Every prime ideal of a product ring comes from one factor

Let $P\in \mathrm{Spec}(R\times S)$. The idempotents $e_1=(1,0)$ and $e_2=(0,1)$ satisfy $e_1{e}_2=0\in P$, so [L1] gives $e_1\in P$ or $e_2\in P$. They cannot both lie in $P$ because then $1=e_1+e_2\in P$. If $e_1\in P$, then every $(r,0)=e_1(r,s)$ lies in $P$, so $P=R\times \mathfrak{q}$ where $\mathfrak{q}=\{s\in S:(0,s)\in P\}$. The same argument with $e_2$ gives the alternative form $\mathfrak{p}\times S$. In either case the factor ideal is prime because products in the factor ring are products in $R\times S$.

Conversely, if $\mathfrak{p}\in \mathrm{Spec}(R)$, then $\mathfrak{p}\times S$ is a prime ideal of $R\times S$: if $(r_1,s_1)(r_2,s_2)\in \mathfrak{p}\times S$, then $r_1{r}_2\in \mathfrak{p}$, so [L1] gives $r_1\in \mathfrak{p}$ or $r_2\in \mathfrak{p}$, which means $(r_1,s_1)$ or $(r_2,s_2)$ lies in $\mathfrak{p}\times S$. The same proof works for $R\times \mathfrak{q}$.

Therefore every prime ideal of the product ring comes from exactly one factor.