partial derivative of a polynomial belongs to a maximal ideal
$begingroup$
If we consider an affine space $mathbb{A}_K^n=mathrm{Spec},K[T_1,cdots,T_m]$ over a field $K$. It's easy to show that $T_xmathbb{A}_K^nsimeq K^n$ where $x$ is a $K$-point corresponding to the maximal ideal of the form $(T_1-x_1,cdots,T_n-x_n)$. But I wonder how to show that $dim T_xmathbb{A}_K^n=n$ (or maybe fail to equal) for a general closed point correspond to a general maximal ideal $mathfrak m$.
I tried to consider the map $mathfrak mto kappa(x)^n,,gmapsto(frac{partial g}{partial T_1}(x),cdots,frac{partial g}{partial T_n}(x))$. If $x$ is $K$-point, it's easy to show that the kernel is $mathfrak m^2$, and induced a bijection: $mathfrak m/mathfrak m^2to kappa(x)^n=K^n$. But in the general case, is that right? I think it's just a injection. This is equivalent to prove the following:
Conjecture: If $ginmathfrak m$ and we have $dfrac{partial g}{partial T_i}inmathfrak m$ for all $i$, then $ginmathfrak m^2$.
If $mathfrak m=(T_1-x_1,cdots,T_n-x_n)$, it is verified by Taylor expansion. But for general maximal ideal, I don't know how to do it.
algebraic-geometry commutative-algebra
asked Nov 24 '18 at 1:30
user8891548user8891548
513
$endgroup$
add a comment |
$begingroup$
If we consider an affine space $mathbb{A}_K^n=mathrm{Spec},K[T_1,cdots,T_m]$ over a field $K$. It's easy to show that $T_xmathbb{A}_K^nsimeq K^n$ where $x$ is a $K$-point corresponding to the maximal ideal of the form $(T_1-x_1,cdots,T_n-x_n)$. But I wonder how to show that $dim T_xmathbb{A}_K^n=n$ (or maybe fail to equal) for a general closed point correspond to a general maximal ideal $mathfrak m$.
I tried to consider the map $mathfrak mto kappa(x)^n,,gmapsto(frac{partial g}{partial T_1}(x),cdots,frac{partial g}{partial T_n}(x))$. If $x$ is $K$-point, it's easy to show that the kernel is $mathfrak m^2$, and induced a bijection: $mathfrak m/mathfrak m^2to kappa(x)^n=K^n$. But in the general case, is that right? I think it's just a injection. This is equivalent to prove the following:
Conjecture: If $ginmathfrak m$ and we have $dfrac{partial g}{partial T_i}inmathfrak m$ for all $i$, then $ginmathfrak m^2$.
If $mathfrak m=(T_1-x_1,cdots,T_n-x_n)$, it is verified by Taylor expansion. But for general maximal ideal, I don't know how to do it.
algebraic-geometry commutative-algebra
asked Nov 24 '18 at 1:30
user8891548user8891548
513
$endgroup$
$begingroup$
@baharampuri The $f$ seems not contained in $mathfrak m$.
$endgroup$
– user8891548
Nov 24 '18 at 7:09
$begingroup$
Sorry missed that.
$endgroup$
– baharampuri
Nov 24 '18 at 7:22
$begingroup$
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
$endgroup$
– KReiser
Nov 24 '18 at 8:06
$begingroup$
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
$endgroup$
– user8891548
Nov 24 '18 at 8:45
add a comment |
$begingroup$
If we consider an affine space $mathbb{A}_K^n=mathrm{Spec},K[T_1,cdots,T_m]$ over a field $K$. It's easy to show that $T_xmathbb{A}_K^nsimeq K^n$ where $x$ is a $K$-point corresponding to the maximal ideal of the form $(T_1-x_1,cdots,T_n-x_n)$. But I wonder how to show that $dim T_xmathbb{A}_K^n=n$ (or maybe fail to equal) for a general closed point correspond to a general maximal ideal $mathfrak m$.
I tried to consider the map $mathfrak mto kappa(x)^n,,gmapsto(frac{partial g}{partial T_1}(x),cdots,frac{partial g}{partial T_n}(x))$. If $x$ is $K$-point, it's easy to show that the kernel is $mathfrak m^2$, and induced a bijection: $mathfrak m/mathfrak m^2to kappa(x)^n=K^n$. But in the general case, is that right? I think it's just a injection. This is equivalent to prove the following:
Conjecture: If $ginmathfrak m$ and we have $dfrac{partial g}{partial T_i}inmathfrak m$ for all $i$, then $ginmathfrak m^2$.
