Unramified Extensions of Discrete Valuation Rings
$begingroup$
Let $i: R subset A$ be an integral extension of discrete valuation rings and denote by $$f: operatorname{Spec}(A) to operatorname{Spec}(R)$$ the corresponding scheme morphism. Since $operatorname{Spec}(R) = {sigma_R, eta_R}$ (resp. $operatorname{Spec}(A) = {sigma_A, eta_A}$ where $sigma$ is the unique closed and $eta$ the unique generic points)
denote
$K_R := kappa(eta_R) = operatorname{Frac}(R)$
$K_A := kappa(eta_A) = operatorname{Frac}(A)$
and
$k_R := kappa(sigma_R)= mathcal{O}_{R, sigma_R}/m_{sigma_R}mathcal{O}_{R, sigma_R}$
$ k_A := kappa(sigma_A)$
we obtain field extensions $K_Rsubset K_A$ and $k_Rsubset k_A$.
I want to show that:
$f$ is a unramified morphism (in sense of scheme morphism) $Leftrightarrow$
$K_A/K_R, k_A/k_R$ are finite separable and $[K_A:K_R]=[k_A: k_R]$.
Source: S. Bosch, Commutative Algebra and Algebraic Geometry, p. 374.
My attempts:
Since there are only two points it seems ok to apply the chararacterization (v) from Thm 3:
So the the finite separability condition for $K_A/K_R, k_A/k_R$ works in both directions.
The only problem lies in the condition $[K_A:K_R]=[k_A: k_R]$ and that $m_{sigma_R}$ generates $m_{sigma_A}$ ($m_{eta_R}$and $m_{eta_A}=0$ so it's ok)
The author's hint was to use exersise 3.1.8 (page 90):
So in our case $K= K_R$ and $L= K_A$. Then the integral closure of $R$ is in $K_A$ a finite $R$-module. How does it help to get $[K_A:K_R]=[k_A: k_R]$?
Btw: In Ex 3.1.8: What would fail with the trace function $Tr_{L/K}$ if $L/K$ wouldn't separable?
algebraic-geometry ring-theory commutative-algebra
edited Dec 28 '18 at 22:41
user26857
39.3k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
6801315
$endgroup$
add a comment |
$begingroup$
Let $i: R subset A$ be an integral extension of discrete valuation rings and denote by $$f: operatorname{Spec}(A) to operatorname{Spec}(R)$$ the corresponding scheme morphism. Since $operatorname{Spec}(R) = {sigma_R, eta_R}$ (resp. $operatorname{Spec}(A) = {sigma_A, eta_A}$ where $sigma$ is the unique closed and $eta$ the unique generic points)
denote
$K_R := kappa(eta_R) = operatorname{Frac}(R)$
$K_A := kappa(eta_A) = operatorname{Frac}(A)$
and
$k_R := kappa(sigma_R)= mathcal{O}_{R, sigma_R}/m_{sigma_R}mathcal{O}_{R, sigma_R}$
$ k_A := kappa(sigma_A)$
we obtain field extensions $K_Rsubset K_A$ and $k_Rsubset k_A$.
I want to show that:
$f$ is a unramified morphism (in sense of scheme morphism) $Leftrightarrow$
$K_A/K_R, k_A/k_R$ are finite separable and $[K_A:K_R]=[k_A: k_R]$.
Source: S. Bosch, Commutative Algebra and Algebraic Geometry, p. 374.
My attempts:
Since there are only two points it seems ok to apply the chararacterization (v) from Thm 3:
So the the finite separability condition for $K_A/K_R, k_A/k_R$ works in both directions.
The only problem lies in the condition $[K_A:K_R]=[k_A: k_R]$ and that $m_{sigma_R}$ generates $m_{sigma_A}$ ($m_{eta_R}$and $m_{eta_A}=0$ so it's ok)
The author's hint was to use exersise 3.1.8 (page 90):
So in our case $K= K_R$ and $L= K_A$. Then the integral closure of $R$ is in $K_A$ a finite $R$-module. How does it help to get $[K_A:K_R]=[k_A: k_R]$?
