This question is based on exercise $2.14$ of chapter $2$ of Hartshorne.

Suppose $\varphi:S\rightarrow T$ is a graded homomorphism of graded (commutative, unital) rings such that $\varphi_d := \varphi|_d$ is an isomorphism for all $d$ sufficiently large. Then I want to show that the natural morphism $ f: $Proj $T \rightarrow $Proj $S$ is an isomorphism by explicitly constructing an inverse.

The morphism is given on spaces by $P \mapsto \varphi^{-1}(P)$ and on sections by composing pointwise with $\varphi_P : S_{\varphi^{-1}(P)}\rightarrow T_P$.

I would like to show that this is an isomorphism by exhibiting an inverse homeomorphism to $f$ and showing the stalk maps are isomorphic. I appreciate that we can cover Proj $S$ with affine pieces and then show that the corresponding maps from the pullbacks to these pieces are isomorphisms, but I would like to know what $f^{-1}$ looks like explicitly. If there is a good way to see what $f^{-1}$ is by chasing through the local method of showing that $f$ is an isomorphism then I would also appreciate an explanation of that.

My candidate for $f^{-1}$ was $P \mapsto \sqrt{(\varphi(P))}=I$, the radical of the ideal generated by $\varphi(P)$. I think that I have shown that $I$ is homogeneous, doesn't contain $T_+$, $\varphi^{-1}(I) = P$ and is almost prime in the sense that if $a,b \in T$ are homogeneous and have degree at least $1$ then $ab \in I \implies a \in I$ or $b \in I$. But I think that in fact $I$ is not in general prime, since the degree $0$ component of $I$ is exactly $\sqrt{(\varphi(P_0))}$ in the ring $T_0$, where $P_0$ is the ideal of $S_0$ given by $S_0 \cap P$ and that for general rings $A$ and $B$, with $\rho:A\rightarrow B$, $P$ prime in $A$ doesn't imply $\sqrt{(\rho(P))}$ prime in $B$.


2 Answers 2


$\newcommand{\proj}[1]{{{\mathrm{proj}}(#1)}}$ $\newcommand{\ideal}[1]{{\mathfrak #1}}$ Let $\phi:S \to T$ be an isomorphism $\phi_d:S_d \to T_d$ for all $d \geqslant d_0$. Then $\proj{S} \cong \proj{T}$ and the question is: Given a homogeneous prime $\ideal{p}$ of $S$, what is the corresponding prime $\ideal{q}$ of $T$, so that $\phi^{-1}(\ideal{q}) = \ideal{p}$.

I claim, that

$$\ideal{q} = (\sqrt{\phi_{\geqslant d_0}(\ideal{p}_{\geqslant d_0}) T_+}: b)$$

where $b \in T_d$ with $d > d_0$ but not in $\sqrt{\phi_{\geqslant d_0}(\ideal{p}_{\geqslant d_0}) T_+}$.

First it is clear, that $\ideal{p}_{\geqslant d_0}$ is a homogeneous ideal of $S$. Next it follows, that $\phi(\ideal{p}_{\geqslant d_0}) T T_+$ is an homogeneous ideal of $T$. It is identical with $\phi(\ideal{p}_{\geqslant d_0}) T_+$. Call that ideal $\ideal{q}''$ and call $\ideal{q}' = \sqrt{\ideal{q}''}$.

Now for $b_1 b_2 \in \ideal{q}'$ and $b_1,b_2 \in T_+$ it follows that $(b_1 b_2)^n \in \ideal{q}''$. Therefore $(b_1 b_2)^{n n'} \in \phi(\ideal{p}_{\geqslant d_0})$ and $b_i^{n n'} \in T_{e_i}$ with $e_i > d_0$, therefore $b_i^{n n'} = \phi(a_i)$ with $a_i \in S_{e_i}$. So we have $\phi(a_1)\phi(a_2) = \phi(a)$ with $a \in \ideal{p}_{e}$ with $e > d_0$. So $a_1 a_2 = a$ and without restriction of generality $a_1 \in \ideal{p}$. So $b_1^{n n'} = \phi(a_1) \in \phi(\ideal{p}_{\geqslant d_0})$ and so $b_1^{n n'} \in \ideal{q''}$, therefore $b_1 \in \sqrt{\ideal{q}''} = \ideal{q}'$.

So the homogeneous ideal $\ideal{q}'$ fulfills a weak primality for $b_1,b_2 \in T_+$. From this follows the strong primality of $\ideal{q} = (\ideal{q}':b)$. For let $b_1 b_2 \in \ideal{q}$ therefore $b_1 b_2 b \in \ideal{q}'$ ($b_1, b_2$ homogeneous in $T$ of arbitrary degree) then $(b_1 b) (b_2 b) \in \ideal{q}'$. Therefore because $b b_i \in T_+$ without restriction of generality $b_1 b \in \ideal{q}'$, that is $b_1 \in \ideal{q}$.

