Lecture 8: Elliptic curves over DVRs
This lecture is devoted to the behavior of elliptic curves over DVRs. The various types of reduction (good, multiplicative, additive) are defined, and their behavior under extension is studied. Then the behavior of torsion points under reduction is discussed. Finally, I prove the NéronOggShafarevich theorem.
Reduction theory of elliptic curves
A good reference for this lecture is Chapter VII of Silverman’s “The arithmetic of elliptic curves” (MR0817210).
Let $R$ be a complete DVR, $\fp$ its maximal ideal, $K$ its field of fractions, $k$ its residue field, and $v$ the valuation with $v(\pi)=1$, for $\pi$ a uniformizer. We are going to study elliptic curves over $K$, and their reduction modulo $\fp$. We assume throughout that $k$ does not have characteristic 2 or 3.
Minimal Weierstrass equations
Let $E/K$ be an elliptic curve given by a Weierstrass equation $y^2=x^3+ax+b$. Recall that the discriminant $\Delta=16(4a^3+27b^2)$ is nonzero. A Weierstrass equation for $E$ is not unique: one can replaces $y$ with $u^3 y$ and $x$ with $u^2 x$, for $u \in K^{\times}$, which has the effect of changing $a$ to $u^{4} a$ and $b$ to $u^{6} b$. We say that a Weierstrass equation is minimal if $a$ and $b$ belong to $R$ and $v(a) \lt 4$ or $v(b) \lt 6$ (this is equivalent to asking that $v(\Delta)$ be minimal). A minimal Weierstrass equation is unique up to a change of variables as above with $u$ a unit. We let $\cE$ be the projective scheme over $R$ defined by a minimal Weierstrass equation. We call this the minimal Weierstrass model for $E$. It is independent of the choice of minimal Weierstrass equation, up to isomorphism (since $u$ must be a unit in any change of variables).
We can now introduce most of the objects we will be interested in:

We let $\ol{E}$ be $\cE_k$, the special fiber of $\cE$. We call this the reduction of $E$ modulo $\fp$. This is an irreducible projective curve over $k$, though possibly singular.

We let $\ol{E}_{\sm}$ be the smooth locus of $\ol{E}$. A basic fact is that $\ol{E}_{\sm}$ is a group variety: the group law can be defined using the secant line construction, as on an elliptic curve.

Since $\cE$ is projective, $\cE(R)=\cE(K)=E(K)$. We therefore have a welldefined map $E(K) \to \ol{E}(k)$, which we call the reduction map.

We let $E_0(K)$ be the subset of $E(K)$ which reduces into $\ol{E}_{\sm}(k)$. Then $E_0(K)$ is a subgroup of $E(K)$, and the reduction map $E_0(K) \to \ol{E}_{\sm}(k)$ is a group homomorphism. In fact, it is surjective by Hensel's lemma.

We let $E_1(K)$ be the kernel of the reduction map $E_0(K) \to \ol{E}_{\sm}(k)$.
Types of reduction
The curve $\ol{E}$ is defined by the equation $y^2=x^3+\ol{a} x+\ol{b}$, where $\ol{a}$ and $\ol{b}$ are the images of $a$ and $b$ in $k$. This curve is an elliptic curve if and only if $\ol{\Delta} \ne 0$, which is equivalent to asking that $\Delta$ be a unit of $R$. If $\ol{E}$ is an elliptic curve, we say that $E$ has good reduction. In this case, $\cE$ is a smooth scheme over $R$ and is naturally a group object in the category of schemes over $R$.
Now suppose $\ol{E}$ is singular, i.e., $\ol{\Delta}=0$. We then say that $E$ has bad reduction. There are two possibilities. If $\ol{a}=\ol{b}=0$ then $\ol{E}$ has a single singularity, at $(0,0)$, and it is a cusp. The smooth locus $\ol{E}_{\sm}$ is isomorphic to $\bG_a$, as a group variety. We therefore say that $E$ has additive reduction. If $\ol{a}$ or $\ol{b}$ is nonzero then both are nonzero (since $\ol{\Delta}=0$), and $E$ has a single singularity, at $(3b/2a, 0)$, and it is a node. The smooth locus $\ol{E}_{\sm}$ is isomorphic (over $\ol{k}$) to $\bG_m$, as a group variety. We therefore say that $E$ has multiplicative reduction.
To summarize:

$E$ has good reduction if and only if $\Delta$ is a unit of $R$.

