# Classifying one dimensional groups (II)

In this post we classify one-dimensional connected group varieties of dimension $1$.

## Motivation

In a previous post I discussed the classification of one-dimensional connected algebraic groups over a field $k$. There an ‘algebraic group’ over $k$ meant a smooth affine group of finite type over $k$.

Since then people have asked me directly and/or via the searches that led them to this blog asked what happens in the non-affine case. I’ll answer that question in this post by classifying one-dimensional connected geometrically reduced group schemes over $k$ (with some mild assumptions on $k$).

The answer, as we will see, essentially breaks in to two cases: the affine case (as discussed in the previous post) and the elliptic curve case. The two are very separate and distinct.

## The general theory

Let us now clarify exactly the assumptions we want and what basic properties we are able to get from them. Namely, we are interested in classifying one-dimensional groups $G$ over $k$ but without more adjectives this will be impossible. So, let discuss what adjectives we want, and what they give us.

Throughout a group variety over $k$ will mean a finite type group scheme over $k$.

### The assumptions

#### Connectivity

The first assumption that we want to assume about $G$ is that it’s connected. The reason for this is silly. If you don’t insist on working with connected groups you have the entire theory of finite groups to contend with. Indeed, for any finite group $G_0$ one can obtain a one-dimensional group (which is even smooth with reductive identity component) as $\mathbb{G}_{m,k}\times \underline{G_0}$ where $\underline{G_0}$ is the constant group. For this reason we will want to assume that $G$ is connected.

Let us note that in general if you don’t want to assume that $G$ is connected then you have a short exact sequence of group varieties

$1\to G^\circ\to G\to \pi_0(G)\to 1$

where $G^\circ$ is the identity component of $G$ and $\pi_0(G)$, the component group of $G$, is a finite etale group scheme. In particular, if $k$ is algebraically closed then $\pi_0(G)$ is essentially the same just an abstract finite group. Thus, one sees that the general non-connected situation comes down to classifying connected groups (which we will essentially do), finite groups (in the form of $\pi_0(G)$, and extensions of the latter by the former.

Let us also make an observation about connected groups, or more generally connected varieties with $k$-points, that will be useful later (this was also in the last point):

Observation 1: Let $k$ be a field and $X$ a finite type $k$-scheme such that $X(k)\ne\varnothing$. Then, $X$ is connected if and only if $X$ is 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. $\blacksquare$

As a corollary we obtain the following:

Corollary 2: Let $k$ be a field and $G$ a group variety over $k$. Then, $G$ is connected if and only if $G$ is geometrically connected.

#### Geometrically reduced

The second assumption we will make is that our group $G$ is geometrically reduced. Recall that if $k$ is perfect this is the same thing as $G$ being reduced. The reason that such assumptions are necessary for a reasonable classification is the prescence of infinitesimal group schemes in characteristic $p>0$.

Namely, recall that a group scheme $G_0$ over $k$ is infinitesimal if $(G_0)_\mathrm{red}$ is the trivial algebraic group.

Example 3: Define $\alpha_{p,k}:=\ker(\mathbb{G}_{a,k}\to \mathbb{G}_{a,k})$ where $F$ is the (relative) Frobenius map on $\mathbb{G}_{a,k}$ which on $R$-points is the additive map $x\mapsto x^p$.  One can see that $\alpha_{k,p}$ is, as a scheme, just $\mathrm{Spec}(k[x]/(x^p))$ and so evidently $(\alpha_{k,p})_{\mathrm{red}}$ is trivial.  So, $\alpha_{k,p}$ is infinitesimal.

Infinitesimal groups can be quite complicated. Their difficulty is somewhat comparable to the theory of finite groups. So, they are something we would like to ignore. For this reason we shall want to assume our groups are reduced. Moreover, to ignore issues with base change it’s also useful to assume that our groups are actually geometrically reduced (there exists reduced groups which are not geometrically reduced–see Examples 1.57 and 1.58 of [Mil]).

Let us note that if we are only interested in characteristic $0$ then this is no real assumption:

Theorem 4(Cartier): Let $k$ be a characteristic $0$ field. Then, every group variety over $k$ is (geometrically) reduced.

Proof: See [Mil, Theorem 3.23]. $\blacksquare$

#### One-dimensional

The assumption that $G$ is one-dimensional is, of course, necessary for a concise classification. Else we have to contend with all of abelian varieties, linear algebraic groups, … Hopefully this is convincing enough.

### The implications

Let $G$ be a group variety over $k$. We see what implications we get if we assume that $G$ is a one-dimensional connected geometrically reduced group variety.

#### Smoothness

In fact, geometrically reducedness for group varieties implies smoothness:

Lemma 5: Let $k$ be a field and $G$ a group variety over $k$. Then, $G$ is geometrically reduced if and only if it’s smooth over $k$.

