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I can interpret the other axioms for a grading via the definition of a group scheme action appropriately (I think):

Let $H = k[x,x^{-1}]$. We are given a coaction $ \rho : A \to A \otimes H$, which is a morphism of $k$ algebras.

  1. The counit axiom implies that the coaction map $\rho: A \to A \otimes H$ is injective (it explicitly gives an inverse), hence the direct sum structure on $A \otimes H$ given by $H = \bigoplus_{ n \in \mathbb{Z}} k_{x^n} \otimes_k A$ pulls back via $\rho^{-1}$ to give a direct sum decomposition of $A$: $A = \bigoplus_{n \in \mathbb{Z}} \rho^{-1} (k_{x^n} \otimes A)$.
  2. That the coaction map is an algebra homomorphism implies that $A_n A_m \subseteq A_{n + m}$.

These appear to be all of the conditions of a $\mathbb{Z}$ grading on the ring $A$ - namely, that $A$ can be written as a direct sum of submodules graded by $\mathbb{Z}$ which multiply in the correct way. But I haven't used the coassociativity. So I suspect that I am being stupid in some way I would appreciate being pointed out.

Or perhaps the point of the coassiativity is to force the coaction to look like $f \to x^n \otimes f$, for $f$ a degree $n$ homogeneous element in the grading induced by $1$ (i.e. living in $\rho^{-1} ( k_{x^n} \otimes_k A)$? However, this seems to come from the conunit axiom anyway, since the counit evaluates $x^n$ to be $1$. ($f$ must be sent to some element in $k_{x^n} \otimes_k A$, i.e. some $ x^n \otimes_k g$, but conuit implies that it must be the case that $g = f$.)

As for the converse - starting with a graded ring, then defining a coaction by sending a degree $n$ homogeneous element to $x^n \otimes f$... it seems that the coassociativity more or less comes for free. So even if one had just a not necessarily associative action of $G_m$ on $\operatorname{Spec}A$ (i.e., a $k$-algebra coaction by $H$ satisfying just the couit axiom), it appears that it becomes automatically associative (it induces a grading, which then induces an associative coaction, which agrees at least on homogeneous elements). But this feels weird...

I'm really confused!

user26857
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Elle Najt
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1 Answers1

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A bit late, but I worked this out just now.

The coaction is not just $f\mapsto x^n\otimes f$ (for some $n$ depending on $f$, or otherwise)! This only applies for homogeneous $f$. (Well, homogeneous according to the very grading we are trying to work out.) In general, $\rho(f)$ is a finite sum $\sum_i x^i\otimes f_i$, which I will write as $\sum_i f_ix^i$ from now on.

The counit axiom says that, for each element $f\in A$, $\rho(f)$, which is a finite sum $\sum_i f_i x^i$ in $A\otimes H$, satisfies $\sum_i f_i=f$ (thanks Daniel).

The coassociativity axiom applied to the same $f$ will give that $\rho(f_i)=f_ix^i$. This is a nontrivial statement because in general, all we know is that $\rho(f_i)$ is some sum $\sum_j (f_i)_j x^j$. But coassociativity says precisely that $\sum_i f_i(x^i\otimes x^i)=\sum_{i,j}(f_i)_j (x^j\otimes x^i)$, and matching terms, this says that $(f_i)_j=0$ for $i\neq j$, and $(f_i)_i=f_i$. So coassociativity is what guarantees that the pieces $f_i$ corresponding to some $f\in A$ are actually homogeneous. (Consistency of the grading, in some sense, if you like.)

Yongyi Chen
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