Timeline for Normalisation of models of elliptic curves in finite extensions and reducedness of fibres
Current License: CC BY-SA 4.0
43 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 22 at 19:39 | comment | added | user267839 | As the model we considering is regular, clarly the components of special fibre correspond to discrete valuations extending valuation of $R$. But it's not clear for me what you mean by a "valuation modulo the valuation of $q$"? | |
| Jun 22 at 19:26 | comment | added | user267839 | #Update: Think I got meanwhile the idea you described in 4th-to-last paragraph. So you give characterization the pts of kernel of the $d$-isogeny should satisfy (= specialization of each "integral pt" of kernel is contained in different comp of special fibre), and in 3rd-to-last pgraph you select the subgroups which may appear as kernel. What I still not got is the part with Tate curve: "The components (of special fibre?) are given by the different possible valuations modulo the valuation of $q$"? Could you give reference elaborating this characterization of comps in terms of Tate curve? | |
| Jun 7 at 10:26 | comment | added | user267839 | typo in 2nd com: of course in last line is meant $\mathcal{D} $ and not $\overline{D}$... | |
| Jun 7 at 10:05 | comment | added | user267839 | each $ x_i:=\overline{\{k_i\}} \cap D_k$ (... the $\overline{\{k_i\}}$ inside $\mathcal{D}$ corresp to the $K_i$ above) lies in different component of $D_k$, right? So it "distributes" the kernel pts injectively on comps of $D_k$. Is this the whole point behind the "So it suffice [...]" part? So which subgroup is allowed to be the kernel of this isogeny, that's it? Or is there additional "dynamic" around "where the torsion points land" going on neccessary to determine the $d$ isogeny we want to characterize? | |
| Jun 7 at 10:01 | comment | added | user267839 | Now actually my question is, when you write that to characterize this unique deg $d$-isogeny with red special fibres it suffice to characterize which subgroups of the torsion points of an elliptic curve with multiplicative reduction map injectively to the components, you asking there which subgroup of tp of $F$ comes combinatorically into question as a candidate to be the kernel of such isogeny subjected to neccessary & sufficient condition that each of it's points intersects $D_k$ in different components, in sense of that for $k_i$ in kernel of $f$ | |
| Jun 7 at 0:13 | comment | added | user267839 | the "lifts" of $x_i$ corresp to elements of "integral kernel", ie components $K_i$ of preimage of $\overline{\{e_E\}}$ correspding to $x_i$ via $x_i = K_i \cap D_k$, right? So the point seems to be that by this each $K_i$ corresponds to unique component of $D_k$, so we see can interpret is as injective map of "components of integral kernel" to components of $D_k$. | |
| Jun 7 at 0:12 | comment | added | user267839 | there exist exactly one cover $D_k \to C_k$ of deg $d$ descending from geom covering (=base change to alg closure of $k$) such that "the identity lifts to a $k$-rational point". Once we have such cover $D_k \to C_k$ with exactly $d$ points $x_1,..., x_d \in D_k$ (with at least one $k$-ratnl) over the "identity" (...I guess by "identity" you mean $e_k = C_k \cap \overline{\{e_E\}}$ (with $e_E$ identity in $E$), as strictly spking $C_k$ has no scheme structure), by Hensel this corresponds to cover $\overline{D} \to \mathcal{C}$, and | |
| Jun 6 at 23:56 | comment | added | user267839 | (I tried to condense the points which still not clear) Re to the second question on why to characterize this unique deg $d$-isogeny with reduced special fibres it suffice to characterize which subgroups of the torsion points of an elliptic curve with multiplicative reduction map injectively to the components: You refer to $7$-th paragraph. There is stated: "The kernel of the isogeny consists of lifts of $d$ points on $d$ different components of the special fiber of $\mathcal{D}$". The upshot so far I understand is the following: In prev paragraph you stated on level of special fibres | |
| Jun 3 at 10:35 | comment | added | Will Sawin | @user267839 A covering $A$ doesn't naturally have the structure of a group (i.e. a fixed identity), it's just a scheme. So when we say $A \to B$ is an isogeny we just mean there exists a choice of group structure that makes it an isogeny. This is the same as a choice of identity so we can just choose $b$ as the identity. For your last question, this is explained in the fourth-to-last paragraph: Any such isogeny must be finite étale on the integral model and thus must be the desired isogeny. | |
