 2012 at 5:55
  • $\begingroup$ Hi AliReza--I use the fact that $\pi$ does not have a square root in $\mathbb{Q}(\pi)$ to see that the map is not surjective. The fact that $\mathbb{Q}(\pi)$ is a purely transcendental extension of $\mathbb{Q}$ allows us to construct the map easily. On the other hand, the extension $\mathbb{R}/\mathbb{Q}$ is rather complicated. $\endgroup$
    – Daniel Litt
    May 2, 2012 at 5:58
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    $\begingroup$ I usually leave only comments rather than answers, to bypass the silly reputation system. But for once, I thought, let me post an actual answer. And what's the result? A bunch of responses about somebody else's answer-posted-as-comment. :) $\endgroup$
    – Sridhar Ramesh
    May 2, 2012 at 6:55
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For any field $K$ of characteristic $p>0$ there is a homomorphism $\phi:K\to K$ given by $\phi(a)=a^p$. This need not be surjective, for example when $K$ is the field $(\mathbb{Z}/p)(t)$ of rational functions over $\mathbb{Z}/p$.

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