In a ring, how do we prove that a * 0 = 0?

Question

In a ring, I was trying to prove that for all $a$, $a0 = 0$.

But I found that this depended on a lemma, that is, for all $a$ and $b$, $a(-b) = -ab = (-a)b$.

I am wondering how to prove these directly from the definition of a ring.

Many thanks!

Answer 1

Proceed like this

• $a0 = a(0+0)$, property of $0$.
• $a0 = a0 + a0$, property of distributivity.
• Thus $a0+ (-a0) = (a0 + a0) +(-a0)$, using existence of additive inverse.
• $a0+ (-a0) = a0 + (a0 + (-a0))$ by associativity.
• $0 = a0 + 0$ by properties of additive inverse.
• Finally $0 = a0$ by property of $0$.

Your lemma is also true, you can now prove it easily:

Just note that $ab +a(-b)= a(b + (-b))= a0= 0$.

Answer 2

$a \cdot 0 = a \cdot (0 + 0) = a \cdot 0 + a \cdot 0$