Relation between the Zariski topology on $k^n$ and $operatorname{MaxSpec}k[x_1,dots,x_n]$
$begingroup$
There is the Zariski topology on the set $k^n$. There is also the Zariski topology on the set of prime ideals of $k[x_1,dots,x_n]$. I was wondering if anyone could explain, on a basic level, without appealing to coordinate rings or tensor products, how these two are related? I do understand that if $k$ is algebraically closed, the points of $k^n$ correspond bijectively to maximal ideals of the polynomial ring. Wouldn't it be natural to define the topology on the set of maximal ideals? Why does one enlarge the set (from maximal ideals to prime ideals) on which the topology is being defined? What is the motivation for the definition of the Zariski topology on the set of prime ideals (how to come up with such definition?) And by the way, are $k^n$ and the set of maximal ideals of $k[x_1,dots,x_n]$ homeomorphic if each of them is given the Zariski topology?
abstract-algebra algebraic-geometry commutative-algebra zariski-topology
asked Jan 21 at 23:10
user437309user437309
695313
$endgroup$
add a comment |
$begingroup$
There is the Zariski topology on the set $k^n$. There is also the Zariski topology on the set of prime ideals of $k[x_1,dots,x_n]$. I was wondering if anyone could explain, on a basic level, without appealing to coordinate rings or tensor products, how these two are related? I do understand that if $k$ is algebraically closed, the points of $k^n$ correspond bijectively to maximal ideals of the polynomial ring. Wouldn't it be natural to define the topology on the set of maximal ideals? Why does one enlarge the set (from maximal ideals to prime ideals) on which the topology is being defined? What is the motivation for the definition of the Zariski topology on the set of prime ideals (how to come up with such definition?) And by the way, are $k^n$ and the set of maximal ideals of $k[x_1,dots,x_n]$ homeomorphic if each of them is given the Zariski topology?
abstract-algebra algebraic-geometry commutative-algebra zariski-topology
asked Jan 21 at 23:10
user437309user437309
695313
$endgroup$
$begingroup$
They are in fact homeomorphic
$endgroup$
– leibnewtz
Jan 22 at 2:07
add a comment |
$begingroup$
There is the Zariski topology on the set $k^n$. There is also the Zariski topology on the set of prime ideals of $k[x_1,dots,x_n]$. I was wondering if anyone could explain, on a basic level, without appealing to coordinate rings or tensor products, how these two are related? I do understand that if $k$ is algebraically closed, the points of $k^n$ correspond bijectively to maximal ideals of the polynomial ring. Wouldn't it be natural to define the topology on the set of maximal ideals? Why does one enlarge the set (from maximal ideals to prime ideals) on which the topology is being defined? What is the motivation for the definition of the Zariski topology on the set of prime ideals (how to come up with such definition?) And by the way, are $k^n$ and the set of maximal ideals of $k[x_1,dots,x_n]$ homeomorphic if each of them is given the Zariski topology?
abstract-algebra algebraic-geometry commutative-algebra zariski-topology
asked Jan 21 at 23:10
user437309user437309
695313
$endgroup$
There is the Zariski topology on the set $k^n$. There is also the Zariski topology on the set of prime ideals of $k[x_1,dots,x_n]$. I was wondering if anyone could explain, on a basic level, without appealing to coordinate rings or tensor products, how these two are related? I do understand that if $k$ is algebraically closed, the points of $k^n$ correspond bijectively to maximal ideals of the polynomial ring. Wouldn't it be natural to define the topology on the set of maximal ideals? Why does one enlarge the set (from maximal ideals to prime ideals) on which the topology is being defined? What is the motivation for the definition of the Zariski topology on the set of prime ideals (how to come up with such definition?) And by the way, are $k^n$ and the set of maximal ideals of $k[x_1,dots,x_n]$ homeomorphic if each of them is given the Zariski topology?