Btw: In Ex 3.1.8: What would fail with the trace function $Tr_{L/K}$ if $L/K$ wouldn't separable?
algebraic-geometry ring-theory commutative-algebra
edited Dec 28 '18 at 22:41
user26857
39.3k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
6801315
$endgroup$
add a comment |
$begingroup$
Let $i: R subset A$ be an integral extension of discrete valuation rings and denote by $$f: operatorname{Spec}(A) to operatorname{Spec}(R)$$ the corresponding scheme morphism. Since $operatorname{Spec}(R) = {sigma_R, eta_R}$ (resp. $operatorname{Spec}(A) = {sigma_A, eta_A}$ where $sigma$ is the unique closed and $eta$ the unique generic points)
denote
$K_R := kappa(eta_R) = operatorname{Frac}(R)$
$K_A := kappa(eta_A) = operatorname{Frac}(A)$
and
$k_R := kappa(sigma_R)= mathcal{O}_{R, sigma_R}/m_{sigma_R}mathcal{O}_{R, sigma_R}$
$ k_A := kappa(sigma_A)$
we obtain field extensions $K_Rsubset K_A$ and $k_Rsubset k_A$.
I want to show that:
$f$ is a unramified morphism (in sense of scheme morphism) $Leftrightarrow$
$K_A/K_R, k_A/k_R$ are finite separable and $[K_A:K_R]=[k_A: k_R]$.
Source: S. Bosch, Commutative Algebra and Algebraic Geometry, p. 374.
My attempts:
Since there are only two points it seems ok to apply the chararacterization (v) from Thm 3:
So the the finite separability condition for $K_A/K_R, k_A/k_R$ works in both directions.
The only problem lies in the condition $[K_A:K_R]=[k_A: k_R]$ and that $m_{sigma_R}$ generates $m_{sigma_A}$ ($m_{eta_R}$and $m_{eta_A}=0$ so it's ok)
The author's hint was to use exersise 3.1.8 (page 90):
So in our case $K= K_R$ and $L= K_A$. Then the integral closure of $R$ is in $K_A$ a finite $R$-module. How does it help to get $[K_A:K_R]=[k_A: k_R]$?
Btw: In Ex 3.1.8: What would fail with the trace function $Tr_{L/K}$ if $L/K$ wouldn't separable?
algebraic-geometry ring-theory commutative-algebra
edited Dec 28 '18 at 22:41
user26857
39.3k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
6801315
$endgroup$
Let $i: R subset A$ be an integral extension of discrete valuation rings and denote by $$f: operatorname{Spec}(A) to operatorname{Spec}(R)$$ the corresponding scheme morphism. Since $operatorname{Spec}(R) = {sigma_R, eta_R}$ (resp. $operatorname{Spec}(A) = {sigma_A, eta_A}$ where $sigma$ is the unique closed and $eta$ the unique generic points)
denote
$K_R := kappa(eta_R) = operatorname{Frac}(R)$
$K_A := kappa(eta_A) = operatorname{Frac}(A)$
and
$k_R := kappa(sigma_R)= mathcal{O}_{R, sigma_R}/m_{sigma_R}mathcal{O}_{R, sigma_R}$
$ k_A := kappa(sigma_A)$
we obtain field extensions $K_Rsubset K_A$ and $k_Rsubset k_A$.
I want to show that:
$f$ is a unramified morphism (in sense of scheme morphism) $Leftrightarrow$
$K_A/K_R, k_A/k_R$ are finite separable and $[K_A:K_R]=[k_A: k_R]$.
Source: S. Bosch, Commutative Algebra and Algebraic Geometry, p. 374.
My attempts:
Since there are only two points it seems ok to apply the chararacterization (v) from Thm 3:
So the the finite separability condition for $K_A/K_R, k_A/k_R$ works in both directions.
The only problem lies in the condition $[K_A:K_R]=[k_A: k_R]$ and that $m_{sigma_R}$ generates $m_{sigma_A}$ ($m_{eta_R}$and $m_{eta_A}=0$ so it's ok)
The author's hint was to use exersise 3.1.8 (page 90):
So in our case $K= K_R$ and $L= K_A$. Then the integral closure of $R$ is in $K_A$ a finite $R$-module. How does it help to get $[K_A:K_R]=[k_A: k_R]$?
Btw: In Ex 3.1.8: What would fail with the trace function $Tr_{L/K}$ if $L/K$ wouldn't separable?
algebraic-geometry ring-theory commutative-algebra
algebraic-geometry ring-theory commutative-algebra
edited Dec 28 '18 at 22:41
user26857
39.3k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
6801315
edited Dec 28 '18 at 22:41
user26857
39.3k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
6801315
edited Dec 28 '18 at 22:41
user26857
39.3k124183
edited Dec 28 '18 at 22:41
user26857
39.3k124183
edited Dec 28 '18 at 22:41
user26857
39.3k124183
39.3k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
6801315
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
6801315
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
6801315
6801315
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: You can show that $A/R$ is in fact finite free with $mathrm{rk}_R A=[K_A:K_R]$. This will create the missing link between the degrees of the two field extensions.
