---
abstract: 'Given a base field $\Bbbk$ of characteristic zero, an arbitrary $\Bbbk$-algebra $R$, a semisimple linear algebraic $\Bbbk$-group $G$, and a $G$-$R$ module $M$ free of finite rank, we prove that the algebra of invariants $(S_{R}^{\bullet}((M^{\vee})^{\oplus m}))^{G_{R}}$ is a finitely presented $R$-algebra. In case $R$ is noetherian, we recover Seshadri's generalization of Nagata's finiteness theorem. Furthermore, if $T$ is an $R$-algebra and $N$ a $T$-module, we prove that the algebra of invariants $(S_{T}^{\bullet}(M^{\vee}\otimes_{T}N))^{G_{T}}$ is also finitely presented as long as $N$ is. If $N$ is an arbitrary $T$-module and $G$ is classical, we show that the algebra of invariants is generated by elements of bounded degree whose ideal of relations is generated by elements with bounded degree as well. The bounds depend on $G$ but not on $N$. Additionally, we prove a Chevalley compactification type theorem for the algebra of invariants.'
address:
- 'Departamento de Matemáticas, Universidad de Salamanca, Spain'
- 'Departamento de Matemáticas, Universidad de León, Spain'
- 'Departamento de Matemáticas, Universidad de Salamanca, Spain'
author:
- 'J. Martín Ovejero'
- 'A. L. Muñoz Castañeda'
- 'F. J. Plaza Martín'
bibliography:
- 'mibiblio2.bib'
title: Finiteness properties of the algebra of invariants
---

[^1]

Introduction
============

The basic problem in classical invariant theory is usually formulated as follows: given a base field $\Bbbk$, a $\Bbbk$-algebraic group $G$, and a finite dimensional $\Bbbk$-representation $G\rightarrow \textrm{GL}(V)$, decompose $T^{r}(V):=V^{\otimes r}$ as a direct sum of irreducible representations. There are two equivalent formulations of this problem: (1) describe $(T^{r}(V)\otimes T^{s}(V^{\vee}))^{G}$ in terms of generators and relations, and (2) describe $(S^{\bullet}(V^{\vee}))^{G}$ in terms of generators and relations. The first formulation appeals directly to representation theory, while the third one refers to the quotient variety $V/G$.

When the group $G$ is classical (i.e., one of the algebraic groups $\textrm{SL}_{n}$, $\textrm{Sp}_{2n}$, $\textrm{O}_n$ or $\textrm{SO}_{n}$), the description of the generators is usually called the First Fundamental Theorem (FFT) while the description of the relations is called the Second Fundamental Theorem (SFT). These problems were faced in this generality by Schur, Weyl and others (see @weyl for instance).

A key fact when describing $(S^{\bullet}(V^{\vee}))^{G}$ in terms of generators and relations when $G$ is classical is that it is a finitely generated $\Bbbk$-algebra. In this regard, the 14th Hilbert problem @hilbert asks whether $(S^{\bullet}(V^{\vee}))^{G}$ is always a finitely generated $\Bbbk$-algebra. A counter example of this statement was given by Nagata @nagata-hilbert. Furthermore, in @nagata, Nagata proved that for any finitely generated $\Bbbk$-algebra $A$ and any (faithful) representation $\rho:G\rightarrow \textrm{Aut}_{\Bbbk\textrm{-alg}}(A)$, the ring of invariants $A^{G}\subset A$ is again a finitely generated $\Bbbk$-algebra provided $G$ is reductive. This fundamental result provides a basis for geometric invariant theory and constructive methods in moduli problems @mumford-git.

Regarding this problem, it is worth pointing out Seshadri's work. Seshadri proved a natural generalization of Nagata's result @seshadri-relative [Theorem 2]. If $S=\textrm{Spec}(R)$, where $R$ is a finitely generated $\Bbbk$-algebra, $G_{R}$ is a reductive group $S$-scheme, $M$ is a finitely generated free $R$-module, and $\rho:G_{R}\rightarrow \textrm{Aut}_{R\textrm{-mod}}(M)$ is a representation, then the algebra of invariants $S^{\bullet}_{R}(M^{\vee})^{G_R}$ is finitely generated over $R$. Furthermore, if $B$ is an $R$-algebra, $\rho': G_{R}\rightarrow \textrm{Aut}_{R\textrm{-alg}}(B)$ another representation and $S^{\bullet}_{R}(M^{\vee})\rightarrow B$ a surjective morphism of $R$-algebras compatible with the two actions, $\rho$ and $\rho'$, then the same result as before holds for $B^{G}$. An important fact in Seshadri's generalization is that $R$ is assumed to be finitely generated over $\Bbbk$.

The goal of this article is to give an analogous result in a relative and not necessarily noetherian context. In analogy with the classical theory, we will study the ring of functions of the module $\textrm{Hom}_{T}(M\otimes_{R}T,N)$ which are $G_{R}$-invariant; that is, the graded algebra $A=(S_{T}^{\bullet}(M^{\vee}\otimes_{T}N))^{G_{R}}$. Here, $M$ and $\rho$ are as in Seshadri's work, $T$ is an $R$-algebra and $N$ a $T$-module. More precisely, our goal is to understand the behavior of $A$ as $N$ varies. We prove that, even in the most general case for $N$, the algebra of invariants satisfies a finiteness condition on the degrees of the generators and relations if the group is classical. This bounding condition is encoded in the notion of partially generated graded algebra, that we introduce and study before proving the main result. Observe that we do not impose $T$ to be a finitely generated $R$-algebra. Thus, we give a step further in generalizing Nagata's finiteness theorem for algebras of invariants. The main result of this work is the following.

\[t:main-fin-pres\] Let $R$ be a $\Bbbk$-algebra, $T$ an $R$-algebra, $M$ a finitely generated free $R$-module and $N$ a $T$-module. Let $G$ be a semisimple linear algebraic $\Bbbk$-group together with a faithful representation $\rho:G_{R}^{\bullet}\hookrightarrow \underline{\emph{Aut}}_{R\textrm{-mod}}(M)$ (meaning a natural transformation between the group-functors). Let us denote by $A$ the graded $T$-algebra $\big( S_{T}^\bullet (M^{\vee} \otimes_{R}N \big)^{G_T}$. Then:

1.  If $G$ is classical, then $A$ is a partially generated graded $T$-algebra.

2.  If $N$ is finitely presented, then $A$ is a finitely presented $T$-algebra.

3.  If $N$ is finitely generated, then $A$ is a finitely generated $T$-algebra.

Assume the representation is defined over $\Bbbk$, $\rho:G^{\bullet}\hookrightarrow \underline{\emph{Aut}}_{\Bbbk}(V)$, and set $M:=V\otimes_{\Bbbk}R$.

1.  If $N$ is a flat $T$-module, then $A$ is a flat $T$-algebra.

Furthermore, if $R\rightarrow T$ is finitely presented (respectively, flat), the same conclusions given in *(1)* and *(2)* (respectively, *(3)*) can be obtained by substituting $T$ by $R$ in the statements.

The proof of Theorem \[t:main-fin-pres\] (2) is mostly based on Theorem \[t:main-affine\], where we prove that $\big( S_{R}^\bullet (M^{\vee})^{\oplus m}\big)^{G_R}$ is a finitely presented $R$-algebra. By classical noetherian reduction arguments one may write $R$ (respectively, $M$) as a directed limit of a directed system $(R_{\lambda})_{\lambda\in I}$ of noetherian $\Bbbk$-algebras (respectively, as a directed limit of a directed system $(M_{\lambda})_{\lambda\in I}$ of finitely generated free $R_{\lambda}$-modules). Theorem \[t:main-affine\] follows by proving that for any group action of $G_R$ on $M$, there exists an index $\lambda$ and a group action of $G_{R_{\lambda}}$ on $M_{\lambda}$ whose base change to $R$ gives the group action we started with. This generalizes Seshadri's Theorem @seshadri-relative [Theorem 2] to the non noetherian case. Theorem \[t:main-fin-pres\] (3) follows from Theorem \[t:main-fin-pres\] (2).

As a consequence of the main result and the fundamental properties of partially generated graded algebras, we obtain the existence of a natural locally closed embedding of the GIT quotient $\underline{\textrm{Hom}}_{T}(M,N) \sslash G:=\textrm{Spec}\big( S_{T}^\bullet (M^{\vee} \otimes_{R}N \big)^{G_T}$ in a projective bundle.

\[lm:cor-chevalley\] Under the hypotheses of Theorem \[t:main-fin-pres\], if either $N$ is finitely presented or $G$ is one of the classical (semisimple) linear algebraic $\Bbbk$-groups ($\emph{SL}_n$, $\emph{Sp}_{2n}$, $\emph{SO}_n$), there is a natural locally closed embedding, $$\label{eq:compact}
\Phi_{M,N,G}:\underline{\emph{Hom}}_{T}(M,N) \sslash G\hookrightarrow {\mathbb P} ( E_d ),$$ where $E_d:= \underset{1\leq \vert \underline{d}\vert \leq d }\oplus  S^{d_1}A_1 \otimes_R \dots \otimes_R  S^{d_t}A_t$ and $A_{i}=(S^{i}(M^{\vee}\otimes_{R}N))^{G_{T}}$, for a given natural number $d$. Furthermore, (1) the natural number $d$ depends on $G$ but not on $N$ if $G$ is classical, and (2) $\Phi_{M,N,G}$ is $\emph{Aut}_{T}(N)$-equivariant in any case.

In case $R=T=\Bbbk$, $N=M=\Bbbk^{n}$ and $\rho$ takes values in $\textrm{SL}(n)$, $\textrm{SL}(n)/G$ is an open subscheme of $\underline{\textrm{Hom}}_{T}(M,N) \sslash G$, so the restriction of $\Phi_{M,N,G}$ gives a locally closed embedding of $\textrm{SL}_{n}/G$ in a projective bundle which is $\textrm{SL}_{n}$-equivariant (acting on the right). Thus, Corollary \[lm:cor-chevalley\] can be seen as a Chevalley compactification type theorem in a more general context.

The article is organized as follows. In Section 2, we introduce some preliminary notions about representations of group schemes. In Section 3 we prove the main technical result needed for proving Theorem \[t:main-fin-pres\]. This result generalizes Seshadri's theorem @seshadri-relative to the non-noetherian case. In Section 4, we deal with Theorem \[t:main-fin-pres\] (2), (3) and (4). In Section 5, we prove Theorem \[t:main-fin-pres\] (1). Before proving the theorem, we introduce the notion of partially generated graded algebra, and give some fundamental properties of this kind of rings that are needed in the proof. Furthermore, we describe how can be $\textrm{Spec}(A)$ embedded in a certain projective space as a locally closed subscheme. This allows to prove Corollary \[lm:cor-chevalley\]. The final section addresses some issues where our results might be applied. Among them, it is worth pointing out that the constructions of moduli spaces of principal bundles based on tensors (e.g., @bhosle [@langer; @munoz1; @schmitt]) may profit particularly from our results and techniques since our work yields new properties (flatness, canonicity, etcetera) of the main objects presented in those papers.

Preliminaries and notation
==========================

We fix once and for all a field of characteristic zero $\Bbbk$. Given a commutative $\Bbbk$-algebra $R$, we denote by $\textrm{Alg}(R)$ the category of $R$-algebras. Let $G$ be a linear algebraic group over $\Bbbk$. For each commutative $\Bbbk$-algebra $R$, we denote by $G_{R}$ the $R$-group scheme obtained from $G$ by base change respect to the structure morphism of $\textrm{Spec}(R)$. The ring of functions of $G_{R}$ will be denoted by $R[G]$, so $G_{R}=\textrm{Spec} (R[G])$, and its associated group-functor by $G^{\bullet}_{R}$. We will drop the subscript if $R=\Bbbk$.

Representations
---------------

Given a commutative $\Bbbk$-algebra $R$ and an $R$-module $M$, we denote by $\underline{\textrm{End}}_{R\textrm{-mod}}(M)$ (respectively, $\underline{\textrm{Aut}}_{R\textrm{-mod}}(M)$) the functor of endomorphisms (respectively, automorphisms) of $M$: $$T\in \textrm{Alg}(R)\leadsto \textrm{End}_{T\textrm{-mod}}(M\otimes_{R}T)$$ (respectively, $\textrm{Aut}_{T\textrm{-mod}}(M\otimes_{R}T))$. If $M$ is a finitely generated free $R$-module, $S^{\bullet}(\textrm{End}_{R\textrm{-mod}}(M))^{\vee}$ is a representative of $\underline{\textrm{End}}_{R\textrm{-mod}}(M)$. Furthermore, we may localize $S^{\bullet}(\textrm{End}_{R\textrm{-mod}}(M)^{\vee})$ by the determinant function, and the functor of points of the corresponding affine $R$-scheme is exactly $\underline{\textrm{Aut}}_{R\textrm{-mod}}(M)$.

