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I am looking for references for the two facts found in the bullet points at the bottom of p. 386 of Diamond & Shurman's "A First Course in Modular Forms"

  • For a field $k$ and a curve $X/k$ of genus $g$, if $M$ is coprime to the characteristic of $k$,

$$\operatorname{Pic}^0(X)[M] \cong \left(\mathbb Z/M\mathbb Z \right)^{2g}$$

  • If a curve $X/\mathbb Q$ has good reduction at prime $p \nmid M$, then the reduction map is injective on $\operatorname{Pic}^0(X)[M]$.
Johnny Apple
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    Unfortunately, this is either easy or hard depending on what you're willing to accept about $\mathrm{Pic}^0$. What do you know about this space? Do you know it's an abelian variety? – Alex Youcis Nov 03 '21 at 03:44
  • No, but I have heard this before. I have read from Bosch's text that Pic^0 as a functor can be represented somehow, but I think the result you are quoting comes from Mumford, yes? If so, I have not read Mumford yet. – Johnny Apple Nov 03 '21 at 04:24
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    I don't know if it comes from Mumford--that seems unlikely, but I don't know who first proved it. It's Theorem 4.4 here: http://virtualmath1.stanford.edu/~conrad/248BPage/handouts/pic.pdf

    With this do you know how to prove this result?

    – Alex Youcis Nov 03 '21 at 04:28
  • I'll read the handout. It looks alot like Bosch. I do not know how to prove the result assuming this. I don't know much about abelian varieties in general, just what little has been mentioned in Silverman's texts. – Johnny Apple Nov 03 '21 at 04:35
  • If you know this fact, and the theory of abelian varieties, the first fact is Theorem 4.2.2 of this: http://virtualmath1.stanford.edu/~conrad/249CS15Page/handouts/abvarnotes.pdf

    The second fact follows from the fact that $\mathrm{Pic}^0(\mathcal{X})$ (if $\mathcal{X}$ is a smooth proper model of $X$) is a model of $\mathrm{Pic}^0(X)$. Since $p\nmid M$ the multiplication map $[M]$ on $\mathrm{Pic}^0(\mathcal{X})$ is smooth (since it's derivative is multiplication by $M$ invertible since we're working over $\mathbb{Z}_{(p)}$) and so its kernel $\mathrm{Pic}^0(\mathcal{X})[M]$ is etale

    – Alex Youcis Nov 03 '21 at 04:40
  • which means that it's geometrically constant which clearly implies the result you want. Unfortunately I don't know a more down-to-erath way to say this that doesn't require a lot more effort. Maybe another user does. – Alex Youcis Nov 03 '21 at 04:41
  • I'm all for this route. Where can I learn more about the solution to the second part? – Johnny Apple Nov 03 '21 at 04:44
  • See the first bullet in the 'Finite Flat Group Schemes' section of this https://ayoucis.wordpress.com/2015/08/10/p-divisible-groups-formal-groups-and-the-serre-tate-theorem/ to see that $\mathrm{Pic}^0(\mathcal{X})[M]$ is finite flat over $\mathbb{Z}_{(p)}$. Use Theorem 2 of op. cit. to show that it's etale. Theorem 1 of https://ayoucis.wordpress.com/2015/07/24/another-basic-viewpoint-on-etale-cohomology/ then says that $\mathrm{Pic}^0(\mathcal{X})[M]$ is etale locally constant, and then things are easy. Since if $\underline{A}$ is a constant finite abelian group scheme over – Alex Youcis Nov 03 '21 at 04:56
  • something like $\mathcal{O}$ (the local ring of $\overline{\mathbb{Z}}$ at some prime over $p$) then almost by definition the natural maps

    $$\underline{A}(\overline{\mathbb{F}}_p)\to \underline{A}(\mathcal{O})\to \underline{A}(\overline{\mathbb{Q}})$$

    are bijective, which gives you what you want. To see that $\mathrm{Pic}^0(\mathcal{X})[M]_\mathcal{O}$ is constant, you just need to use loc. cit. since $\mathcal{O}$ has no non-trivial finite etale covers (e.g. it's normal with algebraically closed generic fiber).

