# In a ring $a \cdot a=a$ then $a+a=0$ [duplicate]

I am stuck in a question for Algebraic Structures:

I need to prove the following:

Let $R$ be a ring. If $a^2=a$ for any $a\in R$, then $a+a=0$ for any $a\in R$.

(of course $+$, $\cdot$ and $0$ regarding $R$)

P.S. This is similar to a different question I asked (if $a \cdot a = 0$ then $a + a = 0$) but I wasn't able to apply the ideas from there. In that question we assumed that $R$ is a unity ring.

## marked as duplicate by Stella Biderman, Namaste, C. Falcon, Daniel W. Farlow, user223391 Mar 21 '17 at 17:47

• Not true in the integers modulo $4$. $1*1=1$, but $1+1=2$ – Mastrem Mar 20 '17 at 20:12
• I think what is meant is: if $R$ is a ring such that $a^2=a$ for all $a$, then $a+a=0$ for all $a\in A$. Is this true? – Aweygan Mar 20 '17 at 20:13
• The statement is unclear to me. Who is $a$ in $a+a=0$ ? Is it all those that satisfy $a*a=a$ ? Is it, if for any $a$ one has $a*a=a$ then for any $a$ one had $a+a=0$ ? – James Well Mar 20 '17 at 20:16
Let $a\in R$ be given. Then $$a+a=(a+a)^2=a\cdot a+a\cdot a+a\cdot a+a\cdot a=a+a+a+a.$$ Adding $-(a+a)$ to both sides yields the result.