A linear representation of $G$ over $R\in \textrm{Alg}(\Bbbk)$, is a pair $(M,\rho)$ where $M$ is an $R$-module, and $\rho:G_{R}^{\bullet}\rightarrow \underline{\textrm{Aut}}_{R\textrm{-mod}}(M)$ is a morphism of functors. In such situation, it is said that $M$ is $G$-$R$ module. If $M$ has an additional structure of $R$-algebra, then a representation is a morphism $\rho$ as above whose image lies in $\underline{\textrm{Aut}}_{R\textrm{-alg}}(M)$. In any case, we say that $G_{R}$ acts on $M$. Note that giving $\rho$ is the same as giving a morphism of groups $\rho_{T}:G_{R}^{\bullet}(T)\rightarrow \textrm{Aut}_{T\textrm{-mod}}(M\otimes_{R}T)$ for each $R$-algebra $T$ in a functorial way (the same holds for algebras). Furthermore, giving the above data is the same as giving a morphism of $R$-modules $$M\rightarrow M\otimes_{R}R[G],$$ which makes $M$ a co-module over the co-algebra $R[G]$. Since $R[G]$ is $R$-flat, giving a morphism $M\rightarrow M\otimes_{R}R[G]$ is, in turn, equivalent to giving a morphism $R$-modules $$\textrm{End}_{R\textrm{-mod}}(M)^{\vee}\rightarrow R[G]=\Bbbk[G]\otimes_{\Bbbk}R.$$ Finally, note that if $M$ is a finitely generated free $R$-module, then there is a canonical isomorphism $M^{\vee}\otimes_{R}T\simeq (M\otimes_{R}T)^{\vee}$. Thus, any representation $\rho:G_{R}^{\bullet}\rightarrow\underline{\textrm{Aut}}_{R\textrm{-mod}}(M)$ induces a representation (the contragradient representation) $\rho^{\vee}:G_{R}^{\bullet}\rightarrow 
\underline{\textrm{Aut}}_{R\textrm{-mod}}(M^{\vee})$ defined by the rule:

$$\begin{split}
\rho_{T}^{\vee}(g): (M\otimes_{R} T)^{\vee} & \rightarrow (M\otimes_{R}T)^{\vee}\\
\varphi &\mapsto  \varphi \circ \rho_{T}(g)
\end{split}$$

The co-module structure induced on $M^{\vee}$ by the contragradient representation is given by $$\label{co-grad}
\begin{split}
M^{\vee} & \rightarrow  \textrm{Hom}_{R}(M,R[G]) \simeq M^{\vee}\otimes_{R}R[G] \\
\delta &\mapsto  (\delta\otimes 1) \circ \rho
\end{split}$$

Invariant exactness and Reynolds operator
-----------------------------------------

Given a commutative $\Bbbk$-algebra $R$, an $R$-module $M$, and a representation $\rho:G_{R}^{\bullet}\rightarrow\underline{\textrm{Aut}}_{R\textrm{-mod}}(M)$ (thus, $M$ is a $G$-$R$ module), we denote by $M^{G_R}$ the $R$-module of invariant elements of $M$.

The group scheme $G_R$ is invariant exact if the functor "taking invariants by $G_R$" transforms short exact sequences of $G$-$R$ modules into short exact sequences of $R$-modules. If $R[G]$ is projective as $R$-module, then $G_R$ is invariant exact if and only if $(R[G])^{\vee}=R\oplus C$ for some $R$-algebra $C$ @sancho [Remark 2.7]. Since $\Bbbk[G]$ is projective and this property is stable under base change, $R[G]$ is projective (as $R$-module) as well for every $\Bbbk$-algebra $R$. Thus, $G_{R}$ is invariant exact for every $\Bbbk$-algebra $R$ because $G$ is.

A Reynolds operator $\mathcal{R}$ for $G_R$ is a projection $$\mathcal{R}(M):M\rightarrow M^{G_{R}}$$ for every $G$-$R$ module $M$ giving the identity when is restricted to $M^{G_{R}}$, and is compatible with morphisms of $G$-$R$ modules. In particular, this makes $M^{G_{R}}$ a direct summand of $M$. If $G_R$ is invariant exact, there is always a Reynolds operator for free $R$-modules (this is a particular case of @sancho [Theorem 2]). Thus, in our situation, there will always be available a Reynolds operator for free $G$-$R$ modules and for any $\Bbbk$-algebra $R$.

Whenever a Reynolds operator is defined, the functor "taking invariants by $G_R$" is compatible with base changes. This follows from the proof @seshadri-relative [Lemma 2] and the fact that $M^{G_{R}}\subset M$ is a direct summand. In particular, the Reynolds operator behaves well under base changes; that is, $$\mathcal{R}(M)\otimes 1=\mathcal{R}(M\otimes_{R}T)$$ for every $R$-algebra $T$ and every free $R$-module $M$.

Further notation
----------------

Through the paper we will use the following notation:

-   $T_{R}^{\bullet}(M)$ ($T_{R}^{n}(M)$) is the ($n$th) tensor algebra (power) of the $R$-module $M$.

-   $S_{R}^{\bullet}(M)$ ($S_{R}^{n}(M)$) is the ($n$th) symmetric algebra (power) of the $R$-module $M$.

-   $\mathbb{P}(M)$ is the homogeneous spectrum $\textrm{Proj} \ S_{R}^{\bullet}(M)$.

-   Given $\rho:G_{R}^{\bullet}\rightarrow \underline{\textrm{Aut}}_{R\textrm{-mod}}(M)$, with $M$ free of finite rank, an $R$-algebra $T$ and a $T$-module $N$, $\underline{\textrm{Hom}}_{T}(M,N) \sslash G$ denotes the (affine) GIT quotient $\textrm{Spec} \big( S_{T}^\bullet (M^{\vee}\otimes_{R}N)\big)^{G_T}$.

Nagata-Seshadri theorem. Non-noetherian case
============================================

The proof of the statements in Theorem \[t:main-fin-pres\] will be based on the particular case where $N$ is a finitely generated free $T$-module. If $R$ is a finitely generated $\Bbbk$-algebra and $N=T=R$, this is precisely Sesadri's theorem @seshadri-relative [Theorem 2]. The argument of the proof of this result will consists of a Noetherian reduction argument following the techniques presented in @ega4-3.

Before stating the main technical result, we recall the fundamental lemmas, and prove others, that will be needed.

Let $R_0$ be a commutative ring, $(R_{\lambda})_{\lambda\in I}$ an inductive system of $R_0$-algebras. If $M_{0}$ is an $R_0$-module, we can set $M_{\lambda}:=M_{0}\otimes_{R_{0}}R_{\lambda}$. Since $M_{\beta}$ and $ M_{\lambda}\otimes_{R_{\lambda}}R_{\beta}$ are canonically isomorphic, there is a canonical morphism $M_{\lambda}\rightarrow M_{\beta}$ for each pair $\lambda\leq \beta \in I$, namely, $m\mapsto m\otimes 1$. Thus, $(M_{\lambda})_{\lambda\in I}$ forms an inductive system. If $N_{0}$ is another $R_{0}$-module, the same discussion yields an inductive system $(N_{\lambda})_{\lambda\in I}$. Note that the tensor product operation $\otimes_{R_{\lambda}}R_{\beta}$ induces a morphism $$\textrm{Hom}_{R_{\lambda}}(M_{\lambda},N_{\lambda})\longrightarrow \textrm{Hom}_{R_{\lambda}}( M_{\beta},N_{\beta})$$ for each pair $\lambda\leq \beta \in I$. Thus, we also have an inductive system $(\textrm{Hom}_{R_{\lambda}}(M_{\lambda},N_{\lambda}))_{\lambda\in I}$.

@ega4-3 [Lemma 93.241.143.245]\[lm:noeth-red-1\] Let $R_0$ be a commutative ring, $(R_{\lambda})_{\lambda\in I}$ an inductive system of $R_0$-algebras, $R=\underset{\longrightarrow}{\lim}\  R_{\lambda}$; let $M_0 , N_0$ be $R_0$-modules, and let us denote $M_{\lambda}=M_{0}\otimes_{R_{0}}R_{\lambda}, N_{\lambda}=N_{0}\otimes_{R_{0}}R_{\lambda}$, $M=M_{0}\otimes_{R_{0}}R=\underset{\longrightarrow}{\lim}\  M_{\lambda}, N=N_{0}\otimes_{R_{0}}R=\underset{\longrightarrow}{\lim}\  N_{\lambda}$. If $M_{0}$ is of finite type (finite presentation), then the canonical homomorphism $$\underset{\longrightarrow}{\lim}\  \emph{Hom}_{R_{\lambda}}(M_{\lambda},N_{\lambda})\rightarrow \emph{Hom}_{R}(M,N)$$ is injective (respectively, bijective).

@ega4-3 [Proposition 8.9.1 (ii)]\[lm:noeth-red-2\] Let $R$ be a commutative ring and $M$ an $R$-module of finite presentation. Then, there exits a noetherian subring $R_{0}\subset R$ and an $R_0$-module of finite presentation such that $M_{0}\otimes_{R_{0}}R\simeq M$.

\[rmk:free\] If $M$ is free, we may assume that $M_0$ is also free (see @ega4-3 [Corollaire 93.241.143.245]).

\[lm:indexes\] Let $R$ be a commutative ring. Let $(R_{\lambda})_{\lambda\in I}$ be the inductive system formed by the noetherian subrings of $R$, so $R=\underset{\longrightarrow}{\lim}\  R_{\lambda}$. Let $R_{0}\in (R_{\lambda})_{\lambda\in I}$. Consider the set of indexes $J=\{\lambda\in I | R_{0}\subset R_{\lambda}\}$. Then, $\underset{\longrightarrow}{\lim}\ R_{\lambda}=R$.

Note that, in this case, for any of both, $I$ and $J$, we have $\underset{\longrightarrow}{\lim}\  R_{\lambda}=\cup R_{\lambda}$. So, to prove the lemma we just need to show that given $R_{\alpha}$ and $R_{\beta}$ such that $R_{0}\subset R_{\alpha}$, the subring $R_{\gamma}$ generated by $R_{\alpha}$ and $R_{\beta}$ is noetherian and $R_{0}\subset R_{\gamma}$, that is, $\gamma\in J$. Let us fix presentations $u_{\alpha}:\mathbb{Z}[x_1,\hdots,x_m]\rightarrow R_{\alpha}\rightarrow 0$ and $u_{\beta}:\mathbb{Z}[y_1,\hdots,y_n]\rightarrow R_{\beta}\rightarrow 0$. Then, $R_{\gamma}$ is given by the image of the morphism of $k$-algebras $$u:\mathbb{Z}[x_1,\hdots,x_m,y_1,\hdots,y_n]\rightarrow R$$ defined by $u(x_{i})=u_{\alpha}(x_{i})$ and $u(y_{j})=u_{\beta}(y_{j})$. Thus, $R_{\gamma}$ is noetherian and, clearly, $R_{\alpha}, R_{\beta}\subset R_{\gamma}$. In particular, $R_{0}\subset R_{\gamma}$.

\[lm:noeth-red-3\] Let $R$ be a commutative ring and $M$ an $R$-module of finite presentation. Then, there exits a noetherian subring $R_{0}\subset R$, a directed system of noetherian $R_{0}$-subalgebras of $R$, $(R_{\lambda})_{\lambda\in J}$, and a directed system of modules of finite presentation $(M_{\lambda})_{\lambda\in J}$ ($M_{\lambda}$ being an $R_{\lambda}$-module) such that $R=\underset{\longrightarrow}{\lim}\  R_{\lambda}$, $M=\underset{\longrightarrow}{\lim}\  M_{\lambda}$ and $\emph{End}_{R}(M)=\underset{\longrightarrow}{\lim}\  \emph{End}_{R_{\lambda}}(M_{\lambda})$.

The ring $R$ can be written as $R=\underset{\longrightarrow}{\lim}\ R_{\lambda}$, where $(R_{\lambda})_{\lambda\in I}$ is the directed system of all noetherian subrings of $R$. Let $R_0$ and $M_0$ be as in Lemma \[lm:noeth-red-2\]. Let $J\subset I$ be the subset of indexes given in Lemma \[lm:indexes\] and denote $M_{\lambda}=M_{0}\otimes_{R_{0}}R_{\lambda}$. From Lemma \[lm:indexes\] it follows that $R=\underset{\longrightarrow}{\lim}\ R_{\lambda}$. This implies that $M=\underset{\longrightarrow}{\lim}\ M_{\lambda}$, since directed limits commutes with tensor product. Now, the result follows directly from Lemma \[lm:noeth-red-1\].

\[prop:noeth-red\] Let $\Bbbk$ be a field, $R$ and $A$ $\Bbbk$-algebras, and $M$ a finitely generated free $R$-module. Let $f:\emph{End}_{R}(M)^{\vee}\rightarrow A\otimes_{\Bbbk}R$ be a morphism of $R$-modules. Then, there exits a noetherian sub-$\Bbbk$-algebra $R_{0}\subset R$, a finitely generated free $R_{0}$-module $M_0$, and a morphism of $R_{0}$-modules $f_{0}:\emph{End}_{R_0}(M_0)^{\vee}\rightarrow A\otimes_{\Bbbk}R_{0}$ such that $f_{0}\otimes 1=f$.

The above lemmas can be stated in terms of $\Bbbk$-algebras instead of just commutative rings. We will use this fact without any further mention to this remark.

