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Let $G$ be an algebraic variety over an algebraically closed field $k$ (any characteristic). Suppose that: (1) the set of $k$-points has the structure of a group. (2) for any $g\in G$ the right-multiplication by $g$ is a morphism of algebraic varieties $G\to G$. (3) the inverse map is a morphism $G\to G$.

Does it imply that $G$ is an algebraic group? (i.e. is the multiplication $G\times G\to G$ a morphism?)

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you might specify if you have in mind characteristic zero or arbitrary characteristic. –  YCor Nov 17 '12 at 18:01
What set are you using to be the elements of the group? –  Will Sawin Nov 17 '12 at 18:03
The multplicative group of quaternions (presented as $\mathbb{C}\times \mathbb{C} \setminus \{0\}$ yields a non-algebraic example of your (1)+(2); yet it fails (3). –  Mikhail Bondarko Nov 17 '12 at 19:02
@Mikhail: In char. $p$ such near-misses (losing (3)) abound. Let $H$ be a linear algebraic group and $H'$ an infinitesimal non-normal closed subgroup scheme of $H$. The quotient scheme $G = H'\backslash H$ is a smooth affine variety and the natural map $H(k) \rightarrow G(k)$ is bijective. If $G(k)$ is thereby equipped with the group structure of $H(k)$ then this is non-algebraic (as otherwise the quotient map $H \rightarrow G$ would be a homomorphism of algebraic groups and so its schematic kernel $H'$ would be normal), and (1) and (2) hold but (3) fails. –  user28172 Nov 17 '12 at 19:25
Since there was some discussion about the extent to which the group structure is needed, the following might be relevant. If $k$ is a countably infinite field, then there is a function $f:k\times k\to k$ that is not a polynomial even though, for each fixed value of $x$ in $k$, $f(x,y)$ is a polynomial function of $y$ and, for each fixed $y\in k$, $f(x,y)$ is a polynomial function of $x$. (There is no such $f$ when $k$ is uncountable.) –  Andreas Blass Nov 18 '12 at 2:21

1 Answer 1

up vote 12 down vote accepted

I predict that in whatever is the situation of motivating interest, you know more: for any algebraically closed field $K/k$ you likewise have a group structure on $G(K)$ functorially in $K/k$ making the translations and inversions by $G(K)$ also be morphisms on $G_K$. Under this additional condition the answer is always affirmative, by using a trick that I believe goes back to Hasse, namely base change to the function field of $G$ (or rather, in our situation, an algebraic closure thereof). This is a Weil-style way of getting at Yoneda's Lemma.

The inversion hypothesis implies that left-translations are also morphisms, and we will use this condition instead of the inversion hypothesis (and deduce at the end that inversion is a morphism too). In fact, in char. 0 we won't even need this condition with the left-translations (or inversion) at all, but we are using a stronger assumption on the existence of group laws on $K$-points for lots of $K/k$. This stronger initial assumption across many $K/k$ bypasses Bondarko's examples in char. 0. The near-miss examples avoiding the inversion condition in char. $p$ in the comments are not ruled out by our stronger hypothesis across all $K/k$, but they are ruled out by the left-translations being morphisms (!), so matters will be more delicate in positive characteristic. We need the left-translation condition in positive characteristic to circumvent some inseparability issues with function fields of $k$-varieties.

Before we begin the actual arguments, note that since $k$ is algebraically closed, so $G(k)$ is Zariski-dense in $G$, for any field extension $F/k$ the subset $G(k)$ inside $G(F) = G_F(F)$ is Zariski-dense $G_F$. Hence, $F$-morphisms among $G_F$, $G_F \times G_F$, etc. are determined by their effect on $k$-points promoted to $F$-points. I mention this at the outset for peace of mind in some later discussions.

The "translation is a morphism" hypothesis ensures that $G$ is smooth, so its connected components are irreducible. I will assume that the variety $G$ is connected, a harmless assumption since (i) for $g \in G^0(k)$ the right-translation morphism by $g$ preserves $G^0$ because it carries $e$ to $g$, (ii) inversion preserves $G^0$ since it fixes $e$, (iii) the morphism property on $G^0$ clearly implies the general case by using the "morphism" property for a few translations by points in the various connected components of $G$. Thus, now $G$ is irreducible and so has a "function field" $k(G)$.

By applying the functoriality in $K/k$ to $k$-automorphisms of $K$, we see via Galois descent that the initial hypothesis for algebraically closed extensions is actually valid for all perfect extensions of $k$ as well. Let $K$ be the perfect closure of $k(G)$, and let $\eta \in G(k(G)) \subset G(K)$ correspond to the generic point. This induces a right-translation morphism $\rho_K:G_K \simeq G_K$ (computing right-translation by $\eta_K$). Since $G$ is finite type over $k$, by expressing $K$ as a directed union of finite purely inseparable extensions of $k(G)$ we obtain a finite purely inseparable extension $F$ of $k(G)$ and an $F$-morphism $\rho_F:G_F \simeq G_F$ that descends $\rho_K$ and clearly computes right-translation by $\eta_F$. The normalization $X$ of $G$ in the finite extension $F/k(G)$ is a $k$-variety finite radiciel over $G$, and $X_{\eta} = {\rm{Spec}}(F)$ over $k(G)$. Thus, by expressing $k(G)$ as the directed union of the coordinate rings of affine opens around $\eta$ in $G$, we find such an open $U$ so that over its preimage $U'$ in $X$ (which is finite radiciel over $U$) there is a $U'$-morphism $\rho_{U'}:G_{U'} \simeq G_{U'}$ extending $\rho_F$.