| Jun 3 at 0:06 | comment | added | user267839 | Furthermore, once having established that the exist a unique degree $d$ isogeny with reduced special fibres in penultimate paragraph you claim that to characterize it, it suffice to analize which subgroups of the torsion points of an elliptic curve with multiplicative reduction map injectively to the components. Why this information completely suffice to to characterize this isogeny? | |
| Jun 3 at 0:06 | comment | added | user267839 | (...modulo typos swapping $R$ and $K$ above in tensorprod) Two points are still not clear to me. Why a finite étale cover of a genus $1$ curve over $K$ where the identity lifts to a $K$ -rational point is an isogeny? Say we have such map $g: A \to B$ and $b \in g^{-1}(e_A)$ is $K$-rational. Clearly we can precompose this map by "shift by $-b$ map and would obtain by such composition an isogeny. Can we from this somehow deduce that $g$ was already an isogeny? Or does this require another approachnto see it? | |
| Jun 2 at 18:57 | comment | added | Will Sawin | @user267839 That all looks right to me. | |
| Jun 2 at 18:45 | comment | added | user267839 | (And then we have chain of equivalences: $f: F \otimes K^{ur} \to E\otimes K^{ur} $ connected (as cover, by geom conn. for ellipt curves definitionally, as above?) **iff** $\mathcal{D} \otimes_K R^{ur} \to \mathcal{C} \otimes_K R^{ur} $ (with $R^{ur}$ integral closure of $R$ in $K^{ur}$) by density of generic fibre, right? iff $D_k \otimes_k \overline{k} \to C_k \otimes_k \overline{k}$ connected due to equivalence of coverings over Henselian local rings & their residue fields + corresp between unramified & residue field extensions. Is this correct equivalence chain? | |
| Jun 2 at 18:44 | comment | added | user267839 | So you mean this boils down to connectedness of base change of $F$ to unramified closure of base $K$, right? And why this holds? Due to just that geometric connectedness of elliptic curves (as part of definition), right? | |
| Jun 2 at 18:06 | comment | added | Will Sawin | @user267839 Yes, by geometrically I mean to pass to the algebraic closure. It's connected over $\overline{k}$ since the original covering is connected over the completion of the maximal unramified extension of $K$. Then classify these covers over an algebraically closed field and descend down. Of course there are other possible approaches here, like to rely more heavily on a classification of isogenies via the Tate curve. | |
| Jun 2 at 17:33 | comment | added | user267839 | equivalence of cov's) must geometrically consist of a loop of $\Bbb P^1$'s winding around that loop multiple times."[...] What do you mean by "geometrically" there? Did you passed at that stage briefly to algebraic closure of $k$ and analyzed the shape the covering should have there firstly? (it seems like that to me because in next sentence you intend to "descend back" to $k$, suggesting that in previous sentence your reasoned on level of alg closure of $k$). | |
| Jun 2 at 17:32 | comment | added | user267839 | There is another step I not fully understand on the construction of the covering of $\mathcal{C}$ (due to the equivalence over Henselian local $R$ this would indeed be equivalent to construct covering over special fibre $C_k$.) You wrote: "An unramified covering is constant over each $\Bbb P^1$ (= a component of the special fibre $C_k$) and thus a geometrically connected unramified covering of $\mathcal{C}$ (...here I assume you switch between cov's over $R$ and $k$ in accordsnce to the mentioned | |
| May 31 at 23:05 | comment | added | Will Sawin | @user267839 If $U$ is the smooth locus of the minimal proper regular model then $U$ is the Neron model so the group law $E \times E \to E$ extends to a map $U \times U \to U$, and it saisfies the group identities since $E \times E \to E$ does. Maybe one can do this without the Neron model. For $C$ a component of a group containing an element $g$, multiplication by $g$ gives an isomorphism from the identity component to $C$. | |
| May 31 at 19:38 | comment | added | user267839 | On the direct proof: Why does the smooth locus of a reduced component of the special fibre have group structure? Also, once knowing this, assume we pick now two reduced components, where one has smooth locus isomorphic to $\Bbb G_m$. Why is then the smooth locus of the other is also consequently $\cong \Bbb G_m$? (...surely, with Kodaira table as you wrote its immediate, but I wondering how one sees it directly without) | |
| May 31 at 19:00 | comment | added | Will Sawin | @user267839 But now calculate the intersection number of the $\mathbb G_m$ component with the fiber, which we know is zero. Its self-intersection is $-2$ and there are two adjacent components, each with intersection number at least $1$, so both adjacent components must have intersection number $1$, which forces them to be reduced (otherwise the intersection number would be divisible by the multiplicity). | |
