Invariant Subschemes of Affine $\IG_m$-Schemes

As we've seen in the last blog post, there is a (contravariant) equivalence between the category of affine $\IG_m$-schemes and the category of ($\IZ$-)graded rings. If $X$ is an affine $\IG_m$-scheme, then a function $f \in \sheaf{O}(X)$ is homogeneous of degree $d$ if and only if $f(tx) = t^d f(x)$ for all $R$-valued points $(t, x) \in_R \IG_m \times X$.

If $X$ is an affine scheme, then a closed subscheme $Z \subset X$ corresponds to an ideal $I(Z) \subset \sheaf{O}(X)$ consisting of those functions that vanish on $Z$. Similarly, an open subscheme $U \subset X$ corresponds to a radical ideal $I \subset \sheaf{O}(X)$. It turns out that $\IG_m$-invariant (to be defined precisely below) closed/open subschemes correspond precisely to homogeneous (radical) ideals of $\sheaf{O}(X)$.

While the case of closed subschemes is straight-forward, the case of open subschemes requires a tiny bit more work. When I first learned about the equivalence between affine $\IG_m$-schemes and graded rings 6 years ago, I couldn't figure it out and I asked a question about it on MathOverflow. It was answered by Will Sawin, using the fact that $\IG_m$ is a smooth group scheme and some facts about smooth morphisms. Since then I found a completely elementary argument in the style of the last blog post, which I want to share here.

Invariant Closed Subschemes

Let $X$ be an affine scheme. If $I \subset \sheaf{O}(X)$ is an ideal, then there is a subfunctor $Z = V(I) \subset X$ defined by $Z(R) = \{x \in_R X \mid f(x) = 0 \;\forall f \in I\}$. It turns out that $Z$ is an affine scheme, the map $i^* \from \sheaf{O}(X) \to \sheaf{O}(Z)$ defined by the embedding $i \from Z \into X$ is surjective and $I$ can be recovered as the kernel $I(Z) := \Ker(i^*)$. A subfunctor of this form is called a closed subscheme of $X$.

The closed subscheme $Z \subset X$ is called ($\IG_m$-)invariant if for all $(t, x) \in_R \IG_m \times X$ with $x \in_R Z$ we have $t \cdot x \in_R Z$. In this case the morphism $\rho \from \IG_m \times X \to X$ restricts to a morphism $\IG_m \times Z \to Z$ equipping $Z$ with the structure of an affine $\IG_m$-scheme such that the embedding $i \from Z \into X$ is a morphism of affine $\IG_m$-schemes. We have seen in the last post that the induced ring homomorphism $i^* \from \sheaf{O}(X) \to \sheaf{O}(Z)$ is hence a morphism of graded rings.

If $f \in \sheaf{O}(X)_d$ is a homogeneous function then the basic closed subscheme $V(f) = V(\langle f \rangle) \subset X$ is invariant. Indeed, if $t \in_R \IG_m$ and $x \in_R V(f)$, then $f(tx) = t^d f(x) = t^d 0 = 0$, so $tx \in_R V(f)$.

Homogeneous Ideals

Let $A$ be a graded ring. An ideal $I \subset A$ is called homogeneous if whenever $f = \sum_{i \in I} f_i \in I$ such that $f_i \in A_i$, then it follows that $f _i \in I$. If we write $I_i := I \cap A_i$, then it follows that $$ I = \bigoplus_{i \in I} I_i. $$ In particular, $I$ is generated by its homogeneous elements.

If $\varphi \from A \to B$ is a homomorphism of graded rings, then it is obvious that the ideal $\Ker(\varphi) \subset A$ is homogeneous.

Proposition 1. Let $X$ be an affine $\IG_m$-scheme. Then a closed sub-scheme $Z \subset X$ with embedding morphism $i \from Z \to X$ is invariant if and only if the ideal $$ I(Z) := \Ker(i^*) \subset \sheaf{O}(X) $$ is invariant.

Proof. It follows from the above two sections that if $Z$ is invariant then $I(Z)$, as the kernel of a homomorphism of graded rings, is invariant.

