Mixing
Fiber Product of Schemas
$begingroup$
I have a question about a equation in the proof of Prop. 3.1.16 in Liu's "Algebraic Geometry" (page 83):
$DeclareMathOperator{Spec}{Spec}$
Why the equation $(X times _Y V) times _V Spec k(y) = f^{-1} (V) _y$ holds, where $Q times _S R$ means the fiber product of $S$-schemes Q and R?
algebraic-geometry schemes
edited Sep 30 '17 at 15:43
Vim
8,18131348
asked Sep 19 '17 at 23:14
KarlPeterKarlPeter
5611316
$endgroup$
add a comment |
$begingroup$
I have a question about a equation in the proof of Prop. 3.1.16 in Liu's "Algebraic Geometry" (page 83):
$DeclareMathOperator{Spec}{Spec}$
Why the equation $(X times _Y V) times _V Spec k(y) = f^{-1} (V) _y$ holds, where $Q times _S R$ means the fiber product of $S$-schemes Q and R?
algebraic-geometry schemes
edited Sep 30 '17 at 15:43
Vim
8,18131348
asked Sep 19 '17 at 23:14
KarlPeterKarlPeter
5611316
$endgroup$
$begingroup$
Please keep questions like this coming. I am also learning algebraic geometry and this is a great way for me to work out the details : )
$endgroup$
– D_S
Sep 20 '17 at 12:54
add a comment |
$begingroup$
I have a question about a equation in the proof of Prop. 3.1.16 in Liu's "Algebraic Geometry" (page 83):
$DeclareMathOperator{Spec}{Spec}$
Why the equation $(X times _Y V) times _V Spec k(y) = f^{-1} (V) _y$ holds, where $Q times _S R$ means the fiber product of $S$-schemes Q and R?
algebraic-geometry schemes
edited Sep 30 '17 at 15:43
Vim
8,18131348
asked Sep 19 '17 at 23:14
KarlPeterKarlPeter
5611316
$endgroup$
I have a question about a equation in the proof of Prop. 3.1.16 in Liu's "Algebraic Geometry" (page 83):
$DeclareMathOperator{Spec}{Spec}$
Why the equation $(X times _Y V) times _V Spec k(y) = f^{-1} (V) _y$ holds, where $Q times _S R$ means the fiber product of $S$-schemes Q and R?
algebraic-geometry schemes
algebraic-geometry schemes
edited Sep 30 '17 at 15:43
Vim
8,18131348
asked Sep 19 '17 at 23:14
KarlPeterKarlPeter
5611316
edited Sep 30 '17 at 15:43
Vim
8,18131348
asked Sep 19 '17 at 23:14
KarlPeterKarlPeter
5611316
edited Sep 30 '17 at 15:43
Vim
8,18131348
edited Sep 30 '17 at 15:43
Vim
8,18131348
edited Sep 30 '17 at 15:43
Vim
8,18131348
8,18131348
asked Sep 19 '17 at 23:14
KarlPeterKarlPeter
5611316
asked Sep 19 '17 at 23:14
KarlPeterKarlPeter
5611316
asked Sep 19 '17 at 23:14
KarlPeterKarlPeter
5611316
5611316
$begingroup$
Please keep questions like this coming. I am also learning algebraic geometry and this is a great way for me to work out the details : )
$endgroup$
– D_S
Sep 20 '17 at 12:54
add a comment |
$begingroup$
Please keep questions like this coming. I am also learning algebraic geometry and this is a great way for me to work out the details : )
$endgroup$
– D_S
Sep 20 '17 at 12:54
$begingroup$
Please keep questions like this coming. I am also learning algebraic geometry and this is a great way for me to work out the details : )
$endgroup$
– D_S
Sep 20 '17 at 12:54
$begingroup$
Please keep questions like this coming. I am also learning algebraic geometry and this is a great way for me to work out the details : )
$endgroup$
– D_S
Sep 20 '17 at 12:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$DeclareMathOperator{Spec}{Spec}$Oh I get what he's saying now. $X times_Y V$ is isomorphic to $f^{-1}(V)$. This implies that $(X times_Y V) times_V Spec kappa(y) cong f^{-1}(V) times_V Spec kappa(y) = f^{-1}(V)_y$. This can be seen in the same way as my previous answer, by showing that $f^{-1}(V)$, together with the maps $i:f^{-1}(V) rightarrow X$ and $f': f^{-1}(V) rightarrow V$, satisfies the required universal property.
