**Given R and S rings, then R \times S can’t be a field**

**Remark** It is good to check out what the structure is from an object. For example, a field consists of elements (\neq 0) which has an inverse, or in other words, is a unit in the field.

**Proof.** Assume by contradiction that R \times S is indeed a field. Then it consists elements of the form (r,s) \in R \times S which has an inverse (r^{-1}, s^{-1}) \in R \times S. That also means that the multiplication of two (r,0),(0,s) \in R \times S has an inverse in R \times S. So we get that (r,0)(0,s) = (0,0) \in R \times S which implies that (r,0) is a zero divisor. That is not possible in a field. Therefore, R \times S can’t be a field, which completes the proof.