Let $R_0$, $(R_{\lambda})_{\lambda\in J}$ and $(M_{\lambda})_{\lambda\in J}$ be as in Lemma \[lm:noeth-red-3\]. By Remark \[rmk:free\], we may assume that $M_\lambda$ is free for every $\lambda$. Note that, under these conditions there is a canonical isomorphism $\textrm{End}_{R_{\alpha}}(M_{\alpha})\otimes_{R_{\alpha}}R_{\beta}=\textrm{End}_{R_{\beta}}(M_{\beta})$ for every $\alpha\leq \beta\in J$. So, $\underset{\longrightarrow}{\lim}\ \textrm{End}_{R_{\lambda}}(M_{\lambda})^{\vee}=\textrm{End}_{R}(M)^{\vee}$. On the other hand, since directed limits commute with tensor product, we may write $A\otimes_{\Bbbk}R=\underset{\longrightarrow}{\lim}\ (A\otimes_{\Bbbk}R_{\lambda})$. Then, $f$ belongs to $$\textrm{Hom}_{R}(\underset{\longrightarrow}{\lim}\  \textrm{End}_{R_{\lambda}}(M_{\lambda})^{\vee},\underset{\longrightarrow}{\lim}\  (A \otimes_{\Bbbk}R_{\lambda})).$$ By Lemma \[lm:noeth-red-1\], the above is isomorphic to $$\underset{\longrightarrow}{\lim}\  \textrm{Hom}_{R}(\textrm{End}_{R_{\lambda}}(M_{\lambda})^{\vee},A \otimes_{\Bbbk}R_{\lambda}),$$ and the result follows.

\[t:main-affine\] Let $R$ be a $\Bbbk$-algebra, $M$ a finitely generated free $R$-module. Let $G$ be a semisimple linear algebraic $\Bbbk$-group together with a faithful representation $\rho:G_R^{\bullet}\hookrightarrow \underline{\emph{Aut}}_{R\textrm{-mod}}(M)$. Then $A:=  \big( S_{R}^\bullet ((M^{\vee})^{\oplus m})\big)^{G_R}$ is an $R$-algebra of finite presentation for all $m>0$.

We claim that $R$ can be assumed to be noetherian. Let $R[G]$ denote the ring of functions of $G_R$; that is, $\textrm{\textrm{Spec}} R[G] = G_R$ and, with this notation, it holds that $\Bbbk[G]\otimes_{\Bbbk} R=R[G]$. The action of $G_R$ on $M$ can be understood as a morphism of $R$-modules $$\tilde \rho: M \longrightarrow R[G]\otimes_R M$$ Let $\rho:\textrm{End}_R (M)^{\vee}\to R[G]=\Bbbk[G]\otimes_{\Bbbk}R$ be the corresponding morphism of $R$-modules. Let $R_0$, $M_0$ and $\rho_0$ be as in Proposition \[prop:noeth-red\]. Denote by $R_1$ the localization of $R_0$ by the multiplicative system given by those $r\in R_0$ which are invertible in $R$. Note that there are inclusions $R_0 \subset R_1 \subset R$. Let $M_1:=M_0\otimes_{R_0} R_1$ and $\rho_1=\rho_0 \otimes 1$ (base change to $R_1$). Note that $R_1$ is noetherian and that $M_1$ is a $R_1$-module free of finite rank.

The next step is to show that $\tilde \rho_1: M_1 \longrightarrow R_1[G]\otimes_{R_1} M_1$ defines an action of $G_{R_1}$ on $M_1$. Namely, we have to check that the following two diagrams $$\label{e:action1}
    \xymatrix{
    M_1 \ar[r]^-{\tilde\rho_1}  \ar[d]^{\tilde\rho_1} & 
    R_1[G]\otimes_{R_1} M_1  \ar[d]^{1\otimes \tilde\rho_1}
    \\
    R_1[G]\otimes_{R_1} M_1  \ar[r]^-{m \otimes 1} & 
    R_1[G]\otimes_{R_1} R_1[G]\otimes_{R_1}  M_1}$$ and $$\label{e:action2}
    \xymatrix@C=22pt{
    M_1 \ar[r]^-{\tilde\rho_1}  \ar[dr]_{\operatorname{Id}} & 
    R_1[G]\otimes_{R_1} M_1  \ar[d]^{e\otimes \tilde\rho_1}
    \\
    & 
    M_1}$$ are commutative, where $m$ (resp. $e$) is the multiplication map (resp. identity element) of $G$. Since $R_1$ is noetherian and $M_1$ is finitely generated, it follows from Nakayama's lemma that the previous two diagrams are commutative if and only if they are so after twisting by $\otimes_{R_1} R_1/{\mathfrak m}_1$ for all maximal ideals ${\mathfrak m}_1 \subset R_1$.

Given a maximal ideal ${\mathfrak m}_1 \subset R_1$, it holds that $ {\mathfrak m}_1  R$ is a proper ideal of $R$; indeed, if $ {\mathfrak m}_1  R =(1)$, the image of $ {\mathfrak m}_1$ would contain an invertible element of $R$, say $r$. Having in mind the multiplicative system considered to construct $R_1$, it would follow that $r$ is invertible in $R_1$ and, thus, $ {\mathfrak m}_1 = R_1$. Hence, consider a maximal ideal ${\mathfrak m}$ of $R$ containing $ {\mathfrak m}_1  R$. Note that $ {\mathfrak m}_1 = {\mathfrak m}\cap  R_1$. Further, since $R_1 / {\mathfrak m}_1  \to  R / {\mathfrak m}$ is faithfully flat, it holds that diagrams and twisted by $\otimes_{R_{1}} R_1/{\mathfrak m}_1$ are commutative if and only if the following diagrams $$\xymatrix{
    M \ar[r]^-{\tilde\rho}  \ar[d]^{\tilde\rho} & 
    R[G]\otimes_{R} M  \ar[d]^{1\otimes \tilde\rho}
    \\
    R[G]\otimes_{R} M  \ar[r]^-{m \otimes 1} & 
    R[G]\otimes_{R} R[G]\otimes_{R}  M}
    \qquad \qquad
        \xymatrix@C=22pt{
    M \ar[r]^-{\tilde\rho}  \ar[dr]_{\operatorname{Id}} & 
    R[G]\otimes_{R} M  \ar[d]^{e\otimes \tilde\rho}
    \\
    & 
    M }$$ are commutative when twisted by $\otimes_{R}R / {\mathfrak m} $. But this is the case since $\rho$ is an action.

By \[co-grad\], the base change of the co-module structure induced on $M_{1}^{\vee}$ by the representation $\rho_1$ is equal to the co-module structure induced on $M^{\vee}$ by $\rho_{1}\otimes 1=\rho$. Now, Lemma \[l:basechangeschur\] implies that $$\label{e:R_2R}
\big(S^\bullet_R (M^{\vee})^{\oplus m}\big)^{G_R} \,=\, \big(S^\bullet_{R_1} (M_{1}^{\vee})^{\oplus m}\big)^{G_{R_1}} \otimes_{R_1} R.$$ Assume that $\big(S^\bullet_{R_1} (M_{1}^{\vee})^{\oplus m}\big)^{G_{R_1}}$ were a finite type $R_1$-algebra. Then it would be of finite presentation since $R_1$ is noetherian. Hence, by equation , $A$ would by a finitely presented $R$-algebra. Thus, it remains to show that, if $R$ is a noetherian ring, then $A$ is a finite type $R$-algebra. But this is @seshadri-relative [Theorem 2].

Algebra of invariants of the module of homomorphisms. Part 1 {#sec:part1}
============================================================

The goal of this section is to prove Theorem \[t:main-fin-pres\] (2), (3) and (4). A key ingredient for proving them, apart from Theorem \[t:main-affine\], is the base change property of the algebra of invariants, $$\label{eq:base-change}
\big( S_{T}^\bullet ((M^{\vee}\otimes_{R}T)^{\oplus m})\big)^{G_{T}}
    \,\simeq \,  \big( S_{R}^\bullet ((M^{\vee})^{\oplus m})\big)^{G_{R}}  \otimes_R T.$$ We give the proof of this result in Appendix (see Lemma \[l:basechangeschur\]), which is based on the Standard Basis Theorem and the base change properties of the Schur modules. This is the main reason for assuming $\textrm{char}(\Bbbk)=0$. It is worth pointing out that a characteristic free proof of this result (\[eq:base-change\]) would allow to drop the assumption on the characteristic of the base field in the main theorem.

\[Theorem \[t:main-fin-pres\] (2)\]\[cor:finitely-presented\] Let $R$ be a $\Bbbk$-algebra, $T$ an $R$-algebra, $M$ a finitely generated free $R$-module of rank $m$ and $N$ a finitely presented $T$-module. Let $G$ be a semisimple linear algebraic $\Bbbk$-group together with a faithful representation $\rho:G_R^{\bullet}\hookrightarrow \underline{\emph{Aut}}_{R\textrm{-mod}}(M)$. Then $A:=  \big( S_{T}^\bullet (M^{\vee}\otimes_{R}N)\big)^{G_T}$ is a $T$-algebra of finite presentation.

The proof consists of successive simplifications. First, consider a finite presentation $N''\to N'\to N\to 0$ where $N''$ and $N'$ are free $T$-modules of finite rank. Note that we have the exact sequence of $T$-modules $$\widetilde N := (M^{\vee}\otimes_R N'' ) \otimes_{T}  S_{T}^\bullet (M^{\vee} \otimes_R N')
    \to 
     S_{T}^\bullet (M^{\vee} \otimes_R N')
    \to 
     S_{T}^\bullet (M^{\vee} \otimes_R N)
     \to 0$$ Considering the $T$-algebras defined by $B:=  \big( S_{T}^\bullet (M^{\vee} \otimes_R N')\big)^{G_T}$, $A:=  \big( S_{T}^\bullet (M^{\vee} \otimes_R N)\big)^{G_T}$ and recalling that $G$ is reductive, it follows that $$\widetilde N^G
    \to 
    B   \to A \to 0$$ is exact. We claim that $\widetilde N^G$ is a finitely generated $C$-module. Indeed, since $\widetilde N$ is a finitely generated module over $S_{T}^\bullet (M \otimes_R N')$, one has a surjective morphism $$\big( S_{T}^\bullet (M \otimes_R N') \big)^{\oplus k} \,\longrightarrow\, \widetilde N$$ and having in mind that $G$ is reductive and the functoriality of the Reynolds operator, it follows a surjective morphism $(B)^{\oplus k} \to \widetilde N^G$ of $B$-modules. Hence, if the statement holds for $B$, then it will hold for $A$ or, what is tantamount, we can assume that $N$ is free of finite rank.

Hence, let $N$ be free of finite rank, say $n$. If we set isomorphisms $N\simeq (T)^{\oplus m}\simeq R^{\oplus m}\otimes_R T$ of $T$-modules and bearing in mind Lemma \[l:basechangeschur\], we deduce that $$A\,:=\,  \big( S_{T}^\bullet (M^{\vee} \otimes_R N)\big)^{G_T}
    \,\simeq \,  \big( S_{R}^\bullet ((M^{\vee})^{\oplus m})\big)^{G_R}  \otimes_R T$$ But in this case, the conclusion follows easily from Theorem \[t:main-affine\].

\[Theorem \[t:main-fin-pres\] (3)\] \[cor:finitely-generated\] Let $R$ be a $\Bbbk$-algebra, $T$ an $R$-algebra, $M$ a finitely generated free $R$-module of rank $m$ and $N$ a finitely generated $T$-module. Let $G$ be a semisimple linear algebraic $\Bbbk$-group together with a faithful representation $\rho:G_R^{\bullet}\hookrightarrow \underline{\emph{Aut}}_{R\textrm{-mod}}(M)$. Then $A:=  \big( S_{T}^\bullet (M^{\vee}\otimes_{R}N)\big)^{G_T}$ is a finitely generated $T$-algebra.

Using the same notation as in Corollary \[cor:finitely-presented\], there exists a surjective morphism $N':=T^{\oplus m}\rightarrow N$. By Corollary \[cor:finitely-presented\],, $B:=\big( S_{T}^\bullet (M^{\vee} \otimes_{R} N')\big)^{G_T}$ is finitely presented $T$-algebra and by construction the induced morphism $B\rightarrow A$ is surjective, hence $B$ is finitely generated.

\[Theorem \[t:main-fin-pres\] (4)\] Let $R$ be a $\Bbbk$-algebra, $T$ an $R$-algebra, $V$ a finite dimensional $\Bbbk$-vector space, and $N$ a flat $T$-module. Let $G$ be a semisimple linear algebraic $\Bbbk$-group together with a faithful representation $\rho:G^{\bullet}\hookrightarrow\underline{\emph{Aut}}_{R\textrm{-mod}}(V)$. Then $A:=  \big( S_{T}^\bullet (V^{\vee}\otimes_{\Bbbk}N)\big)^{G_T}$ is a flat $T$-algebra.

Let us assume for now that $N$ is free and finitely generated. Under this assumption, the Schur modules $\mathbb{S}^{\lambda}_{T}N$ are free and finitely generated for any given partition $\lambda$, and therefore $T$-flat. On the other hand, by Theorem \[l:basechangeschur\], the following isomorphism of $R$-modules holds $$\mathbb{S}^{\lambda}_{R}(V\otimes_{\Bbbk}R)^{G_R}\simeq (\mathbb{S}^{\lambda}_{\Bbbk}V)^{G}\otimes_{\Bbbk}R$$ Since $V$ is a finite dimensional vector space, $\mathbb{S}^{\lambda}_{\Bbbk}V$ is a $\Bbbk$-vector space where the reductive group $G$ acts. Thus, by the Fundamental Theorem of Invariant Theory, $(\mathbb{S}^{\lambda}_{\Bbbk}V)^{G}$ is a finite dimensional vector space over $\Bbbk$. By the base change property for Schur modules stated above, we conclude that $(\mathbb{S}^{\lambda}_{R}(V\otimes_{\Bbbk}R))^{G_R}$ is a free $R$-module, hence $R$-flat.

