Group schemes and graded rings | Arithmetic variety

In this post we will describe how an action of the multiplicative group scheme $\mathbb{G}_m$ on $\text{Spec }R$ defines a $\mathbb{Z}$ -grading of $R$ . A future post may describe how this relates to projective schemes. (I will do all of this using diagrams, but there may be some easier way using the functors of points). All this was taught to me by Mark Haiman in Math 256B (Algebraic Geometry) at UC Berkeley.

Fix a field $k$ ; we will work in the category of $k$ -schemes. Thus $R$ will be a $k$ -algebra, and we will establish a graded $k$ -algebra structure on $R$ . However, none of our arguments change if we just let $k$ be $\mathbb{Z}$ . A group scheme is a group object in the category of $k$ -schemes. A precise definition can be found here. Most importantly, group schemes can act on other schemes. The definition of a group scheme action can be found here. Note that all definitions are given by diagrams (or functor of points). For example, we specify the “identity element” of a group scheme by a map $\text{Spec }k \to G$ , rather than selecting some point in the underlying topological space.

$\mathbb{G}_m$ is defined as $\text{Spec }k[s, t]/(st-1) = \text{Spec }k[t, t^{-1}]$ . (for shorthand, we will write $k[t, t^{-1}]$ as $k[t^\pm]$ ). As a variety, it can be thought of as $k^*$ , the “punctured affine line”. Its group operation is given by a map $\mathbb{G}_m \times_k \mathbb{G}_m \to \mathbb{G}_m$ which corresponds to the $k$ -algebra map $\mu: k[t^\pm] \to k[t^\pm] \otimes_k k[t^\pm] \cong k[t^\pm, u^\pm]$ defined by $t \mapsto tu$ . The identity is given by a map $\text{Spec }k \to \mathbb{G}_m$ corresponding to $i: k[t^\pm] \to k$ defined by $t \mapsto 1$ .

Suppose $\mathbb{G}_m$ acts on $\Spec R$ . The action map $\mathbb{G}_m \times_k \text{Spec }R \to \text{Spec }R$ corresponds to a $k$ -algebra map $\phi: R \to R \otimes_k k[t^\pm] \cong R[t^\pm]$ such that the following diagrams commute:

Associativity:

$\xymatrix{R\ar[r]^{\phi}\ar[d]^{\phi} & {R[t^\pm]} \ar[d]^{id_R\otimes \mu}\\{R[u^\pm]} \ar[r]^{\phi \otimes id_{k[u^\pm]}} & {R[t^\pm, u^\pm]}}$

Identity:

$\xymatrix{R\ar[r]^{\phi}\ar[d]^{id_R} & {R[t^\pm]} \ar[dl]^{i}\\ R}$

For $r \in R$ , write $\phi(r) = \sum_{-\infty}^{\infty} r_it^i \in R[t^\pm]$ , where almost all the $r_i$ are zero. Then the first diagram implies that

$(*)$ if $\phi(r) = r_it^i$ (i.e. the polynomial is just a single monomial), then $\phi(r_i) = r_it^i$ .

This is because, along the top and right arrows, we have $r \mapsto r_it^i \mapsto r_it^iu^i$ and along the left and bottom arrows we have $r \mapsto r_iu^i \mapsto \phi(r_i)u^i$ . Furthermore, the second diagram says that

$(**)$ for all $r$ , $\sum r_i = r$ .

Therefore, letting $R[t^\pm]_d$ stand for the degree $d$ homogenous component of $R[t^\pm]$ (so that it consists of multiples of $t^d$ ), let $R_d := \phi^{-1}(R[t^\pm]_d)$ . Since all the $R[t^\pm]_d$ are disjoint, their preimages are disjoint as well. Furthermore, for an arbitrary element $r$ , we have $r = \sum r_i$ by $(*)$ , and by $(**)$ , we have that each $r_i \in R_i$ . Thus $R = \sum R_i$ as a direct sum.

It remains to show that $R_mR_n \subseteq R_{m+n}$ . But this is easy: if $r_m \in R_m, r_n \in R_n$ , then $\phi(r_mr_n) = \phi(r_m)\phi(r_n) = r_mr_nx^{m+n} \in R[t^\pm]_{m+n}$ , so $r_mr_n \in R_{m+n}$ as desired.

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