# Are open $\mathbb{G}_m$-invariant subschemes of an affine scheme precisely the homogeneous radical ideals of its coordinate ring?

This question is certainly not research level and in fact quite elementary which is why I asked it on math.stackexchange before: math.stackexchange. However it doesn't seem to get much attention there and I thought I would try it here. My questions are regarding actions of a group scheme $$G$$ on a scheme $$X$$. I'm fine with assuming $$G$$ affine. $$\newcommand{\IG}{\mathbb{G}} \newcommand{\pmo}{{\pm 1}} \newcommand{\IZ}{\mathbb{Z}} \newcommand{\Spec}{\mathrm{Spec}} \newcommand{\tensor}{\otimes} \newcommand{\into}{\hookrightarrow} \newcommand{\iso}{\cong} \newcommand{\onto}{\twoheadrightarrow} \newcommand{\sheaf}{\mathcal} \newcommand{\inv}{{-1}}$$

Originally, I was thinking about actions of the multiplicative group $$\IG_m = \Spec(\IZ[T^\pmo])$$ on affine schemes $$X = \Spec(R)$$. The category of affine schemes with a $$\IG_m$$-action is equivalent to the category of $$\IZ$$-graded rings as follows:

To a $$\IZ$$-graded ring $$R$$ associate the action $$\IG_m \times \Spec(R) \to \Spec(R)$$ which is given by the map of rings $$R \to \IZ[T^\pmo] \tensor R$$, $$f = \sum f_d \mapsto \sum f_d \tensor T^d$$, where $$f_d$$ are the homogenous components of $$f$$. Conversely, if the action $$\IG_m \times \Spec(R) \to \Spec(R)$$ is given, call $$f_d \in R$$ homogenous of degree $$d$$ if it is sent to a homogenous element of rank $$d$$ by the morphism $$R \to \IZ[T^\pmo] \tensor R \cong R[T^\pmo]$$, where we regard the latter ring as graded by the powers of $$T$$.

If $$Y \subseteq X$$ is a closed subscheme, I call $$Y$$ invariant under the action if the morphism $$\IG_m \times Y \into \IG_m \times X \to X$$ factors over $$Y \into X$$.

Proposition: The correspondence $$\{\text{Closed subschemes of }X\} \iso \{\text{Ideals of }R\}$$ restricts to a bijective correspondence $$\{\IG_m\text{-invariant closed subschemes of }X\} \iso \{\text{Homogeneous ideals of }R\}.$$

For example if $$Y$$ is $$\IG_m$$-invariant, then by definition the given action restricts to an action of $$\IG_m$$ on $$Y$$ and the inclusion $$Y \into X$$ becomes a morphism of affine schemes with a $$\IG_m$$-action. Hence the surjective map $$R=\sheaf{O}(X) \onto \sheaf{O}(Y)$$ is a morphism of graded rings. But its kernel is $$I$$ which is, hence, homogeneous.

I was trying to prove the corresponding result for open subschemes:

Conjecture 1: The correspondence $$\{\text{Open subschemes of }X\} \iso \{\text{Radical ideals of }R\}$$ restricts to a bijective correspondence $$\{\IG_m\text{-invariant open subschemes of }X\} \iso \{\text{Homogeneous radical ideals of }R\}.$$

It is easy to see that a homogeneous radical ideal defines an invariant open subscheme. First note that the union of invariant open subschemes is again invariant. If the open subschemes $$U_j$$ are defined by $$I_j$$, then $$\bigcup U_j$$ is defined by $$\sqrt{\sum I_j}$$. Hence it suffices to consider $$U = D(f)$$ for $$f$$ a homogeneous element. But then it is easy to see that $$U$$ is invariant. In fact, the restricted action $$\IG_m \times U \to U$$ makes $$U = \Spec(R[f^\inv])$$ into an affine schemes with a $$\IG_m$$-action. The associated grading on $$R[f^\inv]$$ is the natural grading that you would expect on a localisation at a homogeneous element.

However I cannot manage to prove the converse, i.e. if $$I$$ is a radical ideal of $$R$$ such that the open subscheme $$X_I \into X$$ defined by $$I$$ is $$\IG_m$$-invariant then $$I$$ is homogeneous. By the proposition it would suffice to prove

Conjecture 2: If $$G$$ is an (affine) group scheme acting on a scheme $$X$$ then the bijection $$\{\text{Open subschemes of }X\} \iso \{\text{Reduced closed subschemes of }X\}$$ given by “reduced closed complement” and “open complement” restricts to a bijection $$\{G\text{-invariant open subschemes of }X\} \iso \{G\text{-invariant reduced closed subschemes of }X\}.$$

This certainly sounds reasonable if one thinks for example of an action of a topological group on a topological space, where the complement of an invariant subset is clearly invariant again. However I cannot prove it in the context of schemes and I'm not quite sure that it is correct. If $$X = \Spec(R)$$ is reduced and of finite type over an algebraically closed field and $$G = \IG_m$$ (so that $$\IG_m \times X$$ is again reduced and of finite type), then it suffices to consider $$k$$-valued points, so that the reduced complement of an invariant open subscheme is indeed invariant again.