If you want an intuitive reason for why isomorphisms like this should hold, just imagine you're in the category of sets. If $f: X rightarrow Y$ is a map of sets, and $V subseteq Y$, then the fiber product $X times_Y V$ is literally
$$X times_Y V = { (x,v) in X times V : f(x) = v }$$
which you can identify with $f^{-1}(V)$.
You can actually make this intuition precise by giving a different proof of these isomorphisms, using the Yoneda lemma.
edited Jan 29 at 9:45
Martin Sleziak
44.9k10122277
answered Sep 20 '17 at 1:56
D_SD_S
14.1k61653
$endgroup$
$begingroup$
In what way I can apply Yoneda here to prove it? Do you mean by using the consequence of YL that $Hom(-, A)$ commutates with fiber products?
$endgroup$
– KarlPeter
Oct 2 '17 at 22:09
add a comment |
$begingroup$
$DeclareMathOperator{Spec}{Spec}$
I don't know what Liu had in mind, but you can argue directly that $X times_Y Spec kappa(y)$ is isomorphic to $f^{-1}(V) times_V Spec kappa(y)$ as follows. (this is a bit technical and you might be better off just doing this yourself instead of going through what I wrote; I would draw a diagram to follow).
Let $f'$ be the morphism of schemes $f': f^{-1}(V) rightarrow V$ coming from the morphism $f: X rightarrow Y$. Let $iota: Spec kappa(y) rightarrow Y$ be the structural morphism, and let $iota'$ be the corresponding morphism $iota': Spec kappa(y) rightarrow V$. Also, let $i: f^{-1}(V) rightarrow X$ and $j: V rightarrow Y$ be the inclusion morphisms.
Let $pi_1, pi_2$ be the projections on $X times_Y Spec kappa(y)$, and $pi_1', pi_2'$ the projections on $f^{-1}(V) times_V Spec kappa(y)$.
We have morphisms $i circ pi_1'$ and $pi_2'$ from $f^{-1}(V) times_V Spec kappa(y)$ to $X$ and $Spec kappa(y)$, respectively. They are compatible with the structural morphisms $f: X rightarrow Y$ and $iota: Spec kappa(y) rightarrow Y$, since
$$f circ i circ pi_1' = j circ f' circ pi_1' = j circ iota' circ pi_2' = iota circ pi_2'$$
Therefore, by the universal property of $X times_Y Spec kappa(y)$, there exists a unique morphism $Phi: f^{-1}(V) times_V Spec kappa(y) rightarrow X times_Y Spec kappa(y)$ such that $i circ pi_1' = pi_1 circ Phi$ and $pi_2' = pi_2 circ Phi$.
I say $Phi$ is an isomorphism of schemes. By standard universal property arguments, this follows immediately from:
Claim: $f^{-1}(V) times_V Spec kappa(y)$, together with the maps $i circ pi_1'$ and $pi_2'$ to $X$ and $Spec kappa(y)$ respectively, is the fiber product of $X$ and $Spec kappa(y)$ over $Y$.
Proof of claim: Suppose $Z$ is a scheme, and $h_1: Z rightarrow X, h_2: Z rightarrow Spec kappa(y)$ are morphisms such that $f circ h_1 = iota circ h_2$. We need to show that there exists a unique morphism of schemes $h: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ such that $i circ pi_1' circ h = h_1$ and $pi_2' circ h = h_2$.
