Mixing
Find one-dimensional $P,Q$ such that $PQ = P cap Q$ and $P,Q$ not coprime
$begingroup$
I am looking for an example of the following situation:
Let $R = k[x,y,z]$ be the polynomial ring in three indeterminates where $k$ is a field. I want to find two prime ideals $P$ and $Q$ of $R$ which satisfy
$R/P$ and $R/Q$ are Dedekind domains- at least one of $P$ and $Q$ is not principal ,
$P+Q neq R$,
$P cap Q = PQ$.
In geometrical terms: I would like to find two regular affine curves in $mathbb{A}_k^3$ given by prime ideals from which at least one is not principal (Points 1. and 2.). Moreover, they should intersect in at least one point (This is point 3.) and the curve obtained from these two curves (the curve given by $R/PQ$) should be reduced (This is point 4.).
algebraic-geometry commutative-algebra maximal-and-prime-ideals affine-schemes
asked Jan 7 at 13:14
windsheafwindsheaf
612312
$endgroup$
|
show 4 more comments
$begingroup$
I am looking for an example of the following situation:
Let $R = k[x,y,z]$ be the polynomial ring in three indeterminates where $k$ is a field. I want to find two prime ideals $P$ and $Q$ of $R$ which satisfy
$R/P$ and $R/Q$ are Dedekind domains- at least one of $P$ and $Q$ is not principal ,
$P+Q neq R$,
$P cap Q = PQ$.
In geometrical terms: I would like to find two regular affine curves in $mathbb{A}_k^3$ given by prime ideals from which at least one is not principal (Points 1. and 2.). Moreover, they should intersect in at least one point (This is point 3.) and the curve obtained from these two curves (the curve given by $R/PQ$) should be reduced (This is point 4.).
algebraic-geometry commutative-algebra maximal-and-prime-ideals affine-schemes
asked Jan 7 at 13:14
windsheafwindsheaf
612312
$endgroup$
$begingroup$
what about $(x,z)$ and $(y,z)$? EDIT: ignore this comment, I misread part (4) of the question.
$endgroup$
– Jef L
Jan 7 at 15:21
1
$begingroup$
Note that condition 1 implies condition 2 by Krull's principal ideal theorem.
$endgroup$
– Jef L
Jan 7 at 15:24
$begingroup$
@JefL Thanks for your comment. We may assume wlog that the components meet at the origin, i.e. both $P$ and $Q$ are contained in $(x,y,z)$. But then it seems to me that there is "not enough space" for condition 4. to hold. But I cannot make it rigorous.
$endgroup$
– windsheaf
Jan 7 at 15:33
$begingroup$
do you assume $k$ to be algebraically closed? Because otherwise you can't necessarily assume this maximal ideal to be $(x,y,z)$ e.g. it could be something like $(x^2-2,y,z)$ in $mathbb{Q}[x,y,z]$.
$endgroup$
– Jef L
Jan 7 at 15:42
$begingroup$
@JefL I don't see why I assume that the residue field of $(x,y,z)$ is $k$. No, I don't assume $k$ to be algebraically closed. My archetypical ground field is finite.
$endgroup$
– windsheaf
Jan 7 at 15:45
|
show 4 more comments
$begingroup$
I am looking for an example of the following situation:
Let $R = k[x,y,z]$ be the polynomial ring in three indeterminates where $k$ is a field. I want to find two prime ideals $P$ and $Q$ of $R$ which satisfy
$R/P$ and $R/Q$ are Dedekind domains- at least one of $P$ and $Q$ is not principal ,
$P+Q neq R$,
$P cap Q = PQ$.
In geometrical terms: I would like to find two regular affine curves in $mathbb{A}_k^3$ given by prime ideals from which at least one is not principal (Points 1. and 2.). Moreover, they should intersect in at least one point (This is point 3.) and the curve obtained from these two curves (the curve given by $R/PQ$) should be reduced (This is point 4.).
algebraic-geometry commutative-algebra maximal-and-prime-ideals affine-schemes
asked Jan 7 at 13:14
windsheafwindsheaf
612312
$endgroup$
I am looking for an example of the following situation:
Let $R = k[x,y,z]$ be the polynomial ring in three indeterminates where $k$ is a field. I want to find two prime ideals $P$ and $Q$ of $R$ which satisfy
$R/P$ and $R/Q$ are Dedekind domains- at least one of $P$ and $Q$ is not principal ,
$P+Q neq R$,
$P cap Q = PQ$.
