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$\newcommand{\b}[1]{\overline{#1}}$ $\newcommand{\m}[1]{\,(\text{mod }#1)}$ $\def\red{\tilde{E}_{\mathfrak{p}}}$ $\def\p{\mathfrak{p}}$ $\def\at{_{\mathfrak{p}}}$ $\def\comp{K_{\mathfrak{p}}}$

I'm studying elliptic curves of the form $E:y^2=x^3+Ax+B$ over a number field $K$ and I'm trying to understand the concept of reduction at a finite place $\mathfrak{p}$ at $K$.

I use the following notation:

$K\at$: completion of $K$ over the non-archimedian absolute value $|\cdot|\at$

$O\at$: local ring of integers of $K\at$

$m\at$: maximal ideal of $O\at$

$k\at$: residue field $O\at/m\at$

Supposing $y^2=x^3+Ax+B$ is a minimal Weierstrass equation with respect to $\p$, we define the reduction map $E(K)\to \red(k\at)$ (where $\tilde{E}\at:y^2=x^3+\b{A}x+\b{B}$ is the curve over $k\at$) as follows: \begin{align*} E(K)&\to \red(k\at)\\ (x_0:x_1:x_2)&\mapsto (\b{x_0}:\b{x_1}:\b{x_2}) \end{align*} (here we choose $x_0,x_1,x_2\in K\at$ such that $x_i\notin m\at$ for some $i$)

My question is this: since $O\at/m\at\simeq O_K/\p$, why can't we just define the reduction map as $(x:y:z)\mapsto (x\m{\p}:y\m{\p}:z\m{\p})$? Why do we bring up the local field $K\at$?

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You're right that we could define a reduction map directly from $E( K)$ to $E(k_{\mathfrak p})$ without ever going through the local field $K_{\mathfrak p}$.

However, one benefit of considering the curve over the local field is that the reduction map is then (mostly*) surjective. Moreover distinct points over $K$ gives rise to distinct points over the localization $K_{\mathfrak p}$, so we have an injective localization map from $E(K)$. Thus $E(K_\mathfrak p)$ sits as an intermediate piece in the following diagram:

$$E(K) \hookrightarrow E(K_{\mathfrak p}) \twoheadrightarrow E(k_{\mathfrak p}).$$

Considering this extra piece in the middle can help us understand why the map $E(K) \to E(k_\mathfrak p)$ may fail to be injective or surjective.


For example, the projective curve $$ C \, : \, x^2 + y^2 +1 = 0$$ has no points over $\mathbb Q$, but it does have the points $(\pm1,\pm1)$ over $\mathbb F_3$. To understand where these extra points "come from", it helps to consider the 3-adic curve $C(\mathbb Q_3)$. Over the local field, we can find a bijection to the projective line $C(\mathbb Q_3) \cong \mathbb P^1(\mathbb Q_3) $. This descends to $C(\mathbb F_3) \cong \mathbb P^1(\mathbb F_3) $ which explains why we counted 4 points $\mod 3$.


*A caveat about $X(K_\mathfrak p) \to X(k_\mathfrak p)$: this is not always surjective, as I stated earlier. For example if $X$ is defined by the equation $\{x^2 -3y^2 = 0\}$, there exist solutions over $\mathbb F_3$ but none over $\mathbb Q_3$.

The fact that we can usually lift points of $X(k_\mathfrak p)$ to points of $X(K_\mathfrak p)$ is a result of Hensel's lemma, and this has as part of the hypothesis a condition on the derivative of the defining polynomial. Geometrically, what this translates to is that the reduction $X(K_\mathfrak p) \to X(k_\mathfrak p)$ is surjective onto the smooth locus $ X^\mathrm{sm}(k_\mathfrak p)$, i.e. points where the Jacobian matrix of partial derivatives has the correct rank (over $k_\mathfrak p$).

In the specific case of studying elliptic curves, $E$ should be smooth to begin with so $E^\mathrm{sm}(K) = E(K)$, but even so it may be true that at a given prime $\mathfrak p$ the reduction is not smooth i.e. $E^\mathrm{sm}(k_\mathfrak p) \subsetneq E(k_\mathfrak p)$. When this happens we say $E$ has ''bad reduction'' at $\mathfrak p$.

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    $\begingroup$ (the map $C(\mathcal{O}_\mathfrak{p}) \to C(k_\mathfrak{p})$ is understood using Hensel's lemma) $\endgroup$ – reuns Nov 12 '17 at 16:35
  • $\begingroup$ Just another question. Let $E(K)_2=\{P\in E(K)\mid 2P=O\}$ be the $2$-torsion points of $E(K)$. I'm trying to prove that if $E$ has good reduction at $\mathfrak{p}$, then the reduction map $E(K)_2\to E_{\mathfrak{p}}(k_{\mathfrak{p}})$ is injective. Is it possible to prove it without talking about local fields? $\endgroup$ – rmdmc89 Nov 12 '17 at 19:10
  • $\begingroup$ I think we also need to add the condition $ord_{\mathfrak{p}}(2)=0$ for the statement above $\endgroup$ – rmdmc89 Nov 12 '17 at 20:17
  • $\begingroup$ Not sure about that.. I'll let you know if I figure something out! $\endgroup$ – Harry Richman Nov 13 '17 at 17:25

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