The Stacks project

Lemma 27.11.3. With hypotheses and notation as in Lemma 27.11.1 above. Assume $A_ d \to B_ d$ is surjective for all $d \gg 0$. Then

  1. $U(\psi ) = Y$,

  2. $r_\psi : Y \to X$ is a closed immersion, and

  3. the maps $\theta : r_\psi ^*\mathcal{O}_ X(n) \to \mathcal{O}_ Y(n)$ are surjective but not isomorphisms in general (even if $A \to B$ is surjective).

Proof. Part (1) follows from the definition of $U(\psi )$ and the fact that $D_{+}(f) = D_{+}(f^ n)$ for any $n > 0$. For $f \in A_{+}$ homogeneous we see that $A_{(f)} \to B_{(\psi (f))}$ is surjective because any element of $B_{(\psi (f))}$ can be represented by a fraction $b/\psi (f)^ n$ with $n$ arbitrarily large (which forces the degree of $b \in B$ to be large). This proves (2). The same argument shows the map

\[ A_ f \to B_{\psi (f)} \]

is surjective which proves the surjectivity of $\theta $. For an example where this map is not an isomorphism consider the graded ring $A = k[x, y]$ where $k$ is a field and $\deg (x) = 1$, $\deg (y) = 2$. Set $I = (x)$, so that $B = k[y]$. Note that $\mathcal{O}_ Y(1) = 0$ in this case. But it is easy to see that $r_\psi ^*\mathcal{O}_ X(1)$ is not zero. (There are less silly examples.) $\square$


Comments (5)

Comment #1795 by Keenan Kidwell on

I don't understand the justification for surjectivity of . I was thinking the point was that, given a fraction , there is some such that , and by considering the decomposition of (and hence of ) into homogeneous components, one immediately gets that , where , so works. Is this what the remark about degrees is getting at?

Comment #1824 by on

The thing about the degrees is there because we only assume is surjective for sufficiently large. Otherwise what you say is correct. OK?

Comment #1828 by Keenan Kidwell on

Ah, yes, absolutely. I missed that completely. Sorry.

Comment #4970 by Dario WeiƟmann on

typo: should be the pullback of instead.

There are also:

  • 2 comment(s) on Section 27.11: Functoriality of Proj

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