The crucial observation is that, as maps of sets, since the image of $iota$ is the single point $y$ of $Y$, the image of $h_1$ is contained in $f^{-1}{y} subseteq f^{-1}(V)$. A morphism of schemes whose set theoretic image is contained in open subset of a scheme is the same as a morphism of schemes into that open subscheme. So there exists a unique morphism $overline{h_1}: Z rightarrow f^{-1}(V)$ such that $i circ overline{h_1} = h_1$. Now,
$$j circ f' circ overline{h_1} = f circ i circ overline{h_1} = f circ h_1 = iota circ h_2 = j circ iota' circ h_2$$
so $$f' circ overline{h_1} = iota' circ h_2$$
as morphisms of schemes $Z rightarrow f^{-1}(V)$. So by the universal property of $f^{-1}(V) times_V Spec kappa(y)$, there exists ($ast$) a unique morphism $h: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ such that $pi_1' circ h = overline{h_1}$ and $pi_2' circ h = h_2$. Then
$$i circ pi_1' circ h = i circ overline{h_1} = h textrm{ and } pi_2' circ h = h_2 $$
which is what we wanted to show about $h$. To see that $h$ is the unique morphism satisfying this property, suppose $H: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ is another morphism satisfying the same. We can write $i circ pi_1' circ H = i circ overline{h_1}$, and cancel the $i$ to get $pi_1' circ H = overline{h_1}$ and $pi_2' circ H = h_2$. By the uniqueness of $h$ in $ast$, we get $H = h$. $blacksquare$
This reduces the Proposition to the case where $Y$ is affine: once you show that the underlying map of $pi_1'$ is a homeomorphism onto the set of $x in f^{-1}(V)$ such that $f(x) = y$, the fact that $pi_1 circ Phi = pi_1'$ as maps of sets into $X$ shows you that the same is true for $pi_1$.
answered Sep 20 '17 at 1:10
D_SD_S
14.1k61653
$endgroup$
add a comment |
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%2f2436710%2ffiber-product-of-schemas%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$DeclareMathOperator{Spec}{Spec}$Oh I get what he's saying now. $X times_Y V$ is isomorphic to $f^{-1}(V)$. This implies that $(X times_Y V) times_V Spec kappa(y) cong f^{-1}(V) times_V Spec kappa(y) = f^{-1}(V)_y$. This can be seen in the same way as my previous answer, by showing that $f^{-1}(V)$, together with the maps $i:f^{-1}(V) rightarrow X$ and $f': f^{-1}(V) rightarrow V$, satisfies the required universal property.
If you want an intuitive reason for why isomorphisms like this should hold, just imagine you're in the category of sets. If $f: X rightarrow Y$ is a map of sets, and $V subseteq Y$, then the fiber product $X times_Y V$ is literally
$$X times_Y V = { (x,v) in X times V : f(x) = v }$$
which you can identify with $f^{-1}(V)$.
You can actually make this intuition precise by giving a different proof of these isomorphisms, using the Yoneda lemma.
edited Jan 29 at 9:45
Martin Sleziak
44.9k10122277
answered Sep 20 '17 at 1:56
D_SD_S
14.1k61653
$endgroup$
$begingroup$
In what way I can apply Yoneda here to prove it? Do you mean by using the consequence of YL that $Hom(-, A)$ commutates with fiber products?
$endgroup$
– KarlPeter
Oct 2 '17 at 22:09
add a comment |
$begingroup$
$DeclareMathOperator{Spec}{Spec}$Oh I get what he's saying now. $X times_Y V$ is isomorphic to $f^{-1}(V)$. This implies that $(X times_Y V) times_V Spec kappa(y) cong f^{-1}(V) times_V Spec kappa(y) = f^{-1}(V)_y$. This can be seen in the same way as my previous answer, by showing that $f^{-1}(V)$, together with the maps $i:f^{-1}(V) rightarrow X$ and $f': f^{-1}(V) rightarrow V$, satisfies the required universal property.
If you want an intuitive reason for why isomorphisms like this should hold, just imagine you're in the category of sets. If $f: X rightarrow Y$ is a map of sets, and $V subseteq Y$, then the fiber product $X times_Y V$ is literally
$$X times_Y V = { (x,v) in X times V : f(x) = v }$$
which you can identify with $f^{-1}(V)$.
You can actually make this intuition precise by giving a different proof of these isomorphisms, using the Yoneda lemma.
edited Jan 29 at 9:45
Martin Sleziak
44.9k10122277
answered Sep 20 '17 at 1:56
D_SD_S
14.1k61653
$endgroup$
$begingroup$
In what way I can apply Yoneda here to prove it? Do you mean by using the consequence of YL that $Hom(-, A)$ commutates with fiber products?
$endgroup$
– KarlPeter
Oct 2 '17 at 22:09
add a comment |
$begingroup$
$DeclareMathOperator{Spec}{Spec}$Oh I get what he's saying now. $X times_Y V$ is isomorphic to $f^{-1}(V)$. This implies that $(X times_Y V) times_V Spec kappa(y) cong f^{-1}(V) times_V Spec kappa(y) = f^{-1}(V)_y$. This can be seen in the same way as my previous answer, by showing that $f^{-1}(V)$, together with the maps $i:f^{-1}(V) rightarrow X$ and $f': f^{-1}(V) rightarrow V$, satisfies the required universal property.