As to your last remark: If the field extension is not separable then the trace will be zero.
answered Dec 28 '18 at 12:00
asdqasdq
1,8311518
$endgroup$
$begingroup$
The freeness of $A/R$ commes from a structure theorem for modules over Dedekid rings (I never heard about a structure theorem over DVRs). Then $mathrm{rk}_R A=[K_A:K_R]$ boils down to the general property of free extensions. But the link between $ [K_A:K_R]$ and $ [k_A:k_R]$ is still unclear. Is $mathrm{rk}_R$ conserved (why?) if we quotient out someting that is contained in Jacobsen ideal? So some kind of Nakayama argument?
$endgroup$
– KarlPeter
Dec 29 '18 at 4:18
$begingroup$
One way to show this is by observing $A/mathfrak{m}_{sigma_A}=Aotimes R/mathfrak{m}_{sigma_R}$ by using unramifiedness.
$endgroup$
– asdq
Dec 29 '18 at 5:13
$begingroup$
Ah ok, this settle $rk_R(A)=[k_A: k_R]$. One point stays unclear: How do you get the finiteness of $A/R$? SInce after settled this $A$ considered as $R$ module without torsion is indeed finite free so $A = oplus_{rk_R A} R$. Regarding equality $mathrm{rk}_R A=[K_A:K_R]$ there occure following problem. Passing to $K_R$ is the same as localizing by $S := R backslash {0}$ and that respects products. So $AS^{-1} = oplus R S^{-1}= oplus K_R$. But $S$ is not enough to get $AS^{-1} = K_A$. I guess I have to combine it with result from Ex 8.
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Does the information about finiteness of integral closure of $R$ help here?
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Well exercise 8 tells you that $A/R$ is finite since $A$ is precisely the integral closure of $R$ in $K_A$.
$endgroup$
– asdq
Dec 29 '18 at 11:53
|
show 2 more comments
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%2f3054573%2funramified-extensions-of-discrete-valuation-rings%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
$begingroup$
Hint: You can show that $A/R$ is in fact finite free with $mathrm{rk}_R A=[K_A:K_R]$. This will create the missing link between the degrees of the two field extensions.
As to your last remark: If the field extension is not separable then the trace will be zero.
answered Dec 28 '18 at 12:00
asdqasdq
1,8311518
$endgroup$
$begingroup$
The freeness of $A/R$ commes from a structure theorem for modules over Dedekid rings (I never heard about a structure theorem over DVRs). Then $mathrm{rk}_R A=[K_A:K_R]$ boils down to the general property of free extensions. But the link between $ [K_A:K_R]$ and $ [k_A:k_R]$ is still unclear. Is $mathrm{rk}_R$ conserved (why?) if we quotient out someting that is contained in Jacobsen ideal? So some kind of Nakayama argument?
$endgroup$
– KarlPeter
Dec 29 '18 at 4:18
$begingroup$
One way to show this is by observing $A/mathfrak{m}_{sigma_A}=Aotimes R/mathfrak{m}_{sigma_R}$ by using unramifiedness.
$endgroup$
– asdq
Dec 29 '18 at 5:13
$begingroup$
Ah ok, this settle $rk_R(A)=[k_A: k_R]$. One point stays unclear: How do you get the finiteness of $A/R$? SInce after settled this $A$ considered as $R$ module without torsion is indeed finite free so $A = oplus_{rk_R A} R$. Regarding equality $mathrm{rk}_R A=[K_A:K_R]$ there occure following problem. Passing to $K_R$ is the same as localizing by $S := R backslash {0}$ and that respects products. So $AS^{-1} = oplus R S^{-1}= oplus K_R$. But $S$ is not enough to get $AS^{-1} = K_A$. I guess I have to combine it with result from Ex 8.
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Does the information about finiteness of integral closure of $R$ help here?
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Well exercise 8 tells you that $A/R$ is finite since $A$ is precisely the integral closure of $R$ in $K_A$.
$endgroup$
– asdq
Dec 29 '18 at 11:53
|
show 2 more comments
$begingroup$
Hint: You can show that $A/R$ is in fact finite free with $mathrm{rk}_R A=[K_A:K_R]$. This will create the missing link between the degrees of the two field extensions.
As to your last remark: If the field extension is not separable then the trace will be zero.
answered Dec 28 '18 at 12:00
asdqasdq
1,8311518
$endgroup$
$begingroup$
The freeness of $A/R$ commes from a structure theorem for modules over Dedekid rings (I never heard about a structure theorem over DVRs). Then $mathrm{rk}_R A=[K_A:K_R]$ boils down to the general property of free extensions. But the link between $ [K_A:K_R]$ and $ [k_A:k_R]$ is still unclear. Is $mathrm{rk}_R$ conserved (why?) if we quotient out someting that is contained in Jacobsen ideal? So some kind of Nakayama argument?