Proof: If $G$ is smooth over $k$ then it’s geometrically reduced by standard theory. Conversely, suppose that $G$ is geometrically reduced. We need to show that $G_{\overline{k}}$ is regular.  This means that we need to check that the local rings at all points of $G_{\overline{k}}$ are regular. But, by standard theory it suffices to check at only the closed points. By generic smoothness for geometrically reduced varieties we know that there is some point $x\in G(\overline{k})$ such that $\mathcal{O}_{G_{\overline{k}},x}$ is regular. But, for any other closed point $g\in G(\overline{k})$ the left translation by $gx^{-1}$ map $t_{gx^{-1}}:G\to G$ is an automorphism of varieties which carries $x$ to $g$. Thus, $\mathcal{O}_{G_{\overline{k}},g}$ is also regular. $\blacksquare$

#### Separatedness

In fact, separatedness is true for any group:

Lemma 6: Let $k$ be a field and $G$ a group variety over $k$. Then, $G$ is separated.

Proof: It suffices to show that the diagonal $\Delta_G\subseteq G\times_k G$ is closed. But, the map $f:G\times_k G\to G$ given on $R$-points by $(g_1,g_2)\to g_1g_2^{-1}$ is clearly a morphism and $\Delta_G=f^{-1}(e)$ where $e\in G(k)$ is the identity element. Since $e$ is a closed point we see that $G$ is separated as desired. $\blacksquare$

#### Geometric integrality

From our assumptions that $G$ is smooth and geometrically connected we immediately get the following from general theory:

Lemma 7: Let $k$ be a field and let $X$ be geometrically connected smooth finite type $k$-scheme. Then, $X$ is geometrically integral.

Proof: Evidently we may assume that $k$ is algebraically closed. Suppose that $X$ had more than one irreducible component–say that $C_1$ and $C_2$ are distinct irreducible components of $X$. Note then that if $x\in C_1\cap C_2$ then $\mathcal{O}_{X,x}$ is not a domain since $C_1$ and $C_2$ give rise to two distinct minimal primes of $\mathcal{O}_{X,x}$. But, since $X$ is regular we know that $\mathcal{O}_{X,x}$ is a domain, and this is a contradictoin. Thus, the irreducible components of $X$ are disjoint. But, since $X$ is finite type we know it has finitely many irreducible components. This implies that the irreducible components are clopen. Since we assumed that $X$ had more than one irreducible component this contradicts that $X$ is connected. $\blacksquare$

But, in fact, we can also derive this irreducibility from the theory of group schemes:

Lemma 8: Let $k$ be a field and let $G$ be a connected group variety over $k$. Then, $G$ is geometrically irreducible.

Proof: From Observation 1 it suffices to assume that $k$ is algebraically closed. Suppose that $G$ were not irreducible. Then, as in the proof of Lemma 7 there must be two irreducible components $C_1$ and $C_2$ of $G$ that intersect and, in particular, contain a common $k$-point $x$. Note though that $C_1$ is not contained in the union of the other irreducible components (by irreducibility!) and thus there is some $k$-point $y$ of $G$ which is contained within a unique irreducible component. Let $t_{xy^{-1}}:G\to G$ be the left translation by $yx^{-1}$ automorphism. This takes $x$ to $y$. But, since $x$ lies on the intersection of two irreducible components of $G$ and $y$ lies in a unique irreducible component of $G$ this is a contradiction. $\blacksquare$

#### Affineness or projectiveness

One thing that we get for free from our assumptions, in fact just the fact that we’re working with a geometrically connected separated curve, is that our group is either affine or projective. Namely, we have the following:

Proposition 9: Let $k$ be a field and let $X$ be a one-dimensional geometrically connected separated variety over $k$. Then, either $X$ is affine or projective.

The idea of the proof is simple. Namely we will show that $X$ is at least contained in a projective curve and then use the fact that since we’re in one dimensional all proper open subvarieties of a projective curve are affine.

We begin by noting the following observation:

Observation 10: In the proof of Proposition 9 it suffices to assume that $X$ is smooth over $k$.

Remark 11: We are using the definition of projectiveness as in here. This is contrast to the definition that Hartshorne uses (which is called $H$-projective in loc. cit). This makes no real difference when discussing projectiveness of a variety, but does have difference in general. For instance, in Hartshorne’s definition of a projective morphism it needn’t be true that finite maps are projective (e.g. see the remark here).

We will cite the following lemma for simplicity:

Lemma 12: Let $k$ be a field and let $X$ be a connected one-dimensional variety over $k$. Then, $X$ is projective if and only if it’s proper.

Proof: If one assumes that $X$ is smooth and integral (which is the case we’re in for our groups) then this is somewhat easy. Namely, let $\mathrm{Spec}(A)$ be an affine open in $X$ and let $K(X)$ be the common function field between this affine open and $X$. Note that if we embed $\mathrm{Spec}(A)$ in to some $\mathbb{P}^n_k$ we can take the closure, which we’ll denote $Z$, to obtain a connected projective variety. It may be true that $Z$ is singular. But, after taking its normalization $C$ (see below for references for normalization) we get a smooth projective integral curve which is birational to $X$ (since $C$ is birational to $Z$ and $Z$ contains an open that is also an open in $X$). One can then use the valuative criterion for propertness to take the birational map $C-\! - \! \rightarrow X$ and extend it unique to a map $C\to X$ which is necessarily an isomorphism.