abstract-algebra algebraic-geometry commutative-algebra zariski-topology
abstract-algebra algebraic-geometry commutative-algebra zariski-topology
asked Jan 21 at 23:10
user437309user437309
695313
asked Jan 21 at 23:10
user437309user437309
695313
asked Jan 21 at 23:10
user437309user437309
695313
asked Jan 21 at 23:10
user437309user437309
695313
asked Jan 21 at 23:10
user437309user437309
695313
695313
$begingroup$
They are in fact homeomorphic
$endgroup$
– leibnewtz
Jan 22 at 2:07
add a comment |
$begingroup$
They are in fact homeomorphic
$endgroup$
– leibnewtz
Jan 22 at 2:07
$begingroup$
They are in fact homeomorphic
$endgroup$
– leibnewtz
Jan 22 at 2:07
$begingroup$
They are in fact homeomorphic
$endgroup$
– leibnewtz
Jan 22 at 2:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One of the reasons we use the prime spectrum is because of the following: if $varphi:Rto S$ is a ring homomorphism and $mathfrak{p}$ is a prime ideal of $S$, then $varphi^{-1}(mathfrak{p})$ is a prime ideal of $R$. This allows us to view $mathrm{Spec}$ as a functor $mathrm{CommRing^{op}}tomathrm{LocRing}$, where $mathrm{LocRing}$ is the category of locally ringed spaces. One thing we get is an equivalence of categories $mathrm{CommRing^{op}}simeqmathrm{AffSch}$ (the other direction is given by the global sections functor).
It is natural to consider $mathrm{MaxSpec}$, but not all the rings we want to consider are algebraically closed fields. In addition, we lose extra information like (non)closed points.
The motivation for the Zariski topology is sort of extending the Zariski topology from algebraically closed fields to all commutative rings. Given $mathrm{Spec},R$, we can form an open basis of subsets $D(f)$ for $fin R$, $D(f)={mathfrak{p}inmathrm{Spec},R : fnotinmathfrak{p}}$. These $D(f)$ can be thought of where $f$ does not vanish.
As for why $mathrm{MaxSpec},k[x_1,dots,x_n]simeq k^n$, this is one way to state Hilbert's Nullstellensatz.
answered Jan 22 at 0:09
Ryan KeletiRyan Keleti
1469
$endgroup$
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%2f3082536%2frelation-between-the-zariski-topology-on-kn-and-operatornamemaxspeckx-1%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
$begingroup$
One of the reasons we use the prime spectrum is because of the following: if $varphi:Rto S$ is a ring homomorphism and $mathfrak{p}$ is a prime ideal of $S$, then $varphi^{-1}(mathfrak{p})$ is a prime ideal of $R$. This allows us to view $mathrm{Spec}$ as a functor $mathrm{CommRing^{op}}tomathrm{LocRing}$, where $mathrm{LocRing}$ is the category of locally ringed spaces. One thing we get is an equivalence of categories $mathrm{CommRing^{op}}simeqmathrm{AffSch}$ (the other direction is given by the global sections functor).
It is natural to consider $mathrm{MaxSpec}$, but not all the rings we want to consider are algebraically closed fields. In addition, we lose extra information like (non)closed points.
The motivation for the Zariski topology is sort of extending the Zariski topology from algebraically closed fields to all commutative rings. Given $mathrm{Spec},R$, we can form an open basis of subsets $D(f)$ for $fin R$, $D(f)={mathfrak{p}inmathrm{Spec},R : fnotinmathfrak{p}}$. These $D(f)$ can be thought of where $f$ does not vanish.
As for why $mathrm{MaxSpec},k[x_1,dots,x_n]simeq k^n$, this is one way to state Hilbert's Nullstellensatz.
answered Jan 22 at 0:09
Ryan KeletiRyan Keleti
1469
$endgroup$
add a comment |
$begingroup$
One of the reasons we use the prime spectrum is because of the following: if $varphi:Rto S$ is a ring homomorphism and $mathfrak{p}$ is a prime ideal of $S$, then $varphi^{-1}(mathfrak{p})$ is a prime ideal of $R$. This allows us to view $mathrm{Spec}$ as a functor $mathrm{CommRing^{op}}tomathrm{LocRing}$, where $mathrm{LocRing}$ is the category of locally ringed spaces. One thing we get is an equivalence of categories $mathrm{CommRing^{op}}simeqmathrm{AffSch}$ (the other direction is given by the global sections functor).
