Finiteness of the normalization of an algebra over a DVR
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Let $R$ be a DVR, $A$ a finitely generated integral $R$-algebra, and $A'$ the normalization of $A$ in the fraction field of $A$.
Then is $A'$ finite as an $A$-module?
I know that if $R$ is a field, then it's true.
And I know this is true, see here, section 2 in page 8.
But I don't know its proof.
So please show it or suggest some references.
Thank you.
algebraic-geometry commutative-algebra
asked Dec 19 '18 at 5:14
k.j.k.j.
39819
$endgroup$
add a comment |
$begingroup$
Let $R$ be a DVR, $A$ a finitely generated integral $R$-algebra, and $A'$ the normalization of $A$ in the fraction field of $A$.
Then is $A'$ finite as an $A$-module?
I know that if $R$ is a field, then it's true.
And I know this is true, see here, section 2 in page 8.
But I don't know its proof.
So please show it or suggest some references.
Thank you.
algebraic-geometry commutative-algebra
asked Dec 19 '18 at 5:14
k.j.k.j.
39819
$endgroup$
add a comment |
$begingroup$
Let $R$ be a DVR, $A$ a finitely generated integral $R$-algebra, and $A'$ the normalization of $A$ in the fraction field of $A$.
Then is $A'$ finite as an $A$-module?
I know that if $R$ is a field, then it's true.
And I know this is true, see here, section 2 in page 8.
But I don't know its proof.
So please show it or suggest some references.
Thank you.
algebraic-geometry commutative-algebra
asked Dec 19 '18 at 5:14
k.j.k.j.
39819
$endgroup$
Let $R$ be a DVR, $A$ a finitely generated integral $R$-algebra, and $A'$ the normalization of $A$ in the fraction field of $A$.
Then is $A'$ finite as an $A$-module?
I know that if $R$ is a field, then it's true.
And I know this is true, see here, section 2 in page 8.
But I don't know its proof.
So please show it or suggest some references.
Thank you.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Dec 19 '18 at 5:14
k.j.k.j.
39819
asked Dec 19 '18 at 5:14
k.j.k.j.
39819
edited Dec 19 '18 at 5:20
k.j.
asked Dec 19 '18 at 5:14
k.j.k.j.
39819
asked Dec 19 '18 at 5:14
k.j.k.j.
39819
asked Dec 19 '18 at 5:14
k.j.k.j.
39819
39819
add a comment |
add a comment |
1 Answer
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The property you state is related to the property of $R$ being excellent, and more specifically to weaker condition of whether $R$ has the N-2 property or the stronger Nagata property; see [Matsumura, §31] and [Illusie–Laszlo–Orgogozo, Exp. I].
Not all DVR's satisfy your property. We give an example below, due to Nagata. On the other hand, both Dedekind domains of characteristic zero and complete local rings are excellent [Illusie–Laszlo–Orgogozo, Exp. I, Prop. 3.1 and §4], hence satisfy your property.
Example [Nagata, App. A1, Ex. 3]. We give an example of a DVR $R$ and an algebra of finite type $A$ over $R$ such that the normalization $A'$ of $A$ in $operatorname{Frac}(A)$ is not finite over $A$. Let $k$ be a field of characteristic $p > 0$ such that $[k : k^p] = infty$, and consider the ring
$$R = biggl{sum_{i=0}^infty a_ix^i in k[[x]] biggmvert [k^p(a_0,a_1,ldots):k^p] < infty biggr}.$$
This is a DVR by [Nagata, App. A1, (E3.1)]. Let ${b_1,b_2,ldots} subseteq k$ be a sequence of elements in $k$ that are $p$-independent over $k^p$. Set
$$c = sum_{i=0}^infty b_ix^i,$$
and consider the ring $A = R[c]$. Then, the normalization $A'$ of $A$ in $operatorname{Frac}(A)$ is not finite over $A$ by [Nagata, App. A1, (E3.2)].
answered Dec 19 '18 at 12:24
Takumi MurayamaTakumi Murayama
6,34611645
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1 Answer
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1 Answer
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active
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votes
$begingroup$
The property you state is related to the property of $R$ being excellent, and more specifically to weaker condition of whether $R$ has the N-2 property or the stronger Nagata property; see [Matsumura, §31] and [Illusie–Laszlo–Orgogozo, Exp. I].
Not all DVR's satisfy your property. We give an example below, due to Nagata. On the other hand, both Dedekind domains of characteristic zero and complete local rings are excellent [Illusie–Laszlo–Orgogozo, Exp. I, Prop. 3.1 and §4], hence satisfy your property.
Example [Nagata, App. A1, Ex. 3]. We give an example of a DVR $R$ and an algebra of finite type $A$ over $R$ such that the normalization $A'$ of $A$ in $operatorname{Frac}(A)$ is not finite over $A$. Let $k$ be a field of characteristic $p > 0$ such that $[k : k^p] = infty$, and consider the ring
$$R = biggl{sum_{i=0}^infty a_ix^i in k[[x]] biggmvert [k^p(a_0,a_1,ldots):k^p] < infty biggr}.$$
This is a DVR by [Nagata, App. A1, (E3.1)]. Let ${b_1,b_2,ldots} subseteq k$ be a sequence of elements in $k$ that are $p$-independent over $k^p$. Set
$$c = sum_{i=0}^infty b_ix^i,$$
and consider the ring $A = R[c]$. Then, the normalization $A'$ of $A$ in $operatorname{Frac}(A)$ is not finite over $A$ by [Nagata, App. A1, (E3.2)].
answered Dec 19 '18 at 12:24
Takumi MurayamaTakumi Murayama
6,34611645
$endgroup$
add a comment |
$begingroup$
The property you state is related to the property of $R$ being excellent, and more specifically to weaker condition of whether $R$ has the N-2 property or the stronger Nagata property; see [Matsumura, §31] and [Illusie–Laszlo–Orgogozo, Exp. I].
