Does such ring extension exist?
0
$begingroup$
Does there exist a commutative ring extension $Asubsetneq B$ satisfying the following conditions:
$A$ is a normal local domain and $B$ is a regular domain with the same dimension;
$A$ and $B$ have the same fraction field;
$B$ is finitely generated as an $A$-algebra.
commutative-algebra
asked Dec 14 '18 at 2:25
G.-S. ZhouG.-S. Zhou
1288
$endgroup$
add a comment |
0
$begingroup$
Does there exist a commutative ring extension $Asubsetneq B$ satisfying the following conditions:
$A$ is a normal local domain and $B$ is a regular domain with the same dimension;
$A$ and $B$ have the same fraction field;
$B$ is finitely generated as an $A$-algebra.
commutative-algebra
asked Dec 14 '18 at 2:25
G.-S. ZhouG.-S. Zhou
1288
$endgroup$
1
$begingroup$
Hi. You can take $B = k[x,y]$ polynomial ring over a field, and $A = k[x,xy]$. Since $y = xy/x$, they have the same field of fractrions, these two rings are regular (so normal), and $A[y] to B$ gives item 3.
$endgroup$
– Youngsu
Dec 15 '18 at 20:08
$begingroup$
Hi. We need $A$ to be local.
$endgroup$
– G.-S. Zhou
Dec 16 '18 at 2:40
$begingroup$
You can localize $A, B$ at the complement of the maxial ideal $(x,xy)A$.
$endgroup$
– Youngsu
Dec 16 '18 at 7:58
add a comment |
0
0
0
$begingroup$
Does there exist a commutative ring extension $Asubsetneq B$ satisfying the following conditions:
$A$ is a normal local domain and $B$ is a regular domain with the same dimension;
$A$ and $B$ have the same fraction field;
$B$ is finitely generated as an $A$-algebra.
commutative-algebra
asked Dec 14 '18 at 2:25
G.-S. ZhouG.-S. Zhou
1288
$endgroup$
Does there exist a commutative ring extension $Asubsetneq B$ satisfying the following conditions:
$A$ is a normal local domain and $B$ is a regular domain with the same dimension;
$A$ and $B$ have the same fraction field;
$B$ is finitely generated as an $A$-algebra.
commutative-algebra
commutative-algebra
asked Dec 14 '18 at 2:25
G.-S. ZhouG.-S. Zhou
1288
asked Dec 14 '18 at 2:25
G.-S. ZhouG.-S. Zhou
1288
asked Dec 14 '18 at 2:25
G.-S. ZhouG.-S. Zhou
1288
asked Dec 14 '18 at 2:25
G.-S. ZhouG.-S. Zhou
1288
asked Dec 14 '18 at 2:25
G.-S. ZhouG.-S. Zhou
1288
1288
1
$begingroup$
Hi. You can take $B = k[x,y]$ polynomial ring over a field, and $A = k[x,xy]$. Since $y = xy/x$, they have the same field of fractrions, these two rings are regular (so normal), and $A[y] to B$ gives item 3.
$endgroup$
– Youngsu
Dec 15 '18 at 20:08
$begingroup$
Hi. We need $A$ to be local.
$endgroup$
– G.-S. Zhou
Dec 16 '18 at 2:40
$begingroup$
You can localize $A, B$ at the complement of the maxial ideal $(x,xy)A$.
$endgroup$
– Youngsu
Dec 16 '18 at 7:58
add a comment |
1
$begingroup$
Hi. You can take $B = k[x,y]$ polynomial ring over a field, and $A = k[x,xy]$. Since $y = xy/x$, they have the same field of fractrions, these two rings are regular (so normal), and $A[y] to B$ gives item 3.
$endgroup$
– Youngsu
Dec 15 '18 at 20:08
$begingroup$
Hi. We need $A$ to be local.
$endgroup$
– G.-S. Zhou
Dec 16 '18 at 2:40
$begingroup$
You can localize $A, B$ at the complement of the maxial ideal $(x,xy)A$.
$endgroup$
– Youngsu
Dec 16 '18 at 7:58
1
1
$begingroup$
Hi. You can take $B = k[x,y]$ polynomial ring over a field, and $A = k[x,xy]$. Since $y = xy/x$, they have the same field of fractrions, these two rings are regular (so normal), and $A[y] to B$ gives item 3.
$endgroup$
– Youngsu
Dec 15 '18 at 20:08
$begingroup$
Hi. You can take $B = k[x,y]$ polynomial ring over a field, and $A = k[x,xy]$. Since $y = xy/x$, they have the same field of fractrions, these two rings are regular (so normal), and $A[y] to B$ gives item 3.
$endgroup$
– Youngsu
Dec 15 '18 at 20:08
$begingroup$
Hi. We need $A$ to be local.
$endgroup$
– G.-S. Zhou
Dec 16 '18 at 2:40
$begingroup$
Hi. We need $A$ to be local.
$endgroup$
– G.-S. Zhou
Dec 16 '18 at 2:40
$begingroup$
You can localize $A, B$ at the complement of the maxial ideal $(x,xy)A$.
$endgroup$
– Youngsu
Dec 16 '18 at 7:58
$begingroup$
You can localize $A, B$ at the complement of the maxial ideal $(x,xy)A$.
$endgroup$
– Youngsu
Dec 16 '18 at 7:58
add a comment |
0
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1
$begingroup$
Hi. You can take $B = k[x,y]$ polynomial ring over a field, and $A = k[x,xy]$. Since $y = xy/x$, they have the same field of fractrions, these two rings are regular (so normal), and $A[y] to B$ gives item 3.
$endgroup$
– Youngsu
Dec 15 '18 at 20:08
$begingroup$
Hi. We need $A$ to be local.
$endgroup$
– G.-S. Zhou
Dec 16 '18 at 2:40
$begingroup$
You can localize $A, B$ at the complement of the maxial ideal $(x,xy)A$.
$endgroup$
– Youngsu
Dec 16 '18 at 7:58