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.


[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 $\bigcap_\lambda \mathfrak a_\lambda^e=(\bigcap_\lambda \mathfrak a_\lambda)^e$.

  • $\begingroup$ Something which is also worth noting: For a module, being finitely presented and flat is equivalent to being finitely presented and projective. $\endgroup$ – Uriya First Sep 12 '18 at 10:31
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    $\begingroup$ Lam's Prop. 4.44 shows that the "obvious" map $\varepsilon\!:\!(\prod_\lambda S/\mathfrak a_\lambda)\otimes R\to \prod_\lambda (S/\mathfrak a_\lambda\otimes R)$ is bijective if and only if $R$ is finitely presented over $S$, but actually, the statement $\bigcap_\lambda \mathfrak a_\lambda^e=(\bigcap_\lambda \mathfrak a_\lambda)^e$ would follow just from $\varepsilon$ being injective. This "more subtle" question (injectivity of $\varepsilon$) is "left out" in Lam's book, could you point me out where to find it? $\endgroup$ – MonLau Sep 19 '18 at 20:05
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    $\begingroup$ @ManLau Good point. The condition for this is that $R$ is a Mittag--Leffler $S$-module, see stacks.math.columbia.edu/tag/059M and stacks.math.columbia.edu/tag/0599. I am not sure if there is a simple characterization of the flat Mittag--Leffler modules. However, any direct sum of finitely presented modules is Mittag--Leffler. $\endgroup$ – Uriya First Sep 19 '18 at 20:25
  • $\begingroup$ Many thanks. I do not know whether it is simple but for further reference the article "D.Herbera, J.Trlifaj, Almost free modules and Mittag–Leffler conditions" gives a characterization of the flat Mittag-Leffler modules as $\aleph_1$-projective modules. $\endgroup$ – MonLau Sep 20 '18 at 7:13

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