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The Reynolds operator and the ideal theory of the invariant subring

Statement

Assume the Axiom of Choice inherited from the named suppliers. Let $G$ be a complex reductive affine algebraic group and let $X$ be an affine algebraic set with an algebraic $G$ -action (Classical complex affine algebraic actions and rational modules); write $A = C [X]$ and let $R_{X} : A \to A^{G}$ be the Reynolds operator of Complete reducibility and the Reynolds operator for a complex reductive group. Then:

(a) $R_{X}$ is $A^{G}$ -linear, idempotent, and its image is exactly $A^{G}$ ;

(b) for every ideal $I \subseteq A^{G}$ one has $R_{X} (I A) = I$ , and consequently the map $I \mapsto I A$ is injective on ideals of $A^{G}$ and $A^{G}$ is Noetherian whenever $A$ is Noetherian (Left and right Noetherian rings);

(d) if $A$ carries a $G$ -stable grading and an ideal is homogeneous, the same conclusions hold in the graded subalgebra of invariants (Nonnegatively graded rings and modules, homogeneous elements, and twists).

Facts & Assumptions

Given: AC; a complex reductive affine algebraic group $G$ ; an affine algebraic set $X$ with algebraic $G$ -action, $A = C [X]$ , and the Reynolds operator $R_{X} : A \to A^{G}$ of the bridge theorem.

[F1]

Properties of the Reynolds operator. The projection $R_{A} : A \to A^{G}$ is $G$ -equivariant, restricts to the identity of $A^{G}$ , is natural under morphisms of rational $G$ -modules and is $A^{G}$ -linear: $R_{A} (a b) = a R_{A} (b)$ for $a \in A^{G}$ , $b \in A$ (Complete reducibility and the Reynolds operator for a complex reductive group).

[F2]

The coordinate ring is a rational module. If $G$ acts algebraically on an affine algebraic set $X$ , then $C [X]$ with $(g f) (x) = f (g^{- 1} x)$ is a rational $G$ -module on which every finite set of functions lies in a finite-dimensional stable subspace, and the action preserves multiplication and the unit (The coordinate ring of an affine algebraic action is a locally finite rational module, Classical complex affine algebraic actions and rational modules).

[F3]

Noetherian rings. A ring is Noetherian when its left regular module is Noetherian (Left and right Noetherian rings). The ideal-level ascending chain condition used below is the equivalence in [F5].

[F4]

Graded rings. A nonnegatively graded ring is a commutative ring $S = ⨁_{n \geq 0} S_{n}$ with $S_{n} S_{m} \subseteq S_{n + m}$ (Nonnegatively graded rings and modules, homogeneous elements, and twists).

[F5]

Ascending chain condition. A commutative ring is Noetherian if and only if every ascending chain of ideals stabilises (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member, dependent choice included in AC).

Proof

technique · direct

1.1F1F2

Part (a): [F1] gives $R_{X} (a b) = a R_{X} (b)$ for $a \in A^{G}$ , $b \in A$ , so $R_{X}$ is $A^{G}$ -linear; it is idempotent because it restricts to the identity on $A^{G}$ and its values lie in $A^{G}$ , so $R_{X} (R_{X} b) = R_{X} b$ ; and its image is exactly $A^{G}$ , since every invariant is fixed and every value is invariant.

2.1F1F3F5step 1.1

Part (b): let $I \subseteq A^{G}$ be an ideal. The extension $I A$ is $G$ -stable, so by [F1] and step 1.1, $R_{X} (I A) = I R_{X} (A) = I A^{G} = I$ . If $I A = J A$ for ideals of $A^{G}$ , then $I = R_{X} (I A) = R_{X} (J A) = J$ , so $I \mapsto I A$ is injective. An ascending chain $I_{1} \subseteq I_{2} \subseteq \dots$ of ideals of $A^{G}$ gives the ascending chain $I_{n} A$ of ideals of $A$ , which stabilizes when $A$ is Noetherian by the ascending chain condition; applying $R_{X}$ to a stable equality and using $R_{X} (I_{n} A) = I_{n}$ gives $I_{n} = R_{X} (I_{n} A) = R_{X} (I_{n + 1} A) = I_{n + 1}$ , so $A^{G}$ is Noetherian.

2.2F1step 1.1

Part (c): let $φ : A \to B$ be a surjective $G$ -equivariant homomorphism of rational $G$ -algebras. Clearly $φ (A^{G}) \subseteq B^{G}$ . Conversely, if $b \in B^{G}$ , choose $a \in A$ with $φ (a) = b$ ; naturality of the Reynolds operator [F1] gives $b = R_{B} (b) = R_{B} (φ (a)) = φ (R_{A} (a)) \in φ (A^{G})$ . Hence $φ (A^{G}) = B^{G}$ .

3.1F1F4step 2.1step 2.2∎

Part (d): if $A = ⨁_{n} A_{n}$ is a $G$ -stable grading, the degree projections are $G$ -equivariant, so by naturality $R_{X}$ preserves each degree and $A^{G} = ⨁_{n} A_{n}^{G}$ . For a homogeneous ideal $I \subseteq A^{G}$ its extension $I A$ is homogeneous; for an ascending chain of homogeneous ideals $I_{n}$ , all extensions $I_{n} A$ are homogeneous. Thus $R_{X}$ preserves homogeneity, and the computations of steps 1.1, 2.1 and 2.2 apply verbatim: $R_{X} (I A) = I$ , the map $I \mapsto I A$ is injective on homogeneous ideals, and $A^{G}$ is Noetherian when $A$ is; the surjectivity statement of part (c) holds for surjective graded equivariant maps by the same argument.

Remarks

This isolates the three computations used repeatedly in the finite-generation theorem and in the stable-locus theorem: $A^{G}$ is a direct summand as an $A^{G}$ -module, ideal extension is injective, and surjections descend to invariants. They are Brion's steps in the proof of Theorem 1.24(i) and the corresponding properties of the Reynolds operator in Popov–Vinberg.
The Axiom of Choice is inherited from the bridge theorem and the coordinate-ring rationality theorem; the argument itself uses none.

Depends on

Used by

Dependency tree · two levels

46 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Michel Brion, Introduction to actions of algebraic groups, Les cours du CIRM 1 (2010), no. 1, 1-22 (standard reference, not scraped)
V. L. Popov and E. B. Vinberg, Invariant Theory, in Algebraic Geometry IV, Encyclopaedia of Mathematical Sciences 55, Springer 1994 (standard reference, not scraped)