If you want an intuitive reason for why isomorphisms like this should hold, just imagine you're in the category of sets. If $f: X rightarrow Y$ is a map of sets, and $V subseteq Y$, then the fiber product $X times_Y V$ is literally
$$X times_Y V = { (x,v) in X times V : f(x) = v }$$
which you can identify with $f^{-1}(V)$.
You can actually make this intuition precise by giving a different proof of these isomorphisms, using the Yoneda lemma.
edited Jan 29 at 9:45
Martin Sleziak
44.9k10122277
answered Sep 20 '17 at 1:56
D_SD_S
14.1k61653
$endgroup$
$DeclareMathOperator{Spec}{Spec}$Oh I get what he's saying now. $X times_Y V$ is isomorphic to $f^{-1}(V)$. This implies that $(X times_Y V) times_V Spec kappa(y) cong f^{-1}(V) times_V Spec kappa(y) = f^{-1}(V)_y$. This can be seen in the same way as my previous answer, by showing that $f^{-1}(V)$, together with the maps $i:f^{-1}(V) rightarrow X$ and $f': f^{-1}(V) rightarrow V$, satisfies the required universal property.
If you want an intuitive reason for why isomorphisms like this should hold, just imagine you're in the category of sets. If $f: X rightarrow Y$ is a map of sets, and $V subseteq Y$, then the fiber product $X times_Y V$ is literally
$$X times_Y V = { (x,v) in X times V : f(x) = v }$$
which you can identify with $f^{-1}(V)$.
You can actually make this intuition precise by giving a different proof of these isomorphisms, using the Yoneda lemma.
edited Jan 29 at 9:45
Martin Sleziak
44.9k10122277
answered Sep 20 '17 at 1:56
D_SD_S
14.1k61653
edited Jan 29 at 9:45
Martin Sleziak
44.9k10122277
edited Jan 29 at 9:45
Martin Sleziak
44.9k10122277
edited Jan 29 at 9:45
Martin Sleziak
44.9k10122277
44.9k10122277
answered Sep 20 '17 at 1:56
D_SD_S
14.1k61653
answered Sep 20 '17 at 1:56
D_SD_S
14.1k61653
answered Sep 20 '17 at 1:56
D_SD_S
14.1k61653
14.1k61653
$begingroup$
In what way I can apply Yoneda here to prove it? Do you mean by using the consequence of YL that $Hom(-, A)$ commutates with fiber products?
$endgroup$
– KarlPeter
Oct 2 '17 at 22:09
add a comment |
$begingroup$
In what way I can apply Yoneda here to prove it? Do you mean by using the consequence of YL that $Hom(-, A)$ commutates with fiber products?
$endgroup$
– KarlPeter
Oct 2 '17 at 22:09
$begingroup$
In what way I can apply Yoneda here to prove it? Do you mean by using the consequence of YL that $Hom(-, A)$ commutates with fiber products?
$endgroup$
– KarlPeter
Oct 2 '17 at 22:09
$begingroup$
In what way I can apply Yoneda here to prove it? Do you mean by using the consequence of YL that $Hom(-, A)$ commutates with fiber products?
$endgroup$
– KarlPeter
Oct 2 '17 at 22:09
add a comment |
$begingroup$
$DeclareMathOperator{Spec}{Spec}$
I don't know what Liu had in mind, but you can argue directly that $X times_Y Spec kappa(y)$ is isomorphic to $f^{-1}(V) times_V Spec kappa(y)$ as follows. (this is a bit technical and you might be better off just doing this yourself instead of going through what I wrote; I would draw a diagram to follow).
Let $f'$ be the morphism of schemes $f': f^{-1}(V) rightarrow V$ coming from the morphism $f: X rightarrow Y$. Let $iota: Spec kappa(y) rightarrow Y$ be the structural morphism, and let $iota'$ be the corresponding morphism $iota': Spec kappa(y) rightarrow V$. Also, let $i: f^{-1}(V) rightarrow X$ and $j: V rightarrow Y$ be the inclusion morphisms.
Let $pi_1, pi_2$ be the projections on $X times_Y Spec kappa(y)$, and $pi_1', pi_2'$ the projections on $f^{-1}(V) times_V Spec kappa(y)$.