    – Alex Youcis Nov 03 '21 at 04:57
  • Oops, that should read

    $$\underline{A}(\overline{\mathbb{F}}_p)\leftarrow \underline{A}(\mathcal{O})\to \underline{A}(\overline{\mathbb{Q}})$$

    – Alex Youcis Nov 03 '21 at 05:05
  • Hi Alex. I am almost through assembling the results. Where does one find the result that Pic^0 preserves models, mentioned in the 5th comment above? Thank you for your answers! – Johnny Apple Nov 03 '21 at 19:41
  • Can you say a little more about how why your natural maps are bijective by definition because of the given assumptions on $\underline{A}$? Also, what does it mean that a ring has no finite etale covers and has algebraically closed generic point? Do you mean the Spec of that ring? Finally, why does normal + algebraically closed generic fiber imply no nontrivial etale covers? – Johnny Apple Nov 03 '21 at 20:04

1 Answers1

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At the request of the OP, I am going expand on my comments.

Let us first note that if $g$ has genus $0$ then this is basically all moot, since in this case $X_{\overline{k}}$ is $\mathbb{P}^1_{\overline{k}}$ so that $\mathrm{Pic}^0_X$ is trivial.

We then recall the following fact.

Theorem (e.g. see [Conrad, Theorem 4.4]): Let

  • $S$ be a locally Noetherian scheme,
  • and $X\to S$ be smooth proper morphism all of whose fibers are geometrically connected curves of genus $g>0$
  • $e:S\to X$ a section of $X\to S$.

The the functor $\mathrm{Pic}^0_X$ on $S$ schemes taking $T\to S$ to isomorphism classes of pairs $(\mathscr{L},i)$ where

  • $\mathscr{L}$ is a line bundle on $X\times_S T$ which is fiber-by-fiber (rel. $T$) degree $0$,
  • $i\colon e_T^\ast\mathscr{L}\cong \mathcal{O}_T$ is an isomorphism (where $e_T$ is the base change of $S\to X$ to $T$),
  • and two pairs are isomorphic if the line bundles are isomorphic in a way that preserves the trivalizations over $e$,

is representable by an abelian scheme of relative dimension $g$.

Note that you may take objections to my definition of the Picard functor on two counts:

  1. I am assuming that $X\to S$ has a section,
  2. my actual definition of the Picard functor with 'rigifications along $e$' may seem foreign to you.

But, as an answer:

  1. this will not be an issue for us since what you're asking about is an esssentially geometric question, so we are happy to base change to a finite extension to obtain a rational point,
  2. it turns out (e.g. see [Kleinman, Lemma 2.9]) that this definition of the Picard functor essentially agrees with other 'more natural' definitions (e.g. the fppf sheafification of $T\mapsto \mathrm{Pic}^0(X\times T)$).

Also, let us note the following corollary that is useful to answer your questions.

Corollary: Let $\mathcal{O}$ be a DVR and $X$ be a smooth proper curve over $F=\mathrm{Frac}(\mathcal{O})$. Suppose that $X$ has a good reduction, then $\mathrm{Pic}^0_X$ has good reduction.

The proof is simple: the abelian variety over $\mathcal{O}$ modelling $\mathrm{Pic}^0_X$ is $\mathrm{Pic}^0_\mathcal{X}$ for $\mathcal{X}$ a smooth proper model of $X$!

But, truthfully, this explanation elides one small detail that's handled by the following lemma.

Lemma: Any smooth proper model $\mathcal{X}\to\mathrm{Spec}(\mathcal{O})$ of $X$ then its geometric fiber $\mathcal{X}_{\overline{s}}$ (where $\overline{s}$ is the closed geometric point of $\mathrm{Spec}(\mathcal{O})$) is connected of genus $g$.

Proof: The fact that $\mathcal{X}_{\overline{s}}$ is connected can be seen as follows. Note that $\mathcal{X}$ itself is connected: if not then $\mathcal{X}=C_1\sqcup C_2$, so then $X=(C_1)_\eta\sqcup (C_2)_\eta$ ($\eta$ is the generic point of $\mathrm{Spec}(\mathcal{O})$), so since $X$ is connected we deduce that (WLOG) $(C_1)_\eta$ is empty. But, since $C_1$ is an open in $\mathcal{X}$, which is flat over $\mathcal{O}$, it's flat over $\mathcal{O}$ and so this is impossible (just think that locally on $C_1$ this means that it's of the form $\mathrm{Spec}(A)$ with $A_F=0$, but $\mathcal{O}\to A$ is flat, so this is impossible). But, since $\mathcal{X}\to\mathrm{Spec}(\mathcal{O})$ is smooth, and the target is normal, we deduce that $\mathcal{X}$ is normal, and thus since it's also connected it's integral (e.g. see this). The connectedness then follows from general principles related to the Theorem on Formal Functions (e.g. see [Illusie, Corollary 2.23]).