If $mathfrak m=(T_1-x_1,cdots,T_n-x_n)$, it is verified by Taylor expansion. But for general maximal ideal, I don't know how to do it.
algebraic-geometry commutative-algebra
asked Nov 24 '18 at 1:30
user8891548user8891548
513
$endgroup$
If we consider an affine space $mathbb{A}_K^n=mathrm{Spec},K[T_1,cdots,T_m]$ over a field $K$. It's easy to show that $T_xmathbb{A}_K^nsimeq K^n$ where $x$ is a $K$-point corresponding to the maximal ideal of the form $(T_1-x_1,cdots,T_n-x_n)$. But I wonder how to show that $dim T_xmathbb{A}_K^n=n$ (or maybe fail to equal) for a general closed point correspond to a general maximal ideal $mathfrak m$.
I tried to consider the map $mathfrak mto kappa(x)^n,,gmapsto(frac{partial g}{partial T_1}(x),cdots,frac{partial g}{partial T_n}(x))$. If $x$ is $K$-point, it's easy to show that the kernel is $mathfrak m^2$, and induced a bijection: $mathfrak m/mathfrak m^2to kappa(x)^n=K^n$. But in the general case, is that right? I think it's just a injection. This is equivalent to prove the following:
Conjecture: If $ginmathfrak m$ and we have $dfrac{partial g}{partial T_i}inmathfrak m$ for all $i$, then $ginmathfrak m^2$.
If $mathfrak m=(T_1-x_1,cdots,T_n-x_n)$, it is verified by Taylor expansion. But for general maximal ideal, I don't know how to do it.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Nov 24 '18 at 1:30
user8891548user8891548
513
asked Nov 24 '18 at 1:30
user8891548user8891548
513
edited Nov 24 '18 at 6:51
user8891548
asked Nov 24 '18 at 1:30
user8891548user8891548
513
asked Nov 24 '18 at 1:30
user8891548user8891548
513
asked Nov 24 '18 at 1:30
user8891548user8891548
513
513
$begingroup$
@baharampuri The $f$ seems not contained in $mathfrak m$.
$endgroup$
– user8891548
Nov 24 '18 at 7:09
$begingroup$
Sorry missed that.
$endgroup$
– baharampuri
Nov 24 '18 at 7:22
$begingroup$
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
$endgroup$
– KReiser
Nov 24 '18 at 8:06
$begingroup$
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
$endgroup$
– user8891548
Nov 24 '18 at 8:45
add a comment |
$begingroup$
@baharampuri The $f$ seems not contained in $mathfrak m$.
$endgroup$
– user8891548
Nov 24 '18 at 7:09
$begingroup$
Sorry missed that.
$endgroup$
– baharampuri
Nov 24 '18 at 7:22
$begingroup$
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
$endgroup$
– KReiser
Nov 24 '18 at 8:06
$begingroup$
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
$endgroup$
– user8891548
Nov 24 '18 at 8:45
$begingroup$
@baharampuri The $f$ seems not contained in $mathfrak m$.
$endgroup$
– user8891548
Nov 24 '18 at 7:09
$begingroup$
@baharampuri The $f$ seems not contained in $mathfrak m$.
$endgroup$
– user8891548
Nov 24 '18 at 7:09
$begingroup$
Sorry missed that.
$endgroup$
– baharampuri
Nov 24 '18 at 7:22
$begingroup$
Sorry missed that.
$endgroup$
– baharampuri
Nov 24 '18 at 7:22
$begingroup$
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
$endgroup$
– KReiser
Nov 24 '18 at 8:06
$begingroup$
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
$endgroup$
– KReiser
Nov 24 '18 at 8:06
$begingroup$
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
$endgroup$
– user8891548
Nov 24 '18 at 8:45
$begingroup$
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
$endgroup$
– user8891548
Nov 24 '18 at 8:45
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%2f3011065%2fpartial-derivative-of-a-polynomial-belongs-to-a-maximal-ideal%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%2f3011065%2fpartial-derivative-of-a-polynomial-belongs-to-a-maximal-ideal%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
$begingroup$
@baharampuri The $f$ seems not contained in $mathfrak m$.
$endgroup$
– user8891548
Nov 24 '18 at 7:09
$begingroup$
Sorry missed that.
$endgroup$
– baharampuri
Nov 24 '18 at 7:22
$begingroup$
One clarifying question: which field are you considering $T_xBbb A^n_K$ as a vector space over when you ask for its dimension? If you ask over the residue field at $x$, the answer is always $n$, whereas if you ask over the field $K$, the dimension is $n$ times the degree of the residue field as an extension over $K$.
$endgroup$
– KReiser
Nov 24 '18 at 8:06
$begingroup$
@KReiser $T_xmathbb A_K^n$ considered as a $kappa(x)$-vector space. May I ask how to prove that the dimension is $n$?
$endgroup$
– user8891548
Nov 24 '18 at 8:45