If $$G = \IG_m$$ and $$X$$ is affine then this conjecture is equivalent to the first one.

My conjecture is also equivalent to the claim that if $$U \subseteq X$$ is an open $$\IG_m$$-invariant subscheme then $$U$$ is a union of subschemes of the form $$D(f) \subseteq X$$ where $$f$$ in $$R$$ is a homogeneous element. In particular, this would imply (a special case of) the following conjecture:

Conjecture 3: If $$G$$ is an (affine) group scheme acting on a scheme $$X$$ and if $$U \subseteq X$$ is an open $$G$$-invariant subscheme then $$U$$ is a union of $$G$$-invariant affine open subschemes.

Unfortunately I really don't have a good intuition for actions of groups schemes and I would be glad about some clarification.

• In Conj. 3 you mean that $U$ is such a union? Commented Jan 26, 2020 at 8:20
• @MartinBrandenburg Yes, sure. Commented Jan 26, 2020 at 10:35

$$\mathbb G_m \times Y \hookrightarrow \mathbb G_m \times X \to X$$ factors over $$Y \hookrightarrow X$$

can be simplified a bit. A better one for our purposes is

The two subschemes of $$\mathbb G_m \times X$$ defined as the inverse image of $$Y$$ under the right projection $$\mathbb G_m \times X \to X$$ and the action map $$\mathbb G_m \times X \to X$$ are equal.

To see they are equivalent, note that the universal property of a fiber product implies your factorization is equivalent to the claim that the immersion $$\mathbb G_m \times Y \hookrightarrow \mathbb G_m \times X$$ factors through the map from $$\left(\mathbb G_m \times X \right) \times_X Y$$ to $$\mathbb G_m \times X$$. That fiber product is the inverse image of $$Y$$ under the action map, and $$\mathbb G_m \times Y$$ is the inverse image of $$Y$$ under the right projection. So your factorization is equivalent to $$p^* Y \subseteq a^* Y$$, where $$p$$ is projection and $$a$$ is action. But $$p^* Y \subseteq a^* Y$$ is equivalent to $$a^* Y \subseteq p^* Y$$, and therefore equivalent to $$p^* Y =a^* Y$$, because we can swap the projection and action maps using the automorphism of $$\mathbb G_m \times X$$ that sends $$(g,x)$$ to $$(g^{-1}, gx)$$.

Now this one is equivalent for open subschemes and their closed complements because the operation of taking closed complement is compatible with pullback under smooth morphisms.

• Where does this use the special case $\mathbb{G}_m$? Angelo wrote that it is false for general group schemes. Commented Jan 28, 2020 at 9:49
• @MartinBrandenburg Then the maps $\mathbb G_m \times X\to X$ are not smooth, and in particular the inverse image of a reduced closed subscheme can clearly be non-reduced. I think this approach does work for general smooth group schemes. Commented Jan 28, 2020 at 13:24

Your conjecture 2 is false if you don't assume that $$G$$ is reduced (in positive characteristic there are affine group schemes that are not reduced).

As to conjecture 3, it is hopelessly wrong (think of the action of $$\mathrm{GL}_n$$ on $$\mathbb P^{n-1}$$). The only non-trivial case I know is Sumihiro's theorem: a normal algebraic variety with an action of a torus can be covered by invariant affine open subsets. Being normal is essential: consider the standard action of $$\mathbb{G}_{\mathrm m}$$ on $$\mathbb P^1$$, and glue together the origin and the point at infinity.

### Intermediate cases

While the smooth group case has been affirmed by Will Sawin and the most general case has been refuted by Angelo, there is quite some space in between where we can still find more affirmative cases.

#### 1 A Simple Example

You may skip this example and read the further sections if you are only interested in the results. Otherwise, it serves two purposes:

• A more gentle start than abstract schemes (after all, it is not quite a reserach-level question and I'd like to make some allowances for a broader range of readers accordingly)
• A motivating example showing some (but not all) interesting behaviour investigated in general

1.1 Algebraic Setting Consider the ring $$R = \mathbb{F}_2[T]/(T^2-1)$$ as a $$\mathbb{Z}/2\mathbb{Z}$$-graded ring with the grading $$\mathrm{deg}(T) = 1 + 2\mathbb{Z}$$, inducing the ring homomorphism $$\begin{split}\alpha\colon R &\to R[S]/(S^2-1)\\T &\mapsto TS.\end{split}$$.