The fact about genus $g$ is then easy as it follows from the connectedness of the fibers and the constancy of Euler characteristic in flat families (see [Vakil, Theorem 24.7.1]). $\blacksquare$


From all of this, we see that your question really can be interpretted as follows:

Question: Let $A$ be an abelian variety over a field $k$ of dimension $g$.

  1. Why is $A[N](k^\mathrm{sep})\cong (\mathbb{Z}/N\mathbb{Z})^{2g}$ for $N$ invertible in $k$?
  2. Let $A$ be an abelian variety over $\mathbb{Q}$ with good reduction at $p$. Then, if $p\nmid N$, why is the reduction map $A[N](\overline{\mathbb{Q}})\to \mathscr{A}[N](\overline{\mathbb{F}_p})$ injective (here $\mathscr{A}$ is the abelian scheme over $\mathbb{Z}_{(p)}$ modelling $A$--it's Neron model)?

The proof of 1. is not super easy, and definitely requires some legwork. I suggested reading [Conrad 2, Theorem 4.2.2]. Although, some of the ideas already appear in the explanation of 2. below.

To prove 2., note that the map $[N]\colon \mathscr{A}\to\mathscr{A}$ is finite flat (see [Conrad 2, Theorem 4.1.1]), so $\mathscr{A}[N]\to\mathrm{Spec}(\mathbb{Z}_{(p)})$ is finite flat (since it's the base change of $[N]$ along the identity section). So $\mathscr{A}[N]\to\mathrm{Spec}(\mathbb{Z}_{(p)})$ is a finite flat group scheme. In fact, by 1., we know that it's of order $N^{2g}$, which is invertible in $\mathbb{Z}_{(p)}$. So, by the disucssion in the first bullet of 'Finite Flat Group Schemes' section of [Blog] we deduce that $\mathscr{A}[N]\to\mathrm{Spec}(\mathbb{Z}_{(p)})$ is finite etale. Thus, also we have that $\mathscr{A}[N]_{\mathcal{O}}\to\mathrm{Spec}(\mathcal{O})$ is finite etale where $\mathcal{O}$ is the localization of $\overline{\mathbb{Z}}$ (the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$) at a prime above $p$.

Note though that any finite etale morphism $f\colon X\to Y$ is split by a finite etale cover of $Y$. Indeed, one can take a Galois cover $\widetilde{X}\to Y$ of $f$ (see e.g. see Tag 0BN2) dominating $X\to Y$, and then check that $X\times_Y \widetilde{X}\to\widetilde{X}$ is split.

But, note that $\text{Spec}(\mathcal{O})$ has no non-trivial (connected) finite etale covers. Indeed, by Tag 0BQL such a finite etale cover would be the normalization of $\mathrm{Spec}(\mathcal{O})$ in a finite extension of $\mathrm{Frac}(\mathcal{O})=\overline{\mathbb{Q}}$, of which there are no non-trivial ones.

So, $\mathscr{A}[N]_{\mathcal{O}}\to\mathrm{Spec}(\mathcal{O})$ becomes split on a finite etale cover of $\mathrm{Spec}(\mathcal{O})$, but such covers are disjoint unions of isomorphisms, so $\mathscr{A}[N]_{\mathcal{O}}\to\mathrm{Spec}(\mathcal{O})$ must already be split. This means that, in particular, it represents the constant sheaf on $\mathrm{Spec}(\mathcal{O})$ so that for any $\mathcal{O}$-scheme $T$ one has that $\mathscr{A}[N]_{\mathcal{O}}(T)=(\mathscr{A}[N](\mathcal{O}))^{\#(\pi_0(T))}$ (at least for non-degenerate schemes, see this)--but note that since $\mathscr{A}[N](\mathcal{O})=A[N](\mathbb{Q})$ we know from the first part that this group is just $(\mathbb{Z}/N\mathbb{Z})^{2g}$. But, in particular this means that if $T\to T'$ is a morphism of $\mathcal{O}$-schemes, both of which are connected, then the induced map $\mathscr{A}[N]_\mathcal{O}(T')\to\mathscr{A}[N]_\mathcal{O}(T)$ is a bijection.

