# 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? – Martin Brandenburg Jan 26 '20 at 8:20
• @MartinBrandenburg Yes, sure. – Jakob Werner Jan 26 '20 at 10:35

## 2 Answers

Your definition

$$\mathbb G_m \times Y \hookrightarrow \mathbb G_m \times X \to X$$ 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. – Martin Brandenburg Jan 28 '20 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. – Will Sawin Jan 28 '20 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.