A projective variety is irreducible if and only if it does not contain a line
$begingroup$
Let $X$ be a proper closed subset of $mathbb P^2_{mathbb C}$. Is it true that $X$ is irreducible if and only if it does not contain a line $ax+by+cz = 0$? I have never seen or thought about irreducibility this way.
This is claimed in these notes on elliptic curves in the proof that an elliptic curve is irreducible.
algebraic-geometry
asked Dec 16 '18 at 22:49
D_SD_S
13.5k51551
$endgroup$
add a comment |
$begingroup$
Let $X$ be a proper closed subset of $mathbb P^2_{mathbb C}$. Is it true that $X$ is irreducible if and only if it does not contain a line $ax+by+cz = 0$? I have never seen or thought about irreducibility this way.
This is claimed in these notes on elliptic curves in the proof that an elliptic curve is irreducible.
algebraic-geometry
asked Dec 16 '18 at 22:49
D_SD_S
13.5k51551
$endgroup$
4
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18
add a comment |
$begingroup$
Let $X$ be a proper closed subset of $mathbb P^2_{mathbb C}$. Is it true that $X$ is irreducible if and only if it does not contain a line $ax+by+cz = 0$? I have never seen or thought about irreducibility this way.
This is claimed in these notes on elliptic curves in the proof that an elliptic curve is irreducible.
algebraic-geometry
asked Dec 16 '18 at 22:49
D_SD_S
13.5k51551
$endgroup$
Let $X$ be a proper closed subset of $mathbb P^2_{mathbb C}$. Is it true that $X$ is irreducible if and only if it does not contain a line $ax+by+cz = 0$? I have never seen or thought about irreducibility this way.
This is claimed in these notes on elliptic curves in the proof that an elliptic curve is irreducible.
algebraic-geometry
algebraic-geometry
asked Dec 16 '18 at 22:49
D_SD_S
13.5k51551
asked Dec 16 '18 at 22:49
D_SD_S
13.5k51551
asked Dec 16 '18 at 22:49
D_SD_S
13.5k51551
asked Dec 16 '18 at 22:49
D_SD_S
13.5k51551
asked Dec 16 '18 at 22:49
D_SD_S
13.5k51551
13.5k51551
4
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18
add a comment |
4
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18
4
4
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18
add a comment |
0
active
oldest
votes
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%2f3043290%2fa-projective-variety-is-irreducible-if-and-only-if-it-does-not-contain-a-line%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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%2f3043290%2fa-projective-variety-is-irreducible-if-and-only-if-it-does-not-contain-a-line%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
4
$begingroup$
For an elliptic curve (which in this case means a cubic) what you say is correct. The only non-trivial decomposition for 3 is $2+1$ or $1+1+1$. Higher degree curves, this is false.
$endgroup$
– Mohan
Dec 16 '18 at 23:05
$begingroup$
You're saying if $X$ is a projective curve defined by a cubic equation, then the proper irreducible subsets of $X$ can either be lines or quadratics?
$endgroup$
– D_S
Dec 16 '18 at 23:24
$begingroup$
Oh I see..if $X = Z(f)$ and $X_0 = Z(f_0)$ is contained in $X$ for $f, f_0$ homogeneous, then $f_0$ divides $f$.
$endgroup$
– D_S
Dec 17 '18 at 0:18