Interesting invariants of a scheme satisfying additivity in the product
$begingroup$
I am curious if there is an invariant of a scheme satisfying $tau(X times_{k} Y) = tau(X) + tau(Y)$ for $X$ and $Y$ schemes over a field $k$. (Of course, except the dimension). This might seem like a vague question. But, I am trying to prove that a certain product scheme has a property, and the quickest way seems to be relating it to such an invariant, hence the question.
algebraic-geometry schemes
$endgroup$
add a comment |
$begingroup$
I am curious if there is an invariant of a scheme satisfying $tau(X times_{k} Y) = tau(X) + tau(Y)$ for $X$ and $Y$ schemes over a field $k$. (Of course, except the dimension). This might seem like a vague question. But, I am trying to prove that a certain product scheme has a property, and the quickest way seems to be relating it to such an invariant, hence the question.
algebraic-geometry schemes
$endgroup$
$begingroup$
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
$endgroup$
– Minseon Shin
Dec 1 '18 at 20:08
add a comment |
$begingroup$
I am curious if there is an invariant of a scheme satisfying $tau(X times_{k} Y) = tau(X) + tau(Y)$ for $X$ and $Y$ schemes over a field $k$. (Of course, except the dimension). This might seem like a vague question. But, I am trying to prove that a certain product scheme has a property, and the quickest way seems to be relating it to such an invariant, hence the question.
algebraic-geometry schemes
$endgroup$
I am curious if there is an invariant of a scheme satisfying $tau(X times_{k} Y) = tau(X) + tau(Y)$ for $X$ and $Y$ schemes over a field $k$. (Of course, except the dimension). This might seem like a vague question. But, I am trying to prove that a certain product scheme has a property, and the quickest way seems to be relating it to such an invariant, hence the question.
algebraic-geometry schemes
algebraic-geometry schemes
asked Dec 1 '18 at 10:41
user73260user73260
112
112
$begingroup$
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
$endgroup$
– Minseon Shin
Dec 1 '18 at 20:08
add a comment |
$begingroup$
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
$endgroup$
– Minseon Shin
Dec 1 '18 at 20:08
$begingroup$
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
$endgroup$
– Minseon Shin
Dec 1 '18 at 20:08
$begingroup$
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
$endgroup$
– Minseon Shin
Dec 1 '18 at 20:08
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%2f3021213%2finteresting-invariants-of-a-scheme-satisfying-additivity-in-the-product%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%2f3021213%2finteresting-invariants-of-a-scheme-satisfying-additivity-in-the-product%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
$begingroup$
If $X,Y$ are not of finite type, then the dimension may not even be additive, consider $X = Y = k(T)$ in which case $X times_{k} Y$ has dimension 1.
$endgroup$
– Minseon Shin
Dec 1 '18 at 20:08