Classifying One-dimensional Algebraic Groups

In this post, we classify one-dimensional algebraic groups over an algebraically closed field. This post was later extended in this one.


The world of algebraic groups, and even more generally affine group schemes, is a vast and scary place; all kinds of unexpected things can happen. Even in the cases that we best understand, like reductive groups over local fields, we are far from being able to ‘list’ all such possible groups (the best we can do is try and list root data in the reductive case).

That said, the vista isn’t so bleak that in the absolutely simplest of all cases, dimension 1 (connected) algebraic groups over an algebraically closed field, we can’t say anything. In fact, quite the opposite. There are, up to isomorphism, only two such groups: the multiplicative group \mathbf{G}_m, and the additive group \mathbf{G}_a.

This result is very useful, not only in the sense of comfort it gives in the chaotic world of groups (“well, at least I know that case!”), but as a means of limiting the open affine subgroup schemes of some given object. Moreover, in certain nice situations, it will even allow us to classify one-dimensional connected algebraic groups over non-closed fields.

The setup

Before we start the proof, let’s take a moment to set conventions. Let us fix an algebraically closed field k (of arbitrary characteristic). An algebraic group \mathbf{G}/k is smooth affine group scheme over k (i.e. a smooth, closed subgroup scheme of \text{GL}_n for some n).

Our goal is to show that if \mathbf{G}/k is a connected algebraic group of dimension 1, then \mathbf{G} must be one of the two simplest possible group schemes: the multiplicative group \mathbf{G}_m, or the additive group \mathbf{G}_a. Recall their definitions:

The multiplicative group: The multiplicative group \mathbf{G}_m has the obvious underlying scheme structure of the punctured line: D(t)=\mathbb{A}^1_k-\{(0)\}\subseteq \mathbb{A}^1_k. The group structure on \mathbf{G}_m has multiplication map m given by

m:k[t,t^{-1}]\to k[u,v,u^{-1},v^{-1}]:t\mapsto uv

inverse map i given by

i:k[t,t^{-1}]\to k[t,t^{-1}]:t\mapsto t^{-1}

and identity section e

e:k[t,t^{-1}]\to k:t\mapsto 1

Functorially, \mathbb{G}_m represents the group functor \mathbf{Sch}/k\to \mathbf{Ab} given by X\mapsto \mathcal{O}_X(X)^\times.

The additive group: The additive group \mathbf{G}_a has the underlying scheme structure of the line \mathbb{A}^1_k. The addition map a is given by

a:k[t]\to k[u,v]:t\mapsto u+v

inversion map i given by

i:k[t]\to k[t]:t\mapsto -t

and identity section 0 given by

0:k[t]\to k:t\mapsto 0

Functorially, \mathbf{G}_a is the group functor sending a k-scheme X to \mathcal{O}_X(X).

\text{ }

While I am sure that the vast majority of readers were well aware of those definitions, it’s nice to keep them in mind for later, when we do a fine structural analysis of the group structures on the line and punctured line.

The proof

Before we jump right into the formalities of the proof, let’s take a quick second to outline its major steps:

  • Open embed our group \mathbf{G} into an integral smooth projective curve C.
  • Show, that using the group structure of \mathbf{G} that \text{Aut}(C) contains an infinite subset, and, moreover, this subset preserves the k-points of C-\mathbf{G}.
  • Use Hurwitz’s automorphism theorem, along with basic facts about elliptic curves, to deduce that the genus of C can’t be greater than 2, and can’t be 1 respectively. Thus, deduce that C\cong \mathbb{P}^1_k.
  • Note that since \text{Aut}(C), has an infinite subset which stabilizes the set of k-points in C-G, that C-G can contain only 1 or 2 k-points.
  • Deduce that, as a scheme, \mathbf{G} is isomorphic to the line or the punctured line.
  • Show that all group structures on the line and punctured line give algebraic groups isomorphic to either \mathbf{G}_m or \mathbf{G}_a.

Now that we have this rough outline, let’s fill in the details:

Theorem(Main) 1: Let \mathbf{G}/k be a connected algebraic group of dimension 1. Then, \mathbf{G} is isomorphic to either \mathbf{G}_a or \mathbf{G}_m.