We have morphisms $i circ pi_1'$ and $pi_2'$ from $f^{-1}(V) times_V Spec kappa(y)$ to $X$ and $Spec kappa(y)$, respectively. They are compatible with the structural morphisms $f: X rightarrow Y$ and $iota: Spec kappa(y) rightarrow Y$, since
$$f circ i circ pi_1' = j circ f' circ pi_1' = j circ iota' circ pi_2' = iota circ pi_2'$$
Therefore, by the universal property of $X times_Y Spec kappa(y)$, there exists a unique morphism $Phi: f^{-1}(V) times_V Spec kappa(y) rightarrow X times_Y Spec kappa(y)$ such that $i circ pi_1' = pi_1 circ Phi$ and $pi_2' = pi_2 circ Phi$.
I say $Phi$ is an isomorphism of schemes. By standard universal property arguments, this follows immediately from:
Claim: $f^{-1}(V) times_V Spec kappa(y)$, together with the maps $i circ pi_1'$ and $pi_2'$ to $X$ and $Spec kappa(y)$ respectively, is the fiber product of $X$ and $Spec kappa(y)$ over $Y$.
Proof of claim: Suppose $Z$ is a scheme, and $h_1: Z rightarrow X, h_2: Z rightarrow Spec kappa(y)$ are morphisms such that $f circ h_1 = iota circ h_2$. We need to show that there exists a unique morphism of schemes $h: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ such that $i circ pi_1' circ h = h_1$ and $pi_2' circ h = h_2$.
The crucial observation is that, as maps of sets, since the image of $iota$ is the single point $y$ of $Y$, the image of $h_1$ is contained in $f^{-1}{y} subseteq f^{-1}(V)$. A morphism of schemes whose set theoretic image is contained in open subset of a scheme is the same as a morphism of schemes into that open subscheme. So there exists a unique morphism $overline{h_1}: Z rightarrow f^{-1}(V)$ such that $i circ overline{h_1} = h_1$. Now,
$$j circ f' circ overline{h_1} = f circ i circ overline{h_1} = f circ h_1 = iota circ h_2 = j circ iota' circ h_2$$
so $$f' circ overline{h_1} = iota' circ h_2$$
as morphisms of schemes $Z rightarrow f^{-1}(V)$. So by the universal property of $f^{-1}(V) times_V Spec kappa(y)$, there exists ($ast$) a unique morphism $h: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ such that $pi_1' circ h = overline{h_1}$ and $pi_2' circ h = h_2$. Then
$$i circ pi_1' circ h = i circ overline{h_1} = h textrm{ and } pi_2' circ h = h_2 $$
which is what we wanted to show about $h$. To see that $h$ is the unique morphism satisfying this property, suppose $H: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ is another morphism satisfying the same. We can write $i circ pi_1' circ H = i circ overline{h_1}$, and cancel the $i$ to get $pi_1' circ H = overline{h_1}$ and $pi_2' circ H = h_2$. By the uniqueness of $h$ in $ast$, we get $H = h$. $blacksquare$
This reduces the Proposition to the case where $Y$ is affine: once you show that the underlying map of $pi_1'$ is a homeomorphism onto the set of $x in f^{-1}(V)$ such that $f(x) = y$, the fact that $pi_1 circ Phi = pi_1'$ as maps of sets into $X$ shows you that the same is true for $pi_1$.
answered Sep 20 '17 at 1:10
D_SD_S
14.1k61653
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperator{Spec}{Spec}$
I don't know what Liu had in mind, but you can argue directly that $X times_Y Spec kappa(y)$ is isomorphic to $f^{-1}(V) times_V Spec kappa(y)$ as follows. (this is a bit technical and you might be better off just doing this yourself instead of going through what I wrote; I would draw a diagram to follow).
Let $f'$ be the morphism of schemes $f': f^{-1}(V) rightarrow V$ coming from the morphism $f: X rightarrow Y$. Let $iota: Spec kappa(y) rightarrow Y$ be the structural morphism, and let $iota'$ be the corresponding morphism $iota': Spec kappa(y) rightarrow V$. Also, let $i: f^{-1}(V) rightarrow X$ and $j: V rightarrow Y$ be the inclusion morphisms.