$endgroup$
– KarlPeter
Dec 29 '18 at 4:18
$begingroup$
One way to show this is by observing $A/mathfrak{m}_{sigma_A}=Aotimes R/mathfrak{m}_{sigma_R}$ by using unramifiedness.
$endgroup$
– asdq
Dec 29 '18 at 5:13
$begingroup$
Ah ok, this settle $rk_R(A)=[k_A: k_R]$. One point stays unclear: How do you get the finiteness of $A/R$? SInce after settled this $A$ considered as $R$ module without torsion is indeed finite free so $A = oplus_{rk_R A} R$. Regarding equality $mathrm{rk}_R A=[K_A:K_R]$ there occure following problem. Passing to $K_R$ is the same as localizing by $S := R backslash {0}$ and that respects products. So $AS^{-1} = oplus R S^{-1}= oplus K_R$. But $S$ is not enough to get $AS^{-1} = K_A$. I guess I have to combine it with result from Ex 8.
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Does the information about finiteness of integral closure of $R$ help here?
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Well exercise 8 tells you that $A/R$ is finite since $A$ is precisely the integral closure of $R$ in $K_A$.
$endgroup$
– asdq
Dec 29 '18 at 11:53
|
show 2 more comments
$begingroup$
Hint: You can show that $A/R$ is in fact finite free with $mathrm{rk}_R A=[K_A:K_R]$. This will create the missing link between the degrees of the two field extensions.
As to your last remark: If the field extension is not separable then the trace will be zero.
answered Dec 28 '18 at 12:00
asdqasdq
1,8311518
$endgroup$
Hint: You can show that $A/R$ is in fact finite free with $mathrm{rk}_R A=[K_A:K_R]$. This will create the missing link between the degrees of the two field extensions.
As to your last remark: If the field extension is not separable then the trace will be zero.
answered Dec 28 '18 at 12:00
asdqasdq
1,8311518
answered Dec 28 '18 at 12:00
asdqasdq
1,8311518
answered Dec 28 '18 at 12:00
asdqasdq
1,8311518
answered Dec 28 '18 at 12:00
asdqasdq
1,8311518
1,8311518
$begingroup$
The freeness of $A/R$ commes from a structure theorem for modules over Dedekid rings (I never heard about a structure theorem over DVRs). Then $mathrm{rk}_R A=[K_A:K_R]$ boils down to the general property of free extensions. But the link between $ [K_A:K_R]$ and $ [k_A:k_R]$ is still unclear. Is $mathrm{rk}_R$ conserved (why?) if we quotient out someting that is contained in Jacobsen ideal? So some kind of Nakayama argument?
$endgroup$
– KarlPeter
Dec 29 '18 at 4:18
$begingroup$
One way to show this is by observing $A/mathfrak{m}_{sigma_A}=Aotimes R/mathfrak{m}_{sigma_R}$ by using unramifiedness.
$endgroup$
– asdq
Dec 29 '18 at 5:13
$begingroup$
Ah ok, this settle $rk_R(A)=[k_A: k_R]$. One point stays unclear: How do you get the finiteness of $A/R$? SInce after settled this $A$ considered as $R$ module without torsion is indeed finite free so $A = oplus_{rk_R A} R$. Regarding equality $mathrm{rk}_R A=[K_A:K_R]$ there occure following problem. Passing to $K_R$ is the same as localizing by $S := R backslash {0}$ and that respects products. So $AS^{-1} = oplus R S^{-1}= oplus K_R$. But $S$ is not enough to get $AS^{-1} = K_A$. I guess I have to combine it with result from Ex 8.
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Does the information about finiteness of integral closure of $R$ help here?
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Well exercise 8 tells you that $A/R$ is finite since $A$ is precisely the integral closure of $R$ in $K_A$.
$endgroup$
– asdq
Dec 29 '18 at 11:53
|
show 2 more comments
$begingroup$
The freeness of $A/R$ commes from a structure theorem for modules over Dedekid rings (I never heard about a structure theorem over DVRs). Then $mathrm{rk}_R A=[K_A:K_R]$ boils down to the general property of free extensions. But the link between $ [K_A:K_R]$ and $ [k_A:k_R]$ is still unclear. Is $mathrm{rk}_R$ conserved (why?) if we quotient out someting that is contained in Jacobsen ideal? So some kind of Nakayama argument?
$endgroup$
– KarlPeter
Dec 29 '18 at 4:18
$begingroup$
One way to show this is by observing $A/mathfrak{m}_{sigma_A}=Aotimes R/mathfrak{m}_{sigma_R}$ by using unramifiedness.