Proof: Begin by noting that \mathbf{G} is automatically integral. Indeed, since \mathbf{G} is Noetherian, connected, and has local rings which are domains (since it’s regular), the standard trick shows that if it were reducible, and the components intersected, one could push this intersection to the local ring at the intersection point, contradicting the local ring being integral. Now, since \mathbf{G} is integral and regular, we know (Theorem 17.4.2 of Vakil) that we can open embed \mathbf{G} into a smooth integral projective curve C/k.

For each g\in \mathbf{G}(k) we have the translation operator \ell_g:\mathbf{G}\to\mathbf{G}. By the curve-to-extension theorem (cf. Vakil), each translation automorphism extends to a morphism \widetilde{\ell_g}:C\to C. But, since this map is obviously of degree 1, it must be an automorphism of C. Thus, we obtain an injection \mathbf{G}(k)\hookrightarrow \text{Aut}(C). Moreover, since \dim\mathbf{G}=1, and k=\overline{k}, we know that \mathbf{G}(k) is infinite, and thus \text{Aut}(C) is also infinite.

Now, we claim that this implies that the genus of C is 0, so that C\cong\mathbb{P}^1_k. It is a famous theorem of Hurwitz, that if X is a curve over an algebraically closed field, and the genus of X is at least 2, then \text{Aut}(X) is finite. Thus, we certainly know that C has genus at most 2. So, we only need to discount the case that the genus of C is 1.

To do this, let us suppose that C is of genus 1, and choose a k-point x_0.. Note that we have a map \mathbf{G}(k)\to \text{Sym}(C-\mathbf{G}) which respects composition. In particular, note that, since C-\mathbf{G} is finite, \widetilde{\ell_g} is the identity map infinitely often. In particular, there are infinitely many automorphisms of C which fix x_0.

The pair (C,x_0) is an elliptic curve over k, and so there exists a unique k-group scheme structure on C such that x_0 is the identity point. We then obtain infinitely many automorphisms of C which fix x_0, and thus an infinite number of elements of \text{End}((C,x_0)) (here \text{End} denotes self-isogenies). But, it’s a common fact that this set is finite (in fact, it’s of order at most 24–cf. Theorem 10.1 of Silverman). Thus, we arrive at a contradiction.

So, at this point we know that C\cong\mathbb{P}^1_k, and we have an open embedding \mathbf{G}\hookrightarrow C. Moreover, we have an infinite set of automorphisms of C stabilizing the finite set C-\mathbf{G}. But, if C-\mathbf{G} consisted of more than 2-points, then the permutation of C-\mathbf{G} would determine the automorphism of \mathbb{P}^1_k (this is just the classical theory of linear fractional transformations), meaning there could only be finitely many automorphisms coming from \mathbf{G}(k). Thus, C-\mathbf{G} consists of at most two k-points. Since \mathbb{P}^1_k-\{x\}\cong\mathbb{A}^1_k and \mathbb{P}^1_k-\{x,y\}\cong \mathbb{A}^1_k-\{(0)\} for any k-points x,y we may conclude that, as schemes, \mathbf{G} is isomorphic to either \mathbb{A}^1_k or \mathbb{A}^1_k-\{(0)\}.

Thus, it remains to show what the possible algebraic group structures on \mathbb{A}^1_k and \mathbb{A}^1_k-\{(0)\} are.

We begin by showing that, up to isomorphism of algebraic groups, the only algebraic group with underlying scheme structure \mathbb{A}^1_k is \mathbf{G}_a. To see this, let \mathbf{G} be an algebraic group with underlying scheme structure \mathbb{A}^1_k. By changing \mathbf{G} up to an isomorphism, we may assume without loss of generality that \mathbf{G} has identity at 0. Now, it suffices to show that for each x\in \mathbf{G}(k), the translation map \ell_x is just the map k[t]\to k[t]:t\mapsto t+x. To see this, note that \ell_x is an automorphism of \mathbb{A}^1_k (as a scheme) with no fixed points. A quick check shows that it’s of the form t\mapsto t+a for some a\in k. But, since \ell_x(0)=x, we may conclude that a=x.