| May 31 at 18:59 | comment | added | Will Sawin | @user267839 Any source (e.g. Silverman) should do if you want the argument using the structure theory. The point is to just look through the Kodaira classification and observe that only the I_n have any components with smooth locus G_m. (It's crucial here that non-reduced componens have empty smooth locus). For a direct proof, the smooth loci of the reduced components form a group so if one is G_m they are all G_m, and thus one just needs to check each component is reduced. By induction, it suffices to check each component adjacent to a G_m component is reduced. | |
| May 31 at 14:11 | comment | added | user267839 | Just to summarize: Your argumentation in comment from May 27, 15:15 is about the allowed intersection combinatorics of the components of the special fibre, but what is still not clear to me is how to obtain the "=>" direction in the equivalence from your penultimate comment. (ie that the type of regular locus of every component is determined completely by r.l. of only one picked component. That appears from naive point of view rather surprising. | |
| May 31 at 13:42 | comment | added | user267839 | How much is involved in the left-to-right implication? Could you recommend a say "standard source" working this out (...if that's not "immediate" and so there really the whole arsenal of structure theory of elliptic fibrations is involved)? It looks that there is a kind of action present in the backgroup permuting the components (and so mapping regular loci to regular loci, such that it suffice to analyze only the regular locus of only one comp; but I'm wondering if somewhere this implication is rigorously worked out) | |
| May 31 at 13:01 | comment | added | Will Sawin | @user267839 yes, every component. | |
| May 31 at 12:28 | comment | added | user267839 | On this equivalence you stated there for the minimal proper regular model: Presumably, you wanted to write that "smooth locus of one component of the special fiber of the minimal proper regular model isomorphic to $\Bbb G_m$ is equivalent to that the smooth locus of every component(!) of the special fiber is isomorphic to $\Bbb G_m$, or am I misunderstanding you there? (as above we allow the special fibre to nave several components) | |
| May 31 at 0:45 | comment | added | Will Sawin | @user267839 Having the smooth locus of one component of the special fiber of the minimal proper regular model isomorphic to $\mathbb G_m$ is known to be equivalent to having the smooth locus of every fiber isomorphic to $\mathbb G_m$. If asked to give a definition of multiplicative reduction I would probably pick one of those. | |
| May 30 at 23:57 | comment | added | user267839 | fibre of $\mathcal{C}$ can have more than one components? A naive guess: Does it here mean that for every component $C_i$ of special fibre holds that $C_i - \cup_{j \neq i} C_j$ is isomorphic to $\mathbb{G}_m$? And so every component of the special fibre of $\mathcal{C}$ is more or mless by definition a $\mathbb{P}^1$ (since we not allow triple intersections)? | |
| May 30 at 23:55 | comment | added | user267839 | yes, reading the question the term "multiplicative reduction" indeed confused me a bit in this context. I know it from models of elliptic curves where one says that a curve has "multiplicative reduction" if the special fibre of its model is nodal curve, especially irreducible (intuitively the "multiplicative" refers here to that if we remove this node= singular locus, then the complement is $\cong \mathbb{G}_m$). But what does "multiplicative reduction" mean here, where it seemingly allowed that the special | |
| May 27 at 15:15 | comment | added | Will Sawin | @user267839 It probably depends a bit on how you define "multiplicative reduction". One knows various definitions are equivalent because of the structure theory of elliptic fibrations, so one doesn't need to specify a definition. If one knows the special fiber is reduced and topologically a loop of $\mathbb P^1$, then one computes the intersection of the special fiber with one of the $\mathbb P^1$s: On the one hand this is zero because the fiber is movable, and on the other hand this is two minus the intersection multiplicites of the two singularities, so the multiplicities are both $1$. | |
| May 27 at 14:28 | comment | added | user267839 | Can one see this "instantly" (=so say "short" argument) - at least that all components are $\Bbb P^1$'s - or does this require to invoke "deeper" results from structure theory of elliptic fibration and so cannot be seen "immerdiately"? | |