The last thing is to prove $\phi^{-1}(\ideal{q}) = \ideal{p}$. Let $\phi(a) \in \ideal{q}$ ($a$ homogeneous in $S$) that is $\phi(a) b \in \ideal{q'}$. Then $(\phi(a) b)^n \in \ideal{q}''$ and $(\phi(a)b)^{n n'} \in \phi(\ideal{p}_{\geqslant d_0})$. Now $b^{n n'} = \phi(a_1)$ with $a_1 \notin \ideal{p}$ and the right side is $\phi(a_2)$ with $a_2 \in \ideal{p}$. So we have $a^{n n'} a_1 = a_2 \in \ideal{p}$ therefore $a^{n n'} \in \ideal{p}$, therefore $a \in \ideal{p}$ as was to be shown.

P.S. I used in the above, that for $b \in \ideal{q}'' = \phi(\ideal{p}_{\geqslant d_0}) T_+$, $b$ homogeneous, a high power $b^N$ is in $\phi(\ideal{p}_{\geqslant d_0})$. This is obvious from the equation

$$b = b_1 z_1 + \cdots + b_r z_r$$

with $b_i \in \phi(\ideal{p}_{\geqslant d_0})$ and $z_i \in T_+$.

  • $\begingroup$ This is great, thanks a lot! $\endgroup$ Mar 11, 2015 at 14:39
  • $\begingroup$ A related question: It is possible to see the continuity of the inverse of $\mathfrak{p}\mapsto\varphi^{-1}(\mathfrak{p})$ directly from this explicit formula? $\endgroup$
    – Yuxiao Xie
    Jul 24, 2020 at 1:29

In view of the complexity of the accepted answer, I want to sketch an alternate characterization of a point $\mathfrak q\in\operatorname{Proj} T$ lying over a point $\mathfrak p\in\operatorname{Proj} S$. First, as a matter of fact, $\ker\varphi\subset\mathfrak p$ for all $\mathfrak p\in\operatorname{Proj} S$, so (since $\varphi$ is a graded map) we may reduce to considering $S\subset T$ a subring with $S_d=T_d$ for $d\geq d_0$.

Let $\mathfrak p\in\operatorname{Proj} S$. Then $\mathfrak p\not\supset S_+$, and in fact $\exists w\in\bigoplus_{d\geq d_0}T_d$ homogeneous such that $w\notin\mathfrak p$. Put

$$\mathfrak q:=\{a\in T:aw\in\mathfrak p\}.$$

Then one can show that $\mathfrak q\in\operatorname{Proj} T$ and it is immediate that $\mathfrak q\cap S=\mathfrak p$.

So $f^{-1}(\mathfrak p)=\mathfrak q$ (showing injectivity of $f$ is not hard).

  • 1
    $\begingroup$ I must say that it is not at all obvious to me how $\mathfrak{q}$ is an ideal in $T$. Also, perhaps you wanted $w\notin \mathfrak{p}$. $\endgroup$ Mar 27, 2017 at 0:15
  • 2
    $\begingroup$ Thanks I meant $w\notin\mathfrak p$. To see $\mathfrak q$ is an ideal, let $b\in T, a\in\mathfrak q$. Then $bw\in S$ and so $abw^2\in\mathfrak p$. But $w\in S,w\notin\mathfrak p,abw\in S$, and $\mathfrak p$ is prime, so $abw\in\mathfrak p$. To see $\mathfrak q$ is prime, just take $a,b\in T$ not in $\mathfrak q$. Then $abw^2\notin\mathfrak p$ so $abw\notin\mathfrak p$ so $ab\notin\mathfrak q$. Homogeneity is easy. $\endgroup$
    – Tomo
    Jul 18, 2018 at 17:34
  • $\begingroup$ Why $\ker\varphi\subset\mathfrak p$ for all $\mathfrak p\in\operatorname{Proj} S$? I'm not sure about it on the degree $0$ component. $\endgroup$
    – Yuxiao Xie
    Jul 24, 2020 at 0:38
  • $\begingroup$ Dear @Colescu, suppose otherwise, and you’ve found $a\in\ker\varphi$ with $a\not\in\mathfrak p\in\operatorname{Proj} S$. Then as $\mathfrak p$ does not contain all of $S_+$, there is some $b\in S_+, b\not\in\mathfrak p$. Then $ab^n=0$ for some $n$, contradicting the primality of $\mathfrak p$. $\endgroup$
    – Tomo
    Jul 24, 2020 at 17:06
  • $\begingroup$ @Tomo I see. It turns out that the counterexample I had in mind contained all of $S_+$, which is forbidden here. Thanks. :) $\endgroup$
    – Yuxiao Xie
    Jul 25, 2020 at 2:11

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