Similarly, let \mathbf{G} be a group scheme with underlyling scheme structure of \mathbb{A}^1_k-\{(0)\}. By the same argument as before, we may assume without loss of generality that the identity element of \mathbf{G} is at 1. The automorphisms of \mathbf{G}_m are of the form t\mapsto a t^{\pm 1}, for a\in k^\times. Now, let x\in \mathbf{G}(k), as before, we want to show that \ell_x is the automorphism t\mapsto xt. Now, since \ell_x has no fixed points it must be of the form t\mapsto at (the map t\mapsto at^{-1} has the fixed point \sqrt{a}). But, then since it must preserve 1, we see that x=a, and so the conclusion follows. \blacksquare

A simple application

Commutativity of one-dimensional algebraic groups

Before we discuss actual applications, let’s discuss some ‘weaker’ corollaries, which are in and of themselves surprising. If some person came up to you on the street and said “Hey, buddy, is every connected algebraic group of dimension 1 over an algebraically closed field commutative?”, I don’t know what you’d say. It’s not at all obvious this is true, although we could conclude, a posteriori, from the above that this is true. Of course, you could try and prove this Lie theoretically (using the Lie algebra), but Lie theory for algebraic groups gets weird over positive characteristic.

Moreover, we actually get the commutativity of connected algebraic groups of dimension 1, even over non-closed fields:

Theorem 2: Let k be any field (not necessarily algebraically closed!), the any connected one-dimensional algebraic group over k is abelian.

Proof: Note we can check commutativity on \overline{k}-points. Indeed, let \mathbf{G}/k be our connected, one-dimensional algebraic group. We want to show that the following identity of morphisms holds:

\begin{matrix} \mathbf{G}\times_k\mathbf{G} & \xrightarrow{\text{switch}} & \mathbf{G}\times_k \mathbf{G}\\ & \searrow^{m} & \downarrow^{m}\\ & & \mathbf{G}\end{matrix}

where \text{switch} is the obvious switching of coordinates map. Well, since \mathbf{G}\times_k \mathbf{G} and \mathbf{G} are extremely nice (in particular, reduced and separated), we can check this equality over \overline{k}, where the result follows from the main theorem. \blacksquare

Restriction on open group subschemes of curves

The above theorem puts a serious restriction on what schemes can contain algebraic groups as (proper) open subschemes. The particular example I have in mind is the following.

Theorem 3: Suppose that X/k is a smooth, projective, geometrically integral curve (here I am NOT assuming that \overline{k}=k!), and that X is not \mathbb{P}^1_k. Then, no (proper) open subscheme of X can have the structure of a k-group scheme.

Proof: Indeed, suppose that U is such an open subscheme. Then, U_{\overline{k}}\to X_{\overline{k}} is an open subscheme of X_{\overline{k}}, with the structure of a k-group scheme. Note though that U_{\overline{k}} is necessarily affine. Indeed:

Lemma 4: Let X/\bar{k} be a smooth, projective, geometrically integral curve. Then, any proper open subscheme U is affine.

Proof: If U is empty, we’re done. So, assume that U is not empty. Then, U=X-\{p_1,\ldots,p_n\} for some k-points p_i. Consider then the line bundle \mathscr{L}=\mathcal{O}(D), where D=m_1 p_1+\cdots+ m_n p_n for m_i\gg 0.


Now, choosing the m_i sufficiently large, we know that K-D will have negative degree, and so we may assume that \ell(K-D)=0.  So, by choosing the m_i sufficiently large, we can made \ell(D) sufficiently large. In particular, we can find f\in K(X) such that f has poles precisely at the p_i. Thus, we obtain a non-constant rational map f:X\to \mathbb{P}^1_k off of the p_i. Extending this to a morphism f:X\to \mathbb{P}^1_k, we obtain a finite map with X-\{p_1,\ldots,p_n\}=f^{-1}(\mathbb{A}^1_k), and thus X-\{p_1,\ldots,p_n\} is affine. \blacksquare

