Let $R$ and $S$ be two commutative rings with unity. Prove that $R\times S$ is NOT an integral domain.
This is the best I could think of so far, please give me a push in the right direction and correct me.
It suffices to show that $R\times S$ has zero-divisors
Therefore, let $a$ be an element of $R$ and $b$ be an element of $S$, thus $R\times S$ has elements of the form $(a,b)$.
Consider the elements $A=(a,0)$ and $B=(0,b)$, such that $a$ does not equal zero and $b$ does not equal zero. Then $AB = 0$. Therefore $R\times S$ has zero divisors.
Best Answer
Your argument here is perfectly correct, as others have indicated.
It would help the clarity of your proof (and, on a fairly pedantic level, would make your proof more correct) if you chose specific non-zero elements $a,b$. In particular, we could take $a = 1_R$ and $b = 1_S$.
This way, we could always say that we're explicitly using the fact that a ring with unity has both a $1$ and a $0$.