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Consider the problem asked in my quiz of abstract algebra.

Let R be a commutative ring and P be a prime ideal and $I_1$ and $I_2$ are ideals such that $P= I_1 \bigcap I_2$. Then prove that either $P=I_1$ or $P= I_2$.

Attempt: $P=I_1 \bigcap I_2$ implies that $P\subseteq I_1 \bigcap I_2$ , and I proved that in this case $P\subseteq I_1$ or $P\subseteq I_2$. But I am unable to prove the converse that $I_1\subseteq P$ or $I_2\subseteq P$.

I shall be really thankful for your help.

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  • $\begingroup$ Assume $I_2 \subsetneq P$ and that $y\in I_2, y \notin P$. It follows for any $x\in I_1$ that $xy\in I_1 \cap I_2=P$, and hence $x\in P$. This implies $I_1 \subseteq P$. $\endgroup$
    – hm2020
    Commented Oct 14, 2021 at 13:40
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    $\begingroup$ Use \cap instead of \bigcap. $\endgroup$
    – Randall
    Commented Oct 14, 2021 at 13:48

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You need to use the definition of prime ideal.

Suppose $P\neq I_1$. So there exists $a\in I_1$ such that $a\neq P$. Thus for all $x\in I_2$, since $ax\in I_1I_2\subseteq I_1\cap I_2\subseteq P$ we must have $x\in P$. In other words, $I_2\subseteq P$. Therefore $$I_2\subseteq P=I_1\cap I_2\subseteq I_2.$$

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