Mixing
Which primes intersect the submonoid $f^mathbb{N}(1+fA)subset A$?
$begingroup$
Let $A$ be a commutative ring and $fin A,Ivartriangleleft A$. We have obvious submonoids $f^mathbb{N},1+Isubset A$.
- The submonoid $f^mathbb{N}subset A$ is saturated, and its complement is the union of primes not containing $f$.
- The complement of the saturation of the submonoid $1+Isubset A$ is the union of maximal ideals containing $I$.
Question. Which primes of $A$ (do not) intersect the submonoid $f^mathbb{N}(1+fA)$? (here $fA=(f)vartriangleleft A$.)
ring-theory commutative-algebra localization
asked Jan 19 at 1:02
ArrowArrow
5,21431446
$endgroup$
add a comment |
$begingroup$
Let $A$ be a commutative ring and $fin A,Ivartriangleleft A$. We have obvious submonoids $f^mathbb{N},1+Isubset A$.
- The submonoid $f^mathbb{N}subset A$ is saturated, and its complement is the union of primes not containing $f$.
- The complement of the saturation of the submonoid $1+Isubset A$ is the union of maximal ideals containing $I$.
Question. Which primes of $A$ (do not) intersect the submonoid $f^mathbb{N}(1+fA)$? (here $fA=(f)vartriangleleft A$.)
ring-theory commutative-algebra localization
asked Jan 19 at 1:02
ArrowArrow
5,21431446
$endgroup$
$begingroup$
$f^{mathbb{N}}$ is typically not saturated (for instance, consider if $A=mathbb{Z}$ and $f$ is composite).
$endgroup$
– Eric Wofsey
Jan 19 at 2:28
add a comment |
$begingroup$
Let $A$ be a commutative ring and $fin A,Ivartriangleleft A$. We have obvious submonoids $f^mathbb{N},1+Isubset A$.
- The submonoid $f^mathbb{N}subset A$ is saturated, and its complement is the union of primes not containing $f$.
- The complement of the saturation of the submonoid $1+Isubset A$ is the union of maximal ideals containing $I$.
Question. Which primes of $A$ (do not) intersect the submonoid $f^mathbb{N}(1+fA)$? (here $fA=(f)vartriangleleft A$.)
ring-theory commutative-algebra localization
asked Jan 19 at 1:02
ArrowArrow
5,21431446
$endgroup$
Let $A$ be a commutative ring and $fin A,Ivartriangleleft A$. We have obvious submonoids $f^mathbb{N},1+Isubset A$.
- The submonoid $f^mathbb{N}subset A$ is saturated, and its complement is the union of primes not containing $f$.
- The complement of the saturation of the submonoid $1+Isubset A$ is the union of maximal ideals containing $I$.
Question. Which primes of $A$ (do not) intersect the submonoid $f^mathbb{N}(1+fA)$? (here $fA=(f)vartriangleleft A$.)
ring-theory commutative-algebra localization
ring-theory commutative-algebra localization
asked Jan 19 at 1:02
ArrowArrow
5,21431446
asked Jan 19 at 1:02
ArrowArrow
5,21431446
edited Jan 19 at 1:30
Arrow
asked Jan 19 at 1:02
ArrowArrow
5,21431446
asked Jan 19 at 1:02
ArrowArrow
5,21431446
asked Jan 19 at 1:02
ArrowArrow
5,21431446
5,21431446
$begingroup$
$f^{mathbb{N}}$ is typically not saturated (for instance, consider if $A=mathbb{Z}$ and $f$ is composite).
$endgroup$
– Eric Wofsey
Jan 19 at 2:28
add a comment |
$begingroup$
$f^{mathbb{N}}$ is typically not saturated (for instance, consider if $A=mathbb{Z}$ and $f$ is composite).
$endgroup$
– Eric Wofsey
Jan 19 at 2:28
$begingroup$
$f^{mathbb{N}}$ is typically not saturated (for instance, consider if $A=mathbb{Z}$ and $f$ is composite).
$endgroup$
– Eric Wofsey
Jan 19 at 2:28
$begingroup$
$f^{mathbb{N}}$ is typically not saturated (for instance, consider if $A=mathbb{Z}$ and $f$ is composite).
$endgroup$
– Eric Wofsey
Jan 19 at 2:28
add a comment |
1 Answer
1
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oldest
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$begingroup$
Well, a prime $P$ intersects $f^mathbb{N}(1+fA)$ iff it intersects either $f^mathbb{N}$ or $1+fA$. This happens iff either $fin P$ or $f$ is a unit mod $P$.
