This is an exercise out of Herstein which seems pretty straightforward but is eluding me.

Let $R,R'$ be rings and let $\phi:R\to R'$ be a mapping such that, for every $x,y\in R$: $${\rm 1.}\qquad\phi(x+y)=\phi(x)+\phi(y)$$ $${\rm 2.}\qquad\phi(xy)=\phi(x)\phi(y)\qquad{\rm or}\qquad\phi(xy)=\phi(y)\phi(x)$$ I'm to show that one of the conditions holds uniformly. That is, for all $x,y\in R$, $\phi(xy)=\phi(x)\phi(y)$ or, for all $x,y\in R$, $\phi(xy)=\phi(y)\phi(x)$. Of course, this doesn't exclude the possibility that both conditions hold uniformly (which happens when e.g. $R'$ is commutative).

Herstein hints to fix $a\in R$ and to consider the sets $$W_a=\{x\in R\mid\phi(ax)=\phi(a)\phi(x)\},$$ $$V_a=\{x\in R\mid\phi(ax)=\phi(x)\phi(a)\}.$$ I have shown that one of $W_a$ or $V_a$ must be the entire ring $R$. I think I'm missing something simple to complete the argument. Any hints?


I forgot to mention that in my copy of the text, the condition to be shown is stated as "$\phi(xy)=\phi(x)\phi(y)$ or $\phi(x)=\phi(y)\phi(x)$". I am pretty confident that it's a typo, but maybe not?

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    $\begingroup$ In case anyone's curious, it is Topics in Algebra, 2nd ed., chapter 3 supplementary problem #21. $\endgroup$ – Jonathan May 22 '13 at 2:59
  • $\begingroup$ If memory serves me correctly, I thought I used the result of a problem before this one. Unfortunately, I do not have a copy of this text with me to verify this. $\endgroup$ – Karl Kronenfeld May 22 '13 at 3:10
  • $\begingroup$ @user1 thanks. There doesn't seem to be anything relevant in the problems preceding this one. For instance, The exercise immediately prior is to show that $x^4=x$ for all $x\in R$ implies $R$ is commutative. $\endgroup$ – Jonathan May 22 '13 at 3:13
  • $\begingroup$ My memory must be bad (otherwise I would just recall the solution I had then! :)). $\endgroup$ – Karl Kronenfeld May 22 '13 at 3:14

Fact. If a group $G$ is a union of two subgroups $G = G_1 \cup G_2$, then $G=G_1$ or $G=G_2$.

We use this two times. Let $a \in R$. Then $W_a$ and $V_a$ are additive subgroups of $R$ with union $R$. Hence, $W_a=R$ or $V_a=R$. Now, $\{a \in R : W_a=R\}$ and $\{a \in R : V_a=R\}$ are additive subgroups of $R$ with union $R$. Hence, $\{a \in R : W_a=R\}$, i.e. $\phi$ is a homomorphism, or $\{a \in R :V_a=R\}$, i.e. $\phi$ is an anti-homomorphism.

  • $\begingroup$ That is neat! :) $\endgroup$ – Prism May 22 '13 at 15:16
  • $\begingroup$ Great solution, thank you! $\endgroup$ – Jonathan May 22 '13 at 16:10
  • $\begingroup$ Why the downvote? Is there something wrong? $\endgroup$ – Martin Brandenburg Apr 20 '17 at 7:25

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