For the general proper case see this (see Remark 12 for why this phrase ‘$H$-projective’ doesn’t matter here). $\blacksquare$

Proof (Observation 10): By our assumptions we may clearly assume that $k$ is algebraically closed. Then, we know from standard theory that the smoothness of $X$ is equivalent to the regularity of $X$. But, since $X$ is one-dimensional this is equivalent to the normality of $X$. But, note that we have the normalization map $\nu:\widetilde{X}\to X$. This is a surjection, this much is clear. What lies significantly deeper is that, in fact, $\nu$ is finite.  For example, one can see [Vak, Theorem 9.7.3]. At a deeper level what is happening is that varieties are Nagata since locally their functions are finite type over a field which implies Nagata by basic theory. It is then also well-known that Nagata schemes have finite normalization maps.

Regardless, note that $\widetilde{X}$ is regular, thus smooth and so to finish it suffices to prove that $\widetilde{X}$ is affine or projective if and only if $X$ is.

If $X$ is affine then $\widetilde{X}$, being a finite over $X$, is also affine. If $X\to\mathrm{Spec}(k)$ is projective, then $\widetilde{X}\to \mathrm{Spec}(k)$ is projective since finite maps are projective and composition of projective maps are projective.

If $\widetilde{X}$ is affine then $X$ is affine. Indeed, this follows from non-trivial general theory about finite morphisms and affineness (it is really this tag that is most relevant). If $\widetilde{X}$ is projective then $X$ is necessarily proper by standard theory and thus projective by Lemma 12. $\blacksquare$

So, we are now free to assume that $X$ is smooth. It is now not hard to justify our intuitive sketch of proof after the statement of Proposition 9:

Proof(Proposition 9): We may assume that $k$ is algebraically closed.

Let $U$ be an affine open subscheme of $X$. As in the first paragraph of the proof of Lemma 12 we can find an open embedding $U\hookrightarrow C$ where $C$ is smooth projective geometrically connected curve over $k$. By the valuative criterion for properness we can uniquely extend $U\hookrightarrow C$ to a map $X\to C$. We claim that $X\to C$ is an open embedding.

Indeed, let us note that $X\to C$ is certainly quasi-finite since it’s quasi-finite on $U\subseteq X$ (it’s an open embedding) and $X-U$ is finite (since $X$ is one-dimensional). Thus, by Zariski’s main theorem (evidently our maps and schemes are quasi-compact and separated) we can find a factorization

$\begin{matrix} \overline{X} & & \\ {^j}\uparrow & \searrow{^\pi} & \\ X & \rightarrow & C\end{matrix}$

where $j$ is a dense open embedding and $\pi$ is finite. It’s clear we may assume that $\overline{X}$ is integral since both $X$ and $C$ are integral. Moreover, note since the normalization map $\nu:\widetilde{\overline{X}}\to\overline{X}$ is an isomorphism on the smooth locus and finite we can, up to replacing $\overline{X}$ with $\widetilde{\overline{X}}$, assume that $\overline{X}$ is smooth over $k$. Note then that $\pi:\overline{X}\to X$ is a birational map of smooth proper curves, so an isomorphism. Indeed, since $\pi$ is surjective we know that it’s flat (e.g. see [Qin, Proposition 3.9]). Note then that $\pi_\ast\mathcal{O}_{\overline{X}}$ is vector bundle on $C$. So, to show that $\mathcal{O}_C\to \pi_\ast\mathcal{O}_{\widetilde{X}}$ is an isomorphism (which is clearly all we need to do) it suffices to show it’s an isomorphism generically, but this is true since $\pi$ is birational. The claim that $X\to C$ is an open embedding then immediately follows.

Now, since $X$ is an open subscheme of $C$, a smooth projective integral curve, it suffices to show that any such variety is affine or projective. This follows from the lemma following this proof. $\blacksquare$

Lemma 13:Let $k$ be a field and $C$ a smooth geometrically integral projective curve over $k$. Then, any open subscheme $U\subsetneq C$ is affine.

Remark 14: This was also contained in the previous post.

Proof: We may assume that $k$ is algebraically closed. If $U$ is empty, we’re done. So, assume that $U$ is not empty. Then, $U=C-\{p_1,\ldots,p_n\}$ 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 $m_i\gg 0$.

$\ell(D)-\ell(K-D)=m_1+\cdots+m_n+1-g$

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 $f\in K(C)$ such that $f$ has poles precisely at the $p_i$. Thus, we obtain a non-constant rational map $f:C\to \mathbb{P}^1_k$ off of the $p_i$. Extending this to a morphism $f:C\to \mathbb{P}^1_k$, we obtain a finite map with $X-\{p_1,\ldots,p_n\}=f^{-1}(\mathbb{A}^1_k)$, and thus $C-\{p_1,\ldots,p_n\}$ is affine. $\blacksquare$

Remark 15: There is almost certainly an easier way to do the above. For example, if one uses this your life becomes much easier. In the above I don’t really use this proposition (save I use part of it in the proof of the non-smooth case of Lemma 12). The reason I opted to not use it is that while I think this approach is more elementary, it’s somehow trickier than what I did above. A lot of the techniques used above are incredibly common and worth internalizing (albeit maybe overkill).

