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.
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 $$I_2\subseteq P=I_1\cap I_2\subseteq I_2.$$