Ring Properties

Question

If $(S,.,+)$ is a ring with the property that $a^2 = a$ for all a an element of $S$, which of the following must be true, given:

I $a + a = 0$ for all $a\in{ S}$.

II $(a + b)^2 = a^2 + b^2 $ for all $a, b \in{ S}.$

III S is commutative.

• A. III only
• B. I and II only
• C. I and III only
• D. II, and III only
• E. I, II, and III

Help please I know that it is not only III so A. is out.

Answer 1

We have $$(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b=a+b$$ hence we find $$ab=-ba$$ and $$(ab)^2=abab=a(-ab)b=-a^2b^2=-ab=ab=-ba$$ hence $$ab=ba=-ba$$ finally we have $$(a+a)^2=a^2+a^2+a^2+a^2=a+a+a+a=a+a$$ so $$a+a=0$$ and then the three assertions are true.

Answer 2

• I: Let $a ∈ S$. Then $a = a^2 = (-a)^2 = (-a) = -a$, so $a + a = 0$.
• II: Let $a, b ∈ S$. Then $(a+b)^2 = (a+b) = a + b = a^2 + b^2$.
• III: Let $a, b ∈ S$. Then $(a+b)^2 = a^2 + ab + ba + b^2 = (a^2 + b^2) + (ab + ba)$. By II, subtracting $(a+b)^2 = a^2 + b^2$ on both sides yields $ab = -ba$. By I, $ab = ba$.