This question is certainly not research level and in fact quite elementary which is why I asked it on math.stackexchange before: [math.stackexchange](https://math.stackexchange.com/questions/3519251/are-open-mathbbg-m-invariant-subschemes-of-an-affine-scheme-precisely-the-h). 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.



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


