# Wednesday February 26th Example (regarding generalization from last time) : Let $G = \prod^\infty \ZZ/p^n\ZZ$, which is a commutative pro-$p$ group. Then $$\oplus \ZZ_p^n \ZZ ~(?)~\subsetneq G[\tors] \subsetneq G $$ with proper containments, but the direct sum is dense in the direct product, so $G[\tors]$ is proper and dense and thus not closed. ## (Clark-Xarles) Local Bounds for Torsion Points on Abelian Varieties > Recommended reading! Introduces a lot of tools in their most basic form. Let $K/\QQ_p$ be a $p\dash$adic field with $[K: \QQ_p] = d$, $R$ a valuation ring, $\mfm$ a maximal ideal, $\FF_q = R/\mfm$. > Recall previously: finiteness on torsion subgroups in certain cases. > Mordell-Weil works for global fields, but what is that finite group? > No positive results bounding the torsion for a Laurent series field, can be arbitrary large. Let $A/K$ be a $g\dash$dimensional abelian variety and recall \begin{align*} A(K) &\cong_{\mathrm{TopGrp}} (R, +)^g \oplus T \cong \ZZ_p^{dg} \oplus T ,\end{align*} where $\abs{T} < \infty$ We'll prove bounds in three steps: 1. Good reduction 2. Potentially good reduction 3. General reduction and results in the case of *anisotropic reduction*. ## Good Reduction Definition (Good Reduction for Abelian Varieties) : $A/K$ has *good reduction* if there exists a smooth group scheme $A?R$ with generic fiber $A/K$. Noting that $\spec(R) = \theset{(0), (\pi)}$ is a two point space, we have the fibers \begin{tikzcd} A \arrow[dd] & & A/K \arrow[dd] & & A/\FF_q \arrow[dd] \\ & & & & \\ \spec(R) & & (0) & & (\pi) \end{tikzcd} We get a SES $$ 0 \to A' \to A(R) \mapsvia{\text{reduction}} A(\FF_q) \to 0 $$ and the fact that it is surjective is due to **Hensel's lemma**. Here $A' = \mfm^g$ endowed with a formal group law. Since we have a projective variety, we can clear denominators to replace $A(R)$ with $A(K)$ in the above sequence. Note that the quotient is precisely a Mordell-Weil group, so using the Weil conjectures, we can say quite a bit about its size. Thus $A' \injects A(K)$ is an open finite-index subgroup. Moreover, all of the torsion is $p\dash$primary, and $A'[\tors] = A'[p^\infty]$. Define $G[\tors]'$ to be the torsion that is prime to $p$. If the kernel is $p\dash$primary torsion, then the reduction map induces an isomorphism \begin{align*} r: A(K)[\tors]' \mapsvia{\cong} A(\FF_q)[\tors]' .\end{align*} Theorem (Weil, Serre. Improved Weil Bounds) : $\# A(\FF_q) \leq \floor{ (1 + \sqrt q)^2}^g$. (Note that Weil's original bound is just this without the floor function.) Therefore, for $A/K$ and $K/\QQ$ a number field of degree $d$, choose $\mfp_1, \mfp_2$ in $\spec \ZZ_k$ points of good reduction. Then \begin{align*} \# A(K)[\tors] \leq \floor{ \qty{1+p_1^{\frac d 2} }^2 }^g \floor{ \qty{ 1 + p_2^{\frac d 2} }^2 } .\end{align*} > Note: potentially "blah" always means "blah" after some base extension. ## Potentially Good Reduction Definition (Potentially Good Reduction) : Let $K$ be a $p\dash$adic field, then $A/K$ has *potentially good reduction* iff there exists a finite degree extension $L/K$ such that base extending to $L$, $A/L$, has good reduction. Proposition : $\# A(K)[\tors]' \leq \floor{\qty{1 + \sqrt q }^2 }^g$. Remark : If $\endo(A)$ is "large enough", e.g. if $A$ has $X\dash$multiplication for $X = \CC, \HH$ (complex or quaternionic) then $A$ must have potentially good reduction (see Pete's thesis). Proof : Serre-Tate observed that if you have potentially good reduction, it can always be obtained via a totally ramified extension $L/K$. That is, if $S$ is the valuation ring of $L$ and $\mfm_S$ is its maximal ideal, then $S/\mfm_S \cong \FF_q$, so it doesn't change the size of the residue field. Applying step 1 over $L$ yields a bound. Application : If $E/\QQ$ is an elliptic curve with potentially good reduction over $2, 3$ (i.e. the $j\dash$invariant doesn't have 2 or 3 in the denominator). Applying the bound yields \begin{align*} \# E(\QQ)[\tors] &\leq \floor{(1 + \sqrt 2)^2 } \floor{ (1 + \sqrt 3)^2 } \\ &= (5)(7) \\ &= 35 .\end{align*} Note that a much harder theorem yields a tighter bound at $15$, but this is much easier and not much worse of a bound. Taking now $A/\QQ$ with $\dim A = 2$ with the same conditions yields $\# A(\QQ)[\tors]' \leq (35^2) = 1225$. Note that without this minor observation, the known bound is astronomically larger! Theorem (C-Xarles) : Fix $A/\QQ$ with $\dim A = 2$ and everywhere potentially good reduction. Then $$ \# A(\QQ)[\tors] \in [1, 16] \union [18, 20], \union \theset{22, 24, 25, 28, 30, 36, 48, 60, 72} $$ Conversely, $[1, 10] \union \theset{12, 16, 18, 19 ,20, 24, 36}$ are known to occur. > Note that for elliptic curves over $\QQ$, it's $\theset{1,2,4,6}$. Proof (Sketch) : For $K$ a $p\dash$adic field, the absolute unramified extension has galois group $\hat \ZZ$, then extend to $\bar K$ with galois group $H$ to obtain a SES $$ 1 \to H \to g_K \to \hat \ZZ \to 0 $$ Since $\hat \ZZ$ is a free profinite group, thus a projective object in that category of profinite groups, so this sequence splits and $g_K \cong \hat \ZZ \oplus H$. What does this buy us? We can think of the subgroup $H$ now as a quotient, so there is a $K^{\mathrm{tr}} \leq \bar K$ such that $\Aut(K\tr/K) = H$, i.e. we have a compositum \begin{tikzcd} & \bar K \arrow[ldd, "H"'] \arrow[rdd] & \\ & & \\ K^\text{unramified} \arrow[rdd] & & K^{\text{tr}} \arrow[ldd, "H"] \\ & & \\ & K & \end{tikzcd} Thus $\bar K = K^{\text{unramified}} \cross K^{\text{tr}}$. ## General Reduction Let $A/K$ be an abelian variety, $A/R$ the Néron model, which is a smooth finite type group $R\dash$group scheme with generic fiber $A/K$ and $A(K) = A(R)$. There is a reduction map $A' \to A(R) \mapsvia{r} A(\FF_1) \to 0$ given by reduction, which is again surjective by Hensel's lemma. Here the kernel $A'$ is exactly as before, obtained by evaluating a $g\dash$dimensional formal group law on $\mfm^g$. The difference here is that $A(\FF_q)$ is no longer an abelian variety, and instead the group of $\FF_q$ rational points of a commutative algebraic group over $\FF_q$. There is structure theory of commutative algebraic groups, which may be disconnected, so we'll need to consider things like the component group -- but this can be arbitrarily large, which can be an issue.