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Let's say I have a complex projective variety $X\subseteq\mathbb P^n$ with homogeneous coordinate ring $S=\bigoplus_{d\ge 0} S_d$. The localization by some homogeneous $f\in S$ (of nonzero degree) yields a graded ring $S_f=:R=\bigoplus_{d\in\mathbb Z} R_d$. Now, instead of looking at the spectrum of the degree zero part, I could look at $R_{\ge 0}=\bigoplus_{d\ge 0} R_d$. I have an inclusion of graded $\mathbb C$-algebras $S\hookrightarrow R_{\ge 0}$ and both are positively graded, so I may very well consider the morphism $$Y:=\mathrm{Proj}(R_{\ge 0})\xrightarrow{\quad\textstyle\pi\quad}\mathrm{Proj}(S)=X.$$ Since $\mathrm{Spec}(R_0)=X_f$, the variety $Y$ is projective over $X_f$. However, what is the nature of $\pi$? Is it surjective? I am sure this must have interested someone before, but I cannot find anything in the 'standard' literature on algebraic geometry.

More generally, is there any deeper geometric meaning to taking the positive part of a graded ring?

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You should specify that $f$ is homogeneous of nonzero degree. –  Jason Starr Oct 4 '13 at 11:32
    
@JasonStarr: Of course, thanks. –  Jesko Hüttenhain Oct 4 '13 at 11:48
    
To understand the geometric meaning to taking the positive part of a graded ring for defining Proj, you need go back to geometric quotients by algebraic groups. Proj is a quotient of an affine space (without "the origin") by the multiplicative group G_m. Unfortunately, this is not explained in the usual common books. –  Al-Amrani Oct 14 '13 at 11:51

1 Answer 1

up vote 9 down vote accepted

Up to replacing $S$ by the "Veronese subring" $S_{(e)} := \oplus_d S_{de}$, which does not change Proj, you may as well assume that $f$ has degree $1$. Consider the graded $R_0$-algebra homomorphism $u:R_0[t]\to R_{\geq 0}$ by $u(t) = f$.

First, $u$ is surjective. Indeed, for every homogeneous element $g$ in $R_d$, then $h=gf^{-d}$ is in $R_0$. Thus $ht^d$ is an element in $R_0[t]_d$ such that $u(ht^d)$ equals $g$.

Next, $u$ is injective. To prove this, observe that, since $u$ is a graded $R_0$-algebra homomorphism, also $\text{Ker}(u)$ is a homogeneous ideal. Every element in $R_0[t]_d$ is of the form $ht^d$ for some $h\in R_0$. Also, $u(ht^d)$ equals $hf^d$. Since $f$ is not a zero divisor in $R$ (by construction), if $hf^d$ equals $0$, then $h$ equals $0$, hence $ht^d$ equals $0$. Therefore the only homogeneous element in $\text{Ker}(u)$ is $0$, i.e., $u$ is injective.

Since $u$ is an isomorphism of graded $R_0$-algebras, also $\text{Proj}(R_{\geq 0})$ equals $\text{Proj}(R_0[t])$. Of course $\text{Proj}(R_0[t])$ is just $\text{Spec}(R_0)$. Thus $\text{Proj}(R_{\geq 0})$ is canonically isomorphic to $\text{Spec}(R_0)$.

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Ouch, I was 1 min late. Anyway, perfect answer! –  Anton Fonarev Oct 4 '13 at 12:06
    
That's sweet! Thanks. –  Jesko Hüttenhain Oct 4 '13 at 15:25
    
Conclusion:the morphism π in the question is nothing else but the canonical open immersion defined by "f different from 0". Recall that Proj is, by construction, covered by such affine open subsets. –  Al-Amrani Oct 14 '13 at 11:36

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