Not all DVR's satisfy your property. We give an example below, due to Nagata. On the other hand, both Dedekind domains of characteristic zero and complete local rings are excellent [Illusie–Laszlo–Orgogozo, Exp. I, Prop. 3.1 and §4], hence satisfy your property.
Example [Nagata, App. A1, Ex. 3]. We give an example of a DVR $R$ and an algebra of finite type $A$ over $R$ such that the normalization $A'$ of $A$ in $operatorname{Frac}(A)$ is not finite over $A$. Let $k$ be a field of characteristic $p > 0$ such that $[k : k^p] = infty$, and consider the ring
$$R = biggl{sum_{i=0}^infty a_ix^i in k[[x]] biggmvert [k^p(a_0,a_1,ldots):k^p] < infty biggr}.$$
This is a DVR by [Nagata, App. A1, (E3.1)]. Let ${b_1,b_2,ldots} subseteq k$ be a sequence of elements in $k$ that are $p$-independent over $k^p$. Set
$$c = sum_{i=0}^infty b_ix^i,$$
and consider the ring $A = R[c]$. Then, the normalization $A'$ of $A$ in $operatorname{Frac}(A)$ is not finite over $A$ by [Nagata, App. A1, (E3.2)].
answered Dec 19 '18 at 12:24
Takumi MurayamaTakumi Murayama
6,34611645
$endgroup$
add a comment |
$begingroup$
The property you state is related to the property of $R$ being excellent, and more specifically to weaker condition of whether $R$ has the N-2 property or the stronger Nagata property; see [Matsumura, §31] and [Illusie–Laszlo–Orgogozo, Exp. I].
Not all DVR's satisfy your property. We give an example below, due to Nagata. On the other hand, both Dedekind domains of characteristic zero and complete local rings are excellent [Illusie–Laszlo–Orgogozo, Exp. I, Prop. 3.1 and §4], hence satisfy your property.
Example [Nagata, App. A1, Ex. 3]. We give an example of a DVR $R$ and an algebra of finite type $A$ over $R$ such that the normalization $A'$ of $A$ in $operatorname{Frac}(A)$ is not finite over $A$. Let $k$ be a field of characteristic $p > 0$ such that $[k : k^p] = infty$, and consider the ring
$$R = biggl{sum_{i=0}^infty a_ix^i in k[[x]] biggmvert [k^p(a_0,a_1,ldots):k^p] < infty biggr}.$$
This is a DVR by [Nagata, App. A1, (E3.1)]. Let ${b_1,b_2,ldots} subseteq k$ be a sequence of elements in $k$ that are $p$-independent over $k^p$. Set
$$c = sum_{i=0}^infty b_ix^i,$$
and consider the ring $A = R[c]$. Then, the normalization $A'$ of $A$ in $operatorname{Frac}(A)$ is not finite over $A$ by [Nagata, App. A1, (E3.2)].
answered Dec 19 '18 at 12:24
Takumi MurayamaTakumi Murayama
6,34611645
$endgroup$
The property you state is related to the property of $R$ being excellent, and more specifically to weaker condition of whether $R$ has the N-2 property or the stronger Nagata property; see [Matsumura, §31] and [Illusie–Laszlo–Orgogozo, Exp. I].
Not all DVR's satisfy your property. We give an example below, due to Nagata. On the other hand, both Dedekind domains of characteristic zero and complete local rings are excellent [Illusie–Laszlo–Orgogozo, Exp. I, Prop. 3.1 and §4], hence satisfy your property.
Example [Nagata, App. A1, Ex. 3]. We give an example of a DVR $R$ and an algebra of finite type $A$ over $R$ such that the normalization $A'$ of $A$ in $operatorname{Frac}(A)$ is not finite over $A$. Let $k$ be a field of characteristic $p > 0$ such that $[k : k^p] = infty$, and consider the ring
$$R = biggl{sum_{i=0}^infty a_ix^i in k[[x]] biggmvert [k^p(a_0,a_1,ldots):k^p] < infty biggr}.$$
This is a DVR by [Nagata, App. A1, (E3.1)]. Let ${b_1,b_2,ldots} subseteq k$ be a sequence of elements in $k$ that are $p$-independent over $k^p$. Set
$$c = sum_{i=0}^infty b_ix^i,$$
and consider the ring $A = R[c]$. Then, the normalization $A'$ of $A$ in $operatorname{Frac}(A)$ is not finite over $A$ by [Nagata, App. A1, (E3.2)].
answered Dec 19 '18 at 12:24
Takumi MurayamaTakumi Murayama
6,34611645
edited Dec 19 '18 at 13:25
answered Dec 19 '18 at 12:24
Takumi MurayamaTakumi Murayama
6,34611645
answered Dec 19 '18 at 12:24
Takumi MurayamaTakumi Murayama
6,34611645
answered Dec 19 '18 at 12:24
Takumi MurayamaTakumi Murayama
6,34611645
6,34611645
add a comment |
add a comment |
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