2
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Yes; also if $A$ or $B$ has zero divisors themselves, then there are more zero divisors of $A\times B$. $\endgroup$ – vadim123 May 5 '14 at 20:23
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Sure, that makes sense. But how do I state that last part in the general sense? $\endgroup$ – allie May 5 '14 at 20:32
  • $\begingroup$ @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 $\endgroup$ – Belgi May 5 '14 at 20:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.