# Actions of the Multiplicative Group

My qualifying exam is fast approaching and so I’d like to use the next few months to refresh myself on the basics. In particular, I’d like to finally nail down a bunch of concepts that I somehow seem to forget about and then have to re-figure out every two months or so. Thus, I anticipate several very short blog posts in the coming weeks, which will hopefully serve me as a memory tool.

Today’s simple concept is how a group action $\mathrm{G}_m \times \mathrm{Spec} A \to \mathrm{Spec} A$ is the same thing as a grading on $A$. To be clear, this is a map of schemes $g: \mathrm{G}_m \times \mathrm{Spec} A \to \mathrm{Spec} A$ such that:

• If $\mu: \mathrm{G}_m \times \mathrm{G}_m \to \mathrm{G}_m$ is the multiplication map for $\mathrm{G}_m$, then $g \circ (\mu \times id ) =g \circ (id \times g)$.
• $1: \mathrm{Spec} \mathbb{Z} \to \mathrm{G}_m$ is the identity section, then $g \circ (1 \times id)$ is the identity map on $\mathrm{Spec} A$.

We translate these conditions  into maps of global sections. If $g:A\to A[x,x^{-1}]$ takes $\alpha$ to $\sum_i p_i(\alpha) \otimes x^i$, then the condition we get is  $\sum_{i,j} p_j(p_i(\alpha)) \otimes x^i \otimes z^j = \sum_i p_i(\alpha) \otimes x^i \otimes z^i$ . This equality implies that $p_j \circ p_i=0$ if $i \neq j$ and $p_i \circ p_i= p_i$.

The second condition implies that $\sum\limits_i p_i(\alpha)=\alpha$.

Thus, the map $g$ splits $A$ into infinitely many subsets $A_i$ corresponding to the images of the different $p_i$. Since $p_i$ is a homomorphism of additive groups, $A_i$ is a group, and since $g$ is a homomorphism of algebras, $A_i \cdot A_j \subset A_{i+j}$. If $p_i(\alpha)=p_j(\beta)$ for $i \neq j$, then applying $p_i$ to both sides, we get that $p_i(\alpha)=0$. Thus $A_i \cap A_j=0$. Finally $\bigoplus_i A_i=A$ and so $A$ has the structure of a graded ring. Conversely projection maps from a graded ring to its components give maps $p_i$ satisfying the above properties.