1.2 Geometric Setting Consider the group scheme $$G = \mathrm{Spec}(\mathbb{Z}[S]/(S^2-1)) = \mathrm{Spec}(\mathbb{Z}[\mathbb{Z}/2]),$$ the affine scheme $$X = \mathrm{Spec}(R)$$ and the group action $$G \times X = \mathrm{Spec}(R[S]/(S^2-1)) \to \mathrm{Spec}(R) = X$$ given by $$\alpha$$.

1.3 Classical Subobjects When working in the unmodified setting of the question, a somewhat unclean picture emerges:

• Open subschemes of $$X$$:
1. $$\emptyset$$ (invariant)
2. $$X$$ (invariant)
• Closed subschemes of $$X$$:
1. $$X$$ (invariant, non-reduced)
2. $$\emptyset$$ (invariant, reduced)
3. $$\mathrm{Spec}(\mathbb{F}_2)$$ (non-invariant, reduced)
• Ideals of $$R$$:
1. $$(0)$$ (homogeneous, non-radical)
2. $$(1)$$ (homogeneous, radical)
3. $$(T+1)$$ (non-homogeneous, radical)

So far, no real surprises. There are two ideals (resp. closed subschemes) corresponding to the empty open subscheme, but neither of them is homogeneous (resp. invariant) and radical (resp. reduced). However, as indicated by my choice of numbering, there still appears to be a nice correspondence between invariant open subschemes, invariant closed subschemes and homogeneous ideals, except that for a perfect correspondence we'd like to ignore the existence of the nilpotent non-homogeneous element $$X+1$$ so that $$(0)$$ would be a radical. So, we'll do just that.

1.4 Better Subojects When the closed subscheme is only required to be as reduced as an invariant subscheme can be and the ideal is considered to have the radical property only with respect to the homogeneous elements, we instead obtain the following picture:

• Open subschemes of $$X$$:
1. $$\emptyset$$ (invariant)
2. $$X$$ (invariant)
• Closed subschemes of $$X$$:
1. $$X$$ (invariant, maximally reduced)
2. $$\emptyset$$ (invariant, maximally reduced)
3. $$\mathrm{Spec}(\mathbb{F}_2)$$ (non-invariant, whatever)
• Ideals of $$R$$:
1. $$(0)$$ (homogeneous, homogeneously radical)
2. $$(1)$$ (homogeneous, homogeneously radical)
3. $$(T+1)$$ (non-homogeneous, whatever)

#### 2 Diagonalizable and Affine

Consider the diagonalizable group schemes $$\mathrm{D}(M) = \mathrm{Spec}(\mathbb{Z}[M])$$ for Abelian groups $$M$$ (in particular, $$\mathbb{G}_{\mathrm{m}} = \mathrm{D}(\mathbb{Z})$$). Then, a $$\mathrm{D}(M)$$-action on an affine scheme $$X$$ is the same as an $$M$$-graded ring. Restricting ourselves to these, i.e. diagonalizable groups and affine schemes, your first two conjectures are nearly correct and the last one may even be strengthened slightly.

2.1 Definition A homogeneous ideal is called homogeneously radical iff it contains all homogeneous elements of its radical. An invariant closed subscheme $$C$$ is called maximally reduced if any of its invariant closed subschemes on the same points is already identical to $$C$$.

2.2 Lemma When a diagonalizable group acts on an affine scheme, there is a bijective three-way correspondence working exactly like you'd expect between

• maximally reduced invariant closed subschemes and
• invariant open subschemes.

Proof. The equivalence of the first two points is immediate from the correspondence between homogeneous ideals and invariant closed subschemes. The equivalence of the latter two points is a special case of the more general treatment in the next section.

2.3 Corollary From this, we get the following alterations of your three conjectures:

2.4 Fixed Conjecture 1 The assignment $$\{\textrm{ideals of $$R$$}\} \to \{\textrm{open subschemes of $$X$$}\}$$ restricts to a bijective correspondence $$\{\textrm{homogeneously radical homogeneous ideals of $$R$$}\} \cong \{\textrm{invariant open subschemes of $$X$$}\}.$$

2.5 Fixed Conjecture 2 The open-complement assignment $$\{\textrm{closed subschemes of $$X$$}\} \to \{\textrm{open subschemes of $$X$$}\}$$ restricts to a bijective correspondence $$\{\textrm{maximally reduced invariant closed subschemes of $$X$$}\} \cong \{\textrm{invariant open subschemes of $$X$$}\}.$$

2.6 Fixed Conjecture 3 Any invariant open subscheme is the union of invariant principal opens (and thus in particular of invariant affine open subschemes).