Since $V\otimes_{\Bbbk}R$ is a finitely generated free $R$-module, it follows from Lemma \[l:lemaprevio\] that the $k$-degree component of $A$ is $$\big({\operatorname{S}}^{k}_{T}(V^{\vee}\otimes_{\Bbbk}N)\big)^{G_T}\simeq\bigoplus_{\vert\lambda\vert=k}(\mathbb{S}_{R}^{\lambda}(V^{\vee}\otimes_{\Bbbk}R))^{G_R}\otimes_{R}\mathbb{S}^{\lambda}_{T}N.$$ Since the flat property is stable by pullbacks, and the tensor product of two flat modules is again flat, we conclude that for each partition $\lambda$, $$(\mathbb{S}_{R}^{\lambda}(V^{\vee}\otimes_{\Bbbk}R))^{G_R}\otimes_{R}\mathbb{S}^{\lambda}_{T}N\simeq (((\mathbb{S}_{R}^{\lambda}(V^{\vee}\otimes_{\Bbbk}R))^{G_R}\otimes_{R} T)\otimes_{T}\mathbb{S}^{\lambda}_{T}N$$ is a flat $T$-module, and therefore, $\big({\operatorname{S}}^{k}_{T}(V^{\vee}\otimes_{\Bbbk}N)\big)^{G_T}$ is too. So, if $N$ is free and finitely generated we have just proved the theorem, because in this case, $A$ is a direct sum of flat modules.

Consider now the general case. Since $N$ is flat, it follows from Lazard's Theorem @Lazard that it can be expressed as a direct limit of finitely generated free $T$-modules. Since direct limits commute with tensor products and, also, with the formation of the Schur modules we have reduced the problem to the case where $N$ is free and finitely generated.

Algebra of invariants of the module of homomorphisms. Part 2
============================================================

In order to complete the proof of the main theorem, it remains to show that Theorem \[t:main-fin-pres\] (1) holds true.

We introduce the notion of $t$-partially generated graded $R$-algebra ($t$-pgg-algebra). Roughly speaking, a $t$-pgg-algebra $A$ is a graded $R$-algebra which is generated by elements of degree less than or equal to a given natural number $t$, and such that the ideal of relations between the generators satisfies the same condition. Observe that we do not impose a finiteness condition on the number of generators, but only on the number of different degrees of the generators of the algebra, as well as on the degrees of the generators of the ideal of relations.

Partially generated graded algebras
-----------------------------------

Given a ring $R$ and a subset $I\subseteq {\mathbb{Z}}$, a partial algebra over $R$ with support $I$ consists of the data $\{A_i, m_{ij}\}$, where each $A_i$ is an $R$-module and $m_{ij}:A_i\times A_j\to A_{i+j}$, $i,j\in I$ with $i+j\in I$, are bilinear maps of $R$-modules satisfying:

1.  commutativity: $m_{ij}(a_i,a_j) = m_{j,i}(a_j,a_i)$ for all $a_i\in A_i, a_j\in A_j$ and all pairs $i,j$ with $i,j,i+j\in I$;

2.  associativity: $m_{i+j,k} \circ (m_{ij},1) = m_{i,j+k}\circ (1,m_{jk})$ for all triples $i,j,k$ with $i+j,i+k,j+k ,i+j+k\in I$.

If no confusion arises, we will simply write $A\,=\, \oplus_{i\in I} A_i$.

Given two partial algebras over $R$, $A=\oplus_{i\in I} A_i$ and $B=\oplus_{j\in J} B_j$ with $I,J\subseteq {\mathbb{Z}}$, a morphism of partial $R$-algebras $f:A\to B$ is a set of morphisms of $R$-modules $\{f_k:A_k\to B_k\,\vert\, k\in I\cap J\}$ such that $f_{i+j} \circ m_{ij}^A  = m_{ij}^B \circ (f_i, f_j) $ for all $i,j,$ with $i,j,i+j \in I\cap J$.

Note that for any ${\mathbb{Z}}$-graded $R$-algebra, $A=\oplus_{n\in {\mathbb{Z}}} A_n$, and each subset $I\subset\mathbb{Z}$, it holds that $A_I:=\oplus_{i\in I} A_i$ is a partial algebra. Any (non-graded) $R$-algebra $A$ is also a partial algebra (setting $A_0:=A$ and $A_i=(0)$ for all $i\neq 0$).

From now on, unless otherwise stated, all rings will be considered to be commutative; for all graded $R$-algebras it will be assumed that $I={\mathbb{Z}}_{\geq 0}$ and $A_0=R$; and, for all partial $R$-algebras with $0\in I$, it will be assumed that $A_0=R$. For a graded (resp. partial) algebra, we say elements of $A_n$ are homogeneous of degree $n$.

Let us introduce the following notation $$\label{eq:degreen}
    \begin{split}
    [A]_n \,&:=\, \{a\in A\, \vert\, \text{$a$ is homogeneous of degree $n$} \}, \\
    [A]_{\leq t} \,&:=\, \oplus_{i=1}^t [A]_i,
    \end{split}$$ and similarly for graded modules. Further, if no confusion arises, we will simply write $A_{\leq t}$ instead of $[A]_{\leq t}$.

Let $\operatorname{Alg}(R)$ denote the category of $R$-algebras. Given $t>0$ and a graded $R$-algebra of the type $A=\oplus_{n\geq 0} A_n$, there is a morphism of functors on $\operatorname{Alg}(R)$ $$\label{e:Phit}
        \Phi_{\leq t}: {\operatorname{Hom}}_{\text{$R$-alg}}(A,-)  \longrightarrow {\operatorname{Hom}}_{\text{partial $R$-alg}}(A_{\leq t},-)$$ which, for an $R$-algebra $B$, is given by the map $$\label{e:PhitB}
    \begin{split}
    \Phi_{\leq t}(B): {\operatorname{Hom}}_{R\textrm{-alg}}(A,B) & \longrightarrow {\operatorname{Hom}}_{\textrm{partial $R$-alg}}(A_{\leq t},B).\\
    f \,  & \longmapsto  \Phi_{\leq t}(B)(f):=(f_{i}:=f\vert_{A_{i}} )_{i=0,\hdots,t}
    \end{split}$$

\[d:pgg-alg\] A graded $R$-algebra $A=\oplus_{n\geq 0} A_n$ is called partially generated graded algebra (or, for the sake of brevity, pgg-algebra) if $A_0=R$ and there exists $t>0$ such that $\Phi_{\leq t}$ is an isomorphism of functors on $\operatorname{Alg}(R)$. In the latter situation, we say that $A$ is a $t$-pgg-algebra.

Observe that if $\Phi_{\leq t_0}$ is an isomorphism, then so is $\Phi_{\leq t}$ for any $t\geq t_0$.

Recall from the notation $[-]_n$ for the homogeneous elements of degree $n$ of a graded algebra (resp. partial algebra or module). Note, for instance, that $[ \oplus_{n\geq  0} A_n ]_i = A_i$. Note also that the tensor product of graded objects (rings and modules) inherits a canonical graduation given by setting $\operatorname{deg}(a\otimes b):= \operatorname{deg}(a) + \operatorname{deg}(b)$ for homogeneous elements $a,b$ and, in full generality, $$\big[ ( \oplus_{m} A_m ) \otimes ( \oplus_{n} B_n) \big]_i
    \,:=\, \oplus_{j } ( A_j \otimes B_{i-j})$$ This rule can be used to introduce a graduation in symmetric tensor products (of an algebra or a module), since they are submodules of tensor products. In particular, given a graded $R$-module, $( M=\oplus_{n > 0} M_n )$, it holds that the symmetric algebra (over $R$), $S_R^\bullet(M)$, is a graded $R$-algebra whose degree $n$ component is explicitly described as $$\label{e:Sym-part-n}
    \big[S_R^\bullet(M)\big]_n \,=\,
            \underset{ \vert \underline{d}\vert =n }\bigoplus  S^{d_1}M_1 \otimes_R \dots \otimes_R  S^{d_n}M_n$$ where $n>0$, $\underline d=(d_1,\dots, d_n)\in ({\mathbb{Z}}_{\geq 0})^n$. The number $\vert \underline{d}\vert:=\sum_{i=1}^n i \cdot d_i$ will be called the norm of the tuple $\underline d=(d_1,\dots, d_n)\in ({\mathbb{Z}}_{\geq 0})^n$. We set $\big[S_R^\bullet(M)\big]_0:=R$.

Let $( A=\oplus_{n= 0}^t A_n , m_{ij} )$ be a partial $R$-algebra with $A_0=R$. It follows from the relation  that $$\label{e:Sym-alg-par}
    S_R^\bullet(A_{\leq t})       \,=\, S_R^\bullet(\oplus_{n= 1}^t A_n).$$

For a graded $R$-algebra $A=\oplus_{n\geq 0} A_n$ and $t>0$, the natural inclusion $A_{\leq t} \hookrightarrow A$ together with the product of $A$ induces a canonical homogeneous morphism of graded $R$-algebras $$\label{e:multi-sym-A}
    S_R^\bullet(A_{\leq t}) \to A .$$ Note that in the l.h.s. only the additive structure of $A$ as $R$-module is involved. The behavior of this map with respect to the multiplicative structure of $A$ is our next task.

\[l:repre\] Let $R$ be a ring and let $( A=\oplus_{n= 0}^t A_n , m_{ij} )$ be a partial $R$-algebra. On the category $\operatorname{Alg}(R)$, it holds that ${\operatorname{Hom}}_{\emph{partial $R$-alg}}(A,-)$ is a closed subfunctor of ${\operatorname{Hom}}_{\emph{$R$-alg}}(S_R^\bullet(A_{\leq t}),-)$.

It suffices to show that there exists an ideal $I_t\subseteq S_R^\bullet(A_{\leq t})$, depending on $A$ and $t$, such that the quotient map $S_R^\bullet(A_{\leq t}) \to S_R^\bullet(A_{\leq t})/ I_t$ induces an isomorphism $$\label{e:representable}
    {\operatorname{Hom}}_{\text{partial $R$-alg}}(A,-) \, \simeq 
    {\operatorname{Hom}}_{\text{$R$-alg}}(S_R^\bullet(A_{\leq t})/ I_t ,-)$$ The desired $I_t$ is given by the homogeneous ideal $$\label{e:It}
    I_t\,:=\, 
    \big\langle\{ m_{ij}(a_i ,  a_j) - m(a_i \otimes a_j) 
    \,\vert \, a_i \in  A_i,     a_j \in  A_j     \text{ and } 0\leq i,j,i+j\leq t
    \}\big\rangle$$ Note that the above expression makes sense since $m_{ij}(a_i ,  a_j) \in A_{i+j} \subseteq [S_R^\bullet(A_{\leq t})]_{i+j}$ and $$m(a_i \otimes a_j) \in \operatorname{Im} \big( 
     [S_R^\bullet(A_{\leq t})]_{i}\otimes [S_R^\bullet(A_{\leq t})]_{j} \to [S_R^\bullet(A_{\leq t})]_{i+j}
     \big)$$

\[e:k\[V\]\] Let us consider the ring of functions of the vector space $V$, $A:=\Bbbk[V]=S^{\bullet}(V^{\vee})$. The degree $n$ component is $A_n = S^n (V^{\vee})$. The expression reads $$\big[S_R^\bullet(A_{\leq t})\big]_n \,=\,
    \bigoplus_{\vert \underline{d}\vert =n} S^{d_1}(S^1 V^{\vee})  \otimes_R \dots \otimes_R  S^{d_t}(S^t V^{\vee})$$ and, the degree $n$ component of the multiplication map is the sum of the maps $$S^{d_1}(S^1 V^{\vee})  \otimes_R \dots \otimes_R  S^{d_t}(S^t V^{\vee})
    \,\longrightarrow \, 
    S^n V^{\vee}$$ over $\underline{d}$ with $\vert \underline{d}\vert =n$.

\[t:pgg-exact\] Let $R$ be a ring and $A$ a graded $R$-algebra $A=\oplus_{n\geq 0} A_n$. Then, $A$ is a $t$-pgg-algebra if and only if the sequence $$\label{e:pgg-exact}
    0 \to I_t \to S_R^\bullet(A_{\leq t}) \to A \to 0$$ is an exact sequence of graded $R$-modules, where $I_t$ is the homogeneous ideal in . In particular, $A$ is a pgg-algebra if and only if is exact for come $t > 0$.

Recall that we say that $A$ is a $t$-pgg-algebra iff $\Phi_{\leq t}$ (see equation ) is an isomorphism.

Given $A$ and $t>0$, we consider the following composition of morphisms of functors $${\operatorname{Hom}}_{\text{$R$-alg}}(A,-)  
    \overset{\Phi_{\leq t}}\longrightarrow {\operatorname{Hom}}_{\text{partial $R$-alg}}(\oplus_{n=0}^t A_n,-)
    \overset{\sim}\longrightarrow 
    {\operatorname{Hom}}_{\text{$R$-alg}}(S_R^\bullet(A_{\leq t})/ I_t ,-)$$ where the isomorphism on the r.h.s. comes from Lemma \[l:repre\]. If $A$ is a pgg-algebra and $\Phi_{\leq t}$ is an isomorphism for $t$, the above composition is an isomorphism and, by the uniqueness of the representative of a functor (up to isomorphisms), it holds that $A\simeq S_R^\bullet(A_{\leq t})/ I_t$. For the converse, it suffices to reverse these arguments.

Note that the exactness of can be rephrased as follows: $A$ is a $t$-pgg-algebra (i.e. $\Phi_{\leq t}$ is an isomorphism) if and only if $A$ admits a set of generators of degree less than or equal to $t$ and the ideal of relations among these generators is generated by those relations of degree less than or equal to $t$. Indeed, the latter means that $$\label{eq:Itgen}
        I_t\,=\,  [I_t]_{\leq t} \cdot S_R^\bullet( [A]_{\leq t})  \,=\, 
        \big( I_t\cap \big[ S_R^\bullet( [A]_{\leq t}) \big]_{\leq t} \big)  S_R^\bullet( [A]_{\leq t})$$

Examples
--------

### Example 1: finitely presented algebras

The following proposition provide examples of this type of algebras.

\[p:finpres-pgg\] Let $R$ be a ring and let $A=\oplus_{n\geq 0} A_n$ be a graded $R$-algebra. If $A$ is of finite presentation over $R$, then $A$ is a pgg-$R$-algebra.

