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I am studying old GRE problems for a class and think I have a handle on this problem and am just looking for guidance and verification. Here are the 5 choices:

A) $x=-x$ for all $x\in R$.

B) $R$ is commutative

C) $xy+yx=0$ for all $x,y \in R$

D) Both A and C

E) A, B, and C

Since $R$ is a ring, take $x,y \in R$ and look at $(x+y)^2$. We know $x+y$ is in the ring. Thus,

$$(x+y)^2=(x+y)(x+y)=x(x+y)+y(x+y)$$ $$=xx+xy+yx+yy=x+xy+yx+y=x+y$$

So this shows that $xy+yx=0$ so C is definitely true. A would not be necessarily true, since $xy\in R$, we have by the implication, $$xy=-yx\neq -(xy)$$

Therefore, logically, it has to just be C, right?

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    Notice that $1=(-1)^2=-1$. Thus, $x=-x$. By C), $xy=yx$ also. – 2017-01-29

2 Answers 2

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No, (A) holds in fact. For any $x\in R$,

$$1+x=(1+x)^2=1+2x+x^2=1+2x+x$$

which implies $2x=0$, i.e. $x=-x$.

Then (C) still holds, you're correct about that. But use the fact that (A) holds to show that (B) holds as well.

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    See, it's little things like letting $y=1$ that i always struggle with. That makes it so clear; – 2017-01-29
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(A) is true! Try the following: \begin{align*} x^2&=x\\ (-x)^2&=-x\\ (-x)^2&=x^2\\ &=x \end{align*} By lines (2) and (4), it is clear that (A) holds.

Have you given (B) any thought?

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    okay, so once A holds, then B is extended off of that, since $xy=-yx=yx$, right? so all A B and C – 2017-01-29
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    That's exactly right! – 2017-01-29