# $k$-rational points of the automorphism functor of a scheme

Let $$X$$ be a scheme and let $$\operatorname{Aut}_X$$ denote the functor sending a scheme $$T$$ to the set of $$T$$-automorphisms of $$X \times T$$. Assume that $$\operatorname{Aut}_X$$ is representable by a group scheme. I will continue to denote this group scheme by $$\operatorname{Aut}_X$$.

I am reading "Notes on automorphism groups of projective varieties" by Michel Brion.

In it he defines $$\operatorname{Aut}(X):=\operatorname{Aut}_X(k)$$ and then writes that $$\operatorname{Aut}_X$$ is algebraic if and only if $$\operatorname{Aut}(X)$$ is.

The property of being algebraic is something assigned to a group scheme, but, as far as I understand, $$\operatorname{Aut}_X(k)$$ is just a group.

Does $$\operatorname{Aut}_X(k)$$ also have a group scheme structure?

• It seems to me as if Brion is using the same construction as in your other question, though I am not an expert in algebraic groups. Apr 20, 2021 at 22:02

There are several definitions of the notion "group scheme". If a group scheme $$G$$ (over a commutative ring $$k$$) is a representable functor (let $$G:=Spec(R)$$)

$$h_G: Aff(k) \rightarrow Groups$$

where $$C:=Aff(k)$$ is the category of affine schemes over $$k$$ and $$Groups$$ is the category of groups, it follows

$$G(k):=h_G(Spec(k)):=Hom_C(Spec(k), G) \cong Hom_{k-alg}(R,k).$$

Note: When considering a group valued functor $$h_G$$ as above, there are problems since the category $$Aff(k)$$ of arbitrary $$k$$-algebras is not a small category - the objects $$Ob(Aff(k))$$ of $$Aff(k)$$ is not a set. You could view a group scheme as a triple $$(G,m,i)$$ where $$G$$ is a scheme over $$k$$, $$m:G\times_k G \rightarrow G$$ and $$i:G \rightarrow G$$ are morphisms satisfying a set of criteria. Here $$m$$ is the group multiplication and $$i$$ is the inversion. Then you avoid such set theoretical problems. If you are not careful when introducing methods from category theory, you may find you are unknowingly working with "non standard set theory". On page 5 in "Representations of algebraic groups" (Jantzen) you find the following:

"Let us assume $$k$$ to be arbitrary again. In the definitions to follow, we shall be rather careless about the foundations of mathematics. Instead of working with "all" $$k$$-algebras, we should (as in [DG]) take only those in some "universe". We shall leave the appropriate modifications to the interested reader."

If you care about foundations you should either read about the "universe" mentioned in Jatzen's book or instead work with a triple $$(G,m,i)$$ as explained above.

If you dont care, many mathematicians will accept the following method of proof: When giving a lecture on your results you use the "hand-waving method" to prove results. You "wave your hands" and speak about "categories" and "universes".

One of the problems with being a student in modern algebraic geometry is that much of the results presented in the EGA book series is formulated in the functorial language and hence it assumes the notion "universe" as mentioned in Jantzens book. The EGA book series consists of several tousands of pages and as a student you will not have the time to read everything and verify every detail. This is why you - as a student - will be forced to do some "hand-waving" every now and then.

Let us from now "assume all $$k$$-algebras live in a fixed universe".

Question: "Does AutX(k) also have a group scheme structure?"

Answer: If $$G:=GL(V)$$ where $$V:=k\{e_1,..,e_n\}$$ it follows $$R:=k[x_{ij}][t]/(tdet(x_{ij})-1)$$ and you may identify $$G(k)$$ with the set of matrices

$$\{A:=(a_{ij})\in Mat(n,k)\text{ with }det(A)\in k^*\},$$

hence the set $$G(k):=GL(V)(k)$$ may be interpreted as a set of invertible $$n\times n$$ matrices with coefficients in $$k$$ - this set has a group operation.

Hence if a "group scheme" in your language is a representable functor it follows $$Aut(X):=Aut_X(k)$$ is not a group scheme - it is the group of $$k$$-rational points of the group scheme $$Aut_X$$ - this is by definition a "set" and not a "representable functor". If a group scheme is a triple $$(G,m,i)$$ as defined above, it follows $$Aut_X(k)$$ is not a group scheme since $$Aut_X(k)$$ is a set and not such a triple $$(G,m,i)$$.

It seems to me that a group scheme $$G$$ in this setting is "algebraic" iff $$k$$ is an algebraically closed field and $$R$$ is a finitely generated $$k$$-algebra which is an integral domain. It follows $$R$$ is a Hilbert-Jacobson ring and that any prime ideal in $$R$$ equals the intersection of the maximal ideals containing it. The set

$$G(k)\cong Hom_{k-alg}(R,k)$$

may be identified with the maximal ideals in $$R$$ and when you choose a set of generators $$R\cong k[x_1,..,x_n]/\mathfrak{p}$$ you choose an embedding $$V(\mathfrak{p}) \subseteq \mathbb{A}^n_k$$ as a closed subvariety. Here - since $$k$$ is algebraically closed we use the language of Hartshorne Chapter I where an "algebraic variety" is an irreducible algebraic subset of $$k^n$$. If $$V(\mathfrak{p})$$ is irreducible it follows there is - by the Nullstellensatz - an equality of ideals

$$I(V(\mathfrak{p})) = \mathfrak{p},$$

hence in a sense you recover the ring $$R$$ from the set of maximal ideals in $$R$$. There is an isomorphism

$$R \cong k[x_1,..,x_n]/I(V(\mathfrak{p})) .$$

This may explain the statement

Proposition:"$$Aut_X$$ is algebraic if and only if $$Aut(X)$$ is."

Proof: If $$Aut_X:=Spec(R)$$ it follows that if the $$k$$-rational points $$Aut(X) \subseteq k^n$$ is an algebraic subvariety of $$k^n$$ in the above sense, it follows $$R$$ is a finitely generated $$k$$-algebra which is an integral domain, hence $$Aut_X$$ is algebraic. Conversely we define a group scheme $$G:=Spec(R)$$ to be algebraic iff $$R$$ is a finitely generated $$k$$-algebra which is a domain. It follows the "set of $$k$$-rational points" $$G(k) \subseteq k^n$$ may be realized as an irreducible algebraic subset for some $$n\geq 1$$.

We also observe that if we choose an embedding $$G(k) \subseteq k^n$$ as an algebraic variety and an ideal $$I(G(k)) \subseteq k[x_1,..,x_n]$$ we may define $$R(G(k)):=k[x_1,..,x_n]/I(G(k))$$ and then since $$\mathfrak{p}$$ is prime we get

$$R \cong R(G(k))$$

hence we recover $$G:=Spec(R)$$ from $$G(k)$$: There is an isomorphism of affine schemes

$$G \cong G_k:=Spec(R(G(k)))$$

and an isomorphism of group schemes

$$h_G \cong h_{G_k}.$$

Question: "Does AutX(k) also have a group scheme structure?"

Answer: No, $$G(k):=Aut_X(k)$$ does not have a group scheme structure, but there is a way to construct a group scheme $$h_{G_k}$$ from $$G(k)$$ and an isomorphism $$h_G \cong h_{G_k}$$ of group schemes. Hence if the base field $$k$$ is algebraically closed and if the group scheme is reduced, you may recover the group scheme $$Aut_X$$ from the group $$Aut_X(k)$$.