In geometrical terms: I would like to find two regular affine curves in $mathbb{A}_k^3$ given by prime ideals from which at least one is not principal (Points 1. and 2.). Moreover, they should intersect in at least one point (This is point 3.) and the curve obtained from these two curves (the curve given by $R/PQ$) should be reduced (This is point 4.).
algebraic-geometry commutative-algebra maximal-and-prime-ideals affine-schemes
algebraic-geometry commutative-algebra maximal-and-prime-ideals affine-schemes
asked Jan 7 at 13:14
windsheafwindsheaf
612312
asked Jan 7 at 13:14
windsheafwindsheaf
612312
asked Jan 7 at 13:14
windsheafwindsheaf
612312
asked Jan 7 at 13:14
windsheafwindsheaf
612312
asked Jan 7 at 13:14
windsheafwindsheaf
612312
612312
$begingroup$
what about $(x,z)$ and $(y,z)$? EDIT: ignore this comment, I misread part (4) of the question.
$endgroup$
– Jef L
Jan 7 at 15:21
1
$begingroup$
Note that condition 1 implies condition 2 by Krull's principal ideal theorem.
$endgroup$
– Jef L
Jan 7 at 15:24
$begingroup$
@JefL Thanks for your comment. We may assume wlog that the components meet at the origin, i.e. both $P$ and $Q$ are contained in $(x,y,z)$. But then it seems to me that there is "not enough space" for condition 4. to hold. But I cannot make it rigorous.
$endgroup$
– windsheaf
Jan 7 at 15:33
$begingroup$
do you assume $k$ to be algebraically closed? Because otherwise you can't necessarily assume this maximal ideal to be $(x,y,z)$ e.g. it could be something like $(x^2-2,y,z)$ in $mathbb{Q}[x,y,z]$.
$endgroup$
– Jef L
Jan 7 at 15:42
$begingroup$
@JefL I don't see why I assume that the residue field of $(x,y,z)$ is $k$. No, I don't assume $k$ to be algebraically closed. My archetypical ground field is finite.
$endgroup$
– windsheaf
Jan 7 at 15:45
|
show 4 more comments
$begingroup$
what about $(x,z)$ and $(y,z)$? EDIT: ignore this comment, I misread part (4) of the question.
$endgroup$
– Jef L
Jan 7 at 15:21
1
$begingroup$
Note that condition 1 implies condition 2 by Krull's principal ideal theorem.
$endgroup$
– Jef L
Jan 7 at 15:24
$begingroup$
@JefL Thanks for your comment. We may assume wlog that the components meet at the origin, i.e. both $P$ and $Q$ are contained in $(x,y,z)$. But then it seems to me that there is "not enough space" for condition 4. to hold. But I cannot make it rigorous.
$endgroup$
– windsheaf
Jan 7 at 15:33
$begingroup$
do you assume $k$ to be algebraically closed? Because otherwise you can't necessarily assume this maximal ideal to be $(x,y,z)$ e.g. it could be something like $(x^2-2,y,z)$ in $mathbb{Q}[x,y,z]$.
$endgroup$
– Jef L
Jan 7 at 15:42
$begingroup$
@JefL I don't see why I assume that the residue field of $(x,y,z)$ is $k$. No, I don't assume $k$ to be algebraically closed. My archetypical ground field is finite.
$endgroup$
– windsheaf
Jan 7 at 15:45
$begingroup$
what about $(x,z)$ and $(y,z)$? EDIT: ignore this comment, I misread part (4) of the question.
$endgroup$
– Jef L
Jan 7 at 15:21
$begingroup$
what about $(x,z)$ and $(y,z)$? EDIT: ignore this comment, I misread part (4) of the question.
$endgroup$
– Jef L
Jan 7 at 15:21
1
1
$begingroup$
Note that condition 1 implies condition 2 by Krull's principal ideal theorem.
$endgroup$
– Jef L
Jan 7 at 15:24
$begingroup$
Note that condition 1 implies condition 2 by Krull's principal ideal theorem.
$endgroup$
– Jef L
Jan 7 at 15:24
$begingroup$
@JefL Thanks for your comment. We may assume wlog that the components meet at the origin, i.e. both $P$ and $Q$ are contained in $(x,y,z)$. But then it seems to me that there is "not enough space" for condition 4. to hold. But I cannot make it rigorous.
$endgroup$
– windsheaf
Jan 7 at 15:33
$begingroup$
@JefL Thanks for your comment. We may assume wlog that the components meet at the origin, i.e. both $P$ and $Q$ are contained in $(x,y,z)$. But then it seems to me that there is "not enough space" for condition 4. to hold. But I cannot make it rigorous.
