Extension of intersection of ideals Let $f:A \rightarrow B$ be a ring homomorphism and $\mathfrak{a}_1,\mathfrak{a_2}$ be ideals of $A$. Let $\mathfrak{a}^e$ denote the extension of an ideal $\mathfrak{a}$ of $A$ in $B$. An exercise shows that
$$(\mathfrak{a}_1 \cap \mathfrak{a}_2)^e \subseteq \mathfrak{a}_1^e \cap \mathfrak{a}_2^e.$$
It seems that $(\mathfrak{a}_1 \cap \mathfrak{a}_2)^e =\mathfrak{a}_1^e \cap \mathfrak{a}_2^e$ is wrong in general, but I can not find a counter-example. 

Hint. Try a ring extension which is not flat (see here why), e.g. $K[X^2,X^3]\subset K[X]$. (I leave you the pleasure to choose the right ideals.)
