The normalization of an integral domain and its quotient
0
$begingroup$
Let $k$ be a field, $A$ a finitely generated local integral domain over $k$ of Krull dimension $1$, and $A'$ be the normalization of $A$ in the fraction field of $A$.
Then is $A'/A$ a torsion $A$-module, i.e., $A' /A cong bigoplus A/g_i $ for some $g_i in A$?
I want to use it to show that $dim_k A'/A$ is finite and the equation $dim_kA'/fA' = dim_k A/fA$, where $f$ is a non-unit element of $A$.
I've shown that $A'$ is a finite $A$-module.
Thank you very much.
commutative-algebra
edited Dec 17 '18 at 16:05
user26857
39.3k124183
asked Dec 16 '18 at 20:03
k.j.k.j.
39819
$endgroup$
|
show 1 more comment
0
$begingroup$
Let $k$ be a field, $A$ a finitely generated local integral domain over $k$ of Krull dimension $1$, and $A'$ be the normalization of $A$ in the fraction field of $A$.
Then is $A'/A$ a torsion $A$-module, i.e., $A' /A cong bigoplus A/g_i $ for some $g_i in A$?
I want to use it to show that $dim_k A'/A$ is finite and the equation $dim_kA'/fA' = dim_k A/fA$, where $f$ is a non-unit element of $A$.
I've shown that $A'$ is a finite $A$-module.
Thank you very much.
commutative-algebra
edited Dec 17 '18 at 16:05
user26857
39.3k124183
asked Dec 16 '18 at 20:03
k.j.k.j.
39819
$endgroup$
$begingroup$
It is torsion, but not necessarily of the form you suggest. What you say in the second paragraph is true.That is, if $M$ is any finitely generated torsion free $A$-module of rank 1, and $0neq fin A$, then the dimension formula holds for $M/fM$ (which might be zero if $f$ is a unit).
$endgroup$
– Mohan
Dec 16 '18 at 20:41
$begingroup$
@Mohan Thank you for your comment. Would you show me why $A'/A$ is torsion?
$endgroup$
– k.j.
Dec 16 '18 at 20:45
$begingroup$
Any element in $A'$ is of the form $a/b$ with $a,bin A, bneq 0$. So, its image in $A'/A$ is annihilated by $b$. This is true of every element of $A'/A$ and so it is torsion.
$endgroup$
– Mohan
Dec 16 '18 at 20:49
$begingroup$
@Mohan Thank you. I understand that $A'/A$ is annihilated. But if $A'/A$ can't be written as $oplus A/g_i A$, I can't show that $dim_k A'/A$ is finite. Could you show it?
$endgroup$
– k.j.
Dec 16 '18 at 21:15
1
$begingroup$
Since $A'$ and thus $A'/A$ is finitely generated, you can find a single $bneq 0$ such that $b(A'/A)=0$. That means, $A'/A$ is a finitely generated module over $A/bA$. But $A/bA$ has finite dimension.
$endgroup$
– Mohan
Dec 16 '18 at 22:59
|
show 1 more comment
0
0
0
1
$begingroup$
Let $k$ be a field, $A$ a finitely generated local integral domain over $k$ of Krull dimension $1$, and $A'$ be the normalization of $A$ in the fraction field of $A$.
Then is $A'/A$ a torsion $A$-module, i.e., $A' /A cong bigoplus A/g_i $ for some $g_i in A$?
I want to use it to show that $dim_k A'/A$ is finite and the equation $dim_kA'/fA' = dim_k A/fA$, where $f$ is a non-unit element of $A$.
I've shown that $A'$ is a finite $A$-module.
Thank you very much.
commutative-algebra
edited Dec 17 '18 at 16:05
user26857
39.3k124183
asked Dec 16 '18 at 20:03
k.j.k.j.
39819
$endgroup$
Let $k$ be a field, $A$ a finitely generated local integral domain over $k$ of Krull dimension $1$, and $A'$ be the normalization of $A$ in the fraction field of $A$.
Then is $A'/A$ a torsion $A$-module, i.e., $A' /A cong bigoplus A/g_i $ for some $g_i in A$?
I want to use it to show that $dim_k A'/A$ is finite and the equation $dim_kA'/fA' = dim_k A/fA$, where $f$ is a non-unit element of $A$.
I've shown that $A'$ is a finite $A$-module.
Thank you very much.
commutative-algebra
commutative-algebra
edited Dec 17 '18 at 16:05
user26857
39.3k124183
asked Dec 16 '18 at 20:03
k.j.k.j.
39819
edited Dec 17 '18 at 16:05
user26857
39.3k124183
asked Dec 16 '18 at 20:03
k.j.k.j.
39819
edited Dec 17 '18 at 16:05
user26857
39.3k124183
edited Dec 17 '18 at 16:05
user26857
39.3k124183
edited Dec 17 '18 at 16:05
user26857
39.3k124183
39.3k124183
asked Dec 16 '18 at 20:03
k.j.k.j.
39819
asked Dec 16 '18 at 20:03
k.j.k.j.
39819
asked Dec 16 '18 at 20:03
k.j.k.j.
39819
39819
$begingroup$
It is torsion, but not necessarily of the form you suggest. What you say in the second paragraph is true.That is, if $M$ is any finitely generated torsion free $A$-module of rank 1, and $0neq fin A$, then the dimension formula holds for $M/fM$ (which might be zero if $f$ is a unit).
$endgroup$
– Mohan
Dec 16 '18 at 20:41
$begingroup$
@Mohan Thank you for your comment. Would you show me why $A'/A$ is torsion?
$endgroup$
– k.j.
Dec 16 '18 at 20:45
$begingroup$
Any element in $A'$ is of the form $a/b$ with $a,bin A, bneq 0$. So, its image in $A'/A$ is annihilated by $b$. This is true of every element of $A'/A$ and so it is torsion.
$endgroup$
– Mohan
Dec 16 '18 at 20:49
$begingroup$
@Mohan Thank you. I understand that $A'/A$ is annihilated. But if $A'/A$ can't be written as $oplus A/g_i A$, I can't show that $dim_k A'/A$ is finite. Could you show it?
$endgroup$
– k.j.
Dec 16 '18 at 21:15
1
$begingroup$
Since $A'$ and thus $A'/A$ is finitely generated, you can find a single $bneq 0$ such that $b(A'/A)=0$. That means, $A'/A$ is a finitely generated module over $A/bA$. But $A/bA$ has finite dimension.
$endgroup$
– Mohan
Dec 16 '18 at 22:59
|
show 1 more comment
$begingroup$
It is torsion, but not necessarily of the form you suggest. What you say in the second paragraph is true.That is, if $M$ is any finitely generated torsion free $A$-module of rank 1, and $0neq fin A$, then the dimension formula holds for $M/fM$ (which might be zero if $f$ is a unit).
$endgroup$
– Mohan
Dec 16 '18 at 20:41
$begingroup$
@Mohan Thank you for your comment. Would you show me why $A'/A$ is torsion?
$endgroup$
– k.j.
Dec 16 '18 at 20:45
$begingroup$
Any element in $A'$ is of the form $a/b$ with $a,bin A, bneq 0$. So, its image in $A'/A$ is annihilated by $b$. This is true of every element of $A'/A$ and so it is torsion.
$endgroup$
– Mohan
Dec 16 '18 at 20:49
$begingroup$
@Mohan Thank you. I understand that $A'/A$ is annihilated. But if $A'/A$ can't be written as $oplus A/g_i A$, I can't show that $dim_k A'/A$ is finite. Could you show it?
$endgroup$
– k.j.
Dec 16 '18 at 21:15
1
$begingroup$
Since $A'$ and thus $A'/A$ is finitely generated, you can find a single $bneq 0$ such that $b(A'/A)=0$. That means, $A'/A$ is a finitely generated module over $A/bA$. But $A/bA$ has finite dimension.
$endgroup$
– Mohan
Dec 16 '18 at 22:59
$begingroup$
It is torsion, but not necessarily of the form you suggest. What you say in the second paragraph is true.That is, if $M$ is any finitely generated torsion free $A$-module of rank 1, and $0neq fin A$, then the dimension formula holds for $M/fM$ (which might be zero if $f$ is a unit).
$endgroup$
– Mohan
Dec 16 '18 at 20:41
$begingroup$
It is torsion, but not necessarily of the form you suggest. What you say in the second paragraph is true.That is, if $M$ is any finitely generated torsion free $A$-module of rank 1, and $0neq fin A$, then the dimension formula holds for $M/fM$ (which might be zero if $f$ is a unit).
$endgroup$
– Mohan
Dec 16 '18 at 20:41
$begingroup$
@Mohan Thank you for your comment. Would you show me why $A'/A$ is torsion?
$endgroup$
– k.j.
Dec 16 '18 at 20:45
$begingroup$
@Mohan Thank you for your comment. Would you show me why $A'/A$ is torsion?
$endgroup$
– k.j.
Dec 16 '18 at 20:45
$begingroup$
Any element in $A'$ is of the form $a/b$ with $a,bin A, bneq 0$. So, its image in $A'/A$ is annihilated by $b$. This is true of every element of $A'/A$ and so it is torsion.
$endgroup$
– Mohan
Dec 16 '18 at 20:49
$begingroup$
Any element in $A'$ is of the form $a/b$ with $a,bin A, bneq 0$. So, its image in $A'/A$ is annihilated by $b$. This is true of every element of $A'/A$ and so it is torsion.
$endgroup$
– Mohan
Dec 16 '18 at 20:49
$begingroup$
@Mohan Thank you. I understand that $A'/A$ is annihilated. But if $A'/A$ can't be written as $oplus A/g_i A$, I can't show that $dim_k A'/A$ is finite. Could you show it?
$endgroup$
– k.j.
Dec 16 '18 at 21:15
$begingroup$
@Mohan Thank you. I understand that $A'/A$ is annihilated. But if $A'/A$ can't be written as $oplus A/g_i A$, I can't show that $dim_k A'/A$ is finite. Could you show it?
$endgroup$
– k.j.
Dec 16 '18 at 21:15
1
1
$begingroup$
Since $A'$ and thus $A'/A$ is finitely generated, you can find a single $bneq 0$ such that $b(A'/A)=0$. That means, $A'/A$ is a finitely generated module over $A/bA$. But $A/bA$ has finite dimension.
$endgroup$
– Mohan
Dec 16 '18 at 22:59
$begingroup$
Since $A'$ and thus $A'/A$ is finitely generated, you can find a single $bneq 0$ such that $b(A'/A)=0$. That means, $A'/A$ is a finitely generated module over $A/bA$. But $A/bA$ has finite dimension.
$endgroup$
– Mohan
Dec 16 '18 at 22:59
|
show 1 more comment
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$begingroup$
It is torsion, but not necessarily of the form you suggest. What you say in the second paragraph is true.That is, if $M$ is any finitely generated torsion free $A$-module of rank 1, and $0neq fin A$, then the dimension formula holds for $M/fM$ (which might be zero if $f$ is a unit).
$endgroup$
– Mohan
Dec 16 '18 at 20:41
$begingroup$
@Mohan Thank you for your comment. Would you show me why $A'/A$ is torsion?
$endgroup$
– k.j.
Dec 16 '18 at 20:45
$begingroup$
Any element in $A'$ is of the form $a/b$ with $a,bin A, bneq 0$. So, its image in $A'/A$ is annihilated by $b$. This is true of every element of $A'/A$ and so it is torsion.
$endgroup$
– Mohan
Dec 16 '18 at 20:49
$begingroup$
@Mohan Thank you. I understand that $A'/A$ is annihilated. But if $A'/A$ can't be written as $oplus A/g_i A$, I can't show that $dim_k A'/A$ is finite. Could you show it?
$endgroup$
– k.j.
Dec 16 '18 at 21:15
1
$begingroup$
Since $A'$ and thus $A'/A$ is finitely generated, you can find a single $bneq 0$ such that $b(A'/A)=0$. That means, $A'/A$ is a finitely generated module over $A/bA$. But $A/bA$ has finite dimension.
$endgroup$
– Mohan
Dec 16 '18 at 22:59