In a ring, how do we prove that a * 0 = 0? 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!
 A: $a\cdot0=a\cdot0+0$ (Additive identity) 
$=a\cdot0+(a+(-a))$ (Additive inverse) 
$=(a\cdot0+a)+(-a)$ (Associativity) 
$=(a\cdot0+a\cdot1)+(-a)$ (Multiplicative identity) 
$=a\cdot(0+1)+(-a)$ (Distributivity) 
$=a\cdot1+(-a)$ (Additive identity) 
$=a+(-a)$ (Multiplicative identity) 
$=0$ (Additive inverse) 
Then, we can prove that $-a=(-1)\cdot a$: 
If $a+(-1)\cdot a=0$, then $(-1)\cdot a=-a$ (there's a unique additive inverse element) 
$a+(-1)\cdot a=a\cdot1+a\cdot(-1)=a\cdot(1+(-1))=a\cdot0=0$ (we used the first theorem)
So, the first theorem is necessary to prove the second one, but not conversely.
A: $a \cdot 0 = a \cdot (0 + 0) = a \cdot 0 + a \cdot 0$
A: 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 $ab +a(-b)= a(b + (-b))= a0= 0$.