Apply this for the diagram

$$\mathrm{Spec}(\overline{\mathbb{Q}})\to\mathrm{Spec}(\mathcal{O})\leftarrow\mathrm{Spec}(\overline{\mathbb{F}_p})$$

to see that the maps in the diagram

$$\mathscr{A}[N](\overline{\mathbb{Q}})\leftarrow \mathscr{A}[N](\mathcal{O})\to \mathscr{A}[N](\overline{\mathbb{F}_p})$$

are isomorphisms. Since the reduction map is defined as the composition of the inverse of the left map with the right map, the claim follows.

This map is also equivariant for the action of $D$, the decomposition of group in $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ corresponding to our chosen prime (by construction). Here we are having $D$ act on $\mathscr{A}[N](\overline{\mathbb{F}_p})$ via its quotient $D\to\mathrm{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p)$.

In particular, let's fix an extension $L$ of $\mathbb{Q}$. Note that our choice of a prime of $\overline{\mathbb{Z}}$ lying over $p$ induces such a choice for $\mathcal{O}_L$ thus giving rise to a residue field $\ell$ and a decomposition group $D_L\subseteq D$. We then see that we get an isomorphism

$$A[N](\overline{\mathbb{Q}})^{D_L}\xrightarrow{\approx}\mathscr{A}[N](\overline{\mathbb{F}_p})^{\mathrm{Gal}(\overline{\mathbb{F}_p}/\ell)}=\mathscr{A}[N](\ell).$$

Of course, this source group is, in general, smaller than the group $A[N](L)=A[N](\overline{\mathbb{Q}})^{\mathrm{Gal}(\overline{\mathbb{Q}}/L)},$ and so one only gets an injection

$$A[N](L)\to \mathscr{A}[N](\ell)$$

which is not necessarily a surjection. This non-surjectivity is related to the fact that $L$ is not Henselian (with respect to the valuation determined by our lift of $p$)--indeed, if that were true then there would be no difference between $\mathrm{Gal}(\overline{\mathbb{Q}}/L)$ and $D_L$. Antoher way to see this, is that if we had the Henselian property then $A[N](L)=\mathscr{A}[N](\mathcal{O}_L)\to \mathscr{A}[N](\ell)$ would be surjective by Hensel's lemma.

References:

[Blog] https://ayoucis.wordpress.com/2015/08/10/p-divisible-groups-formal-groups-and-the-serre-tate-theorem/

[Conrad] http://virtualmath1.stanford.edu/~conrad/248BPage/handouts/pic.pdf

[Conrad 2] http://virtualmath1.stanford.edu/~conrad/249CS15Page/handouts/abvarnotes.pdf