We are now ready to start classifying one-dimensional connected geometrically reduced group varieties over a field $k$. We will break our discussion in to its two separated, and distinct cases: the affine case and the projective case.

## The projective case

In short, the projective one-dimensional geometrically reduced group varieties are just elliptic curves. This is what we endeavor to prove. But, before we state this precisely let us remind ourselves what an elliptic curve is.

Definition 16: Let $k$ be a field. Then, an elliptic curve over $k$ is a pair $(E,e)$ where $E$ is variety of the form $E=V(F(x,y,z))\subseteq \mathbb{P}^2_k$ is a smooth curve where $F$ has homogenous degree $3$ and $e\in E(k)$.

Of course, we know that elliptic curves over $k$ have a group law. More explicitly, let $p,q\in E(\overline{k})$ be distinct. There exists a unique line (i.e. a closed subscheme of the form $V(L(x,y,z))$ where $L(x,y,z)$ is a degree one homogenous polynomial) $\ell_{p,q}\subseteq \mathbb{P}^2_{\overline{k}}$ such that $\ell_{p,q}$ passes through $p$ and $q$. If $p=q$ we set $\ell_{p,q}$ to be the unique line in $\mathbb{P}^2_{\overline{k}}$ passing through $p$ and which is tangent to $E$ at $p$ (i.e. $\ell_{p,p}$ is the line $V(dF)$ where $dF=F_x(p)x+F_y(p)y+F_z(p)z$).

Note then that by Bezout’s theorem that for any $p,q\in E(\overline{k})$ (possibly with $p=q$) we have that $\ell_{p,q}\cap E$ (by which we rigorously mean the fiber product $\ell_{p,q}\times_{\mathbb{P}^2_{\overline{k}}}E$) is a finite $\overline{k}$-scheme with $3$-dimensional global sections. If $p\ne q$ then it’s easy to see that $\ell_{p,q}\cap E$ is the disjoint union of three $\overline{k}$-points which are $p,q$ and a third point $r$. If $p=q$ then $\ell_{p,p}\cap E$ is the disjoint union of $\mathrm{Spec}(\mathcal{O}_{\mathbb{P}^2_{\overline{k}},p}/(I_p^E,I_p^{\ell_{p,p}}))$ (where for a closed subscheme $Z$ of $\mathbb{P}^2_{\overline{k}}$ we denote by $I^Z_p$ the ideal of $\mathcal{O}_{\mathbb{P}^2_{\overline{k}},p}$ induced by $Z$) and an $\overline{k}$-point $r$. In either case we see that we obtain a third point $r\in E(\overline{k})$ from $p,q$.

Note then the exact same ideas yield associated to $r,e\in E(\overline{k})$ a line $\ell_{r,e}\subseteq \mathbb{P}^2_{\overline{k}}$ and a third point of intersection of the line $\ell_{r,e}$ and $E$. We denote this third $\overline{k}$-point of $E$ by $p+q$.

We then have the following:

Theorem 17: Let $k$ be a field and $E$ an elliptic curve over $k$.

1. The operation $E(\overline{k})\times E(\overline{k})\to E(\overline{k})$ given by $(p,q)\mapsto p+q$ described above endows $E(\overline{k})$ with the structure of a group.
2. There exists a unique group variety structure on $E$ such that the group structure on $E(\overline{k})$ agrees with that from 1.

Proof: The proof of 1. is [Sil, Proposition 2.2]. The unicity of the group scheme structure in 2. is clear. Indeed, let $m:E\times_k E\to E$ be a multiplication map. Since $E(\overline{k})\times E(\overline{k})$ is a dense subset of $E\times_k E$, and all schemes in consideration are separated, the map $m$ is determined by its value on $E(\overline{k})\times E(\overline{k})$. The existence of such $m$ (i.e. the algebraicity of the group operation from 1. and the fact that it’s defined over $k$) follows from [Sil, Group Law Algorithm 2.3]. $\blacksquare$

We are now able to state our desired broad classification in the projective case:

Proposition 18: Let $k$ be a field. Every elliptic cruve $E$ is a one-dimensional geometrically connected proper group variety over $k$. Conversely, if $G$ is a one-dimensional geometrically connected projective group variety over $k$ then $G$ is isomorphic (as a group variety) to an elliptic curve.

