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

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

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

$a.0=a.0+0$ (Additive identity)
$=a.0+(a+(-a))$ (Additive inverse)
$=(a.0+a)+(-a)$ (Associativity)
$=(a.0+a.1)+(-a)$ (Multiplicative identity)
$=a(0+1)+(-a)$ (Distributativity)
$=a.1+(-a)$ (Additive identity)
$=a+(-a)$ (Multiplicative identity)
$=0$ (Additive inverse)

Then, we can prove that -a=(-1)a
If a+(-1)a=0, then (-1)a=-a (there's a unique additive inverse element)
$a+(-1)a=a.1+a(-1)=a(1+(-1))=a.0=0$ (we used the first theorem)

So, the first theorem is necessary to prove the second one, but not conversely.