Prime ideals in a finite direct product of rings

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 $i\neq j$, $P_i=R_i$.

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 $P=\pi_1(P)\times \prod_{i=2}^{n}{R_i}.$

• Can you clarify why $\pi_1 (P)$ is prime in $R_1$? – Future Dec 26 '15 at 3:19
• @Prospect Because of the following: Suppose $xy \in \pi_1(P)$ for some $x,y \in R_1$. Thus there is $(xy,t_1,t_2,...) \in P$. $P$ is prime, thus $(x,t_1,t_2,...) \in P$ or $(y,1,1,...) \in P$. Hence $x \in \pi_1(P)$ or $y \in \pi_1(P)$. – Emolga Feb 1 '16 at 17:41
• So, what is the ring R in this answer? S itself? – neptun Nov 14 '16 at 2:27

Let $R_{1}$ and $R_{2}$ be two commutative rings with unity.

Let $P$ be a prime ideal in $S=R_{1} \times R_{2}$.

Let $P= P_{1} \times P_{2}$ where $P_{1}$ is a ideal in $R_{1}$ and $P_{2}$ is a ideal in $R_{2}$.

Then $S/P \simeq R_{1}/P_{1} \times R_{2}/P_{2}$.

Since product of two integral domains is not an integral domain,

Therefore only one of the $P_{1}$ or $P_{2}$ is a prime ideal and other should be corresponding ring.

• The other ideal should be the corresponding ring. Thanks for pointing that out, I think the proof is complete now. – avCva Mar 16 '17 at 12:43