Let $pi_1, pi_2$ be the projections on $X times_Y Spec kappa(y)$, and $pi_1', pi_2'$ the projections on $f^{-1}(V) times_V Spec kappa(y)$.
We have morphisms $i circ pi_1'$ and $pi_2'$ from $f^{-1}(V) times_V Spec kappa(y)$ to $X$ and $Spec kappa(y)$, respectively. They are compatible with the structural morphisms $f: X rightarrow Y$ and $iota: Spec kappa(y) rightarrow Y$, since
$$f circ i circ pi_1' = j circ f' circ pi_1' = j circ iota' circ pi_2' = iota circ pi_2'$$
Therefore, by the universal property of $X times_Y Spec kappa(y)$, there exists a unique morphism $Phi: f^{-1}(V) times_V Spec kappa(y) rightarrow X times_Y Spec kappa(y)$ such that $i circ pi_1' = pi_1 circ Phi$ and $pi_2' = pi_2 circ Phi$.
I say $Phi$ is an isomorphism of schemes. By standard universal property arguments, this follows immediately from:
Claim: $f^{-1}(V) times_V Spec kappa(y)$, together with the maps $i circ pi_1'$ and $pi_2'$ to $X$ and $Spec kappa(y)$ respectively, is the fiber product of $X$ and $Spec kappa(y)$ over $Y$.
Proof of claim: Suppose $Z$ is a scheme, and $h_1: Z rightarrow X, h_2: Z rightarrow Spec kappa(y)$ are morphisms such that $f circ h_1 = iota circ h_2$. We need to show that there exists a unique morphism of schemes $h: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ such that $i circ pi_1' circ h = h_1$ and $pi_2' circ h = h_2$.
The crucial observation is that, as maps of sets, since the image of $iota$ is the single point $y$ of $Y$, the image of $h_1$ is contained in $f^{-1}{y} subseteq f^{-1}(V)$. A morphism of schemes whose set theoretic image is contained in open subset of a scheme is the same as a morphism of schemes into that open subscheme. So there exists a unique morphism $overline{h_1}: Z rightarrow f^{-1}(V)$ such that $i circ overline{h_1} = h_1$. Now,
$$j circ f' circ overline{h_1} = f circ i circ overline{h_1} = f circ h_1 = iota circ h_2 = j circ iota' circ h_2$$
so $$f' circ overline{h_1} = iota' circ h_2$$
as morphisms of schemes $Z rightarrow f^{-1}(V)$. So by the universal property of $f^{-1}(V) times_V Spec kappa(y)$, there exists ($ast$) a unique morphism $h: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ such that $pi_1' circ h = overline{h_1}$ and $pi_2' circ h = h_2$. Then
$$i circ pi_1' circ h = i circ overline{h_1} = h textrm{ and } pi_2' circ h = h_2 $$
which is what we wanted to show about $h$. To see that $h$ is the unique morphism satisfying this property, suppose $H: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ is another morphism satisfying the same. We can write $i circ pi_1' circ H = i circ overline{h_1}$, and cancel the $i$ to get $pi_1' circ H = overline{h_1}$ and $pi_2' circ H = h_2$. By the uniqueness of $h$ in $ast$, we get $H = h$. $blacksquare$
This reduces the Proposition to the case where $Y$ is affine: once you show that the underlying map of $pi_1'$ is a homeomorphism onto the set of $x in f^{-1}(V)$ such that $f(x) = y$, the fact that $pi_1 circ Phi = pi_1'$ as maps of sets into $X$ shows you that the same is true for $pi_1$.
answered Sep 20 '17 at 1:10
D_SD_S
14.1k61653
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperator{Spec}{Spec}$
I don't know what Liu had in mind, but you can argue directly that $X times_Y Spec kappa(y)$ is isomorphic to $f^{-1}(V) times_V Spec kappa(y)$ as follows. (this is a bit technical and you might be better off just doing this yourself instead of going through what I wrote; I would draw a diagram to follow).
Let $f'$ be the morphism of schemes $f': f^{-1}(V) rightarrow V$ coming from the morphism $f: X rightarrow Y$. Let $iota: Spec kappa(y) rightarrow Y$ be the structural morphism, and let $iota'$ be the corresponding morphism $iota': Spec kappa(y) rightarrow V$. Also, let $i: f^{-1}(V) rightarrow X$ and $j: V rightarrow Y$ be the inclusion morphisms.
