Set of idempotent elements closed under addition: prove $a+a = 0, ab=ba$

Question

The statement of the problem:

If $R$ is a ring and $a \in R$, $a$ is idempotent if $a^2 = a$. If the set of all idempotent elements of $R$ is closed under addition, prove that for any $x, y \in R$ we have $x+x = 0, \space xy = yx$. Hint: start with $(x+x)^2$ and $(x+y)^2$.

What's confusing me here is that the statement seems wrong, as I have no idea how to use the fact that the idempotent elements are closed under addition to prove $x+x = 0$ or $xy = yx$ for any $x, y \in R$, but I can't seem to think of a counterexample. The only example where the set of idempotent elements is closed under addition that I could think of is $\mathbb{Z}_{2}$, and the statement indeed holds for that ring. Is there a way to connect an arbitrary element of a ring with the idempotent ones, or is the statement of the problem wrong?

P.S. The statement is easy to prove if we add the requirement that $x, y$ are themselves idempotent: $(x+x)^2 = 4x^2 = 4x$, and since $x + x$ is also idempotent, $(x+x)^2 = x+x =2x$, so $4x = 2x$, so $x+x+x+x=x+x$, so $x+x=0$.

On the other hand, $(x+y)^2 = x + y = x^2+y^2+xy+yx=x+y+xy+yx$, so $xy+yx=0$, and I get stuck here again because $xy = -yx$, which isn't the same as $xy = yx$.

Answer 1

The observation of the characteristic being $2$ is already well covered by Hagen von Eitzen's solution. This is to address the claim that the ring is also commutative.

What's confusing me here is that the statement seems wrong, as I have no idea how to use the fact that the idempotent elements are closed under addition to prove $x+x = 0$ or $xy = yx$ for any $x, y \in R$, but I can't seem to think of a counterexample.

I think you're right to feel that way.

Consider the free $F_2$ algebra (where $F_2$ is the field of two elements) $F_2\langle x,y\rangle$. Keep in mind that $x$ and $y$ are two indeterminates that do not commute.

We'll say that the "degree" of a polynomial in this algebra is the largest total of exponents of a monomial with nonzero coefficient. So, for example, the degree of $xyxy+x^2y^2x$ is $5$. Clearly if $a,b$ are in the algebra and $deg(b)>0$, then $deg(ab)>deg(a)$. Using this observation, we can see that the idempotents have to have degree $0$, and so the only ones are $\{0,1\}$. (Alternatively, this ring is a domain, so the only idempotents are $\{0,1\}$.)

However, this algebra is not commutative.

Answer 2

As $1$ is idempotent, so is $2:=1+1$. Hence $4=2$ and $2=0$. Consequently $$ x+x=2\cdot x=0\cdot x=0$$ for all $x\in R$.

If $x,y$ are idempotent, then so is their sum, hence $$ x+y=(x+y)^2=x^2+xy+yx+y^2=x+y+xy+yx$$ and using the above, $$xy=-yx= yx.$$ Then also $(xy)^2=xyxy=xxyy=x^2y^2$, i.e., we see that the idempotents are not only closed under addition (and, as $-1=1$, under subtraction), but also under multiplication, i.e., they form a subring $S$ of $R$. Also, $S$ is abelian. Can we show $S=R$ from this?