If $A$ is a prime ideal on $R\times S$, then $A = P \times S$, where P is a prime ideal on R or $A = R\times Q$, where $Q$ is a prime ideal on $S$.

Question

Let $R$ and $S$ be rings with unity. If $A$ is a prime ideal of $R\times S$, then $A = P \times S$, where $P$ is a prime ideal of $R$ or $A = R\times Q$, where $Q$ is a prime ideal on $S$.

Answer 1

First, you should check that every ideal $I \subseteq R \times S$ has the form $I = I_1 \times I_2$ for some ideals $I_1 \subseteq R, I_2 \subseteq S$.

Then, let $\mathfrak{p} \subseteq R\times S$ be a prime ideal. By the previous fact, there are ideals $I_1 \subseteq R, I_2 \subseteq S$, such that $\mathfrak{p} = I_1 \times I_2$. Since $\mathfrak{p}$ is a proper ideal, at least of the factors is proper. In fact, exactly one of them is proper. Suppose both of them are proper ideals. Then $(0,1), (1,0) \notin \mathfrak{p}$ but $(0,0) = (1,0)\cdot (0,1)\in \mathfrak{p}$, which is a contradiction since $\mathfrak{p}$ is prime. What can you tell about the proper ideal, knowing that $\mathfrak{p}$ is prime?