Mixing
Two definitions of a projective morphism
In Hartshorne's Algebraic Geometry, page 103, a morphism $f: X rightarrow Y$ is said to be projective if it factors as a closed immersion $X rightarrow {bf P}^n_Y$ followed by the projection ${bf P}^n_Y rightarrow Y$. As noted there, EGA II, 5.5 has another definition,
namely $f$ is projective if it factors as a closed immersion $X rightarrow {bf P}(cal E)$ followed by the projection map, where $cal E$ is finite-type quasi-coherent ${cal O}_Y$-module.
Hartshorne states without proof nor reference that the two definitions "are equivalent in case $Y$ itself is quasi-projective over an affine scheme".
My question is: does anyone know a proof or a reference for this statement? And if not: is it correct?
algebraic-geometry
asked Nov 20 '18 at 19:05
Visitor
335111
add a comment |
In Hartshorne's Algebraic Geometry, page 103, a morphism $f: X rightarrow Y$ is said to be projective if it factors as a closed immersion $X rightarrow {bf P}^n_Y$ followed by the projection ${bf P}^n_Y rightarrow Y$. As noted there, EGA II, 5.5 has another definition,
namely $f$ is projective if it factors as a closed immersion $X rightarrow {bf P}(cal E)$ followed by the projection map, where $cal E$ is finite-type quasi-coherent ${cal O}_Y$-module.
Hartshorne states without proof nor reference that the two definitions "are equivalent in case $Y$ itself is quasi-projective over an affine scheme".
My question is: does anyone know a proof or a reference for this statement? And if not: is it correct?
algebraic-geometry
asked Nov 20 '18 at 19:05
Visitor
335111
2
It is a good idea to say what is the second definition.
– Sasha
Nov 20 '18 at 19:07
@sasha: You're right. That's done now.
– Visitor
Nov 20 '18 at 20:10
add a comment |
In Hartshorne's Algebraic Geometry, page 103, a morphism $f: X rightarrow Y$ is said to be projective if it factors as a closed immersion $X rightarrow {bf P}^n_Y$ followed by the projection ${bf P}^n_Y rightarrow Y$. As noted there, EGA II, 5.5 has another definition,
namely $f$ is projective if it factors as a closed immersion $X rightarrow {bf P}(cal E)$ followed by the projection map, where $cal E$ is finite-type quasi-coherent ${cal O}_Y$-module.
Hartshorne states without proof nor reference that the two definitions "are equivalent in case $Y$ itself is quasi-projective over an affine scheme".
My question is: does anyone know a proof or a reference for this statement? And if not: is it correct?
algebraic-geometry
asked Nov 20 '18 at 19:05
Visitor
335111
In Hartshorne's Algebraic Geometry, page 103, a morphism $f: X rightarrow Y$ is said to be projective if it factors as a closed immersion $X rightarrow {bf P}^n_Y$ followed by the projection ${bf P}^n_Y rightarrow Y$. As noted there, EGA II, 5.5 has another definition,
namely $f$ is projective if it factors as a closed immersion $X rightarrow {bf P}(cal E)$ followed by the projection map, where $cal E$ is finite-type quasi-coherent ${cal O}_Y$-module.
Hartshorne states without proof nor reference that the two definitions "are equivalent in case $Y$ itself is quasi-projective over an affine scheme".
My question is: does anyone know a proof or a reference for this statement? And if not: is it correct?
algebraic-geometry
algebraic-geometry
asked Nov 20 '18 at 19:05
Visitor
335111
asked Nov 20 '18 at 19:05
Visitor
335111
edited Nov 20 '18 at 20:10
asked Nov 20 '18 at 19:05
Visitor
335111
asked Nov 20 '18 at 19:05
Visitor
335111
asked Nov 20 '18 at 19:05
Visitor
335111
335111
2
It is a good idea to say what is the second definition.
– Sasha
Nov 20 '18 at 19:07
@sasha: You're right. That's done now.
– Visitor
Nov 20 '18 at 20:10
add a comment |
2
It is a good idea to say what is the second definition.
– Sasha
Nov 20 '18 at 19:07
@sasha: You're right. That's done now.
– Visitor
Nov 20 '18 at 20:10
2
2
It is a good idea to say what is the second definition.
– Sasha
Nov 20 '18 at 19:07
It is a good idea to say what is the second definition.
– Sasha
Nov 20 '18 at 19:07
@sasha: You're right. That's done now.
– Visitor
Nov 20 '18 at 20:10
@sasha: You're right. That's done now.
