In this post, we classify one-dimensional algebraic groups over an algebraically closed field.
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 (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 , and the additive group .
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.
Before we start the proof, let’s take a moment to set conventions. Let us fix an algebraically closed field (of arbitrary characteristic). An algebraic group is smooth affine group scheme over (i.e. a smooth, closed subgroup scheme of for some ).
Our goal is to show that if is a connected algebraic group of dimension , then must be one of the two simplest possible group schemes: the multiplicative group , or the additive group . Recall their definitions:
The multiplicative group: The multiplicative group has the obvious underlying scheme structure of the punctured line: . The group structure on has multiplication map given by
inverse map given by
and identity section
Functorially, represents the group functor given by .
The additive group: The additive group has the underlying scheme structure of the line . The addition map is given by
inversion map given by
and identity section given by
Functorially, is the group functor sending a -scheme to .
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.
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 into an integral smooth projective curve .
- Show, that using the group structure of that contains an infinite subset, and, moreover, this subset preserves the -points of .
- Use Hurwitz’s automorphism theorem, along with basic facts about elliptic curves, to deduce that the genus of can’t be greater than , and can’t be respectively. Thus, deduce that .
- Note that since , has an infinite subset which stabilizes the set of -points in , that can contain only or -points.
- Deduce that, as a scheme, 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 or .
Now that we have this rough outline, let’s fill in the details:
Theorem(Main) 1: Let be a connected algebraic group of dimension . Then, is isomorphic to either or .
Proof: Begin by noting that is automatically integral. Indeed, since 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 is integral and regular, we know (Theorem 17.4.2 of Vakil) that we can open embed into a smooth integral projective curve .
For each we have the translation operator . By the curve-to-extension theorem (cf. Vakil), each translation automorphism extends to a morphism . But, since this map is obviously of degree , it must be an automorphism of . Thus, we obtain an injection . Moreover, since , and , we know that is infinite, and thus is also infinite.
Now, we claim that this implies that the genus of is , so that . It is a famous theorem of Hurwitz, that if is a curve over an algebraically closed field, and the genus of is at least , then is finite. Thus, we certainly know that has genus at most . So, we only need to discount the case that the genus of is .
To do this, let us suppose that is of genus , and choose a -point .. Note that we have a map which respects composition. In particular, note that, since is finite, is the identity map infinitely often. In particular, there are infinitely many automorphisms of which fix .
The pair is an elliptic curve over , and so there exists a unique -group scheme structure on such that is the identity point. We then obtain infinitely many automorphisms of which fix , and thus an infinite number of elements of (here denotes self-isogenies). But, it’s a common fact that this set is finite (in fact, it’s of order at most –cf. Theorem 10.1 of Silverman). Thus, we arrive at a contradiction.
So, at this point we know that , and we have an open embedding . Moreover, we have an infinite set of automorphisms of stabilizing the finite set . But, if consisted of more than -points, then the permutation of would determine the automorphism of (this is just the classical theory of linear fractional transformations), meaning there could only be finitely many automorphisms coming from . Thus, consists of at most two -points. Since and for any -points we may conclude that, as schemes, is isomorphic to either or .
Thus, it remains to show what the possible algebraic group structures on and are.
We begin by showing that, up to isomorphism of algebraic groups, the only algebraic group with underlying scheme structure is . To see this, let be an algebraic group with underlying scheme structure . By changing up to an isomorphism, we may assume without loss of generality that has identity at . Now, it suffices to show that for each , the translation map is just the map . To see this, note that is an automorphism of (as a scheme) with no fixed points. A quick check shows that it’s of the form for some . But, since , we may conclude that .
Similarly, let be a group scheme with underlyling scheme structure of . By the same argument as before, we may assume without loss of generality that the identity element of is at . The automorphisms of are of the form , for . Now, let , as before, we want to show that is the automorphism . Now, since has no fixed points it must be of the form (the map has the fixed point ). But, then since it must preserve , we see that , and so the conclusion follows.
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 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 , even over non-closed fields:
Theorem 2: Let be any field (not necessarily algebraically closed!), the any connected one-dimensional algebraic group over is abelian.
