Proof $a+a = 0$ by given $a\times a = a$

Question

I'm wondering how I can proof that for the following Ring $(R; +,-,0,\times,1)$ with $a \times a = a$ that $a + a = 0$.

How can I proof that? I just know about the lemma:

(i) $0(a) = a(0) = 0$

(ii) $(−a)b = −(ab)$

(iii) $(−a)(−b) = ab$

Many thanks in advance!!!

Kind regards

Answer 1

This homework usually assumes that $a^2=a$ holds for all elements $a$. Then we have $$(a+a) = (a+a)^2 = a^2+2a^2+a^2= a+a+a+a \Rightarrow a+a = 0 .$$

Possible duplicte of

Let R be a ring such that $a^2 = a$ , $\forall a$ $\in R$ . Prove that R is commutative.