This question is from the book "Abstract Algebra" by Dummit and Foote.(Exercise: 7.6.4):
Problem: Prove that if $R$ and $S$ are any two non-zero rings then $R \times S$ is never a field.
My problem with this question is that it doesn't feel right. If $R$ and $S$ are rings then $R \times S$ is also so. (The addition and multiplication on $R \times S$ is defined as $(r_1,s_1)+(r_2,s_2)=(r_1+r_2,s_1+s_2)$ and $(r_1,s_1)(r_2,s_2)=(r_1r_2,s_1s_2)$ respectively where $(r_1,s_1),(r_2,s_2) \in R \times S$.) So proving that $R \times S$ is a field reduces to proving commutativity of multiplication, existence of multiplicative indentity and existence of multiplicative inverse for each element of it. Now, if the two rings are fields themselves then multiplication is commutative, the identy is $(1_R,1_S)$ and inverse of an element $(r,s)$ is $(r^{-1},s^{-1})$. So $R \times S$ is also a field.
Am I right? Please help.
Update: The question comes from a section about the Chinese Remainder Theorem. Is it possible to prove this fact using Chinese Remainder Theorem?
Best Answer
Hint: What is the inverse of $(1,0)$?