Let $pi_1, pi_2$ be the projections on $X times_Y Spec kappa(y)$, and $pi_1', pi_2'$ the projections on $f^{-1}(V) times_V Spec kappa(y)$.
We have morphisms $i circ pi_1'$ and $pi_2'$ from $f^{-1}(V) times_V Spec kappa(y)$ to $X$ and $Spec kappa(y)$, respectively. They are compatible with the structural morphisms $f: X rightarrow Y$ and $iota: Spec kappa(y) rightarrow Y$, since
$$f circ i circ pi_1' = j circ f' circ pi_1' = j circ iota' circ pi_2' = iota circ pi_2'$$
Therefore, by the universal property of $X times_Y Spec kappa(y)$, there exists a unique morphism $Phi: f^{-1}(V) times_V Spec kappa(y) rightarrow X times_Y Spec kappa(y)$ such that $i circ pi_1' = pi_1 circ Phi$ and $pi_2' = pi_2 circ Phi$.
I say $Phi$ is an isomorphism of schemes. By standard universal property arguments, this follows immediately from:
Claim: $f^{-1}(V) times_V Spec kappa(y)$, together with the maps $i circ pi_1'$ and $pi_2'$ to $X$ and $Spec kappa(y)$ respectively, is the fiber product of $X$ and $Spec kappa(y)$ over $Y$.
Proof of claim: Suppose $Z$ is a scheme, and $h_1: Z rightarrow X, h_2: Z rightarrow Spec kappa(y)$ are morphisms such that $f circ h_1 = iota circ h_2$. We need to show that there exists a unique morphism of schemes $h: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ such that $i circ pi_1' circ h = h_1$ and $pi_2' circ h = h_2$.
The crucial observation is that, as maps of sets, since the image of $iota$ is the single point $y$ of $Y$, the image of $h_1$ is contained in $f^{-1}{y} subseteq f^{-1}(V)$. A morphism of schemes whose set theoretic image is contained in open subset of a scheme is the same as a morphism of schemes into that open subscheme. So there exists a unique morphism $overline{h_1}: Z rightarrow f^{-1}(V)$ such that $i circ overline{h_1} = h_1$. Now,
$$j circ f' circ overline{h_1} = f circ i circ overline{h_1} = f circ h_1 = iota circ h_2 = j circ iota' circ h_2$$
so $$f' circ overline{h_1} = iota' circ h_2$$
as morphisms of schemes $Z rightarrow f^{-1}(V)$. So by the universal property of $f^{-1}(V) times_V Spec kappa(y)$, there exists ($ast$) a unique morphism $h: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ such that $pi_1' circ h = overline{h_1}$ and $pi_2' circ h = h_2$. Then
$$i circ pi_1' circ h = i circ overline{h_1} = h textrm{ and } pi_2' circ h = h_2 $$
which is what we wanted to show about $h$. To see that $h$ is the unique morphism satisfying this property, suppose $H: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ is another morphism satisfying the same. We can write $i circ pi_1' circ H = i circ overline{h_1}$, and cancel the $i$ to get $pi_1' circ H = overline{h_1}$ and $pi_2' circ H = h_2$. By the uniqueness of $h$ in $ast$, we get $H = h$. $blacksquare$
This reduces the Proposition to the case where $Y$ is affine: once you show that the underlying map of $pi_1'$ is a homeomorphism onto the set of $x in f^{-1}(V)$ such that $f(x) = y$, the fact that $pi_1 circ Phi = pi_1'$ as maps of sets into $X$ shows you that the same is true for $pi_1$.
answered Sep 20 '17 at 1:10
D_SD_S
14.1k61653
$endgroup$
$DeclareMathOperator{Spec}{Spec}$
I don't know what Liu had in mind, but you can argue directly that $X times_Y Spec kappa(y)$ is isomorphic to $f^{-1}(V) times_V Spec kappa(y)$ as follows. (this is a bit technical and you might be better off just doing this yourself instead of going through what I wrote; I would draw a diagram to follow).
Let $f'$ be the morphism of schemes $f': f^{-1}(V) rightarrow V$ coming from the morphism $f: X rightarrow Y$. Let $iota: Spec kappa(y) rightarrow Y$ be the structural morphism, and let $iota'$ be the corresponding morphism $iota': Spec kappa(y) rightarrow V$. Also, let $i: f^{-1}(V) rightarrow X$ and $j: V rightarrow Y$ be the inclusion morphisms.