So, by our classification, we have that U_{\overline{k}} is either \mathbf{G}_a or \mathbf{G}_m. So, if X_{\overline{k}} is birational to \mathbb{P}^1_{\overline{k}} in either case, and so (since X was assumed smooth), we have that X_{\overline{k}}\cong\mathbb{P}^1_{\overline{k}}. Thus, X is a Brauer-Severi variety, and since we assumed that X had a k-point (since U\subseteq X had a k-point, being a group scheme!) we may conclude that X\cong\mathbb{P}^1_k. \blacksquare

Now, this theorem is certainly overkill in most situations, but it does quickly allow one to process/internally verify various statements one often times hears. For example, take the statement

Let E_0:y^2=x^3+Ax+B, and let E be the closure of E_0 inside of \mathbb{P}^2_k. Then, E has a group structure, but E_0 does not.

(this was, if you are curious, paraphrased from Darmon’s Rational Points on Curves). Now, while there may be other ways of verifying this, it also follows immediately from our above analysis.

What goes wrong over non-closed fields

There are various places in the argument for the main theorem that, in an important way, used that our base field was algebraically closed. Just in case you are still not convinced that we couldn’t have tweaked the proof to work in general, we do the most explicit thing we can to convince you. Namely, we give an example of a connected algebraic group \mathbf{G}/\mathbb{F}_q which is neither \mathbf{G}_a nor \mathbf{G}_m.

As we shall see in the next section, we’re better off looking for things which geometrically are \mathbf{G}_m.

To begin, let’s recall the general formalism of Weil restriction. Namely, let L/K be a finite separable extension of fields. Then, for any group scheme \mathbf{G}/L we can form a group functor \left(\mathbf{Sch}/k\right)^\text{op}\to\mathbf{Set}, called the Weil restriction of \mathbf{G}, as follows:

\mathsf{Res}_{L/K}\mathbf{G}(R)=\mathbf{G}(R\otimes_K L)

for every K-algebra R. If the Weil restriction \mathsf{Res}_{L/K}\mathbf{G} is representable, we also call the representing scheme the Weil restriction of \mathbf{G}, and also denote it \mathsf{Res}_{L/K}\mathbf{G}.

It is a general theorem that for L/K finite separable, and an algebraic group \mathbf{G}/L, the Weil restriction \mathsf{Res}_{L/K}\mathbf{G}, is representable by an algebraic group over K. For more information, see Raynaud et. al’s Neron Models. For example, a vast generalization of the above representability statement can be found as Theorem 4 in section 7.6.

Now, the notion of Weil restriction allows us to produce many examples of one-dimensional algebraic groups (obviously over non-closed fields) which are not \mathbf{G}_m or \mathbf{G}_a.

Consider for example the connected algebraic group \mathbf{G}_{m,\mathbb{F}_{q^2}} (I’m now decorating the multiplicative group with what field I’m considering it over). I can then consider the Weil restriction \mathsf{Res}_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}. Note that this is NOT a one-dimensional algebraic group. In fact, the following is fairly easy to show:

Theorem 5: Let L/K be a finite Galois extension, and \mathbf{G}/L an algebraic group. Then:

\displaystyle \mathsf{Res}_{L/K}\mathbf{G}\times_K L=\prod_{\sigma \in\text{Gal}(L/K)}\mathbf{G}^\sigma

where \mathbf{G}^\sigma:=\mathbf{G}\times_{\mathrm{Spec}(L),\sigma}\mathrm{Spec}(L) (here the notation means that in the fiber product the map \mathbf{G}\to\mathrm{Spec}(L) is the usual one and the map \mathrm{Spec}(L)\to \mathrm{Spec}(L) is that induced by \sigma). In particular, if \mathbf{G}=\mathbf{H}_L for some group \mathbf{H}/K we have a canonical isomorphism of L-group schemes \mathbf{G}^\sigma\cong\mathbf{G} and thus the above reduces to

\displaystyle \mathsf{Res}_{L/K}\mathbf{G}\times_K L=\prod_{\sigma \in\text{Gal}(L/K)}\mathbf{G}

In particular, we see that in our specific case, \mathsf{Res}_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}} is two-dimensional.