$endgroup$
– windsheaf
Jan 7 at 15:33
$begingroup$
do you assume $k$ to be algebraically closed? Because otherwise you can't necessarily assume this maximal ideal to be $(x,y,z)$ e.g. it could be something like $(x^2-2,y,z)$ in $mathbb{Q}[x,y,z]$.
$endgroup$
– Jef L
Jan 7 at 15:42
$begingroup$
do you assume $k$ to be algebraically closed? Because otherwise you can't necessarily assume this maximal ideal to be $(x,y,z)$ e.g. it could be something like $(x^2-2,y,z)$ in $mathbb{Q}[x,y,z]$.
$endgroup$
– Jef L
Jan 7 at 15:42
$begingroup$
@JefL I don't see why I assume that the residue field of $(x,y,z)$ is $k$. No, I don't assume $k$ to be algebraically closed. My archetypical ground field is finite.
$endgroup$
– windsheaf
Jan 7 at 15:45
$begingroup$
@JefL I don't see why I assume that the residue field of $(x,y,z)$ is $k$. No, I don't assume $k$ to be algebraically closed. My archetypical ground field is finite.
$endgroup$
– windsheaf
Jan 7 at 15:45
|
show 4 more comments
1 Answer
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$begingroup$
This can never happen. If such an example existed, localizing at a maximal ideal containing $P+Q$, the same holds and so assume that $R$ is a regular local ring of dimension three, $P,Qsubset R$ prime ideals with all your properties, 3) being automatic now. Then $R/P$ is a dvr, so let $z$ be generator of its maximal ideal. Since, $R/Q$ is also a dvr, $Q=(x,y)$ where they form a regular system of parameters in $R$.
First, assume that neither $x$ nor $y$ is zero in $R/P$. Then, $x=uz^m, y=vz^n$ in $R/P$ with $u,v$ units. We may assume without loss of generality $mgeq ngeq 1$. By abuse of notation, we call the lifts of $u,v,z$ to $R$ by the same letters. Notice that $u,v$ are units in $R$ and $z$ is a non-unit. Then, $x'=x-v^{-1}uz^{m-n}y$ goes to zero in $R/P$. Thus, $x'in Pcap Q$. But, $x'$ is part of a regular system of parameter and thus can not be in $PQ$, the latter is contained in the square of the maximal ideal.
The case when $x$ or $y$ is zero in $R/P$ reduces to the above.
answered Jan 7 at 23:27
MohanMohan
11.7k1817
$endgroup$
add a comment |
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$begingroup$
This can never happen. If such an example existed, localizing at a maximal ideal containing $P+Q$, the same holds and so assume that $R$ is a regular local ring of dimension three, $P,Qsubset R$ prime ideals with all your properties, 3) being automatic now. Then $R/P$ is a dvr, so let $z$ be generator of its maximal ideal. Since, $R/Q$ is also a dvr, $Q=(x,y)$ where they form a regular system of parameters in $R$.
First, assume that neither $x$ nor $y$ is zero in $R/P$. Then, $x=uz^m, y=vz^n$ in $R/P$ with $u,v$ units. We may assume without loss of generality $mgeq ngeq 1$. By abuse of notation, we call the lifts of $u,v,z$ to $R$ by the same letters. Notice that $u,v$ are units in $R$ and $z$ is a non-unit. Then, $x'=x-v^{-1}uz^{m-n}y$ goes to zero in $R/P$. Thus, $x'in Pcap Q$. But, $x'$ is part of a regular system of parameter and thus can not be in $PQ$, the latter is contained in the square of the maximal ideal.
The case when $x$ or $y$ is zero in $R/P$ reduces to the above.
answered Jan 7 at 23:27
MohanMohan
11.7k1817
$endgroup$
add a comment |
$begingroup$
This can never happen. If such an example existed, localizing at a maximal ideal containing $P+Q$, the same holds and so assume that $R$ is a regular local ring of dimension three, $P,Qsubset R$ prime ideals with all your properties, 3) being automatic now. Then $R/P$ is a dvr, so let $z$ be generator of its maximal ideal. Since, $R/Q$ is also a dvr, $Q=(x,y)$ where they form a regular system of parameters in $R$.
