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:
- I am assuming that $X\to S$ has a section,
- my actual definition of the Picard functor with 'rigifications along $e$' may seem foreign to you.
But, as an answer:
- 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,
- 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 $g$.
- Why is $A[N](k^\mathrm{sep})\cong (\mathbb{Z}/N\mathbb{Z})^{2g}$ for $N$ invertible in $k$?
- 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
With this do you know how to prove this result?
– Alex Youcis Nov 03 '21 at 04:28The 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$$\underline{A}(\overline{\mathbb{F}}_p)\to \underline{A}(\mathcal{O})\to \underline{A}(\overline{\mathbb{Q}})$$
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$$\underline{A}(\overline{\mathbb{F}}_p)\leftarrow \underline{A}(\mathcal{O})\to \underline{A}(\overline{\mathbb{Q}})$$
– Alex Youcis Nov 03 '21 at 05:05