But, we can cut down on the dimension by finding various subgroup schemes of \mathsf{Res}_{\mathbb{F}_{q^2}\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}. In particular, note that we have a canonical map N:\mathsf{Res}_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}\to \mathbf{G}_{m,\mathbb{F}_q}. Indeed, for each \mathbb{F}_q-algebra R we have the map

\mathsf{Res}_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}(R)=\mathbf{G}_{m,\mathbb{F}_{q^2}}(\mathbb{F}_{q^2}\otimes_{\mathbb{F}_q}R)= (\mathbb{F}_{q^2}\otimes_{\mathbb{F}_q}R)^\times \to R^\times =\mathbf{G}_{m,\mathbb{F}_q}(R)

which is just the usual norm map of algebras, considering the natural algebra map R\to R\otimes_{\mathbb{F}_q}\mathbb{F}_{q^2} (note that the norm map makes sense since R\otimes_{\mathbb{F}_q}\mathbb{F}_{q^2} is a free R-module!).

Let us denote the connected algebraic group \ker N by \mathsf{Res}^1_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}. Note that we have the following short exact sequence of algebraic groups

1\to \mathsf{Res}^1_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}\to \mathsf{Res}_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}\to\mathbf{G}_{m,\mathbb{F}_q}\to 1

Moreover, it’s easy to check that \mathsf{Res}^1_{\mathbb{F}_{q^2}/\mathbb{F}_q} is indeed a connected algebraic group, since (referencing the theorem stated earlier)

(\mathsf{Res}_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}})\times_{\mathbb{F}_q}\mathbb{F}_{q^2}\cong \mathbf{G}_{m,\mathbb{F}_{q^2}}

then one can check directly (by considering the above exact sequence base changed to \mathbb{F}_{q^2})

(\mathsf{Res}^1_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}})\times_{\mathbb{F}_q}\mathbb{F}_{q^2}\cong \mathbf{G}_{m,\mathbb{F}_{q^2}}

Thus, all we have left to check is that \mathsf{Res}^1_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}\not\cong\mathbf{G}_{m,\mathbb{F}_q}. But, this is clear since there are certainly \mathbb{F}_q-algebras R such that R^\times is not isomorphic to the ‘norm one’ elements of (\mathbb{F}_{q^2}\otimes_{\mathbb{F}_q}R)^\times.

For a truly simple example, take R=\mathbb{F}_{q}. Then, we have the following short exact sequence of groups

1\to \mathsf{Res}^1_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}(\mathbb{F}_q)\to \mathbb{F}_{q^2}^\times\to \mathbb{F}_q ^\times\to 1

In particular


whereas \left|\mathbf{G}_{m,\mathbb{F}_q}(\mathbb{F}_q)\right|=q-1.

Twists, and strengthenings of the theorem

It would be nice if we could somehow extend the above theorem to non-algebraically closed fields k. Of course, as the example in the last section shows, this is not always possible in the most naive way. That said, there are several further observations which allow us to tighten the results of theorem, sometimes giving us a very practical, useful list of the possible one-dimensional algebraic groups over k.

Just to lay down some terminology, note that if k is any field (not necessarily algebraically closed!) and \mathbf{G} is a one-dimensional connected algebraic group over k, then \mathbf{G}_{\overline{k}} is either \mathbf{G}_{a,\overline{k}} or \mathbf{G}_{m,\overline{k}}. To see this we merely apply our main theorem and:

Observation: Let X be a finite type k-scheme and suppose that X(k)\ne\varnothing. Then, X is connected if and only if X_{\overline{k}} is connected. In particular, for a (finite type) group scheme over k connected implies geometrically connected.

Proof: Evidently if X_{\overline{k}} is connected then so is X (since we have a continuous surjection X_{\overline{k}}\to X). Suppose now that X is connected. Note that for an extension L/k one has that X_L is disconnected if and only if \mathcal{O}_X(X_L) contains non-trivial idempotents. In particular, since every element of \mathcal{O}_X(X_{\overline{k}})=\mathcal{O}_X(X)\otimes_k \overline{k} lies in \mathcal{O}_X(X_L)=\mathcal{O}_X(X)\otimes_k L (note that we’re applying `flat base change’ here) for some finite extension L/k we see that X_{\overline{k}} disconnected implies that X_L is disconnected for some finite extension L/k.

