The result may be well-known, in general form; but I faced difficulty in a simple case of the result. (Instead of referring to a result, I tried myself in some arguments, but came to some problems, and I want to solve these small problems I faced.)
Lemma: If $P_1,P_2,\ldots,P_k$ are prime ideals in a commutative ring $R$ with unity, and if $P_1\cap \cdots \cap P_k:=P$ is prime, then $P=P_i$ for some $i$.
Proof: Consider the natural map $f:R\rightarrow \frac{R}{P_1} \oplus \cdots \oplus \frac{R}{P_k}$; it is a ring homomorphism, with $\ker f=P_1\cap \cdots \cap P_k=P$ (prime, by assumption).
Thus, $f$ factors through $\overline{f}:\frac{R}{P} \rightarrow \frac{R}{P_1} \oplus \cdots \oplus \frac{R}{P_k}$, and this is injective.
Since $R/P$ is integral domain, so image of $\overline{f}$ should go inside $\frac{R}{P_i}$ for some $i$ (am I right? I couldn't properly justify this!)
Suppose $Im(f)\subseteq \frac{R}{P_1}$.
This forces that $\ker f\supseteq P_2\cup \cdots P_k$.
Thus, $P=P_1\cap P_2\cap \cdots \cap P_k\supset P_2\cap \cdots \cap P_k$. From this, we see that $P_1\supseteq P_2\cap \cdots \cap P_k$.
Question $1$. How should I proceed next?
Question $2$. Can one suggest some clear statements, via the CRT approach above, for the proof of Lemma?
Best Answer
To put it more starkly, suppose each $P_i $ contains an element $x_i$ that is not in $P$.
Notice that $\prod_i x_i\in\bigcap_i P_i=P$.
Therefore, one of the $x_i$ is in $P$. Contradiction.