So, $P$ does not intersect $f^mathbb{N}(1+fA)$ iff $f$ is nonzero and not a unit mod $P$. Equivalently, this means that $P$ can be extended to both prime ideals that contain $f$ and prime ideals that don't. Algebro-geometrically, if $operatorname{Spec}A$ is a variety, then such primes $P$ are the generic points of subvarieties which intersect but are not contained in the vanishing set of $f$.
answered Jan 19 at 2:44
Eric WofseyEric Wofsey
187k14216344
$endgroup$
$begingroup$
Is this set of primes anything like the boundary of $mathbf{V}(f)subsetoperatorname{Spec}A$?
$endgroup$
– Arrow
Jan 19 at 12:51
$begingroup$
It's certainly not the boundary in the Zariski topology; $V(f)$ is closed so it contains its own boundary (whereas none of these primes are in $V(f)$). You could say it's the boundary of $V(f)$ in the reverse specialization order topology, where you say a set is open iff it is specialiization-closed.
$endgroup$
– Eric Wofsey
Jan 19 at 16:15
$begingroup$
I ask because $mathfrak p$ specializes to a closed point of $D(f)$ and to a closed point of $mathbf{V}(f)$, and I imagined this as $mathfrak p$ "floating over the boundary". I would appreciate a corrected picture if you have one! (And thanks for the answer as always.)
$endgroup$
– Arrow
Jan 19 at 21:07
$begingroup$
Your picture is good, but it doesn't correspond to the topological notion of boundary. This is a weird phenomenon that can only happen in non-$T_1$ spaces, so it's not going to be a topological concept that is familiar from nice spaces.
$endgroup$
– Eric Wofsey
Jan 19 at 21:15
$begingroup$
Interestingly, the submonoid $f^mathbb{N}(1+fA)$ intersects any maximal ideals, so the spectrum of its localization (thought of as a subset) does not contain any of the closed points of $operatorname{Spec}A$. Strange stuff...
$endgroup$
– Arrow
Jan 19 at 21:37
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Well, a prime $P$ intersects $f^mathbb{N}(1+fA)$ iff it intersects either $f^mathbb{N}$ or $1+fA$. This happens iff either $fin P$ or $f$ is a unit mod $P$.
So, $P$ does not intersect $f^mathbb{N}(1+fA)$ iff $f$ is nonzero and not a unit mod $P$. Equivalently, this means that $P$ can be extended to both prime ideals that contain $f$ and prime ideals that don't. Algebro-geometrically, if $operatorname{Spec}A$ is a variety, then such primes $P$ are the generic points of subvarieties which intersect but are not contained in the vanishing set of $f$.
answered Jan 19 at 2:44
Eric WofseyEric Wofsey
187k14216344
$endgroup$
$begingroup$
Is this set of primes anything like the boundary of $mathbf{V}(f)subsetoperatorname{Spec}A$?
$endgroup$
– Arrow
Jan 19 at 12:51
$begingroup$
It's certainly not the boundary in the Zariski topology; $V(f)$ is closed so it contains its own boundary (whereas none of these primes are in $V(f)$). You could say it's the boundary of $V(f)$ in the reverse specialization order topology, where you say a set is open iff it is specialiization-closed.
$endgroup$
– Eric Wofsey
Jan 19 at 16:15
$begingroup$
I ask because $mathfrak p$ specializes to a closed point of $D(f)$ and to a closed point of $mathbf{V}(f)$, and I imagined this as $mathfrak p$ "floating over the boundary". I would appreciate a corrected picture if you have one! (And thanks for the answer as always.)
$endgroup$
– Arrow
Jan 19 at 21:07
$begingroup$
Your picture is good, but it doesn't correspond to the topological notion of boundary. This is a weird phenomenon that can only happen in non-$T_1$ spaces, so it's not going to be a topological concept that is familiar from nice spaces.
$endgroup$
– Eric Wofsey
Jan 19 at 21:15
$begingroup$
Interestingly, the submonoid $f^mathbb{N}(1+fA)$ intersects any maximal ideals, so the spectrum of its localization (thought of as a subset) does not contain any of the closed points of $operatorname{Spec}A$. Strange stuff...
$endgroup$
– Arrow
Jan 19 at 21:37
add a comment |
$begingroup$
Well, a prime $P$ intersects $f^mathbb{N}(1+fA)$ iff it intersects either $f^mathbb{N}$ or $1+fA$. This happens iff either $fin P$ or $f$ is a unit mod $P$.
So, $P$ does not intersect $f^mathbb{N}(1+fA)$ iff $f$ is nonzero and not a unit mod $P$. Equivalently, this means that $P$ can be extended to both prime ideals that contain $f$ and prime ideals that don't. Algebro-geometrically, if $operatorname{Spec}A$ is a variety, then such primes $P$ are the generic points of subvarieties which intersect but are not contained in the vanishing set of $f$.
answered Jan 19 at 2:44
Eric WofseyEric Wofsey
187k14216344
$endgroup$
$begingroup$
Is this set of primes anything like the boundary of $mathbf{V}(f)subsetoperatorname{Spec}A$?