That said, note that X_L\to X is finite and flat (since L/k is finite and flat) and thus a clopen map. In particular, since X is connected we see that every connected component of X_L surjects on to X. In particular, fixing x_0\in X(k) we see that every connected component of X_L contains a preimage of x_0. But, since x_0 is a k-point it has only one preimage. Since connected components are disjoint this implies that X_L has only one connected component. The conclusion follows. \blacksquare

Recall that a twist of a group scheme \mathbf{H} over a field k is a group scheme \mathbf{G} for which \mathbf{G}_{\overline{k}}\cong\mathbf{H}_{\overline{k}}. So the last paragraph says, in other words, \mathbf{G} is a twist of either the additive or multiplicative group. We shall call a one-dimensional algebraic group \mathbf{G}/k of additive type if \mathbf{G}_{\overline{k}}\cong \mathbf{G}_{a,\overline{k}}, and of multiplicative type if \mathbf{G}_{\overline{k}}\cong \mathbf{G}_{m,\overline{k}}. Another common terminology for a group of multiplicative type is a one-dimensional torus, but we shall not use that terminology again.

We begin with the theorem which greatly reduces the amount of one-dimensional algebraic groups:

Theorem 6: Let k be a perfect field. Then, the only one-dimensional connected algebraic group \mathbf{G} of additive type is \mathbf{G}_{a,k}.

Proof: Embed \mathbf{G} into a regular curve C/k. Since k is perfect, and geometric regularity (i.e. smoothness) can be checked on perfect closures, we know that C is actually smooth. But, \mathbf{G}_{\overline{k}}=\mathbf{G}_{a,\overline{k}}\subseteq C_{\overline{k}}, and thus, we see that C_{\overline{k}}=\mathbb{P}^1_{\overline{k}} (since they are birational smooth curves). But, note that \mathbb{P}^1_{\overline{k}} removed two points cannot be \mathbb{A}^1_{\overline{k}}, and so \mathbb{P}^1_{\overline{k}}-\mathbf{G}_{\overline{k}} is one point. Moreover, this point p must be a k-point, else the standard short exact sequence

\mathbb{Z}\xrightarrow{n\mapsto n\deg p}\text{Pic}(\mathbb{P}^1_{\overline{k}})\to \text{Pic}(\mathbb{A}^1_{\overline{k}})=0

would be contradicted. But, note that since C_{\overline{k}} is a Brauer-Severi variety (see earlier), with a k-point, it must be \mathbb{P}^1_k. Moreover, since \mathbb{P}^1_k-\mathbf{G} must base-change to the k-point \mathbb{P}^1_{\overline{k}}-\mathbf{G}_{\overline{k}}, \mathbb{P}^1_{\overline{k}}-\mathbf{G}_{\overline{k}} must consist of a single k-point as well. Thus, \mathbf{G}, as a scheme, is \mathbb{A}^1_k. Thus, previous arguments show that it must, in fact, be isomorphic to \mathbf{G}_{a,k} as desired \blacksquare

Remark: Note the important place that this fails for \mathbf{G}_m. In the argument of the main theorem, we proved a statement about \mathbb{A}^1_{\overline{k}} and \mathbb{A}^1_{\overline{k}}-\{(0)\} having only one group structure up to isomorphism. We implied, and this can be easily seen, that this carries over to work over non-closed fields, but only for \mathbb{A}^1_k. For \mathbb{A}^1_{\overline{k}} we used the closedness of k to discount the automorphism \displaystyle t\mapsto \frac{t}{c} as being a possible translation operator.

Thus, over a perfect field, all we have are \mathbf{G}_{a,k}, and groups of multiplicative type (i.e. tori).

There is a nice way of classifying these though. Namely, for any connected algebraic group \mathbf{H}, it’s common knowledge that the (isomorphism classes over k of) twists of \mathbf{H} are just “\text{Aut}_{\mathbf{LAG}}(\mathbf{H}_{\overline{k}})-torsors” (here \mathbf{LAG} denotes the category of linear algebraic groups), and so are classified by


where, here, G_k denotes the absolute Galois group \text{Gal}(\overline{k}/k), and H^1_\text{cont} denotes continuous group cohomology. We are thinking of G_k acting on \text{Aut}(\mathbf{H}_{\overline{k}}) by just acting on the factor of \overline{k}.

