Direct Product of Two Rings and 0-divisors

Question

Let $R = Z_4\times Z_3$ be the direct product of $Z_4\times Z_3$. Find all 0-divisors in $R$.

So far, I have these elements as the product of $Z_4\times Z_3$ : $(0,0)$, $(0,1)$, $(0,2)$, $(0,3)$, $(1,0)$, $(1,1)$, $(1,2)$, $(1,3)$, $(2,0)$, $(2,1)$, $(2,2)$, $(2,3)$

I also know that 0-divisors are the result of taking two non-zero elements in a ring and multiplying them yields the additive identity(0), aka $(Z_6, 2\cdot3 = 0)$.

In this case, would the 0-divisors be $(2,2)$ and $(1,3)$ since they are the 0-divisors for $Z_4$ and $Z_3$?

Answer 1

Notice $Z_{12}\simeq Z_3\times Z_4$ by the Chinese Remainder Theorem. Maybe it would be easier for you to construct this isomorphism, find the zero divisors in $Z_{12}$, and then see what these elements map to in $Z_3\times Z_4$.

In particular, $2,3,4,6,8,9,10$ are the zero divisors of $Z_{12}$. Their images under the isomorphism are the zero divisors in $Z_3\times Z_4$.

Answer 2

Since you've written down all twelve elements from the ring, you can for every element check by hand whether it is a zero divisor.

For example, $(0,1)$ is a zero divisor, because $(0,1)\times(1,0)=(0,0)$ and both $(0,1)$ and $(1,0)$ are nonzero.

On the other hand, $(1,1)$ is not a zero divisor, because $(0,0)$ is the only element which results in $(0,0)$ when multiplied with $(1,1)$.

$(2,2)$ and $(1,3)$ are no zero divisors, do you see why?