$E$ has multiplicative reduction if and only if $\Delta \in \fp$ but $a$ and $b$ are units of $R$.

$E$ has additive reduction if and only if $a$ and $b$ are both in $\fp$.
We say that $E$ has semistable reduction if it has either good or multiplicative reduction. This is equivalent to one of $a$ or $b$ being a unit of $R$.
Behavior of reduction type under extensions
Proposition. Let $K'/K$ be a finite extension. Suppose that either $K'/K$ is unramified or $E$ has semistable reduction over $K$. Then a minimal Weierstrass equation for $E$ over $K$ is still minimal over $K'$. It follows that the reduction type of $E$ over $K$ is the same as that over $K'$.
Proof. Let $v'$ be the valuation on $K'$. First suppose that $K'/K$ is unramified. Then for $x \in K$ we have $v(x)=v'(x)$. Thus if $v(a) \lt 4$ or $v(b) \lt 6$ then $v'(a) \lt 4$ or $v'(b) \lt 6$. Now suppose that $E$ has semistable reduction. Then either $v(a)=0$ or $v(b)=0$, and so $v'(a)=0$ or $v'(b)=0$, which shows that the equation is minimal over $K'$. ◾
Theorem (Semistable reduction theorem). There exists a finite extension $K'/K$ such that $E$ has semistable reduction over $K'$.
Proof. Recall that we can make a change of variables to replace $(a,b)$ with $(a',b')=(u^{4}a, u^{6} b)$. First suppose that $3v(a) \le 2v(b)$. Taking $u=a^{1/4}$, we find that $a'=1$ is a unit and $b'$ is integral, so the new equation is minimal and has semistable reduction. Thus $E$ has semistable reduction over $K'=K(u)$. Now suppose that $3v(a) \ge 2v(b)$. Taking $u=b^{1/6}$, we find that $a'$ is integral and $b'=1$ is a unit, so the new equation is minimal and has semistable reduction. Thus $E$ has semistable reduction over $K'=K(u)$. ◾
Remark. The proof shows that the extension $K'/K$ can always be taken to have degree at most 6.
Combining the above two results, we see that for all sufficiently large extensions $K'/K$, the curve $E_{K'}$ has either good or multiplicative reduction (independent of $K'$). We say that $E$ has potentially good or potentially multiplicative reduction accordingly. There is a simple test to determine which, in terms of the equation for $E$:
Proposition. $E$ has potentially good reduction if and only if $j(E)=1728 (4a)^3/\Delta$ is integral.
Proof. Since the $j$invariant is independent of the model, we may as well assume that we have passed to an extension where $E$ is semistable and we are working with the minimal model. If $E$ has good reduction then $\Delta$ is a unit, and $j(E)$ is integral. If $E$ has multiplicative reduction then $\Delta$ is not a unit but $a$ is, and so $j(E)$ is not integral. ◾
Example. Suppose $E$ is the curve over $\bQ_p$ given by $y^2=x^3+p$. Then $E$ has additive reduction. We have $a=0$ and $b=p$, so $j=0$ is integral, and so $E$ has potentially good reduction. Indeed, changing $y$ to $p^{1/2} y$ and $x$ to $p^{1/3} x$, we find that $E$ is isomorphic to $y^2=x^3+1$ over $\bQ_p(p^{1/6})$, which is still an elliptic curve mod $p$ (since $p \ge 5$).
Reduction of torsion points
We assume in this section that $E$ has good reduction. Since $\cE$ is a proper smooth group over $R$, its $n$torsion $\cE[n]$ is a finite flat group scheme over $R$, for any $n$. We can therefore apply our knowledge of group schemes to its study.
Proposition. Let $G$ be a finite flat group scheme over $R$ whose order is prime to the residue characteristic. Then the reduction map $G(\ol{K}) \to G(\ol{k})$ is an isomorphism of Galois modules. In particular, $G(\ol{K})$ is an unramified Galois module.
Proof. The reduction map is obviously Galois equivariant, so it suffices to show it's a bijection. To do this, we can assume $k$ is algebraically closed. Since the order of $G$ is invertible on the base, $G$ is étale. Thus, if $G=\Spec(A)$, then $A$ is a product of copies of $R$. Clearly then, $G(\ol{K})=G(K)=G(k)$. ◾
Corollary. Suppose $E$ has good reduction and $n$ is prime to the residue characteristic. Then the reduction map $E[n](\ol{K}) \to \ol{E}[n](\ol{k})$ is an isomorphism of Galois modules. In particular, $E[n](\ol{K})$ is an unramified Galois module.
Using Raynaud's theorem, we can say something about the $p$torsion when the residue characteristic is $p$.
Proposition. Suppose $K$ is an extension of $\bQ_p$ with $e \lt p1$. Let $G$ be a finite flat group scheme over $R$. Then the map $G(R) \to G(k)$ is injective.
