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I have some true or false questions and would like to have your help to check on it.

A). in a ring R, if $x^2=x$, $\forall x\in R$, then R is commutative

For (A), when looking at $(x+y)^2$, it has $x+y=(x+y)^2=x^2+xy+yx+y^2$
and then yx+xy=0, and from 2x=4x, therefore 2x=0. how this play a role here?

B) In an integral domain, it $\exists m\in N, s,t, mx=0, \forall x\in R$, then it's a finite integral domain.

I think this one is correct by definition of characteristic for a ring/field.

C) commutative ring with unity has at least two elements, and cancellation holds, then it's an integral domain.

I think it's correct,
first it's a commutative ring with unity,
second, it has at least two element, and there's a property saying that an integral domain must have at least two elements,
and the third, cancellation holds implies it has no zero divisor.
So the three above looks like fit the profile of integral domain.

D) f is a homomorphism from group G to group H,
then f(G) is a normal subgroup in H

I'm not quite sure about this one, kind of remember that normal subg roup of G under homomorphism is normal subgroup of H, but don't know if f(G) will be normal in H

Are these answer or argument correct? Thanks for your help.

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For (A), first remark that $2x=0$ for all $x \in R$, because $2x = (2x)^2 = 4x^2 = 4x$. Then look at $(x+y)^2$.

For (B), it is not necessarily true. For instance, in the polynomial ring $\mathbf F_p[X]$, every element is killed by $p$, but $\mathbf F_p[X]$ is infinite.

(C) is correct.

(D) is false. (Also why do you introduce $K$ if you don't use it?). Take a group $G$ with a non-normal subgroup $M$, and consider the injection of $M$ in $G$...

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  • $\begingroup$ thanks for the hint and correction, I made some correction for (A), but still uncompleted in finishing this question. could you check it for me? And for (D), it's a typo, ha~ $\endgroup$ – Lily Oct 18 '13 at 14:06
  • $\begingroup$ @Lily You have it! Since $2x=0$ for all $x$, in particular you have $yx = -yx$. Combine this with $xy=-yx$, and you see that $xy=yx$. $\endgroup$ – Bruno Joyal Oct 18 '13 at 14:15
  • $\begingroup$ I'm sorry I still cannot follow... Combine 2x=0 and yx=-yx and we can have yx=yx? why? $\endgroup$ – Lily Oct 18 '13 at 14:31
  • $\begingroup$ oh wait!! is that 2x=0, so x+x=0, x=-x for all x, therefore, when we have -yx, (-y)x=yx, and therefore, xy=yx? $\endgroup$ – Lily Oct 18 '13 at 14:32
  • $\begingroup$ @Lily That's it! :) $\endgroup$ – Bruno Joyal Oct 18 '13 at 14:32

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