We have to show that there exists $t\gg 0$ such that $\Phi_{\leq t}$ is an isomorphism. First, let us determine $t$.

Since $A$ is a finitely presented graded $R$-algebra, there exist a finite set of generators $S=\{a_1,\hdots,a_l\}$ such that the kernel of the surjection $$\label{seq:gen}
    \begin{aligned}
    R[x_1,\hdots,x_l] & \rightarrow A \rightarrow 0 
    \\
    x_i \,& \mapsto a_i
    \end{aligned}$$ is a finitely generated ideal $I_S$. Noting that $a_i$ can be assumed to be homogeneous, and setting $\operatorname{deg}(x_i):= \operatorname{deg}(a_i)$, the above map is a morphism of graded $R$-algebras whose kernel is a finitely generated homogeneous ideal; let $\{y_1,\ldots, y_m\}$ be a set homogeneous generators of $I_S$. Then, we define: $$\label{e:syzy}
    \begin{gathered}
    d\,:=\,{\operatorname{max}}\{ \operatorname{deg}(a_i) \,\vert\, i=1,\ldots, l \}, \\
    d^1\,:=\,\operatorname{max}\{\operatorname{deg}(y_i) \,\vert\, i=1,\ldots, m\}, \\
    t\,:=\,  \operatorname{max}\{d,d^1\}.
    \end{gathered}$$

Next, we show that $\Phi_{\leq t}$ is an isomorphism. Let $B$ be an $R$-algebra. In order to show the injectivity of the map $\Phi_{\leq t}(B)$ of equation , consider $f,g:A\rightarrow B$ two morphisms of algebras such that $\Phi_{\leq t}(B)(f)=\Phi_{\leq t}(B)(g)$. Since $f_i =g_i$ for each $i=0,\hdots, t$ and $ t \geq d$, we have $f(a_i)=f_{d_i}(a_i)=g_{d_i}(a_i)=g(a_i)$ for each $i=1,\hdots, l$, so $f=g$.

Let us now prove the surjectivity. The set of generators $S$ allows the construction of the map . Hence, to give a morphism of algebras $f:A\rightarrow B$ is equivalent to give a morphism of algebras $\widehat{f}:R[x_1,\hdots,x_l]\rightarrow B$ such that $\widehat{f}|_{I_{S}}=0$. Consider now a morphism of $(\phi_{i})\in {\operatorname{Hom}}_{\textrm{partial $R$-alg}}(A_{\leq t},B)$ and define $b_i:=\phi_{d_i}(a_i)$ for $i=1,\hdots,l$. The elements $b_1,\hdots,b_l$ determine a morphism of $R$-algebras $\widehat{f}:R[x_1,\hdots,x_l]\rightarrow B$ by the rule $x_i \mapsto b_{i}$. It suffices to check that $\widehat{f}|_{I_{S}}=0$ or, what amounts to the same, that $\widehat{f}(y_i)=0$ for all $i=1,\ldots, m$. Since $y_i$ is an algebraic expression in the $x_j$'s, we may write $y_i=\sum \lambda_{r_1,\hdots,r_l}x_1^{r_1}\cdots x_l^{r_{l}}$ and observe that $$        \phi_{d_1}(a_1)^{r_1}\cdots \phi_{d_{l}}(a_l)^{r_{l}}         =\phi_{r_{1}d_{1}}(a_{1}^{r_{1}})\cdots \phi_{r_{l}d_{l}}(a_{l}^{r_{l}})
                =\phi_{k}(a_{1}^{r_{1}}\cdots a_{l}^{r_{l}})
        \, .
        $$ where we denote $k:=\operatorname{deg}(y_i)$ and $k \leq d^1 \leq t$.

Then, $$        \widehat{f}(y_i)         = \sum \lambda_{r_1,\hdots,r_l}b_1^{r_1}\cdots b_l^{r_{l}}
                = \sum \lambda_{r_1,\hdots,r_l}\phi_{d_1}(a_1)^{r_1}\cdots \phi_{d_{l}}(a_l)^{r_{l}}
        =\phi_{k}(y_i)= 0 \, . 
        $$ and, being $\phi_{k}$ is a morphism of $R$-modules, it follows that $\widehat{f}$ factorizes through a morphism of $R$-algebras $f:A\to B$ and that $\Phi_{\leq t}(f)=(\phi_i)$.

Indeed, in case $A$ is the algebra of polynomial invariants by the natural action of one of the classical groups, we will be able to determine $t$ more precisely thanks to the results of @char-free.

\[c:FFTSFT-SL\] Let $V$ be a vector space of finite dimension. For any natural number $m>0$, let us consider the graded algebra $$A\,:=\, \Bbbk[(V)^{\oplus m}]^{G} \,=\, \big( S_\Bbbk^\bullet ((V^{\vee})^{\oplus m})\big)^G$$ Then, the morphism $\Phi_{\leq t}$ is an isomorphism in the following cases

1.  for $G=\emph{SL}(V)$ and $t=2\dim(V)$;

2.  for $V$ a $\Bbbk$-vector space together with a symplectic form $\eta:V\times V\rightarrow \Bbbk$, $G= \operatorname{Sp}(V)$ and $t=2+\dim(V)$;

3.  for $V$ a $\Bbbk$-vector space together with a non-degenerate symmetric bilinear form $\beta:V\times V\rightarrow \Bbbk$, $G= \operatorname{SO}(V)$ and $t=2+\dim(V)$;

4.  for $V$ a $\Bbbk$-vector space together with a non-degenerate symmetric bilinear form $\beta:V\times V\rightarrow \Bbbk$, $G= \operatorname{O}(V)$ and $t=2+\dim(V)$.

In particular, $A$ is a pgg-algebra for all $m$.

$(1)$ Recall that the ring of invariants coincides with the ring of constants, i.e. $A=\Bbbk$, for $m<n:=\dim(V)$. So it remains to study the case $m\geq n$. Recalling @char-free [§3], we know that $A$ is generated by $\binom{m}{n}$ elements of degree $n$ and that the relations among them are generated by quadratic relations. Hence $d=n$ and $d^1=2n$, as it was to be proven.

$(2)$ The First Fundamental Theorem for the symplectic group $\operatorname{Sp}(V)$ asserts that the ring of invariants is generated by the following degree $2$ functions $$\langle i,j\rangle:V^{\oplus m}\rightarrow\Bbbk\text{,}\hspace{0.5cm}\langle i,j\rangle(v_{1},...,v_{m}):=\eta(v_{i},v_{j})$$ for $1\leq i< j\leq m$. Thus, letting $\dim(V)=2n$ we get that $A$ is generated by $\binom{m}{2n}$ elements of $\bigwedge^{2}V^{\oplus m}$; that is, $d=2$. The SFT for the symplectic group establishes that if $m\leq 2n$, the elements $\langle i,j\rangle$ are algebraically independent, and, on the other hand, if $m>2n$, the ideal of relations among the generators is generated by the Pfaffians of the principal minors of size $2n+2$ of the skew symmetric matrix whose $i, j$-th entry equals $\langle i,j\rangle$ (see @char-free [§4]). Therefore, ${d}^{1}=2n+2$ and $t=\operatorname{max}\{d,d^1\}=\operatorname{max}\{2,2n+2\}=2(n+1)$.

$(3)$ The First Fundamental Theorem for the special orthogonal group asserts that the ring of invariants is generated by the degree $2$ functions $$\langle i,j\rangle:V^{\oplus m}\rightarrow\Bbbk\text{,}\hspace{0.5cm}\langle i,j\rangle(v_{1},...,v_{m}):=\beta(v_{i},v_{j})$$ for $1\leq i< j\leq m$; and by the degree $n:=\dim(V)$ functions $$[i_{1},...,i_{n}]:V^{\oplus m}\rightarrow\bigwedge^{n}V\simeq\Bbbk\text{,}\hspace{0.5cm}[i_{1},...,i_{n}](v_{1},...,v_{m}):=v_{i_{1}}\wedge\cdots \wedge v_{i_{n}}$$ for $1\leq i_{1}<\cdots <i_{n}\leq m$.

By the Second Fundamental Theorem for $\operatorname{SO}(V)$, the ideal of relations among generators is generated by the $(n+1)\times (n+1)$ minors of the $m\times m$ symmetric matrix whose $i,j$-th entry equals $\langle i, j\rangle$, and, hence, $d^1=2(n+1)$ (see @char-free [§5]).

$(4)$ Analogously, for the orthogonal group $O(V)$, the ring of invariants is generated by the previously defined functions $\langle i,j \rangle$ for $1\leq i< j\leq m$, and the ideal of relations is generated by the $(n+1)\times (n+1)$ minors of the $m\times m$ symmetric matrix whose $i,j$-th entry equals $\langle i, j\rangle$ (where $n:=\dim(V)$ as above). Therefore, $d=2$ and $d^1=2(n+1)$ and $t=2(n+1)$ (see @char-free [§5]).

### Example 2: The infinite-dimensional Grassmann manifold {#ex:sato}

Recall from @Sato that the infinite-dimensional Grassmann manifold of $\Bbbk(\!(z)\!)$, denoted by $\operatorname{Gr}$, is defined as the subset of points of an infinite dimensional projective space whose homogeneous coordinates satisfy the set of all Plücker equations. Alternatively, if one introduces $\operatorname{Gr}$ using scheme theory, then the determinant bundle yields the Plücker embedding which is a closed immersion into the projective spectrum of the graded algebra $\Bbbk[\{x_S\}]$ where $S$ runs over the set of Young diagrams. From this point of view, it has been shown in @Plaza that the ideal $I\subset \Bbbk[\{x_S\}]$ defining the infinite-dimensional Grassmann manifold is generated by the set of all Plücker equations, which are quadratic relations in the variables $x_S$. Accordingly, it holds that $\mathcal{O}_{ \operatorname{Gr}} =  \Bbbk[\{x_S\}]/ I $ is a $2$-pgg-algebra.

Fundamental properties
----------------------

We study in this section some basic operations with pgg-algebras that keep the pgg property, like base changes, tensor products, quotients by ideals. We also state a semicontinuity theorem for finitely presented algebras regarding the natural number $t$ for which the algebra is pgg.

\[c:pgg-basechange\] If $A$ is a pgg-$R$-algebra and $R\to R'$ be a morphism of rings, then $A_{R'}:=A\otimes_{R} R'$ is a pgg-$R'$-algebra.

Tensoring by $R'$, we obtain the exact sequence $$I_t \otimes_R R' \to S_{R'}^\bullet(A_{\leq t}\otimes_R R') \to A\otimes_R R' \to 0$$ Setting $A':=A\otimes_R R'$ and bearing in mind that $S_R^\bullet(A_{\leq t})\otimes_R R'\simeq S_{R'}^\bullet(A_{\leq t}\otimes_R R') \simeq S_{R'}^\bullet(A'_{\leq t}) $, the above sequence yields a surjection $$\xymatrix{
     I_t \otimes_R R' 
    \ar@{->>}[r]  & K:= \operatorname{Ker}\big(S_{R'}^\bullet(A'_{\leq t}) \to A' \big)
    }$$ which shows that $K$ is generated by elements of degree $jeffery57@example.net. The claim is proven.

\[p:tensor-pgg\] Let $R$ be a ring and $A$ (resp. $B$) be a $t_A$-pgg-$R$-algebra such that is an isomorphism for $t_A$ (resp. $t_B$-pgg-$R$-algebra). Then, it holds that $A\otimes_R B$ is a $t$-pgg-$R$-algebra where $t=t_A + t_B$.

By Theorem \[t:pgg-exact\], $A$ (resp. $B$) fits into the exact sequence . Let $I_{t_A}^A$ (resp. $I_{t_B}^B$) denote its kernel. It is straightforward that we have the following commutative diagram of graded $R$-modules $$\xymatrix{
    & I_{t_A}^A\otimes B  + A\otimes  I_{t_B}^B \ar[r] \ar@{-->}[d] &
    S^\bullet_R(A_{\leq t_A}) \otimes S^\bullet_R(B_{\leq t_B}) \ar[r] \ar@{-->>}[d] &
    A\otimes B \ar[r] \ar@{=}[d] & 0 \\
    0\ar[r] & K \ar[r] & S^\bullet_R \big( [ A\otimes B]_{\leq t}) \ar[r] &
    A\otimes B \ar[r] & 0 }$$ where $t:= t_A + t_B$, $K$ is the kernel, the vertical map in the middle is the canonical multiplication induced by $A_{\leq t_A} \otimes B_{\leq t_B}\to [ A\otimes B]_{\leq t}$ and the vertical map on the l.h.s. is the one obtained by factorization. Looking at each degree of the symmetric products of the central terms and having in mind that $A$ (resp. $B$) is generated by homogeneous elements of degree less than or equal to $t_A$ (resp. $t_B$), the surjectivity of the vertical map among symmetric algebras follows. The Snake's Lemma shows that $ I_{t_A}^A\otimes B  + A\otimes  I_{t_B}^B\to K$ is surjective. Noting that the image of $$I_{t_A}^A\otimes B  + A\otimes  I_{t_B}^B \,\to\, S^\bullet_R \big( [A\otimes B]_{\leq t})$$ is generated by homogeneous elements of degree less than or equal to $t$, the claim is proved.

\[p:epi-pgg\] Let $R$ be a ring and $A$ be a graded $R$-algebra. Let $J\subset A$ be a homogenous ideal generated by its components of degree less than or equal to $t_J$. If $A$ be a $t_A$-pgg-$R$-algebra, then $A/J$ is a $t$-pgg-$R$-algebra for $t:= \operatorname{max}(t_A,t_J)$.

