Let $\varphi\!:\!S\to R$ be a homomorphisms of $K$-algebras for some field $K$. Let $\{a_{\lambda}\}_{\lambda}$ be a family of ideals of $S$.

Is there some "natural" assumption on $\varphi$ to guaranty that $$ \cap_{\lambda} a_{\lambda}^e = (\cap_\lambda a_\lambda)^e, $$ where $\_^e$ denotes the extension to $R$.

The inclusion $\cap_{\lambda} a_{\lambda}^e \supseteq (\cap_\lambda a_\lambda)^e$ is always satisfied. Which conditions on $\varphi$ (as general as possible) guaranty that the converse inclusion is also satisfied?

For example, if $R= S\otimes_K T$ for some $K$-algebra $T$ and $\varphi(s)=s\otimes 1$. Let's check it.

The ring $R$ is a free $S$ module via $\varphi$. Moreover, given a $K$-basis $\{t_l\}_l$ of $T$, the set $\{1\otimes t_l\}_l$ is a $S$-base of $R$. So, given an ideal $I\subseteq S$ and $r\in R$ with $r=\sum_l s_l(1\otimes t_l)$, then $r\in I^e$ if and only if $s_l\in I$ for all $l$. Hence, if $r\in \cap_\lambda a_\lambda^e$, then $s_l\in a_\lambda$ for all $l$ and $\lambda$, that is $s_l\in\cap_\lambda a_\lambda$ for all $l$ and then $r\in (\cap_\lambda a_\lambda)^e$.

From the geometric point of view it is clear that the corresponding map $f\!:\! X\to Y$, where $X=Spec(R)$ and $Y=Spec(S)$, has to be surjective. Also flatness looks a reasonable assumption and then $\varphi$ is faithfully flat. But with this two assumptions on $\varphi$, I am not able to fine neither a proof that $\cap_{\lambda} a_{\lambda}^e \subseteq (\cap_\lambda a_\lambda)^e$ nor a counterexample. Hence, I am not sure whether "faithfully flat" is the assumption on $\varphi$ that I am looking for or not, but I think it is. (In the example $R=S\otimes T$, $\varphi$ is faithfully flat).

Facts about faithfully flat homomorphisms that could be useful are:

- $\varphi$ is injective, so $S$ is a subring of $R$.
- For every ideal $I\subseteq S$, $I=I^e\cap S$.
- For every prime ideal $p\subseteq R$ the ideal $p\cap S$ is a prime ideal of $S$.

Any suggestion or comment would be highly appreciated.