$endgroup$
– Arrow
Jan 19 at 12:51
$begingroup$
It's certainly not the boundary in the Zariski topology; $V(f)$ is closed so it contains its own boundary (whereas none of these primes are in $V(f)$). You could say it's the boundary of $V(f)$ in the reverse specialization order topology, where you say a set is open iff it is specialiization-closed.
$endgroup$
– Eric Wofsey
Jan 19 at 16:15
$begingroup$
I ask because $mathfrak p$ specializes to a closed point of $D(f)$ and to a closed point of $mathbf{V}(f)$, and I imagined this as $mathfrak p$ "floating over the boundary". I would appreciate a corrected picture if you have one! (And thanks for the answer as always.)
$endgroup$
– Arrow
Jan 19 at 21:07
$begingroup$
Your picture is good, but it doesn't correspond to the topological notion of boundary. This is a weird phenomenon that can only happen in non-$T_1$ spaces, so it's not going to be a topological concept that is familiar from nice spaces.
$endgroup$
– Eric Wofsey
Jan 19 at 21:15
$begingroup$
Interestingly, the submonoid $f^mathbb{N}(1+fA)$ intersects any maximal ideals, so the spectrum of its localization (thought of as a subset) does not contain any of the closed points of $operatorname{Spec}A$. Strange stuff...
$endgroup$
– Arrow
Jan 19 at 21:37
add a comment |
$begingroup$
Well, a prime $P$ intersects $f^mathbb{N}(1+fA)$ iff it intersects either $f^mathbb{N}$ or $1+fA$. This happens iff either $fin P$ or $f$ is a unit mod $P$.
So, $P$ does not intersect $f^mathbb{N}(1+fA)$ iff $f$ is nonzero and not a unit mod $P$. Equivalently, this means that $P$ can be extended to both prime ideals that contain $f$ and prime ideals that don't. Algebro-geometrically, if $operatorname{Spec}A$ is a variety, then such primes $P$ are the generic points of subvarieties which intersect but are not contained in the vanishing set of $f$.
answered Jan 19 at 2:44
Eric WofseyEric Wofsey
187k14216344
$endgroup$
Well, a prime $P$ intersects $f^mathbb{N}(1+fA)$ iff it intersects either $f^mathbb{N}$ or $1+fA$. This happens iff either $fin P$ or $f$ is a unit mod $P$.
So, $P$ does not intersect $f^mathbb{N}(1+fA)$ iff $f$ is nonzero and not a unit mod $P$. Equivalently, this means that $P$ can be extended to both prime ideals that contain $f$ and prime ideals that don't. Algebro-geometrically, if $operatorname{Spec}A$ is a variety, then such primes $P$ are the generic points of subvarieties which intersect but are not contained in the vanishing set of $f$.
answered Jan 19 at 2:44
Eric WofseyEric Wofsey
187k14216344
edited Jan 20 at 21:01
answered Jan 19 at 2:44
Eric WofseyEric Wofsey
187k14216344
answered Jan 19 at 2:44
Eric WofseyEric Wofsey
187k14216344
answered Jan 19 at 2:44
Eric WofseyEric Wofsey
187k14216344
187k14216344
$begingroup$
Is this set of primes anything like the boundary of $mathbf{V}(f)subsetoperatorname{Spec}A$?
$endgroup$
– Arrow
Jan 19 at 12:51
$begingroup$
It's certainly not the boundary in the Zariski topology; $V(f)$ is closed so it contains its own boundary (whereas none of these primes are in $V(f)$). You could say it's the boundary of $V(f)$ in the reverse specialization order topology, where you say a set is open iff it is specialiization-closed.
$endgroup$
– Eric Wofsey
Jan 19 at 16:15
$begingroup$
I ask because $mathfrak p$ specializes to a closed point of $D(f)$ and to a closed point of $mathbf{V}(f)$, and I imagined this as $mathfrak p$ "floating over the boundary". I would appreciate a corrected picture if you have one! (And thanks for the answer as always.)
$endgroup$
– Arrow
Jan 19 at 21:07
$begingroup$
Your picture is good, but it doesn't correspond to the topological notion of boundary. This is a weird phenomenon that can only happen in non-$T_1$ spaces, so it's not going to be a topological concept that is familiar from nice spaces.
$endgroup$
– Eric Wofsey
Jan 19 at 21:15
$begingroup$
Interestingly, the submonoid $f^mathbb{N}(1+fA)$ intersects any maximal ideals, so the spectrum of its localization (thought of as a subset) does not contain any of the closed points of $operatorname{Spec}A$. Strange stuff...