Proof: Let us first observe that elliptic curves are certainly one-dimensional, smooth, and projective (by definition). Thus, it remains to show why they are connected. There a multitude ways to prove this (Bezout’s lemma, [Vak, Exercise 11.3.F],…) but we list here a cohomological one. It suffices to show that $\mathcal{O}(E)$ is a one-dimensional $k$ space since this then forces the ring $\mathcal{O}(E)$ to be $k$ and thus to have no non-trivial idempotents. But, note the ideal sheaf $\mathcal{I}(E)$ of $E$ is evidently $\mathcal{O}(-3)$ and, in particular, is a line bundle. We have the short exact sequence of sheaves

$0\to \mathcal{I}(E)\to \mathcal{O}_{\mathbb{P}^2_k}\to i_\ast \mathcal{O}_E\to 0$

where $i:E\hookrightarrow \mathbb{P}^2_k$ is the tautological closed embedding. We then get a long exact sequence on cohomology group that contains the portion

$0\to \mathcal{I}(E)(\mathbb{P}^2_k)\to \mathcal{O}_{\mathbb{P}^2_k}(\mathbb{P}^2_k)\to (i_\ast \mathcal{O}_E)(\mathbb{P}^2_k)\to H^1(\mathbb{P}^2_k,\mathcal{I}(E))$

But, evidently $H^1(\mathbb{P}^2_k,\mathcal{I}(E))=0$ (e.g. see [Vak, Theorem 18.1.3]) and equally evident is $\mathcal{I}(E)(\mathbb{P}^2_k)=\mathcal{O}(-3)(\mathbb{P}^2_k)=0$ and thus we see that

$k=\mathcal{O}_{\mathbb{P}^2_k}(\mathbb{P}^2_k)\cong (i_\ast \mathcal{O}_E)(\mathbb{P}^2_k)=\mathcal{O}_E(E)$

as desired.

Conversely, suppose that $G$ is a one-dimensional connected geometrically reduced group variety. By the Rigidity Lemma (which immediately succeeds this proof) it suffices to show that there exists an isomorphism of pointed varieties $(G,e_G)\cong (E,e_E)$ where $E$ is an elliptic curve and $e_G$ and $e_E$ are the respective identity elements. Of course, it suffices to show that there exists an isomorphism of varieties $G\cong E$ since we could postcompose with translation by an appropriate element of $E(k)$ to guarantee that $e_G$ mapsto $e_E$.

We begin by noting that $G$ necessarily has genus $1$. There are several ways to see this, but we proceed with the one that is simplest. Namely, it’s not hard to show that any group variety is parallelizable (i.e. has cotangent bundle which is free) by using the theory of invariant differentials (e.g. see [BLR, §4.2 Corollary 3]). In particular, we see that $\Omega^1_{G/k}\cong \mathcal{O}_G$ and thus $\deg(\Omega^1_{G/k})=0$. But, $\deg(\Omega^1_{G/k})=2g-2$ (e.g. see [Har, Example 1.3.3]). Thus, $g=1$ as desired.

We now need to show that $G$ is isomorphic to an actual variety of the form $V(F(x,y,z))\subseteq \mathbb{P}^2_k$ where $F(x,y,z)$ is a smooth cubic function. To do this let us begin by noting that since $\deg(\mathcal{O}(3e))=3\geqslant 2(1)+1$ we have by standard theory (e.g. see [Vak, Conclusion 19.2.11]) that $\mathcal{O}(3e)$ is very ample. Let us note also that, by Riemann-Roch, we have that

$h^0(\mathcal{O}(3e))-h^0(\Omega^1_{G/k}\otimes \mathcal{O}(-3e))=3+1-1=3$

but since $\Omega^1_{G/k}\otimes \mathcal{O}(-3e)=\mathcal{O}(-3e)$, which has negative degree, the left-hand side of the above simplifies to $h^0(\mathcal{O}(3e))$. So, $h^0(\mathcal{O}(3e))=3$. Thus, we see $\mathcal{O}(3e)$ determines a closed embedding $G\hookrightarrow \mathbb{P}(\mathcal{O}(3e)(G))\cong \mathbb{P}^2_k$. So, $G$ is isomorphic to a closed subscheme of $\mathbb{P}^2_k$. Since we know that $\dim G=\dim \mathbb{P}^2_k-1$ we have that $G$ corresponds to a height one homogenous prime in $k[x,y,z]$ which is automatically principal. Thus, $G$ (or more precisely the image of $G$) is actually a hypersurface, say $G=V(F(x,y,z))$.

Thus, it remains to show that $\deg F=3$. But, let us note that if $\deg F=d$ then the ideal sheaf of $G$ is $\mathcal{O}(-d)$. Using the long exact sequence in cohomology for

$0\to \mathcal{O}(-d)\to \mathcal{O}_{\mathbb{P}^2_k}\to i_\ast \mathcal{O}_G\to 0$

(as we’ve already discussed in this proof) we see that (again using [Vak, Theorem 18.3.1]) that

$H^1(\mathbb{P}^2_k,i_\ast\mathcal{O}_G)\cong H^2(\mathbb{P}^2_k,\mathcal{O}(-d))$

But, note that since $i$ is a closed embedding we know from standard theory (e.g. [Har, Exercise 8.2]) that