While this seems very scary, it’s actually reasonably manageable in most specific instances. For example, let’s see what happens when we take \mathbb{H} to be \mathbf{G}_{m,k}, so that this should be classifying groups of multiplicative type. Well, \text{Aut}(\mathbf{G}_{m,\overline{k}})=\mathbb{Z}/2\mathbb{Z}, with the actual automorphisms being t\mapsto t^{\pm 1}. Moreover, since neither of these automorphisms has any constant coefficients, G_k acts trivially. Thus, we see that

\left\{\begin{matrix}\text{Groups of}\\ \text{multiplicative}\\\text{type}\end{matrix}\right\}\cong H^1_\text{cont}\left(G_k,\mathbb{Z}/2\mathbb{Z}\right)=\text{Hom}_\text{cont}(G_k,\mathbb{Z}/2\mathbb{Z})=\left\{\begin{matrix}\text{Field extensions}\\ \text{of degrees}\\ \text{one or two}\end{matrix}\right\}

where the second to last step was because \mathbb{Z}/2\mathbb{Z} was a trivial G_k-module and the last step is by Kummer theory.

Note, while this certainly seems a bit strange at first, it is not at all to be unexpected. Indeed, consider our remark following the last theorem. The impediment to proving that the groups of multiplicative type over k were just \mathbf{G}_{m,k} came down to the automorphisms \displaystyle t\mapsto \frac{t}{c} possibly being fixed point free, which might give us different group structures on \mathbb{A}^1_k-\{(0)\}. But, the automorphisms of those form with no fixed points correspond precisely to c\in k with \sqrt{c}\notin k, or, in other words, extensions of k of degree 2!

As an example of this theory, we can see that there are precisely two groups of multiplicative type over \mathbb{F}_q, corresponding to the trivial extension, and to the extension \mathbb{F}_{q^2}. Or, if one doesn’t want to use Kummer theory, we can stop at the step before, and calculate


But, since \widehat{\mathbb{Z}} is procyclic, generated by \text{Frob}_q, any continuous homomorphism is determined by where \text{Frob}_q goes. Thus, there are two such continuous homomorphisms \widehat{\mathbb{Z}}\to\mathbb{Z}/2\mathbb{Z}. Moreover, from the last section, we know precisely what these two connected algebraic groups must be: the multiplicative group \mathbf{G}_{m,\mathbb{F}_q} and \mathsf{Res}^1_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}.

So, to sum up this last part, together with our theorem of algebraic groups of additive type over perfect fields, we have the following tidy theorem:

Theorem 7: The only connected one-dimensional algebraic groups over \mathbb{F}_q are \mathbf{G}_{a,\mathbb{F}_q}, \mathbf{G}_{m,\mathbb{F}_q}, and \mathsf{Res}^1_{\mathbb{F}_{q^2}/\mathbb{F}_q}\mathbf{G}_{m,\mathbb{F}_{q^2}}.

Reduction types of elliptic curves

Using the classification of the last section, we can give a more conceptual understanding of the various ‘reduction types’ of an elliptic curve E/K where K/\mathbb{Q}_p is a p-adic local field. Let us say that k is the residue field of K, a finite extension of \mathbb{F}_p.

Now, there are various ways of defining the reduction \overline{E} over k. We shall take the low-brow approach here. Namely, to any Weierstrass equation f(x,y)=0 for E/K we can consider a minimal Weierstrass equation f_{\mathrm{min}}\in \mathcal{O}_K[x,y] which is defined to be a Weierstrass equation with coefficients in \mathcal{O}_K whose discriminant has minimal valuation. We shall denote the curve associated to f_\mathrm{min} by \mathcal{W}. Note that \mathcal{W} is a curve contained in \mathbb{P}^2_{\mathcal{O}_K}.