First, assume that neither $x$ nor $y$ is zero in $R/P$. Then, $x=uz^m, y=vz^n$ in $R/P$ with $u,v$ units. We may assume without loss of generality $mgeq ngeq 1$. By abuse of notation, we call the lifts of $u,v,z$ to $R$ by the same letters. Notice that $u,v$ are units in $R$ and $z$ is a non-unit. Then, $x'=x-v^{-1}uz^{m-n}y$ goes to zero in $R/P$. Thus, $x'in Pcap Q$. But, $x'$ is part of a regular system of parameter and thus can not be in $PQ$, the latter is contained in the square of the maximal ideal.
The case when $x$ or $y$ is zero in $R/P$ reduces to the above.
answered Jan 7 at 23:27
MohanMohan
11.7k1817
$endgroup$
add a comment |
$begingroup$
This can never happen. If such an example existed, localizing at a maximal ideal containing $P+Q$, the same holds and so assume that $R$ is a regular local ring of dimension three, $P,Qsubset R$ prime ideals with all your properties, 3) being automatic now. Then $R/P$ is a dvr, so let $z$ be generator of its maximal ideal. Since, $R/Q$ is also a dvr, $Q=(x,y)$ where they form a regular system of parameters in $R$.
First, assume that neither $x$ nor $y$ is zero in $R/P$. Then, $x=uz^m, y=vz^n$ in $R/P$ with $u,v$ units. We may assume without loss of generality $mgeq ngeq 1$. By abuse of notation, we call the lifts of $u,v,z$ to $R$ by the same letters. Notice that $u,v$ are units in $R$ and $z$ is a non-unit. Then, $x'=x-v^{-1}uz^{m-n}y$ goes to zero in $R/P$. Thus, $x'in Pcap Q$. But, $x'$ is part of a regular system of parameter and thus can not be in $PQ$, the latter is contained in the square of the maximal ideal.
The case when $x$ or $y$ is zero in $R/P$ reduces to the above.
answered Jan 7 at 23:27
MohanMohan
11.7k1817
$endgroup$
This can never happen. If such an example existed, localizing at a maximal ideal containing $P+Q$, the same holds and so assume that $R$ is a regular local ring of dimension three, $P,Qsubset R$ prime ideals with all your properties, 3) being automatic now. Then $R/P$ is a dvr, so let $z$ be generator of its maximal ideal. Since, $R/Q$ is also a dvr, $Q=(x,y)$ where they form a regular system of parameters in $R$.
First, assume that neither $x$ nor $y$ is zero in $R/P$. Then, $x=uz^m, y=vz^n$ in $R/P$ with $u,v$ units. We may assume without loss of generality $mgeq ngeq 1$. By abuse of notation, we call the lifts of $u,v,z$ to $R$ by the same letters. Notice that $u,v$ are units in $R$ and $z$ is a non-unit. Then, $x'=x-v^{-1}uz^{m-n}y$ goes to zero in $R/P$. Thus, $x'in Pcap Q$. But, $x'$ is part of a regular system of parameter and thus can not be in $PQ$, the latter is contained in the square of the maximal ideal.
The case when $x$ or $y$ is zero in $R/P$ reduces to the above.
answered Jan 7 at 23:27
MohanMohan
11.7k1817
answered Jan 7 at 23:27
MohanMohan
11.7k1817
answered Jan 7 at 23:27
MohanMohan
11.7k1817
answered Jan 7 at 23:27
MohanMohan
11.7k1817
11.7k1817
add a comment |
add a comment |
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$begingroup$
what about $(x,z)$ and $(y,z)$? EDIT: ignore this comment, I misread part (4) of the question.
$endgroup$
– Jef L
Jan 7 at 15:21
1
$begingroup$
Note that condition 1 implies condition 2 by Krull's principal ideal theorem.
$endgroup$
– Jef L
Jan 7 at 15:24
$begingroup$
@JefL Thanks for your comment. We may assume wlog that the components meet at the origin, i.e. both $P$ and $Q$ are contained in $(x,y,z)$. But then it seems to me that there is "not enough space" for condition 4. to hold. But I cannot make it rigorous.
$endgroup$
– windsheaf
Jan 7 at 15:33
$begingroup$
do you assume $k$ to be algebraically closed? Because otherwise you can't necessarily assume this maximal ideal to be $(x,y,z)$ e.g. it could be something like $(x^2-2,y,z)$ in $mathbb{Q}[x,y,z]$.
$endgroup$
– Jef L
Jan 7 at 15:42
$begingroup$
@JefL I don't see why I assume that the residue field of $(x,y,z)$ is $k$. No, I don't assume $k$ to be algebraically closed. My archetypical ground field is finite.
$endgroup$
– windsheaf
Jan 7 at 15:45