Timeline for Existence of a reduced fiber implies generically reduced (Exercise III-74 from Geometry of Schemes)
Current License: CC BY-SA 4.0
18 events
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| Nov 4, 2023 at 0:39 | comment | added | user267839 | But even if we add this flatness assumption in (#2) (indeed, in that case we get that $B_m \cong A_m^q$ as $A$- modules (!, that's the crucial problem) for free. But how we can then deduce from this that $B_m$ is reduced? Note, the $A$-mod iso $B_m \cong A_m^q$ not "sees" the ring structure of $B$, which carries the information about reducedness. Do you see how to remedy this problem in this "elementary" approach in (#2) under additional assumption $f$ to be flat? | |
| Nov 3, 2023 at 11:05 | comment | added | Libli | @user267839 : You are right. Thanks for pointing out this! There is a gap in the proof (and probably in the statement). One needs to assume that the hypothesis holds for generic $y \in Y$. Then we can localize at a generic point on $Y$ where $f$ is flat. | |
| Oct 26, 2023 at 16:58 | comment | added | user267839 | An update: the equality $\operatorname{Ker}(\Phi) = \mathfrak{m}.\operatorname{Ker}(\Phi)$ would surely hold if we in (#2) additionally assume that $B_{\mathfrak{m}}$ is flat over $A_{\mathfrak{m}}$ because of tag/00HL, but you not assumed it there, right? So I not see how else - without this additional flatness assumption - there the inclusion $\operatorname{Ker}(\Phi) \subset \mathfrak{m}.\operatorname{Ker}(\Phi)$ should follow. Could you sketch the idea? | |
| Oct 26, 2023 at 14:17 | comment | added | user267839 | There is still a detail in the added proof (#2) I not completely understand: Why holds the inclusion $\operatorname{Ker}(\Phi) \subset \mathfrak{m}.\operatorname{Ker}(\Phi) $? We know that $\operatorname{Ker}(\Phi)$ considered as $A_{\mathfrak{m}}$-module is finitely generated (since base ring Noetherian) and torsion free, but why this implies the quoted inclusion? | |
| Sep 4, 2023 at 8:43 | history | edited | Libli | CC BY-SA 4.0 |
Added two (porbably) more elementary proofs of the result the OP was hoping for.
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| Aug 31, 2023 at 9:23 | comment | added | Libli | Let us continue this discussion in chat. | |
| Aug 30, 2023 at 20:31 | comment | added | user267839 | It looks all in all rather doable just by a simple calculation - even without explicitly using notion of regularity, smoothness etc, so very basic - but I not see how to finish this part. Do you have an idea? | |
| Aug 30, 2023 at 20:30 | comment | added | user267839 | So the local version of the problem becomes, why this $R$ is reduced if we choose $U$ "small enough", under assumption that $R \otimes (A/ \mathfrak {m})=R/ \mathfrak {m} \cdot R $ is product of 4 copies of $K$ as assumed. And I' m wondering if it's possible to deduce directly that $R$ is reduced, by assuming there exist some nilpotent $n \in R$ and concluding directly that it must be zero. What we immediately see that such $ n$ must be contained in the ideal $\mathfrak {m} \cdot R $. If we could somehow show that it must be in the image of $A$, since $A$ is reduced, and we win. | |
| Aug 30, 2023 at 20:26 | comment | added | user267839 | Let now $\phi:A \to R$ be a flat, finite ring map (in framework of the exercise $R$ can be easily constructed explicitly by choosing a ring $S$ corresponding to some open affine $U \subset \mathcal{C}_1 \cap \mathcal{C}_2 \subset \mathbb{P}^2_B$, which by construction is a $K[a,...,l]$ algebra, and then set $R:= S \otimes_{K[a,...,l]} A$. | |
| Aug 30, 2023 at 20:24 | comment | added | user267839 | Yes, this also definitely works and also shows a much stronger statement about $X$ (smooth > regular > reduced). What I had in mind by (let me say very very) elemenary , was something like this: Say we have a regular local ring $(A, \mathfrak {m})$ ( obtained from $K[a,...,l]$ by localizing at a prime $ m$ , over which the fiber of $p_1$ is reduced & consists of 4 points ( ie whos existence one shows in the quoted exercise and which corresponds to your $ b$ from last comment). | |
| Aug 30, 2023 at 17:18 | comment | added | Libli | The module $\Omega_B$ being locally free, the cotangent exat sequence shows that $\Omega_X$ is locally free at $x_1$. We deduce that there exists an open neighborhood $V$ of $x_1$ in $X$ such that restriction of $\Omega_X$ to this open is locally free. And so $X$ is smooth at each points of $U$. | |
| Aug 30, 2023 at 17:17 | comment | added | Libli | As for an elementary argument (which would somehow obscure the intuition you had in your metaquestion), this could go as follows. Let $b \in B$ such that the fiber of $p_1$ over $b$ is 4 reduced points say $x_i$. Then $\Omega_{X/B}$, module of relative differentials of $X$ over $B$, is locally free say at $x_1$. Thus there exists an open neighborhood of $x_1$ in $X$ such that the restriction of $\Omega_{X/B}$ to $U$ is locally free. Since the fiber of $p_1$ over $b$ is smooth of the correct dimension, we deduce that the cotangent map $\Omega_B \rightarrow \Omega_X$ is injective at $x_1$. | |
| Aug 30, 2023 at 16:57 | comment | added | Libli | I don't think the points being $K$-rational play such an important role. As Karl Schwede commented below your question, for this type of problem, you really want to deal with geometric fibers. | |
| Aug 30, 2023 at 15:50 | comment | added | user267839 | deduced with "elementary" tools, without exploitation of these Serre's $R_k, S_k$ characterizations. Do you have an ad hoc idea? | |
| Aug 30, 2023 at 15:48 | comment | added | user267839 | essentially this completely solves the posted problem, but one detail still makes me ponder: in (a) one shows an even stonger statement about a fiber: it is not only a finite union of points with reduced scheme structure, but moreover every point is moreover $K$-rational. Also, in the quoted book the authors not explicitly introduced this technical result you stated above. Do you maybe know an "elementary" way to conclude the generical reducedness directly, without exploiting this result? All in all this is an introductory book, and it's interesting to know if the claim can also | |
| Aug 30, 2023 at 13:52 | vote | accept | user267839 | ||
| Aug 29, 2023 at 18:32 | history | edited | Libli | CC BY-SA 4.0 |
deleted 11 characters in body
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| Aug 29, 2023 at 18:25 | history | answered | Libli | CC BY-SA 4.0 |