Let $R$ be an integral domain and $I,J$ be ideals such that $IJ$ is a principal ideal . Then is it true that $I$ is finitely generated ?
I was thinking like if $IJ=(a)$ , where $a=\sum_{i=1}^{k} x_iy_i$ , $x_i\in I , y_i \in J$ , then we might have $I=(x_1,...,x_k)$ , but I am not sure and I cannot proceed further . Please help , Thanks in advance
Here is how to continue your argument.
Suppose $IJ=(a)$, where $a=\sum_{i=1}^{k} x_iy_i$ , $x_i\in I , y_i \in J$. Let $I_0=(x_1,\ldots,x_k)$ and $J_0=(y_1,\ldots,y_k)$. We must have $I_0\subseteq I$ and $J_0\subseteq J$, and the goal is to show that we also have $I\subseteq I_0$.
Choose any $x\in I$. Since $xy_i\in IJ=(a)$ for $1\leq i\leq k$ there must exist $r_i\in R$ such that $xy_i=r_ia$ for $1\leq i\leq k$. Multiplying by $x_i$ yields $xx_iy_i=r_ix_ia$. Summing this equality as $i$ ranges from $1$ to $k$ yields $xa=x(\sum x_iy_i) = (\sum r_ix_i)a$. From the domain property, we may cancel $a$ to obtain $x=\sum_{i=1}^k r_ix_i\in I_0$. This proves $I\subseteq I_0$, as required.