Proof: Note we can check commutativity on -points. Indeed, let be our connected, one-dimensional algebraic group. We want to show that the following identity of morphisms holds:
where is the obvious switching of coordinates map. Well, since and are extremely nice (in particular, reduced and separated), we can check this equality over , where the result follows from the main theorem.
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 is a smooth, projective, geometrically integral curve (here I am NOT assuming that !), and that is not . Then, no (proper) open subscheme of can have the structure of a -group scheme.
Proof: Indeed, suppose that is such an open subscheme. Then, is an open subscheme of , with the structure of a -group scheme. Note though that is necessarily affine. Indeed:
Lemma 4: Let be a smooth, projective, geometrically integral curve. Then, any proper open subscheme is affine.
Proof: If is empty, we’re done. So, assume that is not empty. Then, for some -points . Consider then the line bundle , where for .
Now, choosing the sufficiently large, we know that will have negative degree, and so we may assume that . So, by choosing the sufficiently large, we can made sufficiently large. In particular, we can find such that has poles precisely at the . Thus, we obtain a non-constant rational map off of the . Extending this to a morphism , we obtain a finite map with , and thus is affine.
So, by our classification, we have that is either or . So, if is birational to in either case, and so (since was assumed smooth), we have that . Thus, is a Brauer-Severi variety, and since we assumed that had a -point (since had a -point, being a group scheme!) we may conclude that .
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 , and let be the closure of inside of . Then, has a group structure, but 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 which is neither nor .
As we shall see in the next section, we’re better off looking for things which geometrically are .
To begin, let’s recall the general formalism of Weil restriction. Namely, let be a finite separable extension of fields. Then, for any group scheme we can form a group functor , called the Weil restriction of , as follows:
for every -algebra . If the Weil restriction is representable, we also call the representing scheme the Weil restriction of , and also denote it .
It is a general theorem that for finite separable, and an algebraic group , the Weil restriction , is representable by an algebraic group over . 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 or .
Consider for example the connected algebraic group (I’m now decorating the multiplicative group with what field I’m considering it over). I can then consider the Weil restriction . Note that this is NOT a one-dimensional algebraic group. In fact, the following is fairly easy to show:
Theorem 5: Let be a finite Galois extension, and an algebraic group. Then:
In particular, we see that in our specific case, is two-dimensional.
But, we can cut down on the dimension by finding various subschemes of . In particular, note that we have a canonical map . Indeed, for each -algebra we have the map
which is just the usual norm map of algebras, considering the natural algebra map (note that the norm map makes sense since is a free -module!).
Let us denote the connected algebraic group by . Note that we have the following short exact sequence of algebraic groups
Moreover, it’s easy to check that is indeed a connected algebraic group, since if (referencing the theorem stated earlier)
Thus, all we have left to check is that . But, this is clear since there are certainly -algebras such that is not isomorphic to the ‘norm one’ elements of .
For a truly simple example, take . Then, we have the following short exact sequence of groups
Twists, and strengthenings of the theorem
It would be nice if we could somehow extend the above theorem to non-algebraically closed fields . Of course, as the example in the last section shows, this is not always possible. 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 .
Just to lay down some terminology, note that if is any field (not necessarily algebraically closed!) and is a one-dimensional connected algebraic group over , then is either or . (Note: I’m using the common fact that if a -scheme has a -point, then connected implies geometrically connected).
Recall that a twist of a group scheme over a field is a group scheme for which . So the last paragraph says, in other words, is a twist of either the additive or multiplicative group. We shall call a one-dimensional algebraic group of additive type if , and of multiplicative type if . 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 be a perfect field. Then, the only one-dimensional connected algebraic group of additive type is .
Proof: Embed into a regular curve . Since is perfect, and geometric regularity (i.e. smoothness) can be checked on perfect closures, we know that is actually smooth. But, , and thus, we see that (since they are birational smooth curves). But, note that removed two points cannot be , and so is one point. Moreover, this point must be a -point, else the standard short exact sequence
would be contradicted. But, note that since is a Brauer-Severi variety (see earlier), with a -point, it must be . Moreover, since must base-change to the -point , must consist of a single -point as well. Thus, , as a scheme, is . Thus, previous arguments show that it must, in fact, be isomorphic to as desired
Remark: Note the important place that this fails for . In the argument of the main theorem, we proved a statement about and 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 . For we used the closedness of to discount the automorphism as being a possible translation operator.
