5
$\begingroup$

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 $B$.

$\endgroup$
3
$\begingroup$

$\newcommand{\proj}[1]{{{\mathrm{proj}}(#1)}}$ $\newcommand{\ideal}[1]{{\mathfrak #1}}$ 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 $z_i \in T_+$.

$\endgroup$
  • $\begingroup$ This is great, thanks a lot! $\endgroup$ – Tom Oldfield Mar 11 '15 at 14:39
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ I must say that it is not at all obvious to me how $\mathfrak{q}$ is an ideal in $T$. Also, perhaps you wanted $w\notin \mathfrak{p}$. $\endgroup$ – user2902293 Mar 27 '17 at 0:15
  • 1
    $\begingroup$ 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. $\endgroup$ – Owen Barrett Jul 18 '18 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.