Zero, the Additive Identity, as the Multiplicative Annihilator In the structures I have encountered so far, I have always seen a zero, which is usually defined as the additive identity. For example:

$\exists 0 \in \mathbb{Z}$ s.t. $\forall a \in \mathbb{Z}, a + 0 = 0+a = a$

It just so happens to be that whenever the need arose, $0$ also served as the multiplicative annihilator, i.e. where $X$ is some commutative ring: $\forall a \in X, a \cdot 0 = 0$, as proven below:

$0=0$, so $0+0 = 0$, so $a\cdot(0+0) = a \cdot 0$, so $a\cdot 0 + a \cdot 0 = a \cdot 0$, so $0 = a \cdot 0$.

My question is whether the zero always serves as the multiplicative annihilator as well, whether this is actually part of its definition (or an always-implicit corollary), or if it is possible to have a zero that does not serve as a multiplicative annihilator.
 A: It is clear from the OP's argument:
$0a = (0 + 0)a = 0a + 0a \Rightarrow 0a = 0 \tag{1}$
that $0$ being a multiplicative annihilator depends on two things:  i.)  the fact that zero is the identity element for the "$+$" operation; and (ii.)  the fact that multiplication distributes over addition in the sense that $a(b + c) = a(b + c)$.  So, if you want the zero element to not be a multiplicative annihilator, you'll have to bust one of these postulates.  Well, busting the additive identity won't work and leave us with an Abelian group under "$+$"; thus we'd have to let go of distributivity, the axiom which ties addition and multiplication together.  If that postulate is eliminated or severely altered, all I can say is:
We'll be left with something VERY un-ring like!
So, as long as we want rings to be rings, as it were, we have to accept $0$ as multiplicative annihilator that it is.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: It is a theorem on rings (all rings) that $x0=0$ for all $x$. This is not part of the definition, it follows from simpler axioms.
$$
\begin{align}
x0&=x0+0=x(0+0)=x0+x0\\
x0&=x0+x0\\
x0&=0
\end{align}
$$
The proof for $0x=0$ is similar.