Set $t:=\operatorname{max}(t_A,t_J)$ and $B:=A/J$. Using , we may build the following commutative diagram of $R$-modules $$\xymatrix{
    0 \ar[r] &  I_{t}^A \ar[r] \ar@{-->}[d] &  S_R^\bullet(A_{\leq t})  \ar[r] \ar@{-->>}[d] & A \ar[r] \ar@{->>}[d] & 0 
    \\
    0 \ar[r] &  K \ar[r]  &  S_R^\bullet(B_{\leq t})  \ar[r]^-p & B \ar@{-->}[r] & 0 }$$ where $K$ is the kernel of $p:S_R^\bullet(B_{\leq t}) \to B$. Since $A\to B$ is a surjective map of graded $R$-modules, it follows that the dashed vertical arrow in the middle is surjective and $p$ is surjective too.

It remains to check that $K$ is generated by its components of degree less than or equal to $t$. The Snake's Lemma yields $$I_{t}^A \to K \to L \to 0$$ where $L$ is a quotient of $J$. But, noting that $I_{t}^A$ is generated as $S_R^\bullet(A_{\leq t})$-module by its components of degree less than or equal to $t$, that $J$ is generated as $A$-module by its components of degree less than or equal to $t$, it then follows that $K$ is generated as $S_R^\bullet(B_{\leq t})$-module by its components of degree less than or equal to $t$.

\[p:point-global-pgg\] Let $R$ be a ring and $A$ be a graded $R$-algebra of finite presentation. Then, the function $$\begin{aligned}
    \widehat{t} \colon S=\emph{Spec} R & \longrightarrow\, {\mathbb{Z}}\\
    x & \mapsto \, \widehat{t}(x):= \operatorname{min}\{\, t  \text{ s.t. $\Phi_{\leq t}$ is an isomorphism for $A\otimes_{R} \Bbbk(x)$\,}\}
    \end{aligned}$$ is upper semicontinuous.

Let $x\in S$ and set $t:=\widehat{t}(x)$. Consider the exact sequence $$\ker (\mu) \overset{\iota}\longrightarrow S_R^{\bullet}(A_{\leq t}) \overset{\mu} \longrightarrow A$$ For $y\in S$, let $\mu(y)$ denote $$S_R^{\bullet}(A_{\leq t})\otimes_{R}\Bbbk(y) 
    = S_{\Bbbk(y)}^{\bullet}( [ A\otimes_{R}\Bbbk(y) ]_{\leq t})\,
    \overset{\mu(y)}\longrightarrow \,A\otimes_{R}\Bbbk(y) \,
    \longrightarrow \, 0$$ and, for $i\in {\mathbb{Z}}$, set $$[\mu(y)]_i : [S_R^{\bullet}(A_{\leq t})\otimes_{R}\Bbbk(y) ]_i \, 
    \longrightarrow \,[A\otimes_{R}\Bbbk(y)]_i$$ Since $[\mu]_i$ is a morphism between finite type $R$-modules, it follows that $$V_i:=\{y \in S \,\vert\,\text{ $[\mu(y)]_i$ is surjective}\}$$ is Zariski open in $S$. Since $A$ is generated by elements of degree less than or equal to $t$, it is easy to check that $$U\,:=\, \bigcap_{i=1}^t V_i \,=\, \bigcap_{i\geq 1} V_i \,=\,  \{y \in S \,\vert\, \text{ $\mu(y)$ is surjective}\}$$ and, thus, $U$ an open neighborhood of $x$.

Next, we will show that there is an open subset $U'\subseteq U$ such that $\widehat{t}(y)\leq t$ for all $y\in U'$. Being $A$ an $R$-algebra of finite presentation, it follows that $[A]_i$ is an $R$-module of finite presentation and $[S_R^{\bullet}(A_{\leq t})]_i$ is a finite type $R$-module. Then, by @stacks-project [[Lemma 01BP](https:/\!/stacks.math.columbia.edu/tag/01BP)], $[\ker (\mu)]_i$ is a finite type $R_U$-module. Setting $$U'\,:=\,
    \{ y \in U \,\vert \, \mu_{\leq t}: [\ker (\mu)]_{\leq t} \to [I_t]_{\leq t} \text{ is surjective}\}$$ it follows that the function $\widehat{t}$ is upper semi-continuos.

Projective cones and embeddings
-------------------------------

When a graded $R$-algebra $A=\oplus_{i}A_{i}$ is generated by elements of degree one, one can embed $\textrm{Spec}(A)$ in the projective bundle $\mathbb{P}(R\oplus A_{1})$ through an open immersion followed by a closed immersion (see @ega2). We will state an analogous result for pgg-algebras. Let us recall two constructions which will inspire our proof. The first one taken from @ega2 [Chp II, §8] goes as follows. For an arbitrary graded algebra $A=\oplus_{n\geq 0} A_n$, let us consider an indeterminate $T$ with $\deg T=1$. Then, the morphism of $A$-algebras $$A[T] \,:=\, A \otimes_R R[T]    \,\longrightarrow\, A$$ sending $T$ to $1$ factorizes through the homogeneous localization of $A[T]$ with respect to the multiplicative system generated by $T$, which will be denoted by $A[T]_T^+$; that is, $$A[T] \,\to  \, A[T]_T^+   \,\overset{\sim}\to\, A$$ where the first map is a localization and the second one maps $T\mapsto 1$. The latter is an isomorphism whose inverse sends an homogenous element $a \in A_n$ to $\frac{a}{T^n}$. Accordingly, we obtain an open embedding $${\operatorname{Spec}}A \, \hookrightarrow\, {\operatorname{Proj}}A[T]$$ whose image is dense.

For the second one, based in the $d$-uple or Veronese embedding, we recall from @ega2 [Proposition 3.1.8] that for any graded algebra $A=\oplus_{n\geq 0} A_n$ and any $d>0$, there is a canonical isomorphism ${\operatorname{Proj}}A \simeq {\operatorname{Proj}}A^{(d)}$ where $$A^{(d)} \,:=\, \oplus_{n\geq 0} [ A^{(d)}] _n
    \qquad \text{ where }
     [ A^{(d)}] _n\,:=\, [A]_{nd}\, .$$

\[t:spec-proj\] Let $A$ be a $t_A$-pgg-algebra. Let $d > t_A$ and $t:=t_A+1$. Then, the morphism ${\operatorname{Spec}}A\to{\operatorname{Spec}}R$ factorizes canonically by a locally closed embedding of $R$-schemes $$\Phi:{\operatorname{Spec}}A \,\hookrightarrow\, 
    {\mathbb P} ( E_d ),$$ where $E_d := \underset{1\leq \vert \underline{d}\vert \leq d }\oplus  S^{d_1}A_1 \otimes_R \dots \otimes_R  S^{d_t}A_t$, for certain natural number $d$. The morphism $\Phi$ satisfies two properties: (1) the closure of the image is isomorphic to ${\operatorname{Proj}}(A\otimes_R R[T])$, and (2) there is a natural action of $\mathcal{G}:= \emph{Aut}_{\emph{graded-}\Bbbk\emph{-alg}}(A)$ on ${\mathbb P} ( E_d )$, and $\Phi$ is $\mathcal{G}$-equivariant.

\[rmk:spec-proj\] Note that when $A$ is finitely presented, the module $E_d$ is finitely presented as well, so $\mathbb{P} ( E_d )$ is a projective bundle in the usual sense. Note also that any subgroup $G\subset \mathcal{G}$ satisfies property (2).

Recall that $B:=R[T]$ with $\operatorname{deg}(T)=1$ is a pgg-algebra and that, further, $\Phi_{\leq t_B}$ is an isomorphism for $t_B=1$, $I^B_{t_B}=(0)$ and $S_R^\bullet(B_{\leq t_B})= S_R^\bullet(R[T]_{\leq 1})= S_R^\bullet(\langle T\rangle)= R[T]$ (see ).

For $A$ as in the statement, Proposition \[p:tensor-pgg\] implies that $\bar A=A \otimes_R B =A \otimes_R R[T] $ is also a pgg-algebra. Further, The proof of the Proposition \[p:tensor-pgg\] yields the exact sequence $$0 \to I^A_{t_A} \otimes_R R[T] \to 
    S_R^\bullet(A_{\leq t_A}) \otimes_R R[T] 
     \to A\otimes_R R[T] \to 0$$ Letting $\bar I:=  I^A_{t_A} \otimes_R R[T]$, the above sequence and Theorem \[t:pgg-exact\] show that $\bar A$ is a $t$-pgg-algebra for $t:= t_{\bar A}=t_A+t_B=t_A+1$. Furthermore, there is an isomorphism

$$\bar A^{(d)}  \, \simeq \, 
    \big(S_R^\bullet(\bar A_{\leq t})\big)^{(d)} / \bar  I^{(d)}  ,$$ which composed with the canonical surjection $$S_R^\bullet \big( [ S_R^\bullet  \bar A_{\leq t}]_d) \twoheadrightarrow S_R^\bullet \big( [ S_R^\bullet  \bar A_{\leq t}]_d / [\bar I]_d\big)  \twoheadrightarrow \big(S_R^\bullet(\bar A_{\leq t})\big)^{(d)} / \bar  I^{(d)}$$ gives a surjection $$S_R^\bullet \big( [ S_R^\bullet  \bar A_{\leq t}]_d) \rightarrow \bar A^{(d)}\rightarrow 0.$$ Having in mind that $\operatorname{deg}(T)=1$ and equations  and , it holds that $$[ S_R^\bullet  \bar A_{\leq t}]_d
    \,=\,
    \underset{ \vert \underline{d}\vert = d }\oplus  S^{d_1}\bar A_1 \otimes_R \dots \otimes_R  S^{d_t}\bar A_t$$ Furthermore, the canonical map of $R$-modules $[ S_R^\bullet  \bar A_{\leq t}]_d\to \bar A$ factorizes as follows $$[ S_R^\bullet  \bar A_{\leq t}]_d
    \, \twoheadrightarrow\, 
         \underset{0\leq  \vert \underline{d}\vert \leq d }\oplus  S^{d_1}A_1 \otimes_R \dots \otimes_R  S^{d_t}A_t \otimes_R \langle T^{d- \vert \underline{d}\vert}\rangle
         \,\to \, \bar A$$ Now, the map of graded $R$-algebras given by the multiplication of $A$ and $T\mapsto 1$ $$\label{CA}
     C\,:=\, S_R^\bullet \Big(  
         \underset{1\leq  \vert \underline{d}\vert \leq d }\oplus  S^{d_1}A_1 \otimes_R \dots \otimes_R  S^{d_t}A_t \otimes_R \langle T^{d- \vert \underline{d}\vert}\rangle
         \Big) 
         \,\longrightarrow \, A$$ is surjective and factorizes by the homogeneous localization by the multiplicative system $M$ generated by all elements of the form $1\otimes\ldots\otimes 1\otimes T^i \otimes 1 \ldots \otimes 1,$ that is, $C_M^+$. Hence, considering the spectrum and the projective spectrum, it holds that the composition $${\operatorname{Spec}}A \, \hookrightarrow\, {\operatorname{Spec}}C^+_M \, \hookrightarrow\, {\operatorname{Proj}}C \,\simeq \,  
    {\mathbb{P}}\big( \underset{1\leq  \vert \underline{d}\vert \leq d }\oplus  S^{d_1}A_1 \otimes_R \dots \otimes_R  S^{d_t}A_t\big)$$ is the desired map since the first one is a closed immersion and the second one is an open embedding. From the construction, it is clear that the closure of ${\operatorname{Spec}}A$ by the previous map is ${\operatorname{Proj}}\bar A$. The property (1) follows from the construction of $\Phi$. Regarding property (2), note that there is a natural action $\mathcal{G}\rightarrow \textrm{Aut}_{R\textrm{-mod}}(A_i)$ on each homogeneous component defined by $g\in\mathcal{G}\mapsto g|_{A_i}\in \textrm{Aut}_{R\textrm{-mod}}(A_i)$. This action induces an action of $\mathcal{G}$ on $\mathbb{P}(E_d)$ in a natural way. To prove that $\Phi$ is $\mathcal{G}$-equivariant is enough to show that (\[CA\]) is (note that $\mathcal{G}$ acts on the variable $T$ trivially). By the very definition of the action of $\mathcal{G}$ on $C$, proving the previous statement is reduced to proving that $A_i \hookrightarrow A$ is $\mathcal{G}$-equivariant, which is trivial.

Partial finiteness of the algebra of invariants for classical groups and projective embedding
---------------------------------------------------------------------------------------------

The proof of Theorem \[t:main-fin-pres\] (1) rests on two facts. One on hand, every $T$-module is the inductive limit of finitely presented modules, and, on the other hand, the inductive limit of $t$-pgg algebras is again $t$-pgg. In order to use this last fact, we have to show first that the algebra of invariants, $N$ being finitely presented, is $t$-pgg for a given natural number $t$. In the next lemma, we prove that this is the case for classical groups.

\[lm:classical\] Let $R$ be a $\Bbbk$-algebra, $T$ an $R$-algebra, $M$ a finitely generated free $R$-module of rank $m$ and $N$ a $T$-module of finite presentation. Let $G$ be one the of the classical (semisimple) linear algebraic $\Bbbk$-groups ($\emph{SL}_n$, $\emph{Sp}_{2n}$, $\emph{SO}_n$) together with a faithful representation $\rho:G_R^{\bullet}\hookrightarrow \underline{\emph{Aut}}_{R\textrm{-mod}}(M)$. Then, there exists a natural number $t\in\mathbb{N}$ that may depend on $\rho$ but not on $N$ such that $A:=  \big( S_{T}^\bullet (M^{\vee}\otimes_{R}N)\big)^{G_T}$ is a $t$-pgg $T$-algebra.

