My question is relatively simple, assume that $X$ is an affine variety such that its coordinate ring $A:=\Bbbk[X]:=H^0(\mathcal O_X,X)$ is $\Lambda$-graded for some monoid $\Lambda$.

Now if $\Lambda=(\mathbb Z,+)$ and the grading is in some sense 'compatible' with the one on the polynomial ring after embedding $X$ into affine space, then this would imply that $X$ is a cone, i.e. cut out by homogeneous polynomials. More precisely, assume $\phi:\Bbbk[X_0,\ldots,X_n]\twoheadrightarrow A$ induces a closed immersion $X\hookrightarrow\mathbb A^n$. Then if $\phi$ is a morphism of graded algebras, we know that $\ker(\phi)=I(X)$ is generated by homogeneous polynomials. However, I would like to be able to see this kind of thing from the grading on $A$ in a more intrinsic manner, i.e. without talking about compatibility to some specific embedding into affine space.

Is there a result that makes this vague notion of compatibility precise? What other kind of geometrical information can I obtain from properties of a grading on $A$? I'd be very thankful for any kind of insight on this, admittedly, very broad matter.


1 Answer 1


(Nitpick: "is graded" is a mild type error. A grading is a structure, not a property.)

One geometric interpretation is as follows. Suppose $A$ is an abelian group which acts on $X$. Then $k[X]$ naturally becomes a linear representation of $A$. If this action splits up into a direct sum of one-dimensional representations, then $k[X]$ naturally becomes graded by the group of characters $A \to k^{\times}$. In the previous we may also internalize $A$ to an algebraic group and then talk about the algebraic characters. For example, if $A = k^{\times}$ then the algebraic characters $k^{\times} \to k^{\times}$ are precisely the power maps $x \mapsto x^n$, and if $k^{\times}$ acts on $X$ then $k[X]$ naturally acquires a $\mathbb{Z}$-grading.

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    $\begingroup$ Your nitpick goes somewhat against common usage. A graded something is a something which has been endowed with a grading, and something is graded if it has been endowed with a grading. You can see the same mechanism with split short exact sequence, splitting of a short exact sequence, say. $\endgroup$ Commented Aug 15, 2013 at 16:06
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    $\begingroup$ First of all, thanks for your answer. However, it's not quite what I was looking for (yet): In fact, what you said is my motivation to ask the question. Having $k^\times$ act on $X$ gives me a $\mathbb Z$-grading indeed, but this grading does not have to come from a polynomial ring, or are you saying that it does? Does every affine $k^\times$ variety admit a $k^\times$-morphism that is an immersion into some $\mathbb A^n$? $\endgroup$ Commented Aug 15, 2013 at 16:11

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