11
$\begingroup$

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

$\endgroup$
12
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Can you clarify why $\pi_1 (P)$ is prime in $R_1$? $\endgroup$ – Future Dec 26 '15 at 3:19
  • 3
    $\begingroup$ @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 $y \in \pi_1(P)$. $\endgroup$ – Emolga Feb 1 '16 at 17:41
  • $\begingroup$ So, what is the ring R in this answer? S itself? $\endgroup$ – neptun Nov 14 '16 at 2:27
2
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.