$k$-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 &quot;Notes on automorphism groups of projective varieties&quot; 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?
There are several definitions of the notion &quot;group scheme&quot;. 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 &quot;non standard set theory&quot;. On page 5 in &quot;Representations of algebraic groups&quot; (Jantzen) you find the following:
&quot;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 &quot;all&quot; $k$-algebras, we should (as in [DG]) take only those in some &quot;universe&quot;. We shall leave the appropriate modifications to the interested reader.&quot;
If you care about foundations you should either read about the &quot;universe&quot; 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 &quot;hand-waving method&quot; to prove results. You &quot;wave your hands&quot; and speak about &quot;categories&quot; and &quot;universes&quot;.
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 &quot;universe&quot; 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 &quot;hand-waving&quot; every now and then.
Let us from now &quot;assume all $k$-algebras live in a fixed universe&quot;.
Question: &quot;Does AutX(k) also have a group scheme structure?&quot;
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 &quot;group scheme&quot; 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 &quot;set&quot; and not a &quot;representable functor&quot;. 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 &quot;algebraic&quot; 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 &quot;algebraic variety&quot; 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:&quot;$Aut_X$ is algebraic if and only if $Aut(X)$ is.&quot;
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 &quot;set of $k$-rational points&quot; $G(k) \subseteq k^n$ 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: &quot;Does AutX(k) also have a group scheme structure?&quot;
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 $Aut_X(k)$.
There are several definitions of the notion &quot;group scheme&quot;. 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 &quot;non standard set theory&quot;. On page 5 in &quot;Representations of algebraic groups&quot; (Jantzen) you find the following:
&quot;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 &quot;all&quot; $k$-algebras, we should (as in [DG]) take only those in some &quot;universe&quot;. We shall leave the appropriate modifications to the interested reader.&quot;
If you care about foundations you should either read about the &quot;universe&quot; 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 &quot;hand-waving method&quot; to prove results. You &quot;wave your hands&quot; and speak about &quot;categories&quot; and &quot;universes&quot;.
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 &quot;universe&quot; 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 &quot;hand-waving&quot; every now and then.
Let us from now &quot;assume all $k$-algebras live in a fixed universe&quot;.
Question: &quot;Does AutX(k) also have a group scheme structure?&quot;
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 &quot;group scheme&quot; 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 &quot;set&quot; and not a &quot;representable functor&quot;. 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 &quot;algebraic&quot; 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 &quot;algebraic variety&quot; 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:&quot;$Aut_X$ is algebraic if and only if $Aut(X)$ is.&quot;
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 &quot;set of $k$-rational points&quot; $G(k) \subseteq k^n$ 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: &quot;Does AutX(k) also have a group scheme structure?&quot;
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 $Aut_X(k)$.