One then defines the reduction of E, denoted \overline{E}, to be \mathcal{W}_k, a curve over k. One then begins to partition the set of E into classes based on properties of \overline{E}. The most crude sorting of this type is made by saying that E has good reduction if \overline{E} is an elliptic curve (i.e. whether \mathrm{disc}(f_\mathrm{min})\in\mathcal{O}_K^\times) and that it has bad reduction otherwise.

But, a further reduction is made based off of the smooth locus of \overline{E}. Before we define this, let us recall the following simple lemma:

Lemma 8: For any field L a cubic curve C\subseteq\mathbb{P}^2_L can have at most one singularity, and it’s defined over L.

Proof: Suppose first that L=\overline{L}, and that C had two singularities p,q. Take a line \ell passing through both p and q. Consider then the intersection \ell\cap C (thought of as the intersection of divisors in the sense of intersection theory). Note that each of \ell\cap p and \ell\cap q cannot have intersection multiplicity 1, else p and q would be smooth points of C (since their maximal ideal at those points would be singly generated).

Thus, we may conclude that

\displaystyle \ell\cdot C=\sum_{x\in \ell\cap C}m_x(\ell,C)\geqslant 2+2=4

But, by Bezout’s theorem we have that \ell\cdot C=3 which is a contradiction.

Assume now that L\ne\overline{L}. If C had two singular points they would stay singular over \overline{L} contradicting the previous case. Similarly, even if C had one singular point, but it was not defined over L, then C_{\overline{L}} would pick up at least two singular points over C_{\overline{L}}. \blacksquare

So, from this we see that if \overline{E} has bad reduction, then it has just one singular point. Working a bit harder, one can show that either \overline{E} is essentially ‘cuspidal’ (i.e. isomorphic to y^2=x^3) or ‘nodal’ (i.e. isomorphic to y^2=x^3-x). In the former case we say that E has additive reduction, and in the former case we say that E has multiplicative reduction.

We make the classification even more granular by the following sort of vague statement. Namely, imagine a nodal curve. Then, approaching the singular point there are two ‘branches’, and thus the singular point has two distinct tangent lines at the nodal point. We say that E has split multplicative reduction if these tangent lines are defined over k, and non-split multiplicative reduction otherwise.

The more rigorous statement is that geometrically \overline{E} is a nodal curve. Then, E has split multiplicative reduction if it is actually a nodal curve over k, and non-split if its only nodal over some finite extension (one can show that it actually always happens over the quadratic extension of k).

To me, while these definitions literally make sense, they are a bit cumbersome, and counterintuitive. How did one think of these definitions? What is really happening here? Using the last section we can make better definitions of these terms and, in fact, justify the definitions above (e.g. why geometrically \overline{E} is either cusidpidal or nodal).

Namely, let us consider the reduction \overline{E}, and define the smooth locus of \overline{E}, denoted \overline{E}^\mathrm{sm}, to be the set of smooth points of \overline{E}. By Lemma 8 we know that \overline{E}^\mathrm{sm} is either \overline{E} or \overline{E}-\{p\} for some p\in\overline{E}(k). Now, using the exact same chord-tangent construction in the classical theory of elliptic curves, one can show that \overline{E}^\mathrm{sm} possesses the structure of an algebraic group.

In particular, by Lemma 4, we know that \overline{E} is a connected affine algebraic group over k, and thus by Theorem 7 we know that \overline{E}^\mathrm{sm} is either \mathbf{G}_{a,k}, \mathbf{G}_{m,k}, or \mathsf{Res}^1_{k'/k}\mathbf{G}_{m,k} (where k'/k is the quadratic extension).  Thus, we have the following definition/theorem:

Theorem/Definition 9: Let E/K be an elliptic curve. Then, E has good (resp. additive, resp. split multiplicative, resp. non-split multiplicative) reduction if \overline{E}^\mathrm{sm} is an elliptic curve (resp. \mathbf{G}_{a,k}, resp. \mathbf{G}_{m,k}, resp. \mathsf{Res}^1_{k'/k}\mathbf{G}_{m,k'}).

This is, in my opinion, the best version of the definitions of these terms. But, one can do a pretty simple case analysis to see that this definition is equivalent to the previous ones.


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