From Corollary \[cor:finitely-presented\] and Proposition \[p:finpres-pgg\], it follows that $A$ is $t$-pgg for some $t\in\mathbb{N}$. Now, the result follows from Corollary \[c:FFTSFT-SL\] and Proposition \[p:point-global-pgg\].

\[Theorem \[t:main-fin-pres\] (1)\]\[th:final\] Let $R$ be a $\Bbbk$-algebra, $T$ an $R$-algebra $M$, a finitely generated free $R$-module of rank $m$ and $N$ a $T$-module. Let $G$ be one the of the classical (semisimple) linear algebraic $\Bbbk$-groups ($\emph{SL}_n$, $\emph{Sp}_{2n}$, $\emph{SO}_n$) together with a faithful representation $\rho:G_R^{\bullet}\hookrightarrow \underline{\emph{Aut}}_{R\textrm{-mod}}(M)$. Then, there exists a natural number $t\in\mathbb{N}$ that may depend on $\rho$ but not on $N$ such that $A:=  \big( S_{T}^\bullet (M^{\vee}\otimes_{R}N)\big)^{G_T}$ is a $t$-pgg $T$-algebra.

Every $T$-module $N$ is a directed limit of a directed system of finitely presented $T$-modules $(N_{\lambda},\theta_{\lambda\mu})$, $N=\underset{\longrightarrow}{\lim}\  N_{\lambda}$ @shannon [Lemma 2]. Now, for each $\lambda$, $\rho:G_{R}^{\bullet}\hookrightarrow \underline{\textrm{Aut}}_{R\textrm{-mod}}(M)$ induces a representation $\rho_{\lambda}:G_{T}^{\bullet}\hookrightarrow \underline{\textrm{Aut}}_{R\textrm{-mod}}(M^{\vee}\otimes_{R}N_{\lambda})$. Consider the algebra of invariants $\big( S_{T}^\bullet (M^{\vee}\otimes_{R}N_{\lambda})\big)^{G_{T}}$. Let us prove that $\underset{\longrightarrow}{\lim} \ \big( S_{T}^\bullet (M^{\vee}\otimes_{R}N_{\lambda})\big)^{G_{T}}\simeq \big( S_{T}^\bullet (M^{\vee}\otimes_{R}N)\big)^{G_{T}}$. In order to do so, we have to demonstrate that the functor $\underset{\longrightarrow}{\lim}(-)$ conmutes with $S^{\bullet}(-)$ and $(-)^{G}$.

\(1) The directed system $(N_{\lambda},\theta_{\lambda\mu})$ induces a directed system of algebras $(S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\lambda}), u_{\lambda\mu}:S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\lambda})\rightarrow S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_\mu))$. Furthermore, the morphisms $\theta_\lambda: N_{\lambda}\rightarrow N$ determined by $(N_{\lambda},\theta_{\lambda\mu})$ induce morphisms $u_{\lambda}:S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\lambda})\rightarrow S^{\bullet}_{T}(M^{\vee}\otimes_{R}N)$ which are compatible with $u_{\lambda\mu}$. Therefore, we have a morphism $$\underset{\longrightarrow}{\lim} \  S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\lambda})\rightarrow S^{\bullet}_{T}(M^{\vee}\otimes_{R}N).$$ On the other hand, note that there is a canonical injective morphism $M^{\vee}\otimes_{R}N\hookrightarrow \underset{\longrightarrow}{\lim} \ S_{T}^\bullet (M^{\vee}\otimes_{R}N_{\lambda})$. By the universal property of the symmetric algebra, this induces a morphism of algebras $$S_{T}^\bullet (M^{\vee}\otimes_{R}N)\rightarrow \underset{\longrightarrow}{\lim} \ S_{T}^\bullet (M^{\vee}\otimes_{R}N_{\lambda}).$$ It is clear that both morphisms are inverses to each other, so we have an isomorphism $$\underset{\longrightarrow}{\lim} \  S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\lambda})\simeq S^{\bullet}_{T}(M^{\vee}\otimes_{R}N).$$

\(2) It is easy to see that $u_{\lambda\mu}: S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\lambda})\rightarrow S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_\mu)$ is $G_{T}$-equivariant for each pair $\lambda,\mu$. Therefore, the restriction to $(S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\lambda}))^{G_{T}}$ takes values in $(S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\mu}))^{G_{T}}$. We denote by $v_{\lambda\mu}$ the restriction of $u_{\lambda\mu}$ to the algebra of invariants. Thus, we have a directed system $((S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\lambda}))^{G_{T}},v_{\lambda\mu})$. The same argument as before shows that there is a morphism $$v_{\lambda}: (S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\lambda}))^{G_{T}} \rightarrow (S^{\bullet}_{T}(M^{\vee}\otimes_{R}N))^{G_{T}}$$ for each $\lambda$. Clearly, these morphisms are compatible with $v_{\lambda\mu}$. Thus, we have a morphism of $T$-algebras $$v: \underset{\longrightarrow}{\lim} \  (S^{\bullet}_{T}(M^{\vee}\otimes_{R}N_{\lambda}))^{G_{T}} \rightarrow (S^{\bullet}_{T}(M^{\vee}\otimes_{R}N))^{G_{T}}.$$ Consider now the composition of the canonical isomorphism of $T$-modules $M^{\vee}\otimes_{R}N\simeq \underset{\longrightarrow}{\lim} \ (M^{\vee}\otimes_{R}N_{\lambda})$ with the directed limit of the Reynolds operators $\underset{\longrightarrow}{\lim} \ (M^{\vee}\otimes_{R}N_{\lambda})\rightarrow \underset{\longrightarrow}{\lim} \ (M^{\vee}\otimes_{R}N_{\lambda})^{G_{T}}$. Since there is a canonical injective morphism $\underset{\longrightarrow}{\lim} \ (M^{\vee}\otimes_{R}N_{\lambda})^{G_{T}}\hookrightarrow \underset{\longrightarrow}{\lim} \ (S_{T}^{\bullet}(M^{\vee}\otimes_{R}N_{\lambda}))^{G_{T}}$, we have a morphism of $T$-modules $$M^{\vee}\otimes_{R}N \rightarrow \underset{\longrightarrow}{\lim} \ (S_{T}^{\bullet}(M^{\vee}\otimes_{R}N_{\lambda}))^{G_{T}}$$ and, therefore, a morphism of $T$-algebras $$S_{T}^{\bullet}(M^{\vee}\otimes_{R}N )\rightarrow \underset{\longrightarrow}{\lim} \ (S_{T}^{\bullet}(M^{\vee}\otimes_{R}N_{\lambda}))^{G_{T}}.$$ The restriction to the subalgebra of invariants gives a morphism $$v':(S_{T}^{\bullet}(M^{\vee}\otimes_{R}N ))^{G_{T}}\rightarrow \underset{\longrightarrow}{\lim} \ (S_{T}^{\bullet}(M^{\vee}\otimes_{R}N_{\lambda}))^{G_{T}}.$$ It is easy to see that $v$ and $v'$ are inverses to each other.

Finally, due to Lemma \[lm:classical\], it remains to show that the directed limit of $t$-pgg algebras is again $t$-pgg. Let us denote by $A_{\lambda}$ the $T$-algebra of $G$-invariants $(S_{T}^{\bullet}(M^{\vee}\otimes_{R}N_{\lambda}))^{G_{T}}$. Since $\underset{\longrightarrow}{\lim} \ (-)$ is an exact functor, we have an exact sequence $$\xymatrix{
0 \ar[r] & \underset{\longrightarrow}{\lim} \  I_{\lambda,t}\ar[r] & \underset{\longrightarrow}{\lim} \  S_{T}^{\bullet}(A_{\lambda,\leq t}) \ar[r] & \underset{\longrightarrow}{\lim} \  A_{\lambda} \ar[r] & 0.
}$$ By $(2)$, $\underset{\longrightarrow}{\lim} \  A_{\lambda} \simeq A$ and by $(1)$, $\underset{\longrightarrow}{\lim} \  S_{T}^{\bullet}(A_{\lambda,\leq t}) \simeq S_{T}^{\bullet}(A_{\leq t})$. Finally, it is clear that $\underset{\longrightarrow}{\lim} \  I_{\lambda,t}$ is isomorphic to an ideal of $S_{T}^{\bullet}(A_{\leq t})$ generated by elements of degree less than or equal to $t$, so the result follows from Theorem \[t:pgg-exact\].

\[Corollary \[lm:cor-chevalley\]\] Under the hypotheses of Theorem \[t:main-fin-pres\], if either $N$ is finitely presented or $G$ is one of the classical (semisimple) linear algebraic $\Bbbk$-groups ($\emph{SL}_n$, $\emph{Sp}_n$, $\emph{SO}_n$), there is a natural locally closed embedding, $$\Phi_{M,N,G}:\underline{\emph{Hom}}_{T}(M,N) \sslash G\hookrightarrow {\mathbb P} ( E_d ),$$ where $E_d:= \underset{1\leq \vert \underline{d}\vert \leq d }\oplus  S^{d_1}A_1 \otimes_R \dots \otimes_R  S^{d_t}A_t$ and $A_{i}=(S^{i}(M^{\vee}\otimes_{R}N))^{G_{T}}$, for a given natural number $d$. Furthermore, (1) the natural number $d$ depends on $G$ but not on $N$ if $G$ is classical, and (2) $\Phi_{M,N,G}$ is $\emph{Aut}_{T}(N)$-equivariant in any case.

If $N$ is finitely presented, the result follows from Corollary \[cor:finitely-presented\], Proposition \[p:finpres-pgg\] and Theorem \[t:spec-proj\]. If $G$ is classical, we conclude by Theorem \[th:final\] and Theorem \[t:spec-proj\]. To prove (2), note that the action of $\textrm{Aut}_{T}(N)$ on $N$ induces an action $$\alpha:\textrm{Aut}_{T}(N)\hookrightarrow \textrm{Aut}_{\textrm{graded-}R\textrm{-alg}}(S_{T}^\bullet (M^{\vee}\otimes_{R}N)).$$ In fact, the induced action of $\textrm{Aut}_{T}(N)$ on $S^i (M^{\vee}\otimes_{R}N)$ is compatible with the action of $G_{T}$ (see Lemma \[l:lemaprevio\]). Therefore, $\alpha$ induces another action, $$\alpha^{G}:\textrm{Aut}_{T}(N)\hookrightarrow \textrm{Aut}_{\textrm{graded-}R\textrm{-alg}}(\big(S_{T}^\bullet (M^{\vee}\otimes_{R}N)\big)^{G_{T}}).$$ Now, the result follows from Theorem \[t:spec-proj\] and Remark \[rmk:spec-proj\].

In case $G$ is not classical, we know by Lemma \[l:fieldcase\] and Proposition \[p:point-global-pgg\] that there is a bound of the degree of the generators of $\big( S_{T}^\bullet (M^{\vee}\otimes_{R}N)\big)^{G_T}$ not depending on $N$ (provided its rank is sufficiently high). However, to conclude that $A$ is pgg we still need to know that the degrees of the generators of the ideal of relations are also bounded by a natural number not depending on $N$. Regarding this problem for $G$ not classical, not much is known, apart from the case $G$ finite @DERKSEN.

Comments
========

It is worth making some comments on the results and the techniques used in this work.

First, we have assumed that the base ring $R$ is a $\Bbbk$-algebra, $\Bbbk$ being a field of characteristic zero. As we mentioned at the beginning of Section \[sec:part1\], we impose this condition for the base change of the invariants of the symmetric algebra to hold. Although the base change property for Schur modules is satisfied independently on the characteristic of the base ring, the Standard Basis Theorem holds only (to the best of the authors' knowledge) when the base ring is a $\Bbbk$-algebra of characteristic zero. It would be of interest knowing to what extent this result can be generalized.

Second, we have chosen as a representation of $G$ an $R$-module $M$ which is free. The reason for imposing this property is to introduce the problem analogously as is done in classical invariant theory. A natural problem now is to let $M$ vary; that is, to study the algebra of invariants $(S^{\bullet}_{T}(M\otimes_{R}N))^{G_{T}}$, $M$ being a more general $G$-$R$-module. Note that when working with the non-dual formalism, it might be convenient to start from a power-reductive $R$-group scheme @Franjou [@power-red]. On the other hand, letting $M$ be a directed limit of free modules, $G$ a directed limit of group schemes, and $(\rho_{\lambda})_{\lambda\in I}$ a directed system of representations would allow extending the results presented in @fioresi, concerning an infinite-dimensional version of the FFT. Observe that, in that case, the resulting algebra of invariants is precisely the algebra of functions of the infinite-dimensional Grassmannian, which we already know is $2$-pgg (see Example \[ex:sato\]).