– Visitor
Nov 20 '18 at 20:10
add a comment |
1 Answer
1
active
oldest
votes
Definitely, if $X$ is "Hartshorne-projective" it is also "EGA-projective" (take $mathcal{E}$ to be a free bundle of rank $n+1$). The opposite direction is true if any coherent sheaf is globally generated by a finite-dimensional vector space of sections after some line bundle twist; indeed, if $V$ generates $mathcal{E} otimes L$ then the surjection $V otimes mathcal{O}_Y to mathcal{E} otimes L$ induces a closed embedding
$$
mathbb{P}(mathcal{E}) = mathbb{P}(mathcal{E} otimes L) to mathbb{P}(V otimes mathcal{O}_Y) = mathbb{P}^n_Y.
$$
So, for instance if $Y$ satisfies reasonable finiteness conditions and admits an ample line bundle, the definitions are equivalent.
answered Nov 20 '18 at 20:37
Sasha
4,353139
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006750%2ftwo-definitions-of-a-projective-morphism%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Definitely, if $X$ is "Hartshorne-projective" it is also "EGA-projective" (take $mathcal{E}$ to be a free bundle of rank $n+1$). The opposite direction is true if any coherent sheaf is globally generated by a finite-dimensional vector space of sections after some line bundle twist; indeed, if $V$ generates $mathcal{E} otimes L$ then the surjection $V otimes mathcal{O}_Y to mathcal{E} otimes L$ induces a closed embedding
$$
mathbb{P}(mathcal{E}) = mathbb{P}(mathcal{E} otimes L) to mathbb{P}(V otimes mathcal{O}_Y) = mathbb{P}^n_Y.
$$
So, for instance if $Y$ satisfies reasonable finiteness conditions and admits an ample line bundle, the definitions are equivalent.
answered Nov 20 '18 at 20:37
Sasha
4,353139
add a comment |
Definitely, if $X$ is "Hartshorne-projective" it is also "EGA-projective" (take $mathcal{E}$ to be a free bundle of rank $n+1$). The opposite direction is true if any coherent sheaf is globally generated by a finite-dimensional vector space of sections after some line bundle twist; indeed, if $V$ generates $mathcal{E} otimes L$ then the surjection $V otimes mathcal{O}_Y to mathcal{E} otimes L$ induces a closed embedding
$$
mathbb{P}(mathcal{E}) = mathbb{P}(mathcal{E} otimes L) to mathbb{P}(V otimes mathcal{O}_Y) = mathbb{P}^n_Y.
$$
So, for instance if $Y$ satisfies reasonable finiteness conditions and admits an ample line bundle, the definitions are equivalent.
answered Nov 20 '18 at 20:37
Sasha
4,353139
add a comment |
Definitely, if $X$ is "Hartshorne-projective" it is also "EGA-projective" (take $mathcal{E}$ to be a free bundle of rank $n+1$). The opposite direction is true if any coherent sheaf is globally generated by a finite-dimensional vector space of sections after some line bundle twist; indeed, if $V$ generates $mathcal{E} otimes L$ then the surjection $V otimes mathcal{O}_Y to mathcal{E} otimes L$ induces a closed embedding
$$
mathbb{P}(mathcal{E}) = mathbb{P}(mathcal{E} otimes L) to mathbb{P}(V otimes mathcal{O}_Y) = mathbb{P}^n_Y.
$$
So, for instance if $Y$ satisfies reasonable finiteness conditions and admits an ample line bundle, the definitions are equivalent.
answered Nov 20 '18 at 20:37
Sasha
4,353139
Definitely, if $X$ is "Hartshorne-projective" it is also "EGA-projective" (take $mathcal{E}$ to be a free bundle of rank $n+1$). The opposite direction is true if any coherent sheaf is globally generated by a finite-dimensional vector space of sections after some line bundle twist; indeed, if $V$ generates $mathcal{E} otimes L$ then the surjection $V otimes mathcal{O}_Y to mathcal{E} otimes L$ induces a closed embedding
$$
mathbb{P}(mathcal{E}) = mathbb{P}(mathcal{E} otimes L) to mathbb{P}(V otimes mathcal{O}_Y) = mathbb{P}^n_Y.
$$
So, for instance if $Y$ satisfies reasonable finiteness conditions and admits an ample line bundle, the definitions are equivalent.
answered Nov 20 '18 at 20:37
Sasha
4,353139
answered Nov 20 '18 at 20:37
Sasha
4,353139
answered Nov 20 '18 at 20:37
Sasha
4,353139
answered Nov 20 '18 at 20:37
Sasha
4,353139
4,353139
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006750%2ftwo-definitions-of-a-projective-morphism%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
It is a good idea to say what is the second definition.
– Sasha
Nov 20 '18 at 19:07
@sasha: You're right. That's done now.
– Visitor
Nov 20 '18 at 20:10