Conversely, assume that $I(Z)$ is homogeneous. Since it is then generated by homogeneous elements, we have $$ Z = \bigcap_{f \in I \text{ hom.}} V(f). $$ The $V(f)$ are invariant sub-schemes by the above and it is obvious that the intersection of invariant closed sub-schemes is invariant. ∎

Invariant Open Subschemes

Let $X$ be an affine scheme. Recall that a point $x \in_R X$ leads to a character $\widetilde{x} \from \sheaf{O}(X) \to R$, $f \mapsto f(x)$. Let $I \subset \sheaf{O}(X)$ be an ideal. Define a subfunctor $U \defby D(I) \subset X$ by $$ x \in_R U \iff \langle \widetilde{x}(I) \rangle = \langle 1 \rangle \text { in }R. $$ Note that if $I$ is generated by a set $S$, then $$ x \in_R U \iff 1 \in \langle \widetilde{x}(f) \mid f \in S\} = \langle f(x) \mid f \in S\rangle. $$ A sub-functor of this form is called an open subscheme of $X$. One can show that the mapping $I \mapsto D(I)$ induces a bijection between radical ideals of $\sheaf{O}(X)$ and open subschemes of $S$.

The open subscheme $U \subset X$ is called invariant if for all $(t, x) \in_R \IG_m \times X$ with $x \in_R U$ we have $t \cdot x \in_R U$.

If $\rho \from \IG_m \times X \to X$ denotes the action morphism and $\pi_2 \from \IG_m \times X \to X$ denotes the second projection, then $U \subset X$ is invariant if and only if $\pi_2^{-1}(U) \subset \rho^{-1}(U)$. (In fact it automatically follows that $\pi_2^{-1}(U) = \rho^{-1}(U)$.)

If $f \in \sheaf{O}(X)_d$ is a homogeneous function of degree $d$ then the basic open subscheme $D(f) := D(\langle f \rangle)$ is invariant. Indeed, if $t \in_R \IG_m$ and $x \in_R D(f)$ (i.e. $f(x) \in_R \IG_m \subset \IA^1$), then $f(tx) = t^d f(x) \in_R \IG_m$, so $tx \in_R D(f)$.

Proposition. Let $X$ be an affine $\IG_m$-scheme and let $I \subset \sheaf{O}(X)$ be a radical ideal. Then the radical ideal $I$ is homogeneous if and only if the open subscheme $D(I) \subset X$ is invariant.

Proof. Suppose first that $I$ is homogeneous. In particular, it is generated by homogeneous elements. Then for $x \in_R X$ we have $x \in_R D(I)$ if and only if $1 \in \langle f(x) \mid f \in I, f \text{ hom.}\rangle$ in $R$. Suppose that $x \in_R D(I)$ and $t \in_R \IG_m$. By assumption, there exist finitely many homogeneous elements $f_\lambda$, $\lambda \in \Lambda$ and coefficients $a_\lambda \in R$ such that $\sum_{\lambda \in \Lambda} a_\lambda f_\lambda(x) = 1$. Say that $f_\lambda$ is homogeneous of degree $d_\lambda$. Then we have $1 = \sum_{\lambda \in \Lambda} a_\lambda t^{-d_\lambda} t^{d_\lambda} f_\lambda(x) = \sum_{\lambda \in \Lambda} a_\lambda t^{-d_\lambda} f_\lambda(tx)$, whence proving that $1 \in \langle f(tx) \mid f \in I, f \text{ hom.}\rangle$. This shows that $D(I)$ is invariant.

Conversely assume that $U = D(I)$ is invariant. Consider the action morphism $\rho \from \IG_m \times X \to X$ as well as the second projection $\pi_2 \from \IG_m \times X \to X$. By the above paragraph we have $\pi_2^{-1}(U) = \rho^{-1}(U)$. Given $f = \sum_{i \in \IZ} f_i \in I$ with $f_i \in \sheaf{O}(X)_i$ we have $D(f) \subset U$ and therefore $D(\rho^*(f)) = \rho^{-1}(D(f)) \subset \rho^{-1}(U) = \pi_2^{-1}(U)$. The open subscheme $\pi_2^{-1}(U) \subset \IG_m \times X$ is defined by the radical ideal $I[T^{\pm1}] \subset \sheaf{O}(X)[T^{\pm1}] \cong \sheaf{O}(\IG_m \times X)$, whence we have $\rho^*(f) \in I[T^{\pm1}]$ under this identification. However by Lemma 3 of last time, under the identification of $\sheaf{O}(X)[T^{\pm1}]$ with $\sheaf{O}(\IG_m \times X)$, $\rho^*(f)$ corresponds to the element $\sum_{i \in I} T^i f_i \in \sheaf{O}(X)[T^{\pm1}]$. So the fact that $\rho^*(f) \in I[T^{\pm1}]$ shows exactly that $f_i \in I$, i.e. that $I$ is homogeneous. ∎