Additionally, Theorem \[t:main-affine\] has important consequences in certain moduli problems. Let $\mathcal{U}$ be a $\Bbbk$-scheme (finite-dimensional or not), $X$ a projective curve over $\Bbbk$, $\mathcal{E}$ a locally free sheaf of rank $n$ on $\mathcal{U}_{X}:=\mathcal{U}\times X$, $G$ a semisimple linear algebraic $\Bbbk$-group $G$, and $\rho: G\hookrightarrow \textrm{SL}(V)$ a faithful representation of dimension $n$. Then, Theorem \[t:main-affine\] says that the algebra of invariants $(S^{\bullet}_{\mathcal{O}_{\mathcal{U}_X}}(\mathcal{M}^{\vee}\otimes_{\mathcal{O}_{\mathcal{U}_X}}\mathcal{E}))^{G}$ (where $\mathcal{M}:=V^{\vee}\otimes_{\Bbbk}\mathcal{O}_{\mathcal{U}_X}$) is locally of finite presentation. Note that this is important for ensuring the existence of the scheme of sections, $\textrm{Sec}\rightarrow \mathcal{U}$, of $\underline{\textrm{Hom}}_{\mathcal{U}_{X}}(\mathcal{M},\mathcal{E}) \sslash G \rightarrow \mathcal{U}_{X}$ @ry in case $\mathcal{U}$ is not noetherian. If $\mathcal{U}$ is the moduli space of locally free sheaves on $X$ with formal trivializations (see @alvarez [@mulase; @plaza-kp]) and $\mathcal{E}$ is the universal locally free sheaf, $\textrm{Sec}\rightarrow \mathcal{U}$ is very closely related with the moduli space of principal $G$-bundles on $X$ with formal trivializations (see @bhosle [@langer; @munoz1; @schmitt] for the formalism of singular principal bundles). Although the functor defined by pairs given by a principal $G$-bundle and a formal trivialization is known to be representable by an infinite-dimensional scheme @Frenkel [Proposition 4.1.5], the mentioned approach would lead to a global description of it. Indeed, such description would allow to determine a system of infinitely many differential equations describing the moduli space inside certain infinite-dimensional Grassmannian. We plan to work this problem out in a forthcoming paper.

Finally, Corollary \[lm:cor-chevalley\] may be useful to study lifting properties of principal $G$-bundles over curves (see @Rama for a detailed exposition when the curve is the projective line) if a faithful representation $\rho:G\hookrightarrow \textrm{SL}(V)$ is given. More precisely, let $S=\textrm{Spec}(R)$ be the spectrum of a $\Bbbk$-discrete valuation ring, $\pi:\mathcal{X}\rightarrow S$ a flat family of (polarized) stable curves of genus $g$, the generic fiber $\mathcal{X}_{\eta}$ being smooth, and $\textrm{SPB}(\rho)^{ss}$ the moduli space of semistable singular principal $G$-bundles on $\mathcal{X}$ @munoz1. One can ask for those singular principal $G$-bundles $(\mathcal{F}_0,\tau_0)\in \textrm{SPB}(\rho)^{ss}_{0}$ that arise as deformations of semistable principal $G$-bundles on $\mathcal{X}_{\eta}$. Since $\tau_0$ belongs to the scheme of sections, $\textrm{Sec}_0$, of $\underline{\textrm{Hom}}_{\mathcal{X}_0}(\mathcal{M}_0,\mathcal{F}_0) \sslash G \rightarrow \mathcal{X}_0$ (here $\mathcal{M}:=V^{\vee}\otimes_{\Bbbk}\mathcal{O}_{\mathcal{X}_0}$), the locally closed embedding (\[eq:compact\]) allows to translate the lifting problem of pairs $(\mathcal{F}_0,\tau_0)$ into a lifting problem of sections of certain projective bundle. Note that, in this situation, the embedding (\[eq:compact\]) is globally well-defined. Understanding these lifting properties would allow to understand some geometric properties of certain compactifications of the universal moduli space of principal $G$-bundles over $\overline{\mathcal{M}_g} $@munoz2 [@munoz3].

Schur modules
=============

We begin recalling from @akin and @Procesi [Chp. 9, §2] the definition, as well as some properties, for Schur modules.

For the definition of Schur modules, we consider a commutative ring $R$, a free $R$-module $M$, a natural number $p\in\mathbb{N}$ and a partition $\lambda=(\lambda_1,\hdots,\lambda_q)$ of $p$; that is, $\lambda_1\geq \lambda_2 \geq \dots \geq \lambda_q >0$ with $\vert \lambda\vert:=\sum_{i=1}^q \lambda_i=p$. Let us denote by $\mu=(\mu_{1},\dots,\mu_{t})$ the transpose partition. The natural embeddings of $R$-modules $a_{\mu_{s}}:\bigwedge^{\mu_{s}} M \hookrightarrow \operatorname{T}^{\mu_{s}}M$ yield $$a_{\mu}=\otimes_{s}a_{\mu_{s}} \,:\, 
    \bigwedge^{\mu_{1}}M \otimes \hdots \otimes \bigwedge^{\mu_{t}}M
    \,\hookrightarrow \, 
    \operatorname{T}^{\mu_{1}}M \otimes \hdots \otimes \operatorname{T}^{\mu_{t}}M
    = \operatorname{T}^{p}M\text{.}$$

One also has the following morphism of $R$-modules $$s_{\lambda} \,:\, \operatorname{T}^{p}M = \operatorname{T}^{\lambda_{1}}M \otimes \hdots \otimes \operatorname{T}^{\lambda_q}M 
    \,\longrightarrow \, {\operatorname{S}}^{\lambda_{1}}M\otimes \hdots \otimes {\operatorname{S}}^{\lambda_{q}}M$$

The Schur module (or Weil module) associated to the data $(M,\lambda)$ is defined as the image of $e_{\lambda}:=s_{\lambda}\circ a_{\mu}$, and it will be denoted by $\mathbb{S}^{\lambda}_{R}M$.

By convention, the unique partition of $p=0$ will be denoted by $(0)$ and we define $\mathbb{S}^{(0)}_{R}R=R $.

Schur modules satisfy the following properties:

1.  If $\lambda$ is the partition consisting of a single row of length $p$, $\lambda=(1,\overset{p}\ldots, 1)$, then $\mathbb{S}^{\lambda}_{R}M=\bigwedge^{p}M$;

2.  if $\lambda=(p)$, then $\mathbb{S}^{\lambda}_{R}M={\operatorname{S}}^{p}M$; and, in particular, $\mathbb{S}^{(p)}_{R}R=R$;

3.  if $M$ is a free $R$-module of rank $m$, then $\mathbb{S}^{\lambda}_{R}M=0$ for every $\lambda$ with height greater than $m$, ${\operatorname{ht}}(\lambda)>m$;

4.  $\mathbb{S}^{\lambda}_{R}R=0$ for all $\lambda\neq (0),  (p)$.

5.  The Schur module $\mathbb{S}^{\lambda}_{R}M$ can be defined for an arbitrary $R$-module;

6.  $\mathbb{S}^{\lambda}_{R}M$ is compatible with base change of the ring $R$;

7.  $\mathbb{S}^{\lambda}_{R}M$ is functorial in $M$;

8.  $\mathbb{S}^{\lambda}_{R}M$ is free if $M$ is free.

This properties are well-known. See @akin and @wey for instance.

For the sake of clarity, when the base ring is fixed and no confusion arises, we just write $\mathbb{S}^{\lambda}M$. From now on, we will restrict to the case where base rings are commutative algebras over a field of characteristic zero, $\Bbbk$. Let us recall the following generalization of the Cauchy's formula, which is fundamental in our approach.

\[The Standard Basis Theorem, <span>@akin [Theorem III.1.4]</span>\]\[decomposition\] Let $R$ be a commutative ring of characteristic zero and $M,N$ finitely generated free modules. Then, there is an isomorphism. $$\label{e:sbt}
    {\operatorname{S}}^{k}(M\otimes_{R} N)\simeq \bigoplus_{|\lambda|=k}\left(\mathbb{S}^{\lambda}M\otimes_{R} \mathbb{S}^{\lambda}N\right).$$ of $\underline{\emph{Aut}}_R(M)\times \underline{\emph{Aut}}_R(N)$ modules.

It is worth pointing out that, since $\mathbb{S}^{\lambda}_{R}M=0$ for ${\operatorname{ht}}(\lambda)>\operatorname{rk}M$, the direct sum in  runs over those $\lambda$ with ${\operatorname{ht}}(\lambda)\leq \operatorname{min}(m,n)$. This remark also applies in forthcoming statements.

\[l:basechangeschur\] Let $R$ and $R'$ be two commutative $\Bbbk$-algebras of characteristic zero, $R\rightarrow R'$ be a morphism of $\Bbbk$-algebras, $M$ be a finitely generated free $R$-module. Further, let $G$ be a semisimple linear algebraic $\Bbbk$-group together with $\rho:G_{R}^{\bullet}\hookrightarrow \underline{\emph{Aut}}_R(M)$ a faithful representation. Then, there is an isomorphism: $$\big( \mathbb{S}^{\lambda}_{R'}( M\otimes_{R}R')\big)^{G_{R'}}\simeq 
    (\mathbb{S}^{\lambda}_{R} M)^{G_{R}}\otimes_{R}R'\text{.}$$

Let $\mathcal{R}$ denote the Reynolds operator corresponding to the action of $G_R$ on $\mathbb{S}^{\lambda}_{R}M$. Recall that $\mathcal{R}$, which is unique since $G$ is reductive, can be defined as a retraction of the natural inclusion $i:(\mathbb{S}^{\lambda}_{R}M)^{G_R}\hookrightarrow\mathbb{S}^{\lambda}_{R}M.$ Note that: $$i\circ\mathcal{R}:\mathbb{S}^{\lambda}_{R}M\rightarrow\mathbb{S}^{\lambda}_{R}M$$ is an $G_R$-equivariant map of $R$-modules. Recall also that the Reynolds operators are compatible with base changes; in particular, $\mathcal{R}\otimes 1$ is the Reynolds operator of $(\mathbb{S}^{\lambda}_{R}M)\otimes_{R}R'$.\
Bearing in mind that taking invariants is exact since $G$ is reductive, and that Schur modules are compatible with base changes, it follows that: $$\big(\mathbb{S}^{\lambda}_{R'}(M\otimes_{R}R')\big)^{G_{R'}}=\text{Im}((i\otimes 1)\circ(\mathcal{R}\otimes1))\simeq\text{Im}(i\circ \mathcal{R})\otimes 1=(\mathbb{S}^{\lambda}_{R}M)^{G_{R}}\otimes_{R}R'\text{.}$$

\[l:lemaprevio\] Let $R$ and $R'$ be two commutative $\Bbbk$-algebras of characteristic zero, $R\rightarrow R'$ be a morphism of $\Bbbk$-algebras, $M$ be a finitely generated free $R$-module, and assume, further, that $N$ is a finitely generated free $R'$-module. Then, for each $k\in\mathbb{N}$, (\[e:sbt\]) induces an isomorphism: $$\label{e:lemaprevio}
        \big({\operatorname{S}}^{k}_{R'}(M\otimes_{R}N)\big)^{G_{R'}}\simeq \bigoplus_{\vert\lambda\vert=k}\big(\mathbb{S}^{\lambda}_{R}M\big)^{G_R}\otimes_{R}\mathbb{S}^{\lambda}_{R'}N\text{.}$$ Furthermore, for a finitely generated free submodule $N'$ of $N$, it holds $$\label{e:Procesi633}
    \Big( 
    \big(\mathbb{S}^{\lambda}_{R}M\big)^{G_{R}}\otimes_{R}\mathbb{S}^{\lambda}_{R'}N
    \Big)
    \cap S^\bullet_{R'}(M\otimes_R N') \,=\, 
    \big(\mathbb{S}^{\lambda}_{R}M\big)^{G_{R}}\otimes_{R}\mathbb{S}^{\lambda}_{R'}N'$$

The Standard Basis Theorem \[decomposition\] implies that: $$\big({\operatorname{S}}^{k}_{R'}(M\otimes_{R}N)\big)^{G_{R'}}\simeq\bigoplus_{\vert\lambda\vert=k}\Big(\mathbb{S}^{\lambda}_{R'}(M\otimes_{R}R')\otimes_{R'}\mathbb{S}^{\lambda}_{R'}N\Big)^{G_{R'}}\text{.}$$ Noting that the above isomorphism is an isomorphism of $\textrm{Aut}_{R'}(M\otimes_{R}R')\times\textrm{Aut}_{R'}(N)$-modules, $G$ acts on $M$, that $G$ acts trivially on $N$, and that $\mathbb{S}^{\lambda}_{R'}N$ is a finitely generated free $R'$-module, it follows that the r.h.s of the previous identity is isomorphic to: $$\bigoplus_{\vert\lambda\vert=k}\Big(\mathbb{S}^{\lambda}_{R'}(M\otimes_{R}R')\Big)^{G_{R'}}\otimes_{R'}\mathbb{S}^{\lambda}_{R'}N\text{.}$$ Applying Lemma \[l:basechangeschur\], one gets . Finally, equation is a straightforward generalization of @Procesi [Chap. 6, Eq. 6.3.3].

\[l:fieldcase\] Let $M$ be a $\Bbbk$-vector space of dimension $n$ and $G$ be a linear semisimple algebraic $\Bbbk$-group together with a faithful representation $\rho:G_R\hookrightarrow \emph{SL}(M)$. Then, for $m\geq n$, the ring of invariants $ \big( S_{\Bbbk}^\bullet (M^{\oplus m})\big)^{G}$ is generated, under the action of $\emph{GL}_m(\Bbbk)$, by the invariants of $n$ copies of $M$; that is, by $ \big( S_{\Bbbk}^\bullet (M^{\oplus n})\big)^{G}$. Further, there exists a constant, $d$, depending on $M$, $G$ and $\rho$ but not on $m$ such that $\big( S_{\Bbbk}^\bullet (M^{\oplus m})\big)^{G}$ is generated by its components of degree less than or equal to $d$.

The first part is @Procesi [Chap. 11, Thm 1]. Regarding the bound, it is enough to note that the action of $\textrm{GL}_m(\Bbbk)$ preserves the degree of elements of $(\mathbb{S}^{\lambda}_{\Bbbk}M\big)^{G}\otimes \mathbb{S}^{\lambda}_{\Bbbk} (\Bbbk^{\oplus n})$.

[^1]: The authors are supported by grant PGC2018-099599-B-I00 of MINECO

{'source': 'arXiv_out/2202'}