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
39.4k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
4641315
$endgroup$
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
39.4k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
4641315
$endgroup$
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
39.4k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
4641315
$endgroup$
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
39.4k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
4641315
edited Dec 28 '18 at 22:41
user26857
39.4k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
4641315
edited Dec 28 '18 at 22:41
user26857
39.4k124183
edited Dec 28 '18 at 22:41
user26857
39.4k124183
edited Dec 28 '18 at 22:41
user26857
39.4k124183
39.4k124183
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
4641315
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
4641315
asked Dec 28 '18 at 4:47
KarlPeterKarlPeter
4641315
4641315
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
1,8681518
$endgroup$
$begingroup$
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
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$begingroup$
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
1,8681518
$endgroup$
$begingroup$
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
1,8681518
$endgroup$
$begingroup$
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
1,8681518
$endgroup$
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
1,8681518
answered Dec 28 '18 at 12:00
asdqasdq
1,8681518
answered Dec 28 '18 at 12:00
asdqasdq
1,8681518
answered Dec 28 '18 at 12:00
asdqasdq
1,8681518
1,8681518
$begingroup$
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
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