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.


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