Let $pi_1, pi_2$ be the projections on $X times_Y Spec kappa(y)$, and $pi_1', pi_2'$ the projections on $f^{-1}(V) times_V Spec kappa(y)$.
We have morphisms $i circ pi_1'$ and $pi_2'$ from $f^{-1}(V) times_V Spec kappa(y)$ to $X$ and $Spec kappa(y)$, respectively. They are compatible with the structural morphisms $f: X rightarrow Y$ and $iota: Spec kappa(y) rightarrow Y$, since
$$f circ i circ pi_1' = j circ f' circ pi_1' = j circ iota' circ pi_2' = iota circ pi_2'$$
Therefore, by the universal property of $X times_Y Spec kappa(y)$, there exists a unique morphism $Phi: f^{-1}(V) times_V Spec kappa(y) rightarrow X times_Y Spec kappa(y)$ such that $i circ pi_1' = pi_1 circ Phi$ and $pi_2' = pi_2 circ Phi$.
I say $Phi$ is an isomorphism of schemes. By standard universal property arguments, this follows immediately from:
Claim: $f^{-1}(V) times_V Spec kappa(y)$, together with the maps $i circ pi_1'$ and $pi_2'$ to $X$ and $Spec kappa(y)$ respectively, is the fiber product of $X$ and $Spec kappa(y)$ over $Y$.
Proof of claim: Suppose $Z$ is a scheme, and $h_1: Z rightarrow X, h_2: Z rightarrow Spec kappa(y)$ are morphisms such that $f circ h_1 = iota circ h_2$. We need to show that there exists a unique morphism of schemes $h: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ such that $i circ pi_1' circ h = h_1$ and $pi_2' circ h = h_2$.
The crucial observation is that, as maps of sets, since the image of $iota$ is the single point $y$ of $Y$, the image of $h_1$ is contained in $f^{-1}{y} subseteq f^{-1}(V)$. A morphism of schemes whose set theoretic image is contained in open subset of a scheme is the same as a morphism of schemes into that open subscheme. So there exists a unique morphism $overline{h_1}: Z rightarrow f^{-1}(V)$ such that $i circ overline{h_1} = h_1$. Now,
$$j circ f' circ overline{h_1} = f circ i circ overline{h_1} = f circ h_1 = iota circ h_2 = j circ iota' circ h_2$$
so $$f' circ overline{h_1} = iota' circ h_2$$
as morphisms of schemes $Z rightarrow f^{-1}(V)$. So by the universal property of $f^{-1}(V) times_V Spec kappa(y)$, there exists ($ast$) a unique morphism $h: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ such that $pi_1' circ h = overline{h_1}$ and $pi_2' circ h = h_2$. Then
$$i circ pi_1' circ h = i circ overline{h_1} = h textrm{ and } pi_2' circ h = h_2 $$
which is what we wanted to show about $h$. To see that $h$ is the unique morphism satisfying this property, suppose $H: Z rightarrow f^{-1}(V) times_V Spec kappa(y)$ is another morphism satisfying the same. We can write $i circ pi_1' circ H = i circ overline{h_1}$, and cancel the $i$ to get $pi_1' circ H = overline{h_1}$ and $pi_2' circ H = h_2$. By the uniqueness of $h$ in $ast$, we get $H = h$. $blacksquare$
This reduces the Proposition to the case where $Y$ is affine: once you show that the underlying map of $pi_1'$ is a homeomorphism onto the set of $x in f^{-1}(V)$ such that $f(x) = y$, the fact that $pi_1 circ Phi = pi_1'$ as maps of sets into $X$ shows you that the same is true for $pi_1$.
answered Sep 20 '17 at 1:10
D_SD_S
14.1k61653
edited Sep 20 '17 at 1:17
answered Sep 20 '17 at 1:10
D_SD_S
14.1k61653
answered Sep 20 '17 at 1:10
D_SD_S
14.1k61653
answered Sep 20 '17 at 1:10
D_SD_S
14.1k61653
14.1k61653
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%2f2436710%2ffiber-product-of-schemas%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
$begingroup$
Please keep questions like this coming. I am also learning algebraic geometry and this is a great way for me to work out the details : )
$endgroup$
– D_S
Sep 20 '17 at 12:54