[Illusie] http://staff.ustc.edu.cn/~yiouyang/Illusie.pdf

[Kleinman] https://arxiv.org/pdf/math/0504020.pdf

[Vakil] http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf

Alex Youcis
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  • This is incredibly helpful. Three small questions: 1) I assume by $\mathcal{A}[N]$ you mean the $N$-torsion. If so, what does the subscript $\mathcal{O}$ indicate? 2) What does it mean for a finite etale morphism to be split? Does that mean what it means for topological covering maps? 3) What is $\pi_0$ in your final argument? – Johnny Apple Nov 04 '21 at 02:23
  • @JohnnyApple 1) sort of the thrust of the above is to try and attack the problem by realizing that the two sets $AN$ and $\mathscr{A}N$ are actually the points of schemes $A[N]$ and $\mathscr{A}[N]$, and so we can use algebraic geometry to attack things. So, $\mathscr{A}[N]$ is the $N$-torsion scheme of $\mathscr{A}$, and $\mathscr{A}[N]_{\mathcal{O}}$ is its base change to $\mathcal{O}$. – Alex Youcis Nov 04 '21 at 03:04
  • a finite etale cover $X\to Y$ is split if there exists an isomorphism $X\cong \bigsqcup_i Y$ for some index set which respects the maps to $Y$ (i.e. if as a $Y$-scheme it's a disjoint union of copies of the base). So yes, basically the same as in topology.
  • – Alex Youcis Nov 04 '21 at 03:05
  • $\pi_0(T)$ is the set of connected components of $T$. Truth be told, this claim I made only holds true if $T$ is locally connected (e.g. if it's locally topologically Noetherian)--see this post for why the two may different for pathological schemes: https://math.stackexchange.com/questions/3178359/constant-sheaves-on-the-%c3%a9tale-site-of-a-scheme
  • – Alex Youcis Nov 04 '21 at 03:06
  • @JohnnyApple I was naughty and used notation that probably confused you. I often times think of a constant sheaf as $\underline{A}$ for some abstract abelian group $A$. So, I used $A$ before when this is obviously horrible notation since $A$ is our abelian variety... Please reread that part now. – Alex Youcis Nov 04 '21 at 03:19
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    This is a fantastic answer. Thank you for thoroughly collecting all these resources and results, and organizing them so concisely, and for your prompt replies to my comments. – Johnny Apple Nov 04 '21 at 04:29
  • Hi Alex. Maybe I missed it, but there is one small detail I am wondering about: why ic Pic^0 of $\mathcal X$ a model for Pic^0(X)? Where can I learn more about this? – Johnny Apple Nov 04 '21 at 17:58
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    @JohnnyApple If $Y$ is a $\mathbb{Z}{(p)}$-scheme, then it produces a functor $\mathcal{F}$ on $\mathbb{Z}{(p)}$-schemes by Yoneda. The base change $Y_\mathbb{Q}$ produces a functor $\mathcal{G}$ on $\mathbb{Q}$-schemes by Yoneda. The relationship is that $\mathcal{G}$ is the restriction of $\mathcal{F}$ to $\mathbb{Q}$-schemes where we are thinking of $\mathbb{Q}$-schemes a subcategory of $\mathbb{Z}$-schemes via taking a $\mathbb{Q}$-scheme to the composition $T\to\mathrm{Spec}(\mathbb{Q})$ to $T\to\mathrm{Spec}(\mathbb{Q})\to\mathrm{Spec}(\mathbb{Z})$ – Alex Youcis Nov 04 '21 at 23:48
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    Now, by definition, $\mathrm{Pic}^0_\mathcal{X}(T)$ is something involving line bundles on $\mathcal{X}\times_{\mathrm{Spec}(\mathbb{Z})}T$. If $T$ is a $\mathbb{Q}$-scheme, then $\mathcal{X}\times_{\mathrm{Spec}(\mathbb{Z})}T\cong (\mathcal{X}\times_{\mathrm{Spec}(\mathbb{Z})}\mathrm{Spec}(\mathbb{Q}))\times_{\mathrm{Spec}(\mathbb{Q})}T$. But, this is equal to $X\times_{\mathrm{Spec}(\mathbb{Q})} T$! And so line bundles on $\mathcal{X}\times_{\mathrm{Spec}(\mathbb{Z})}T$ with some data are the same thing as line bundles on $X\times_{\mathrm{Spec}(\mathbb{Q})} T$ – Alex Youcis Nov 04 '21 at 23:50
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    with some data, but the latter is exactly $\mathrm{Pic}^0_X$. So, $\mathrm{Pic}^0_{\mathcal{X}}$'s functor restricted to $\mathbb{Q}$-schemes gives $\mathrm{Pic}^0_X$'s functor. But, by what I claimed above, this restriction also gives $\mathrm{Pic}^0_{\mathcal{X}}\times_{\mathrm{Spec}(\mathbb{Z})}\mathrm{Spec}(\mathbb{Q})$'s functor. So, by Yoneda's lemma, we deduce that $\mathrm{Pic}^0_X\cong \mathrm{Pic}^0_{\mathcal{X}}\times_{\mathrm{Spec}(\mathbb{Z})}\mathrm{Spec}(\mathbb{Q})$.

    EDIT: Oops, please replace all instances of $\mathbb{Z}$ above with $\mathbb{Z}_{(p)}$.

    – Alex Youcis Nov 04 '21 at 23:52
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    Note that there is some minor confusing yoga in the above because I am conflating $\mathrm{Pic}^0_\mathcal{X}$ as a scheme, and the functor they represent...hopefully this is clear enough. – Alex Youcis Nov 04 '21 at 23:55