Let $A$ be a commutative ring and let $\mathfrak{a}$, $\mathfrak{b}$ be ideals in $A$.
I am asked the following question:
Show that $\mathfrak{a} \cap \mathfrak{b}$ is the largest ideal of $A$ contained in the ideals $\mathfrak{a}$ and $\mathfrak{b}.$
Proof:
Let $x \in \mathfrak{a} \cap \mathfrak{b}$ $\implies x \in \mathfrak{a}$ and $x \in \mathfrak{b}$ and let $r \in A$. Then $x \in \mathfrak{a}$ and also $rx \in \mathfrak{a}$ since $\mathfrak{a}$ is an ideal. Since $x \in \mathfrak{b}$, we also have that $rx \in \mathfrak{b}$ since $\mathfrak{b}$ is an ideal. This implies that $rx \in \mathfrak{a} \cap \mathfrak{b}$.
So I have showed that $\mathfrak{a} \cap \mathfrak{b}$ is an ideal, but how do I show that $\mathfrak{a} \cap \mathfrak{b}$ is the largest ideal contained in both $\mathfrak{a}$ and $\mathfrak{b}$?
Thank you for your help!!!
Best Answer
Let $\;I\;$ be an ideal contained in both $\;\mathfrak a\,,\mathfrak b\;$ . Then it is also contained in their intersection $\;\mathfrak a\cap\mathfrak b\;$ , and this proves the maximality.