$endgroup$
– Arrow
Jan 19 at 21:37
add a comment |
$begingroup$
Is this set of primes anything like the boundary of $mathbf{V}(f)subsetoperatorname{Spec}A$?
$endgroup$
– Arrow
Jan 19 at 12:51
$begingroup$
It's certainly not the boundary in the Zariski topology; $V(f)$ is closed so it contains its own boundary (whereas none of these primes are in $V(f)$). You could say it's the boundary of $V(f)$ in the reverse specialization order topology, where you say a set is open iff it is specialiization-closed.
$endgroup$
– Eric Wofsey
Jan 19 at 16:15
$begingroup$
I ask because $mathfrak p$ specializes to a closed point of $D(f)$ and to a closed point of $mathbf{V}(f)$, and I imagined this as $mathfrak p$ "floating over the boundary". I would appreciate a corrected picture if you have one! (And thanks for the answer as always.)
$endgroup$
– Arrow
Jan 19 at 21:07
$begingroup$
Your picture is good, but it doesn't correspond to the topological notion of boundary. This is a weird phenomenon that can only happen in non-$T_1$ spaces, so it's not going to be a topological concept that is familiar from nice spaces.
$endgroup$
– Eric Wofsey
Jan 19 at 21:15
$begingroup$
Interestingly, the submonoid $f^mathbb{N}(1+fA)$ intersects any maximal ideals, so the spectrum of its localization (thought of as a subset) does not contain any of the closed points of $operatorname{Spec}A$. Strange stuff...
$endgroup$
– Arrow
Jan 19 at 21:37
$begingroup$
Is this set of primes anything like the boundary of $mathbf{V}(f)subsetoperatorname{Spec}A$?
$endgroup$
– Arrow
Jan 19 at 12:51
$begingroup$
Is this set of primes anything like the boundary of $mathbf{V}(f)subsetoperatorname{Spec}A$?
$endgroup$
– Arrow
Jan 19 at 12:51
$begingroup$
It's certainly not the boundary in the Zariski topology; $V(f)$ is closed so it contains its own boundary (whereas none of these primes are in $V(f)$). You could say it's the boundary of $V(f)$ in the reverse specialization order topology, where you say a set is open iff it is specialiization-closed.
$endgroup$
– Eric Wofsey
Jan 19 at 16:15
$begingroup$
It's certainly not the boundary in the Zariski topology; $V(f)$ is closed so it contains its own boundary (whereas none of these primes are in $V(f)$). You could say it's the boundary of $V(f)$ in the reverse specialization order topology, where you say a set is open iff it is specialiization-closed.
$endgroup$
– Eric Wofsey
Jan 19 at 16:15
$begingroup$
I ask because $mathfrak p$ specializes to a closed point of $D(f)$ and to a closed point of $mathbf{V}(f)$, and I imagined this as $mathfrak p$ "floating over the boundary". I would appreciate a corrected picture if you have one! (And thanks for the answer as always.)
$endgroup$
– Arrow
Jan 19 at 21:07
$begingroup$
I ask because $mathfrak p$ specializes to a closed point of $D(f)$ and to a closed point of $mathbf{V}(f)$, and I imagined this as $mathfrak p$ "floating over the boundary". I would appreciate a corrected picture if you have one! (And thanks for the answer as always.)
$endgroup$
– Arrow
Jan 19 at 21:07
$begingroup$
Your picture is good, but it doesn't correspond to the topological notion of boundary. This is a weird phenomenon that can only happen in non-$T_1$ spaces, so it's not going to be a topological concept that is familiar from nice spaces.
$endgroup$
– Eric Wofsey
Jan 19 at 21:15
$begingroup$
Your picture is good, but it doesn't correspond to the topological notion of boundary. This is a weird phenomenon that can only happen in non-$T_1$ spaces, so it's not going to be a topological concept that is familiar from nice spaces.
$endgroup$
– Eric Wofsey
Jan 19 at 21:15
$begingroup$
Interestingly, the submonoid $f^mathbb{N}(1+fA)$ intersects any maximal ideals, so the spectrum of its localization (thought of as a subset) does not contain any of the closed points of $operatorname{Spec}A$. Strange stuff...
$endgroup$
– Arrow
Jan 19 at 21:37
$begingroup$
Interestingly, the submonoid $f^mathbb{N}(1+fA)$ intersects any maximal ideals, so the spectrum of its localization (thought of as a subset) does not contain any of the closed points of $operatorname{Spec}A$. Strange stuff...
$endgroup$
– Arrow
Jan 19 at 21:37
add a comment |
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$begingroup$
$f^{mathbb{N}}$ is typically not saturated (for instance, consider if $A=mathbb{Z}$ and $f$ is composite).
$endgroup$
– Eric Wofsey
Jan 19 at 2:28