# Prove that if R and S are nonzero rings then $R\times S$ is never a field.

This question has been asked on here before but I'm looking for some additional insights. The question comes from the section of the Dummit and Foote textbook on the Chinese Remainder Theorem. All of the proofs that I have seen so far don't use the CRT. Is there any way to prove this using the CRT?

I was thinking of trying to show that there exists a ring A with comaximal ideals I, J such that by the Chinese Remainder Theorem $R\times S \cong A/(IJ)$. Then because $A/(IJ)$ is not a field, $R\times S$ is not a field.

• $R\times S$ is not even a domain: the elements $(0,1)$ and $(1,0)$ are non-zero and have zero product. – Mariano Suárez-Álvarez May 10 '13 at 22:28
• I know that RxS has zero divisors. I'm just wondering whether it's possible to prove this using the Chinese Remainder Theorem. – Danny May 10 '13 at 22:30
• Take $A=R\times S$, $I=R\times 0$ and $J=0\times S$. – Mariano Suárez-Álvarez May 10 '13 at 22:32
• Proving this thatway feels so wrong that should not be done in public! :-) (In particular, I am pretty sure you cannot prove $A/(IJ)\cong R\times S$ in that example without using the fact that $R\times S$ is not a domain, and then using this to prove that it is not a field is beyond weird) – Mariano Suárez-Álvarez May 10 '13 at 22:34
• I agree 100% with Mariano. It is absolutely trivial that $R \times S$ is not a field, what's the purpose in proving this with CRT? And of course it is placed perfectly in the section about CRT because CRT computes products, and it is good to know their properties. – Martin Brandenburg May 10 '13 at 22:40

2) In $\,R\times S\;$ , take $\,I:= R\times\{0\}\;,\;\;J=\{0\}\times S\,$