The map $U'(k) \rightarrow U(k)$ is bijective, so for each $g \in U(k)$ let $g' \in U'(k)$ be the unique point over $g$. Consider the specialization $\rho_{g'}:G \simeq G$ of $\rho_{U'}$ over $g'$. I claim that this is the right-translation by $g$. To see this, pick $h \in G(k)$ and consider the morphism of left-translation $\ell_h:G \simeq G$ by $h$. This carries $\eta$ to another $k(G)$-point of $G$ (over $k$) that "spreads out" to a $U$-point of $G$ whose specialization at $g \in U(k)$ is $\ell_h(g) = hg$ (so if we work instead with the $U'$-point over this via the canonical $U' \rightarrow U$ then its specialization at $g' \in U'(k)$ is also $\ell_h(g) = hg$). Now inside the group $G(K)$ we have $\ell_h(\eta_K) = \rho_K(h_K)$, so $\rho_{U'}(h_{U'})$ is the $U'$-point obtained from applying $\ell_h$ to the canonical $U'$-point of $G$ (spreading out $\eta_F$) because this comparison of $U'$-points may be checked by working at the generic point of $U'$ (and this generic point is dominated by the canonical $K$-point, over which we have the equality $\ell_h(\eta_K) = \rho_K(h_K)$). Specializing this equality of $U'$-points at the point $g' \in U'(k)$ over $g \in U(k) \subset G(k)$ gives the equality of $k$-points $\rho_{g'}(h) = \ell_h(g) = hg$, as desired.

To summarize, we have constructed a $U'$-morphism $\rho:G_{U'} \simeq G_{U'}$ whose specialization at any $g' \in U'(k)$ is right-translation on $G$ by the corresponding point $g \in U(k)$. Suppose ${\rm{char}}(k) = 0$, so $U' = U$ and the composite morphism $${\rm{pr}}_1 \circ \rho:G \times U \rightarrow G$$ on $k$-points is the group law restricted to $U(k)$ in the 2nd variable. For any $g \in G(k)$ the right translation morphism $\rho_g$ carries $U$ to another open subscheme $\rho_g(U)$, and I claim that these opens cover $G$. It suffices to check the covering property on $k$-points, so for $h \in G(k)$ we seek $g \in G(k)$ such that $h \in \rho_g(U)$, or equivalently $hg^{-1} \in U(k)$. Pick any $u \in U(k)$ and let $g = u^{-1}h$. This proves the covering property, and the multiplication map $G(k) \times U(k)g \rightarrow G(k)$ arises from a morphism, namely the composition of ${\rm{pr}}_1 \circ \rho$ and the morphism $\rho_g$. We conclude (in char. 0) that the group law on $G(k)$ is induced by a morphism $G \times G \rightarrow G$.

Now we can deduce in char. 0 that inversion is a morphism (so we have a group variety), even though we never used the left-translation or inversion conditions. Indeed, the "universal right-translation" $G \times G \rightarrow G \times G$ defined by $(x,y) \mapsto (xy,y)$ between fppf $G$-schemes (via ${\rm{pr}}_2$) is a scheme isomorphism between fibers over all points in $G(k)$ and therefore is a scheme isomorphism (by fibral isomorphism criteria, adapted to the peculiarities of $k$-points when $k$ is alg. closed). This yields the morphism property for inversion for free! This was a char-free argument, but it required the composition law to be a morphism. Anyway, char. 0 is now settled.

Assume ${\rm{char}}(k) = p > 0$, so for sufficiently large $n \ge 0$ the finite flat $n$-fold relative Frobenius morphism $G^{(1/p^n)} \rightarrow G$ of $G^{(1/p^n)}$ dominates $U'$ over $U$ (namely, pick $n \ge 0$ so that the finite purely inseparable extension $F/k(G)$ is contained inside $k(G)^{1/p^n}$). Since the initial choice of $F$ could be replaced with a finite purely inseparable extension at the outset if we wish, we may therefore assume that $U'$ is open inside $G^{(1/p^n)}$. Using a covering and translation argument similar to characteristic 0, we arrive at a morphism $$m_r:G \times G^{(1/p^n)} \rightarrow G$$ that recovers the given group law on $k$-points via the natural identification of $G^{(1/p^n)}(k)$ with $G(k)$ (for some $n \ge 0$). Our problem is precisely to show that $m_r$ factors (in the sense of morphisms of varieties) through the $n$-fold relative Frobenius morphism in the 2nd variable.

Now we shall use that left-translations are morphisms too. At the cost of increasing our $n$ if necessary, we can run through the same arguments with "left" instead of "right" to arrive at another morphism $$m_{\ell}:G^{(1/p^n)} \times G \rightarrow G$$ which recovers the given group law on $k$-points (with the same $n$).

Returning to our task of checking that $m_r$ factors through the appropriate iterated Frobenius in the 2nd variable, since that iterated Frobenius is fppf (as $G$ is smooth!) we conclude via fppf descent that it is harmless to check the existence of such a factorization after precomposing $m_r$ with an fppf morphism in the first variable. So let's compose with the same iterated Frobenius in the first variable, arriving at a morphism $$G^{(1/p^n)} \times G^{(1/p^n)} \rightarrow G$$ that recovers the composition law of $G(k)$ on $k$-points. It suffices to show that this latter map factors through the $n$-fold Frobenius in its 2nd variable, but this factorization is clear: it is the composition of $m_{\ell}$ with that iterated Frobenius (as we may check by computing on $k$-points, since we're working with reduced $k$-schemes of finite type). QED

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This is a very nice answer ! –  Olivier Benoist Nov 18 '12 at 2:41

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