Proof. Let $\Gamma$ be the group $G(R)$, regarded as a constant group scheme over $R$. There is a natural map $\Gamma \to G$ of group schemes over $R$, inducing the identity on $R$points. Let $\ol{\Gamma}$ be the schemetheoretic image of this map in $G$, which is a closed subgroup of $G$. (One can also describe $\ol{\Gamma}$ as the schemetheoretic closure of $G(K)$ in $G$.) Since the map $\Gamma \to \ol{\Gamma}$ is an isomorphism on the generic fibers, Raynaud's theorem implies that it is an isomorphism. It follows that $\Gamma_k \to G_k$ is injective; since $\Gamma(R) \to \Gamma(k)$ is bijective (as $\Gamma$ is constant), the composite $G(R)=\Gamma(R) \to G(k)$ is injective. ◾
Remark. In the above situation, the reduction map need not be surjective. For example, let $G$ be the Kummer extension of $\bZ/p\bZ$ by $\mu_p$ corresponding to $a \in R$. If $A$ is a connected $R$algebra, then $G(A)$ is the set of pairs $(i, z)$, where $i \in \bZ/p\bZ$ and $z \in A$ satisfies $z^p=a^i$. If $R$ does not contain a primitive $p$th root of unity or a $p$th root of $a$ then $G(R)=0$. But if $k$ is perfect then $G_k$ is the trivial extension (since $a$ has a $p$th root), so $G(k)=\bZ/p\bZ$.
Remark. Without the assumption on $e$, the reduction map need not be injective. For example, take $G=\mu_p$ and suppose $K$ contains the $p$th roots of unity. Then $G(R)=\mu_p(K)$ has order $p$ but $G(k)$ is the trivial group.
Corollary. Suppose $E$ has good reduction and maintain the same assumptions on $K$. Then the reduction map $E[n](K) \to \ol{E}[n](k)$ is injective.
The kernel of reduction
We now study the group $E_1(K)$, the kernel of the homomorphism $E_0(K) \to \ol{E}_{\sm}(k)$. Since points on $E_1(k)$ are $\fp$adically close to the identity, the point at infinity, it makes sense to switch coordinates so that the identity is at $(0,0)$. The projective equation for $E$ is
$$ ZY^2=X^3+aZ^2X+bZ^3 $$We usually put $x=X/Z$ and $y=Y/Z$. We now put $u=X/Y$ and $v=Z/Y$ to obtain the equation
$$ v=u^3+auv^2+bv^3. $$The point at infinity in projective coordinates is $[0:1:0]$, and thus corresponds to $(u,v)=(0,0)$. The set $E_1(K)$ is given by the set of solutions to the above equation with $u$ and $v$ in $\fp$.
Let $F(u,v)$ be the right side of the above equation, so that the equation reads $v=F(u,v)$. We can then plug this expression for $v$ into the right side to find $v=F(u,F(u,v))$. Continuing in this way, we find $v=\phi(u)$, where $\phi(u)$ is the iterate $F(u,F(u,F(u, \ldots)))$. It is not difficult to see that $\phi(u)$ is a power series in $u$ with coefficients in $R$. Note that, because $R$ is complete, if $u$ is an element of $\fp$ then $\phi(u)$ is a welldefined element of $R$, and in fact $\fp$ since $\phi(0)=0$. It is now an easy exercise to show:
Proposition. The map $\fp \to E_1(K)$ sending $u$ to $(u, \phi(u))$ is a bijection of sets taking 0 to the identity element of $E_1(K)$.
Using this bijection, we can transfer the group structure on $E_1(K)$ to a group structure on $\fp$, which we denote by $\oplus$. It is not hard to show that $\oplus$ is given by a power series over $R$, i.e., there exists a power series $G \in R [\![ s,t ]\!]$ such that $s \oplus t = G(s,t)$. Since 0 is the identity element, we have $G(s,0)=G(0,s)=s$, and so $G(s,t)=s+t+\cdots$, where $\cdots$ are higher order terms. It follows that $\fp^n$ is a subgroup of $\fp$ under $\oplus$. Let $E_n(K)$ be the corresponding subgroup of $E_1(K)$. Clearly then, we have an isomorphism
$$ E_n(K)/E_{n+1}(K) = \fp^n/\fp^{n+1} = k. $$We have thus proved:
Proposition. The group $E_1(K)$ has a decreasing filtration $\{E_n(K)\}_{n \ge 1}$ such that $E_n(K)/E_{n+1}(K)$ is isomorphic to $k$.
Corollary. Suppose $n$ is prime to the residue characteristic. Then the map $E_0(K)[n] \to E_{\sm}(k)[n]$ is injective.
Proof. The kernel is a subgroup of $E_1(K)$ killed by $n$, and therefore 0. ◾
Corollary. Suppose $k$ is finite of characteristic $p$. Then $E_1(K)$ is a pro$p$ group.
The quotient of E(K) by E_0(K)
We have the following important result:
Theorem. The group $E(K)/E_0(K)$ is finite. In fact, if $E$ has split multiplicative reduction (i.e., $\ol{E}_{\sm}$ is isomorphic to $\bG_m$ over $k$) then this group is cyclic of order $v(j)$; otherwise, it has cardinality at most 4.
We will not prove this theorem. Some remarks:

The finiteness statement follows immediately from the existence of Néron models. The more precise statement about the structure and cardinality of the group follows from the classification of Néron models. We will discuss these topics in the next lecture.

If $k$ is finite then the finiteness statement is easy, for $E(K)$ is then a compact group and $E_0(K)$ is an open subgroup; thus $E(K)/E_0(K)$ is both discrete and compact, and thus finite.

One can prove the theorem without Néron models through a casebycase analysis. For instance, suppose $v(a)=1$ and $v(b) \ge 2$. If $P=(x,y)$ is a point of $E(K)$ then
$$ x(2P) = \frac{x^42ax^28bx+a^2}{4(x^3+ax+b)}. $$Thus if $(x,y)$ reduces to the singular point $(0,0)$, i.e., $v(x) \ge 1$, then the valuation of the numerator is equal to 2, while the valuation of the denominator is at least 2; thus $v(x(2P)) \le 0$, and so $2p$ does not reduce to the singular point. This shows that $E(K)/E_0(K)$ is killed by 2. In fact, one can show that the sum of any two points reducing to the singular point does not reduce to the singular point, and so $E(K)/E_0(K)=\bZ/2\bZ$.
The NéronOggShafarevich criterion
Let $G_K$ be the absolute Galois group of $K$ and $I_K$ the inertia subgroup.
Theorem. Let $\ell$ be a prime different from the residue characteristic. Then:

$E$ has good reduction if and only if $I_K$ acts trivially on $T_{\ell}(E)$.