It is natural to consider $mathrm{MaxSpec}$, but not all the rings we want to consider are algebraically closed fields. In addition, we lose extra information like (non)closed points.
The motivation for the Zariski topology is sort of extending the Zariski topology from algebraically closed fields to all commutative rings. Given $mathrm{Spec},R$, we can form an open basis of subsets $D(f)$ for $fin R$, $D(f)={mathfrak{p}inmathrm{Spec},R : fnotinmathfrak{p}}$. These $D(f)$ can be thought of where $f$ does not vanish.
As for why $mathrm{MaxSpec},k[x_1,dots,x_n]simeq k^n$, this is one way to state Hilbert's Nullstellensatz.
answered Jan 22 at 0:09
Ryan KeletiRyan Keleti
1469
$endgroup$
add a comment |
$begingroup$
One of the reasons we use the prime spectrum is because of the following: if $varphi:Rto S$ is a ring homomorphism and $mathfrak{p}$ is a prime ideal of $S$, then $varphi^{-1}(mathfrak{p})$ is a prime ideal of $R$. This allows us to view $mathrm{Spec}$ as a functor $mathrm{CommRing^{op}}tomathrm{LocRing}$, where $mathrm{LocRing}$ is the category of locally ringed spaces. One thing we get is an equivalence of categories $mathrm{CommRing^{op}}simeqmathrm{AffSch}$ (the other direction is given by the global sections functor).
It is natural to consider $mathrm{MaxSpec}$, but not all the rings we want to consider are algebraically closed fields. In addition, we lose extra information like (non)closed points.
The motivation for the Zariski topology is sort of extending the Zariski topology from algebraically closed fields to all commutative rings. Given $mathrm{Spec},R$, we can form an open basis of subsets $D(f)$ for $fin R$, $D(f)={mathfrak{p}inmathrm{Spec},R : fnotinmathfrak{p}}$. These $D(f)$ can be thought of where $f$ does not vanish.
As for why $mathrm{MaxSpec},k[x_1,dots,x_n]simeq k^n$, this is one way to state Hilbert's Nullstellensatz.
answered Jan 22 at 0:09
Ryan KeletiRyan Keleti
1469
$endgroup$
One of the reasons we use the prime spectrum is because of the following: if $varphi:Rto S$ is a ring homomorphism and $mathfrak{p}$ is a prime ideal of $S$, then $varphi^{-1}(mathfrak{p})$ is a prime ideal of $R$. This allows us to view $mathrm{Spec}$ as a functor $mathrm{CommRing^{op}}tomathrm{LocRing}$, where $mathrm{LocRing}$ is the category of locally ringed spaces. One thing we get is an equivalence of categories $mathrm{CommRing^{op}}simeqmathrm{AffSch}$ (the other direction is given by the global sections functor).
It is natural to consider $mathrm{MaxSpec}$, but not all the rings we want to consider are algebraically closed fields. In addition, we lose extra information like (non)closed points.
The motivation for the Zariski topology is sort of extending the Zariski topology from algebraically closed fields to all commutative rings. Given $mathrm{Spec},R$, we can form an open basis of subsets $D(f)$ for $fin R$, $D(f)={mathfrak{p}inmathrm{Spec},R : fnotinmathfrak{p}}$. These $D(f)$ can be thought of where $f$ does not vanish.
As for why $mathrm{MaxSpec},k[x_1,dots,x_n]simeq k^n$, this is one way to state Hilbert's Nullstellensatz.
answered Jan 22 at 0:09
Ryan KeletiRyan Keleti
1469
answered Jan 22 at 0:09
Ryan KeletiRyan Keleti
1469
answered Jan 22 at 0:09
Ryan KeletiRyan Keleti
1469
answered Jan 22 at 0:09
Ryan KeletiRyan Keleti
1469
1469
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.
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%2f3082536%2frelation-between-the-zariski-topology-on-kn-and-operatornamemaxspeckx-1%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
$begingroup$
They are in fact homeomorphic
$endgroup$
– leibnewtz
Jan 22 at 2:07