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

Well, in the integers we have $1\times 1 =1$ but $1+1\neq 0$. Did you omit some assumption about $R$? — lulu, Nov 21 '19 at 15:22
Are you claiming that $0$ and $1$ are the only elements? Can you prove that $0 + 0 = 0$ and $1+1= 0$? — fleablood, Nov 21 '19 at 15:30

score 1 · Accepted Answer · answered Nov 21 '19 at 15:26

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.

answered Nov 21 '19 at 15:26

Dietrich Burde

140,055

Thank you very much!!! Now I know how I should have looked at the exercise... – Default Nov 21 '19 at 15:31
Could you maybe explain me the steps from: $a^2$+2$a^2$+$a^2$ =a+a+a+a⇒a+a=0. – Default Nov 21 '19 at 15:39
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Do it termwise. First summand is $a^2=a$. Second term is $2a^2=2a=a+a$, etc. – Dietrich Burde Nov 21 '19 at 16:12

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

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