# Zero Divisors in Direct Product of Rings [duplicate]

Suppose that $R = S × T$ is a direct product of rings with $S$ and $T$ each having at least two elements. Prove that $R$ has zero divisors.

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(0,1)(1,0)=(0,0). – anon Nov 9 '12 at 3:13
The key is $1\ne 0$ because both $|S|$ and $|T|$ are at least 2. – peoplepower Nov 9 '12 at 3:16

For any $a\ne 0$, $(a,0)\cdot (0,a)=(0,0)$.

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Hmmm...what about the elements $\,(1,0)\,\,,\,\,(0,1)\,$?

Of course if one, or both, of the rings have no unit you can choose any non-zero elements instead of $\,1\,$

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What does it mean for $R$ to have zero divisors? That means it has nonzero elements that multiply to zero. How do you write elements of $R$? as ordered pairs of elements from $S$ and $T$.

So you are asking if there are any non-zero solutions to

$$(s_1, t_1) (s_2, t_2) = (0, 0)$$

such that $(s_1, t_1) \neq (0,0)$ and $(s_2, t_2) \neq (0,0)$.

That is, you want to find values $s_1, s_2 \in S$ and $t_1, t_2 \in T$ such that all of the following hold:

• $s_1 s_2 = 0$
• $t_1 t_2 = 0$
• Either $s_1 \neq 0$, $t_1 \neq 0$, or both
• Either $s_2 \neq 0$, $t_2 \neq 0$, or both.

Now, solve!

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