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I am trying to prove that given R (an integral domain) it is not true that then RxR is an integral domain: We know that for the ring Zp for any prime p, Zp is an integral domain because it is a ring with unity 1 not equal to 0, and it has no zero divisors. The ring Zn for any n not prime is not an integral domain because it does have zero divisors. From here would it be sufficient to show that given R= Z5, we know that Z5 must be an integral domain because 5 is prime. However RxR would give us Z25 which is not an integral domain because n=25 and is not prime???

Kaitlyn
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    No, it's not true that $Z_5 \times Z_5 \cong Z_{25}$. – Jair Taylor Dec 18 '14 at 20:51
  • Note that you have to prove that $R \times R$ is not an integral domain for $\it{any}$ integral domain $R$, not just for a specific example like $\mathbb{Z}_5$. – Arthur Dec 18 '14 at 20:54

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This follows from $(1,0)(0,1)=(0,0)$, i.e. $R\times R$ has zero divisors (in an integral domain $0\neq 1$). You only mention a special case, which is also more immediate from the one line argument I gave.

Mister Benjamin Dover
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