Thus, over a perfect field, all we have are , and groups of multiplicative type (i.e. tori).
There is a nice way of classifying these though. Namely, for any connected algebraic group , it’s common knowledge that the (isomorphism classes over of) twists of are just “-torsors” (here denotes the category of linear algebraic groups), and so are classified by
where, here, denotes the absolute Galois group , and denotes continuous group cohomology. We are thinking of acting on by just acting on the factor of .
While this seems very scary, it’s actually reasonably manageable in most specific instances. For example, let’s see what happens when we take to be , so that this should be classifying groups of multiplicative type. Well, , with the actual automorphisms being . Moreover, since neither of these automorphisms has any constant coefficients, acts trivially. Thus, we see that
where the second to last step was because was a trivial -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 were just came down to the automorphisms possibly being fixed point free, which might give us different group structures on . But, the automorphisms of those form with no fixed points correspond precisely to with , or, in other words, extensions of of degree !
As an example of this theory, we can see that there are precisely two groups of multiplicative type over , corresponding to the trivial extension, and to the extension . Or, if one doesn’t want to use Kummer theory, we can stop at the step before, and calculate
But, since is procyclic, generated by , any continuous homomorphism is determined by where goes. Thus, there are two such continuous homomorphisms . Moreover, from the last section, we know precisely what these two connected algebraic groups must be: the multiplicative group and .
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 are , , and .
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 where is a -adic local field. Let us say that is the residue field of , a finite extension of .
Now, there are various ways of defining the reduction over . We shall take the low-brow approach here. Namely, to any Weierstrass equation for we can consider a minimal Weierstrass equation which is defined to be a Weierstrass equation with coefficients in whose discriminant has minimal valuation. We shall denote the curve associated to by . Note that is a curve contained in .
One then defines the reduction of , denoted , to be , a curve over . One then begins to partition the set of into classes based on properties of . The most crude sorting of this type is made by saying that has good reduction if is an elliptic curve (i.e. whether ) and that it has bad reduction otherwise.
But, a further reduction is made based off of the smooth locus of . Before we define this, let us recall the following simple lemma:
Lemma 8: For any field a cubic curve can have at most one singularity, and it’s defined over .
Proof: Suppose first that , and that had two singularities . Take a line passing through both and . Consider then the intersection (thought of as the intersection of divisors in the sense of intersection theory). Note that each of and cannot have intersection multiplicity , else and would be smooth points of (since their maximal ideal at those points would be singly generated).
Thus, we may conclude that
But, by Bezout’s theorem we have that which is a contradiction.
Assume now that . If had two singular points they would stay singular over contradicting the previous case. Similarly, even if had one singular point, but it was not defined over , then would pick up at least two singular points over .
So, from this we see that if has bad reduction, then it has just one singular point. Working a bit harder, one can show that either is essentially ‘cuspidal’ (i.e. isomorphic to ) or ‘nodal’ (i.e. isomorphic to ). In the former case we say that has additive reduction, and in the former case we say that 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 has split multplicative reduction if these tangent lines are defined over , and non-split multiplicative reduction otherwise.
The more rigorous statement is that geometrically is a nodal curve. Then, has split multiplicative reduction if it is actually a nodal curve over , and non-split if its only nodal over some finite extension (one can show that it actually always happens over the quadratic extension of ).
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 is either cusidpidal or nodal).
Namely, let us consider the reduction , and define the smooth locus of , denoted , to be the set of smooth points of . By Lemma 8 we know that is either or for some . Now, using the exact same chord-tangent construction in the classical theory of elliptic curves, one can show that possesses the structure of an algebraic group.
In particular, by Lemma 4, we know that is a connected affine algebraic group over , and thus by Theorem 7 we know that is either , , or (where is the quadratic extension). Thus, we have the following definition/theorem:
Theorem/Definition 9: Let be an elliptic curve. Then, has good (resp. additive, resp. split multiplicative, resp. non-split multiplicative) reduction if is an elliptic curve (resp. , resp. , resp. ).
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.