This is Exercise 1.2.6(a) in Liu, Algebraic Geometry and Arithmetic Curves
- Let $B$ be a flat $A$-algebra. Show that for any finite family $\{I_\lambda\}_{\lambda\in \Lambda}$ of ideals of $A$, we have $\cap_{\lambda\in\Lambda}(I_\lambda B)=(\cap_{\lambda\in \Lambda}I_\lambda)B$.
How can I prove that? Also does the following holds:
Let $B$ be a flat $A$-algebra. Does there exist an infinite family $\{I_\lambda\}_{\lambda\in \Lambda}$ of ideals of $A$ such that $\cap_{\lambda\in\Lambda}(I_\lambda B)\ne(\cap_{\lambda\in \Lambda}I_\lambda)B$?
Let $B$ be an $A$-algebra. Does there exist a finite family $\{I_\lambda\}_{\lambda\in \Lambda}$ of ideals of $A$ such that $\cap_{\lambda\in\Lambda}(I_\lambda B)\ne(\cap_{\lambda\in \Lambda}I_\lambda)B$?
Let $B$ be a flat $A$-algebra and let $I_1,\dots,I_n$ be a finite set of ideals of $A.$ Then clearly, $(\cap_{i=1}^n I_i)B \subseteq \cap_{i=1}^n (I_iB).$ To prove the other inclusion, consider the following exact sequence with natural maps: $$0 \to \cap_{i=1}^n I_i \to A \to \oplus_{i=1}^n (A/I_i).$$ Since $B$ is $A$-flat, tensoring by $B$ will remain exact. So we have an exact sequence $$0 \to (\cap_{i=1}^n I_i)B \to B \to \oplus_{i=1}^n (B/I_iB).$$ Now the kernel of the map $B \to \oplus_{i=1}^n (B/I_iB)$ is $\cap_{i=1}^n(I_iB).$ This shows that $\cap_{i=1}^n (I_iB) \subseteq (\cap_{i=1}^n I_i)B.$
Let $A = \mathbb Z, B = \mathbb Q.$ Let for each $n \in \mathbb N, I_n = (2^n)$ be the ideal in $A$ generated by the element $2^n.$ Then $\cap_{n\in\mathbb N} I_n = (0) \Rightarrow (\cap_{n\in\mathbb N} I_n)B = (0).$ On the other hand $I_nB = B, \forall n \in \mathbb N.$ So $\cap_{n\in\mathbb N} (I_nB) = B.$