$H^1(\mathbb{P}^2_k,i_\ast\mathcal{O}_G)\cong H^1(G,\mathcal{O}_G)$

and the right-hand side is one dimensional since $G$ has genus $1$. Thus, in conclusion we see that $h^2(\mathcal{O}(-d))=1$. But, $h^2(\mathcal{O}(-d))=\frac{1}{2}(d-1)(d-2)$ (e.g. see again [Vak, THeorem 18.3.1]). Thus, $d=3$ as desired.  $\blacksquare$

Lemma 19(The Rigidity Lemma): Let $k$ be a field. Let $X$ and $Y$ be geometrically integral schemes of finite type over $k$ and $Z$ a separated $k$-scheme. Let $f:X\times_k Y\to Z$ be a morphismof $k$-schemes, and assume further that

• $X$ is proper.
• For some algebraically closed extension $K/k$ there exists some $y_0\in Y(K)$ such that the restriction $f_{y_0}:X_K\to Z_K$ to $X_K\times \{y_0\}$ is a constant map to some $z_0\in Z(K)$.

Then $f$ is independent of $X$; i.e. there exists a unique morphism of $k$-schemes $g:Y\to Z$ such that $f(x,y)=g(y)$.

In particular, if $G_i$ are smooth geometrically integral proper $k$-varieties with identity elements $e_i$ then any morphism of pointed $k$-varieties $f:(G_1,e_1)\to (G_2,e_2)$ is a morphism of group varieties.

Proof: See [Con, Theorem 1.7.1] for the first statement. The second statement follows from the first by noting that the map $G_1\times_k G_1\to G_2$ given on points by $(g_1,g_2)\mapsto f(g_1g_2)f(g_1)^{-1}f(g_2^{-1})$ is constant on $G_1\times \{e_1\}$ and thus is trivial. This implies that $f$ is a group morphism. $\blacksquare$

## The affine case

This was taken care of in the previous post. In particular, we have the following:

Theorem 20: Let $k$ be a field and let $G$ be an affine one-dimensional connected geometrically reduced group variety over $k$. Then, $G_{\overline{k}}\cong \mathbb{G}_{m,\overline{k}}$ or $G_{\overline{k}}\cong \mathbb{G}_{a,\overline{k}}$. If $\mathrm{char}(k)=0$ then the assumption that $G$ is geometrically reduced is unnecessary.

## The two cases combined

Combining our two cases we arrive that the following:

Theorem 21: Let $k$ be a field and let $G$ be a one-dimensional connected geometrically reduced group variety over $k$. Then, $G$ one of the following holds:

1. $G$ is an elliptic curve.
2. $G_{\overline{k}}\cong \mathbb{G}_{m,\overline{k}}$
3. $G_{\overline{k}}\cong \mathbb{G}_{a,\overline{k}}$

If $\mathrm{char}(k)=0$ then the assumption that $G$ is geometrically reduced is unnecessary.

# A finer result

Theorem 21 is lacking in two orthogonal ways. The proper case gives us a nice classification over $k$ but it’s pretty inexplicit (how do we explicitly parameterize elliptic curves up to isomorphism?). The  affine case is very explicit, but only works over $\overline{k}$. Our goal now is to remedy both of these issues in (essentially) full generality. In particular, the main assumption we will often make is that $k$ is perfect.

## The proper case

We begin by trying to understand how to explicitly parameterize elliptic curves over $k$. We begin with a well-known and simple first case:

Proposition 22: Let $k$ be an algebraically closed field. Then, associated to every elliptic curve $E$ over $k$ is an element $j(E)\in k$. This integer $j(E)$ only depends on the isomorphism class of $E$ and induces a bijection

$j:\left\{\begin{matrix}\text{Elliptic curves}\\ \text{over}\, k\end{matrix}\right\}/\text{iso}.\to k$

The element $j(E)\in k$ from Proposition 22 is the so-called $j$-invariant of $E$. Its definition can be given purely in terms of a defining equation for $E$ as in [Sil, §3.1]. The proof of Proposition 22 is then the contents of [Sil, Proposition III.1.4.(b)]. If we want to explicitly construct an inverse for the bijection in Proposition $22$ one associates to $j_0\in k$ the elliptic curve $V(F(x,y,z))\subseteq \mathbb{P}^2_{\overline{k}}$ with $F(x,y,z)$ the polynomial

$\displaystyle y^2z+xyz-x^3+\frac{36}{j_0-1728}xz^2+\frac{1}{j_0-1728}z^3$

at least if $j_0\ne 0,1728$. If $j_0=0$ one can take the elliptic curve with equation

$y^2z+yz^2-x^3$

and if $j_0=1728$ one can take the elliptic curve with equation

$y^2z-x^3-xz^2$

which covers all cases. We denote these explicit elliptic curves as $E_{j_0}$ for any $j_0\in \overline{k}$

In fact, from the simple observation that $E_{j_0}$ has a model over $k$ (that with the same equation) we actually deduce the following strengthening of Proposition 22:

Proposition 23: Let $k$ be a field. Then, the map

$j:\left\{\begin{matrix}\text{Elliptic curves}\\ \text{over}\, k\end{matrix}\right\}/\text{iso}.\to k$

is a surjection with fibers the sets

$T_{j_0}:=\left\{\begin{matrix}\text{Elliptic curves }E\text{ over }k\\ \text{such that }E_{\overline{k}}\cong (E_{j_0})_{\overline{k}}\end{matrix}\right\}/\text{iso.}$

Thus it really remains to explicitly understand the sets $T_{j_0}$. These sets $T_{j_0}$ are exactly the sets of ‘twists’ of $E_{j_0}$ (e.g. see the discussion of twists in the notes of this post). For our sanity, we now assume that $k$ is perfect so that $\overline{k}/k$ is Galois. We then have the following:

Lemma 24: Let $k$ be a perfect field. Then, there is a natural bijection

$T_{j_0}\xrightarrow{\approx} H^1_\text{cont.}(\mathrm{Gal}(\overline{k}/k),\mathrm{Aut}((E_{j_0})_{\overline{k}}))$

Proof: For a down-to-earth proof see [Sil, §X.5]. The high level proof is as follows. Let $\mathcal{M}_{1,1}$ be the category fibered in groupoids over the small etale site for $\mathrm{Spec}(k)$. This is a stack (e.g. see [Ols, Theorem 13.1.2]). Thus, by Theorem 5.1 of the notes from this post the claim essentially follows. Namely, the only other reduction is the observation that by standard theory one has that

$H^1(\mathrm{Spec}(k)_{\mathrm{et}},\mathrm{Aut}(E_{j_0}))\cong H^1_\text{cont.}(\mathrm{Gal}(\overline{k}/k),\mathrm{Aut}((E_{j_0})_{\overline{k}})$

from where the conclusion follows. $\blacksquare$

Remark 25: I don’t truthfully know what happens over non-perfect fields. I suspect that as long as the characteristic $k$ is not $2$ or $3$ then the fact that $\mathrm{Aut}(E)$ is smooth should imply that all isomorphisms which occur over $\overline{k}$ actually occur over $k^\mathrm{sep}$ in which case Lemma 24 is still valid. Even if this is true, I don’t know what can happen in the case of characteristic $2$ or $3$. Please feel free to enlighten me if you know the answer.

So, all that remains to do to completely classify elliptic curves over $k$, at least in the case when $k$ is perfect, is to calculate the groups $H^1_\text{cont.}(\mathrm{Gal}(\overline{k}/k),\mathrm{Aut}((E_{j_0})_{\overline{k}})$ for all $j_0\in k$. This turns out to be quite simple in the case when the characteristic of $k$ is not $2$ or $3$. Indeed, we have the following:

Lemma 26: Let $k$ be a perfect field of characteristic not $2$ or $3$. Then,

$\mathrm{Aut}(E_{j_0})\cong \begin{cases}\mu_2 &\mbox{if}\quad j_0\ne 0,1728\\ \mu_4 & \mbox{if}\quad j_0=1728\\ \mu_6 & \mbox{if}\quad j_0=0\end{cases}$

where this is isomorphisms as group schemes over $k$.

Proof: This is precisely [Sil, III.10.2]. $\blacksquare$

Now, the Galois cohomology of $\mu_n$ is very simple in general. Namely, by Hilbert’s theorem 90 we know that in we have a canonical bijection $H^1_\text{cont.}(\mathrm{Gal}(\overline{k}/k),\mu_n(\overline{k}))$ and $k^\times/(k^\times)^n$. In particular, we see that we have bijections

$T_{j_0}\cong\begin{cases} k^\times/(k^\times)^2 & \mbox{if}\quad j_0\ne 0,1728\\ k^\times/(k^\times)^4 & \mbox{if}\quad j_0=1728\\ k^\times/(k^\times)^6 &\mbox{if}\quad j_0=0\end{cases}$

Thus, we can enhance Proposition 22 to the following:

Theorem 27: Let $k$ be a perfect field of characteristic not $2$ nor $3$. Then, there is an explicit bijection

$\left\{\begin{matrix}\text{Elliptic curves}\\ \text{over}\, k\end{matrix}\right\}/\text{iso}.\to (k\times k^\times)/\sim$

where $(j_0,D)\sim (j_0',D')\in k\times k^\times$ if $j_0=j_0'$ and $D/D'\in (k^\times)^{n(j_0)}$ where

$n(j_0)=\begin{cases} 2 & \mbox{if}\quad j_0\ne 0,1728\\ 4 & \mbox{if}\quad j_0=1728\\ 6 & \mbox{if}\quad j_0=0\end{cases}$

The explictness in the statement of Theorem 27 means that there is an explicit inverse (since the map itself just sends $E$ to $(j(E),D(E))$ where $D(E)$ is the element of $k^\times/(k^\times)^{n(j(E))}$ that comes from our discussion). For an explicit description see [Sil, Corollary 5.4.3]. We mention though, since it’s the most common and most simple case, that if $j_0\ne 0,1728$ then as soon as one writes $E_{j_0}\cong V(F(x,y,z))$ where $F(x,y,z)$ is the homogenization of a polynomial of the form $y^2-x^3-ax-b$ (which is always possible since we’re in characteristic different from $2$ or $3$) then the element $T_{j_0}$ that corresponds to $(j_0,d)$ is the elliptic curve whose affine equation is $dy^2-x^3-ax-b$–this is the so-called quadratic twist of $E_{j_0}$.