#### 3 General Case

I have not yet thought about the adaptations which would enable Conjecture 1 to be considered for general schemes and Conjecture 3 has been refuted by Angelo. However, in light of the diagonalizable and affine case it seems straightforward to salvage Conjecture 2. Throughout this section, let $$G$$ be a group scheme acting on a scheme $$X$$ via $$\alpha\colon G \times X \to X$$.

3.1 Lemma When a quasi-compact flat group scheme acts on a scheme, the “open complement” assigment restricts to a bijection of

• maximally reduced invariant closed subschemes and
• invariant open subschemes.

Proof. The remainder of this section.

3.2 Definition Given a closed subscheme $$C$$ of $$X$$, its invariant hull is the scheme-theoretic image of $$\alpha^*C$$ under the projection $$\pi\colon G \times X \to X$$.

3.3 Lemma Let $$C \subseteq X$$ be a closed subscheme and $$C'$$ its invariant hull. Then $$C \subseteq C'$$.

Proof. Using the unit of $$G$$, the inclusion of $$C$$ into $$X$$ factors through both the projection and the action with a single morphism $$f\colon C \to G \times X$$. Consequently, seen as $$\alpha\circ f$$ the inclusion of $$C$$ into $$X$$ factors through $$\alpha^*C$$ and seen as $$\pi\circ f$$ it factors through the image $$C'$$ of $$\alpha^*C$$ under $$\pi$$.

3.4 Lemma Let $$C \subseteq X$$ be a closed subscheme, $$C'$$ its invariant hull and $$D \subseteq X$$ a $$G$$-invariant closed subscheme containing $$C$$. Then $$C' \subseteq D$$.

Proof. We have $$\alpha^*C \subseteq \alpha^*D = \pi^*D$$ by monotonicity and invariance, yielding the result.

3.5 Lemma Let $$C \subseteq X$$ be a closed subscheme and $$C'$$ its invariant hull. If $$G$$ is flat and qc, $$C'$$ is invariant.

Proof. Since scheme-theoretic images of qc morphisms commute with flat base change, $$\pi^*C'$$ is the image of $$D = G \times \alpha^*C \subseteq G \times G \times C$$ under the projection $$\pi_{\blacktriangle\!\triangledown\!\blacktriangle}$$ eliding the middle component and $$\alpha^*C'$$ is the image of $$E = {\underbrace{\langle\pi_{\vartriangle\!\blacktriangledown\!\vartriangle}, \alpha\pi_{\blacktriangle\!\triangledown\!\blacktriangle}\rangle}_g}^*\alpha^*C$$ under $$\pi_{\blacktriangle\!\triangledown\!\blacktriangle}$$. As $$f = \langle\pi_{\blacktriangle\!\triangledown\!\vartriangle}, \mu\langle\pi_{\vartriangle\!\blacktriangledown\!\vartriangle}, \iota\pi_{\blacktriangle\!\triangledown\!\vartriangle}\rangle, \pi_{\vartriangle\!\triangledown\!\blacktriangle}\rangle$$ restricts to an isomorphism $$D \to E$$ over $$G \times X$$, they agree. (This boils down to a calculation that $$\alpha\pi_{\blacktriangle\!\triangledown\!\vartriangle} = \alpha gf$$ and thus $$D = \pi_{\blacktriangle\!\triangledown\!\vartriangle}^*\alpha^*C = (\alpha\pi_{\blacktriangle\!\triangledown\!\vartriangle})^*C = (\alpha gf)^*C = f^*E,$$ which is tedious in this formalism but obvious when spelled out on points.)

3.6 Lemma Let $$C \subseteq X$$ be a closed subscheme, $$C'$$ its invariant hull and $$U \subseteq X$$ a $$G$$-invariant open subscheme not meeting $$C$$. If $$G$$ is quasi-compact, then $$U$$ doesn't meet $$C'$$, either.

Proof. We show the contrapositive. As $$G$$ is qc, so is $$\pi$$, meaning that by [02JQ] for any field-valued point $$p$$ in $$C'$$ there is a point $$q$$ of $$\alpha^*C$$ such that $$p$$ is a specialization of $$\pi(q)$$. We note that if $$p$$ lies in both $$C'$$ and $$U$$ then so does $$\pi(q)$$, meaning $$q$$ lies in both $$\alpha^*C$$ and $$\pi^*U = \alpha^*U$$. Therefore, $$\alpha(q)$$ is a common point of $$C$$ and $$U$$.

3.7 Corollary The claim of 3.1 Lemma holds.

Proof. Given an invariant open subscheme, consider its reduced closed complement $$C$$ and its invariant hull $$C'$$. Now, $$C'$$ is the maximally reduced (3.4 Lemma) invariant (3.5 Lemma) closed subscheme on the same support as $$C$$ (3.3 Lemma and 3.6 Lemma), providing the inverse assignment.