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:


1. $\varphi$ is injective, so $S$ is a subring of $R$.
2. For every ideal $I\subseteq S$, $I=I^e\cap S$.
3. 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.



Homomorphisms of ring extending nicely ideal intersections
[Note: I am assuming that you are considering commutative algebras. However, everything below also works in the non-commutative case if you use right (left) ideals, and assume that $R$ is finitely presented and flat as a *left* (*right*) $S$-module.]


Given an $S$-algebra $R$, the assumption that $R$ is **flat and finitely presented** as an $S$-module guarantees that $\bigcap\_\lambda \mathfrak a\_\lambda^e=(\bigcap\_\lambda \mathfrak a\_\lambda)^e$ for any family of ideals $\{\mathfrak a\_\lambda\}$ in $S$. 


The assumption that $R$ is **flat** over $S$ guarantees the same thing for finite families of ideals $\{\mathfrak a\_\lambda\}$. 
[**Edit:** However, it may fail for infinite families. For example, consider $S=\mathbb{Z}$, $R=\mathbb{Q}$ and the ideals $\{n\mathbb{Z}\}\_{n\in \mathbb{N}}$ (for a faithfully flat coutnerexample, take $R=\mathbb{Q}\times\mathbb{Z}$ and $\{n\mathbb{Z}\times 0\}\_{n\in \mathbb{N}}$).]


*Proof.* Consider the exact sequence 
$$0\to \bigcap\_\lambda\mathfrak a\_\lambda \to S\xrightarrow{f}
\prod\_\lambda S/\mathfrak a\_\lambda$$
in which the first map is the inclusion and $f$ is defined by $f(s)=(s+\mathfrak a\_\lambda)\_\lambda$.
Since $R$ is flat over $S$, by tensoring with $R$, we get an exact sequence
$$0\to (\bigcap\_\lambda\mathfrak a\_\lambda)\otimes R \to S\otimes R \xrightarrow{f\otimes\mathrm{id}\_R}
(\prod\_\lambda S/\mathfrak a\_\lambda)\otimes R$$
(all tensors are taken over $S$).


The middle term $S\otimes R$ is isomorphic to $R$ via $s\otimes r\mapsto \varphi(s)r$. In addition, since $\{\mathfrak a\_\lambda\}$ is finite or $R$ is finitely presented over $S$, the right term is isomorphic to $\prod\_\lambda ((S/\mathfrak a\_\lambda)\otimes R)$ via the "obvious" map (see Lam's "Lectures and Modules and Rings", Proposition 4.44). As $(S/\mathfrak a\_\lambda)\otimes R\cong R/a\_\lambda^e$, the sequence becomes:
$$0\to (\bigcap\_\lambda\mathfrak a\_\lambda)\otimes R \xrightarrow{s\otimes r\mapsto \varphi(s)r} R\to 
\prod\_\lambda R/\mathfrak a\_\lambda^e.$$
The exactness at $R$ means that $\bigcap\_\lambda \mathfrak a\_\lambda^e=(\bigcap\_\lambda \mathfrak a\_\lambda)^e$.


