If $R$ and $S$ are rings then $R \times S$ is never a field (or domain) 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?
 A: Let us prove that $R\times S$ is not a field if $R$ and $S$ are any two non-zero rings. The following steps lead to a solution:
(1) Note that $(1_R,0_S)\in R\times S$ where $1_R$ is the multiplicative identity of $R$ and $0_S$ is the additive identity of $S$.
(2) If $T$ is a ring, if $0_T$ is the additive identity of $T$, and if $t\in T$, prove that $t0_T=0_T=0_Tt$. (Hint: use the distributive law and write $0_T=0_T+0_T$.)
(3) Prove that $(1_R,1_S)$ is the multiplicative identity of $R\times S$ where $1_R$ and $1_S$ are the multiplicative identities of $R$ and $S$, respectively.
(4) If $(r,s)\in R\times S$, prove that $(r,s)\cdot (1_R,0_S)=(r,0)$.
(5) Finally, prove that there does not exist $(r,s)\in R\times S$ such that $(r,s)\cdot (1_R,0_S)=(1_R,1_S)$. Deduce that the element $(1_R,0_S)\in R\times S$ has no multiplicative inverse. (Hint: note carefully where you use the fact that $R$ and $S$ are non-zero rings.)
I hope this helps!
A: Hint $ $ Ring factorizations correspond to idempotents, i.e. nontrivial product rings contain idempotents $\rm\:e = e^2\:$ that are nontrivial $\rm\: e\ne 0,1,\:$ e.g. $\rm\ e = (0,1)\in R\times S\:.\:$  But a field or domain has only trivial idempotents since $\rm\ e\: (1-e) = 0\ \Rightarrow\ e = 0,1.\:$ Note that the nontrivial idempotent $(0,1)$ may be viewed as constructed by diagonalizing out of the set of trivial idempotents $(0,0),\: (1,1)\:.$
Conversely a nontrivial idempotent $\rm\:e\:$ yields a nontrivial factorization $\rm\:R \cong e\:R \times (1-e)\:R\:$ (the Peirce decomposition). This decomposition extends to any finite set of idempotents having sum $1$ and being  mutually orthogonal, i.e. $\rm\:e_i\, e_j = 0\ \ if\ \ i\ne j\:.$ For example, the Chinese Remainder Theorem (CRT) may be viewed in this manner.
A: Hint: What is the inverse of $(1,0)$?
A: Since $R$ is nonzero, there is an $r\in R$, $r\neq 0$. Since $S$ is nonzero, there exists $s\in S$, $s\neq 0$.
Since $r\neq 0$, then $(r,0_S)\in R\times S$ is not zero; and since $s\neq 0$, then $(0_R,s)\in R\times S$ is not zero.
What is the product of $(r,0_S)$ and $(0_R,s)$? 
Can that happen if nonzero elements are cancellable?
Are nonzero elements cancellable in a division ring?
(This shows a slightly stronger statement: a direct product of nontrivial rings always has zero divisors.)
Alternatively. Prove that $R\times\{0\}$ is a proper ideal of $R\times S$ (here you use that $R\neq 0$ and $S\neq 0$). Then remember that fields are simple. 
