Divisors of zero in $A \times B$ Describe the divisors of zero in ring $A  \times B$. 

So I know the definition of a zero divisor is: 
In a ring, a nonzero element $a$ is called a divisor of zero if there is a nonzero element $b$ in the ring such that the product $ab=0$ or $ba=0$
I just don't really understand. Are the divisors of zero in $A \times B$ in the form of $(0,a) $ or $(b,0)$? I don't quite understand
 A: We can go straight from your definition of a zero-divisor.  In $A \times B$, a zero divisor is any two non-zero elements that multiply to give zero.  Note that in this new ring, "zero" is the element $(0, 0)$.  
So the elements $(a, b)$ and $(c, d)$ are a zero divisor pair if $(a, b) \cdot (c, d) = (0, 0)$.
If $A$ or $B$ originally had zero divisors, then an easy way to generate zero divisors for $A \times B$ is as follows.  Say $a, c \in A$ is a zero divisor pair and $b, d \in B$ is also a zero divisor pair.  Then the following are zero divisor pairs in $A \times B$:
$$(a, 0)(c, 0) = (0, 0)$$
$$(a, b)(c, d) = (0, 0)$$
$$(0, b)(0, d) = (0, 0)$$
Furthermore, given the way multiplication has been defined in $A \times B$, any elements of the form $(x, 0)$ and $(0, y)$ are also zero-divisors.
A: Yes, you are on the right track: all $(a,0)$ and $(0,b)$ will be zero divisors.
Observing that $A,B$ themselves can have zero divisors (as the comment says), it suggests that the elements $(a,y)$ and $(x,b)$ will be the zero divisors in $A\times B$ where $x\in A$ and $y\in B$ denote zero divisors.
A: Consider 
$$
G=A\times B
$$
And to avoid trivial cases assume that $|A|,|B|>1$
Then it is true that any element of the form $(a,0)$ where $a\neq0$
is a zero divisor. 
This is since $(0,b)\in G$ and 
$$
(a,0)\cdot(0,b)=(0,0)
$$
This also shows that any element of the form $(0,b)$ is a zero divisor.
But those are not all the zero divisors. For example consider 
$$
(2,2)\in Z_{4}\times Z_{4}
$$
which is a zero divisor since 
$$
(2,2)\cdot(2,2)=(0,0)
$$