$E$ has semistable reduction if and only if $I_K$ acts unipotently on $T_{\ell}(E)$.
Proof. First, note that $I_K$ acts trivially on $T_{\ell}(E)$ if and only if it does so on $E[\ell^n](\ol{K})$ for all $n$. Thus, if $E$ has good reduction then $I_K$ acts trivially on $T_{\ell}(E)$ by what we've already shown. Conversely, suppose $I_K$ acts trivially on $T_{\ell}(E)$. Thus all $\ell^n$ torsion points belong to $E(K^{\un})$. Let $d$ be the order of $E(K^{\un})/E_0(K^{\un})$, which is finite. Then $E_0(K^{\un})[\ell^n]$ is the kernel of the map $E(K^{\un})[\ell^n] \to E(K^{\un})/E_0(K^{\un})$, and thus has cardinality at least $\ell^{2n}/d$. Since the reduction map $E_0(K^{\un}) \to \ol{E}_{\sm}(\ol{k})$ is injective on $\ell$power torsion, it follows that $\ol{E}_{\sm}(\ol{k})[\ell^n]$ has cardinality at least $\ell^{2n}/d$. But this is not true for $\bG_m$ (where the cardinality is $\ell^n$) or $\bG_a$ (where the cardinality is 1), and so $E$ cannot have multiplicative or additive reduction. Thus $E$ has good reduction.
Now suppose that $I_K$ acts unipotently on $T_{\ell}(E)$. It thus fixes some vector in $T_{\ell}(E)$, which implies that $E(K^{\un})[\ell^n]$ has cardinality at least $\ell^n$. Arguing as in the previous paragraph, we see that $\ol{E}_{\sm}$ cannot be $\bG_a$, and so $E$ has semistable reduction.
Finally, suppose that $E$ has semistable reduction. The multiplicationby$\ell^n$ map on the smooth locus $\cE_{\sm}$ of $\cE$ is flat, and so $\cE_{\sm}[\ell^n]$ is a flat group scheme over $R$. Let $G$ be the schemetheoretic closure in $\cE_{\sm}[\ell^n]$ of the set of $\ol{K}$points which extend to $\ol{R}$points. Then $G$ is finite and flat, and $G_k=\ol{E}_{\sm}[\ell^n]$. Since $G$ has $\ell$power order, it is étale, and so $G(K^{\un})=\ol{E}_{\sm}[\ell^n](\ol{k})$, which contains $\bZ/\ell^n \bZ$ (since $E$ is semistable). Thus $E[\ell^n](K^{\un})$ contains $\bZ/\ell^n \bZ$ for all $n$, which shows that $I_K$ fixes a vector in $T_{\ell}(E)$. Since the determinant of $T_{\ell}(E)$ is the $\ell$cyclotomic character, which is trivial on $I_K$, the result follows. ◾
Corollary. If $I_K$ acts trivially (or unipotently) on one $T_{\ell}(E)$ then it does so on all of them.
Corollary. $E$ has potentially good reduction if and only if $I_K$ acts through a finite quotient on $T_{\ell}(E)$.
Corollary. Isogenous curves have the same reduction type.
Proof. If $E$ and $E'$ are isogenous then $T_{\ell}(E)[1/\ell]$ and $T_{\ell}(E)[1/\ell]$ are isomorphic $\bQ_{\ell}$ representations of $I_K$. ◾