Prime ideals in a finite direct product of rings

Question

Let $S=\prod_{i=1}^{n}{R_i}$ where each $R_i$ is a commutative ring with identity. The prime ideals of $S$ are of the form $\prod_{i=1}^{n}{P_i}$ where for some $j$, $P_j$ is a prime ideal of $R_j$ and for $i\neq j$, $P_i=R_i$.

Answer 1

It is clear that any ideal of $S$ of the form stated above is prime. Let $P$ be a prime ideal of $S$. For $1\leq k\leq n$, let $e_k$ be the element of $S$ whose $k$th coordinate is $1$ and all other coordinates are $0$. $P$ is proper, so some $e_j$ (say $e_1$) is not in $P$. For $k\neq 1$ we have $e_{1}e_k=0\in P$, so $e_k\in P$. Thus $0\times \prod_{i=2}^{n}{R_i}\subseteq P$. Let $\pi_1\colon S\to R_1$ be the canonical projection. Then $\pi_1(P)$ is a prime ideal of $R_1$ and $P=\pi_1(P)\times \prod_{i=2}^{n}{R_i}.$