| May 27 at 14:28 | comment | added | user267839 | nice, so properness of the model (resp it's base changes) goes exactly here technically in to assure finiteness (=quasi finite + proper), otherwise we have in general no purity. Another point: two paragraphs lower you mention that the special fiber of $\mathcal{C}$ is a loop of the lines $\Bbb P^1$, so all comps intersect each other with multiplicity one. Why is that the case? | |
| May 27 at 14:13 | comment | added | Will Sawin | @user267839 The reducedness of $C_k$ holds since $\mathcal C$ is the minimal proper regular model of an elliptic curve with multiplicative reduction. (This was mentioned in the question). The reducedness at every stalk follows from étaleness at every stalk which uses purity of the branch locus (finite plus étale in codimension 1 implies étale everywhere). | |
| May 27 at 14:05 | comment | added | user267839 | etaleness "transfers" reducedness to $D_k$ (resp the comp $D_k^1$) at $d$. Is this the argument you are using there? If yes, why this then implies already that the whole component $D_k^1$ must be already reduced, ie at every stalk and not only the generic one? (...if I elaborated your argument on reducedness at $d$ up to now correctly) | |
| May 27 at 14:01 | comment | added | user267839 | oh sorry, I overlooked the "its" in " [...] if the covering $\mathcal{D} \to \mathcal{C}$ is unramified at its generic point." But then, say $d$ is the genertic point of component $\overline{ \{d \}}=D_k^1 \subset D_k$ of the special fiber and assume we know that to local map $O_{C,c} \to O_{D,d}$ ($c$ image of $d$ wrt $\mathcal{D} \to \mathcal{C}$ ) is unramified. Then $D_k \to C_k $ (base change to special fibres) is also unramified in $d$. So at $d$ $D_k^1 \subset D_k \to C_k$ is even etale and once we know that $C_k$ is at $c$ reduced (wo we know it?) the | |
| May 27 at 10:51 | comment | added | Will Sawin | @user267839 I'm restricting the map to the generic point of the special fiber, so it's not just the original map. The local ring in question is not over $K$. | |
| May 27 at 8:03 | comment | added | user267839 | the special fiber $D_k$ non trivially. May I assume that the issue is that if $p \in D_k^1$ is a point where the latter would be nonreduced, there exist a closed $q \in F$ such that $p \in \overline{q}$ (inside $\mathcal{D}$), such that $F \to E$ would be not unramified in $q$? -> contradiction) Is this the whole joke behind implication that unramifiedness over generic point of $\mathcal D \to \mathcal C$ implies reducedness of every component of the special fiber of $\mathcal D $, or did I misunderstood the point and there is a different argument involved? | |
| May 27 at 7:29 | comment | added | user267839 | but let's stick on separable case, I'm not sure if I understood the geometric picture correctly. So assume that the restriction of $\mathcal D \to \mathcal C$ to generic point - that's just the original map $f:F \to E$, right? - is unramified, so as you wrote uniformizer maps to uniformizer in local rings - but everyting over base $K$. The claim is that any component of special fiber $D_k$ is reduced, so over base $k =O_K/m$. For sake of reaching contradiction, let $D^1_k$ nonreduced. Any closed point of $F$ ($=D_K$) gives rise (by taking closure) to a "horizontal" divisor intersecting | |
| May 27 at 1:34 | comment | added | Will Sawin | @user267839 Thanks for pointing this out, there is a subtlety here. Restricted to the generic point, the map $\mathcal D \to \mathcal C$ is a map of regular local rings. This is unramified if and only if the inverse image of a uniformizer is a uniformizer and not a higher power of a uniformizer (i.e. the inverse image of the special fiber is reduced) and the extension on residue fields is separable. The inseparable case can only happen if the degree is divisible by $p$. I think in fact the inseparable case doesn't happen but the argument needs to proceed a different way. | |
| May 27 at 1:31 | history | edited | Will Sawin | CC BY-SA 4.0 |
added 187 characters in body
|
| May 26 at 22:00 | comment | added | user267839 | a question regarding the map $\mathcal D \to \mathcal C$: Why a component of the special fiber of $\mathcal D$ is reduced iff $\mathcal D \to \mathcal C$ is unramified at generic point, once we know that the special fiber $\mathcal D$ is inverse image of special fiber of $\mathcal C$? | |
| May 22 at 10:44 | comment | added | David Hubbard | Hi Will, thanks for the very complete answer! Apologies, my intended meaning was that $\text{ker}(f)$ has its geometric points defined over $K$ i.e. $\text{Gal}(\bar{K}/K)$ acts on the set of geometric points trivially, but of course what I have written above seems to only ask that $\text{ker}(f)$ is a $K$-scheme. | |
| May 22 at 10:41 | vote | accept | David Hubbard | ||
| May 21 at 16:45 | history | answered | Will Sawin | CC BY-SA 4.0 |