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I am trying to figure out the proof of theorem in this link.

Theorem: Any prime ideal of a finite direct product ring $R=\prod_{i=1}^nR_i$ is of the form $P=\prod_{i=1}^nI_i$, where $I_j$ is a prime ideal of $R_j$ for some $j$ and $I_i=R_i$, for $i\neq j$?

I am considering the case $n=2$. That is, the prime ideals of $R \times S$ are of the form $P\times S$ or $R\times Q$ where $P$ is prime ideal of $R$ and $Q$ is prime ideal of $S$.

I want to show $P \times S$ is prime ideal of $R \times S$ by using the following fact $R/I=(\prod R_i)/(\prod I_i)\cong\prod R_i/I_i$. It is enough to show that quotient ring is integral domain.

Hence, $(R \times S) / (P \times S) \cong (R/P) \times (S/S) = (R/P) \times 0_S$. It is clear that $R/P$ is integral domain but why $(R/P) \times 0_S$ is an integral domain?

Pick two elements such that $(r_1P,0_S)(r_2P,0_S)=(P,0) \implies (r_1r_2P,0_S)=(P,0_S)\to r_1r_2 \in P \implies_{\text{P is prime ideal}}$ either $r_1 \in P$ or $r_2 \in P$. So we get $(r_1P,0_s)=(P,0)$. So, quotient ring is integral domain. Is my proof okay? Any better proof suggestions is welcomed.

Elise9
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  • At the end you probably mean that either $(r_1P, 0)=(P,0)$ or $(r_2P, 0)=(P,0)$. This is correct. Though it's easier to simply note that $R/P$ and $(R/P)\times 0_S$ are very naturally isomorphic, and being an integral domain is a property that is preserved under isomorphisms. – Mark Aug 04 '24 at 19:33
  • Of course! How did I miss that? Certainly, in general, any group $G$ is contained in direct product $G \times {1}$. In that case, $\varphi: R/P \to (R/P)\times 0_S$ by $rP \mapsto (rP,0_S)$ Clearly, injective and surjective. – Elise9 Aug 04 '24 at 19:41

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