$endgroup$
– asdq
Dec 29 '18 at 5:13
$begingroup$
Ah ok, this settle $rk_R(A)=[k_A: k_R]$. One point stays unclear: How do you get the finiteness of $A/R$? SInce after settled this $A$ considered as $R$ module without torsion is indeed finite free so $A = oplus_{rk_R A} R$. Regarding equality $mathrm{rk}_R A=[K_A:K_R]$ there occure following problem. Passing to $K_R$ is the same as localizing by $S := R backslash {0}$ and that respects products. So $AS^{-1} = oplus R S^{-1}= oplus K_R$. But $S$ is not enough to get $AS^{-1} = K_A$. I guess I have to combine it with result from Ex 8.
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Does the information about finiteness of integral closure of $R$ help here?
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Well exercise 8 tells you that $A/R$ is finite since $A$ is precisely the integral closure of $R$ in $K_A$.
$endgroup$
– asdq
Dec 29 '18 at 11:53
$begingroup$
The freeness of $A/R$ commes from a structure theorem for modules over Dedekid rings (I never heard about a structure theorem over DVRs). Then $mathrm{rk}_R A=[K_A:K_R]$ boils down to the general property of free extensions. But the link between $ [K_A:K_R]$ and $ [k_A:k_R]$ is still unclear. Is $mathrm{rk}_R$ conserved (why?) if we quotient out someting that is contained in Jacobsen ideal? So some kind of Nakayama argument?
$endgroup$
– KarlPeter
Dec 29 '18 at 4:18
$begingroup$
The freeness of $A/R$ commes from a structure theorem for modules over Dedekid rings (I never heard about a structure theorem over DVRs). Then $mathrm{rk}_R A=[K_A:K_R]$ boils down to the general property of free extensions. But the link between $ [K_A:K_R]$ and $ [k_A:k_R]$ is still unclear. Is $mathrm{rk}_R$ conserved (why?) if we quotient out someting that is contained in Jacobsen ideal? So some kind of Nakayama argument?
$endgroup$
– KarlPeter
Dec 29 '18 at 4:18
$begingroup$
One way to show this is by observing $A/mathfrak{m}_{sigma_A}=Aotimes R/mathfrak{m}_{sigma_R}$ by using unramifiedness.
$endgroup$
– asdq
Dec 29 '18 at 5:13
$begingroup$
One way to show this is by observing $A/mathfrak{m}_{sigma_A}=Aotimes R/mathfrak{m}_{sigma_R}$ by using unramifiedness.
$endgroup$
– asdq
Dec 29 '18 at 5:13
$begingroup$
Ah ok, this settle $rk_R(A)=[k_A: k_R]$. One point stays unclear: How do you get the finiteness of $A/R$? SInce after settled this $A$ considered as $R$ module without torsion is indeed finite free so $A = oplus_{rk_R A} R$. Regarding equality $mathrm{rk}_R A=[K_A:K_R]$ there occure following problem. Passing to $K_R$ is the same as localizing by $S := R backslash {0}$ and that respects products. So $AS^{-1} = oplus R S^{-1}= oplus K_R$. But $S$ is not enough to get $AS^{-1} = K_A$. I guess I have to combine it with result from Ex 8.
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Ah ok, this settle $rk_R(A)=[k_A: k_R]$. One point stays unclear: How do you get the finiteness of $A/R$? SInce after settled this $A$ considered as $R$ module without torsion is indeed finite free so $A = oplus_{rk_R A} R$. Regarding equality $mathrm{rk}_R A=[K_A:K_R]$ there occure following problem. Passing to $K_R$ is the same as localizing by $S := R backslash {0}$ and that respects products. So $AS^{-1} = oplus R S^{-1}= oplus K_R$. But $S$ is not enough to get $AS^{-1} = K_A$. I guess I have to combine it with result from Ex 8.
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Does the information about finiteness of integral closure of $R$ help here?
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Does the information about finiteness of integral closure of $R$ help here?
$endgroup$
– KarlPeter
Dec 29 '18 at 5:54
$begingroup$
Well exercise 8 tells you that $A/R$ is finite since $A$ is precisely the integral closure of $R$ in $K_A$.
$endgroup$
– asdq
Dec 29 '18 at 11:53
$begingroup$
Well exercise 8 tells you that $A/R$ is finite since $A$ is precisely the integral closure of $R$ in $K_A$.
$endgroup$
– asdq
Dec 29 '18 at 11:53
|
show 2 more comments
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%2f3054573%2funramified-extensions-of-discrete-valuation-rings%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