## The affine case

The affine case was covered in the previous post, so we just summarize the results here. Namely, we have the following:

Theorem 28: Let $k$ be a perfect field of characteristic not $2$. Then, a one-dimensional connected affine geometrically reduced group variety over $k$ is either $\mathbb{G}_{a,k}$ or a torus. Moreover, there is a bijection

$\left\{\begin{matrix}\text{One dimensional}\\ \text{tori over}\, k\end{matrix}\right\}/\text{iso}.\to k^\times/(k^\times)^2$

such that that identity element of $k^\times/(k^\times)^2$ corresponds to $\mathbb{G}_{m,k}$ and $d\in k^\times/(k^\times)^2$ non-identity corresponds to the torus

$\mathsf{Res}^1_{k(\sqrt{d})/k}\mathbb{G}_{m,k(\sqrt{d})}:=\ker\left(\mathsf{Res}_{k(\sqrt{d})/k}\mathbb{G}_{m,k(\sqrt{d})}\xrightarrow{\mathrm{Nm}}\mathbb{G}_{m,k}\right)$

## Putting all together

Summarizing everything above we get the following:

Theorem 29:Let $k$ be a field. Then every one-dimensional connected geometrically reduced group variety over $k$ is either affine or an elliptic curve. If $\mathrm{char}(k)=0$ then the geometrically reduced assumption is automatically satisfied.

Moreover, we have the following parameterization of these two families:

1. If $k$ is perfect and of characteristic not $2$ or $3$ then there is an explicit bijection

$\left\{\begin{matrix}\text{Elliptic curves}\\ \text{over}\, k\end{matrix}\right\}/\text{iso}.\to (k\times k^\times)/\sim$

where $(j_0,D)\sim (j_0',D')\in k\times k^\times$ if $j_0=j_0'$ and $D/D'\in (k^\times)^{n(j_0)}$ where

$n(j_0)=\begin{cases} 2 & \mbox{if}\quad j_0\ne 0,1728\\ 4 & \mbox{if}\quad j_0=1728\\ 6 & \mbox{if}\quad j_0=0\end{cases}$

2. If $k$ is perfect and of characteristic not $2$ then an affine $G$ is either $\mathbb{G}_{a,k}$ or a one-dimensional torus. Moreover, there is a bijection

$\left\{\begin{matrix}\text{One dimensional}\\ \text{tori over}\, k\end{matrix}\right\}/\text{iso}.\to k^\times/(k^\times)^2$

such that that identity element of $k^\times/(k^\times)^2$ corresponds to $\mathbb{G}_{m,k}$ and $d\in k^\times/(k^\times)^2$ non-identity corresponds to the torus

$\mathsf{Res}^1_{k(\sqrt{d})/k}\mathbb{G}_{m,k(\sqrt{d})}:=\ker\left(\mathsf{Res}_{k(\sqrt{d})/k}\mathbb{G}_{m,k(\sqrt{d})}\xrightarrow{\mathrm{Nm}}\mathbb{G}_{m,k}\right)$

Remark 30: One can almost certainly remove the perfectness hypotheses from 1. in Theorem 29. Indeed, let $E_1$ and $E_2$ be elliptic curves over $k$. We need only show that if $(E_1)_{\overline{k}}\cong (E_2)_{\overline{k}}$ then $(E_1)_{k^{\mathrm{sep}}}\cong (E_2)_{k^{\mathrm{sep}}}$. But, note that $\mathrm{Isom}(E_1,E_2)$ is an fppf torsor for $\mathrm{Aut}(E_1)$. But, $\mathrm{Aut}(E_1)$ is $\mu_n$ for $n\in\{2,4,6\}$ by Lemma 26. Its simple to show that any fppf torsor for $\mu_n$ is representable and is smooth since $\mu_n$ is. So, $\mathrm{Isom}(E_1,E_2)$ is a representable by a smooth (in fact etale) $k$-scheme over $\mathrm{Spec}(k)$. So, it evidently has a $k^\mathrm{sep}$ point from where the conclusion follows.

# References

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[Mil] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

[Ols] Olsson, M., 2016. Algebraic spaces and stacks (Vol. 62). American Mathematical Soc.

[Qin] Liu, Q., 2002. Algebraic geometry and arithmetic curves (Vol. 6). Oxford University Press on Demand.

[Sil] Silverman, J.H., 2009. The arithmetic of elliptic curves (Vol. 106). Springer Science & Business Media.

[Vak] Vakil, R., 2017. The Rising Sea: Foundations of Algebraic Geometry (Ver. Nov. 18 2017). http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf