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

• Yes; also if $A$ or $B$ has zero divisors themselves, then there are more zero divisors of $A\times B$. – vadim123 May 5 '14 at 20:23

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.

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.

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 $$(2,2)\cdot(2,2)=(0,0)$$

• Sure, that makes sense. But how do I state that last part in the general sense? – allie May 5 '14 at 20:32
• @allie - If $a\in A$ is a zero divisor and $0\neq b\in B$ then $(a,b)$ is a zero divisor since there is $0\neq r\in A$ s.t $a\cdot r=0$ thus $(a,b)\cdot(r,0)=(0,0)$. Similarly the pairs $(a,b)$ are zero divisors, where $b\in B$ is a zero divisor – Belgi May 5 '14 at 20:35