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Asymptotic base loci in positive characteristic
Mircea Mustata
University of Michigan
Chulalongkorn UniversityDecember 22, 2011
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 1
/ 25
Asymptotic base loci: definition and basic properties
Let X be a projective scheme over an algebraically closed field k .Later, we will assume that char(k) = p > 0.
If L is a line bundle on X and s ∈ Γ(X , L), then Z (s) denotes thezero-locus of s.
The base-locus of L is
Bs(L) =⋂
s∈Γ(X ,L)
Z (s).
This is interesting, since L defines a map to a projective space onX r Bs(L)red. Of course, it is quite hard to make general statementsabout base loci of line bundles.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 2
/ 25
Asymptotic base loci: definition and basic properties
Let X be a projective scheme over an algebraically closed field k .Later, we will assume that char(k) = p > 0.
If L is a line bundle on X and s ∈ Γ(X , L), then Z (s) denotes thezero-locus of s.
The base-locus of L is
Bs(L) =⋂
s∈Γ(X ,L)
Z (s).
This is interesting, since L defines a map to a projective space onX r Bs(L)red. Of course, it is quite hard to make general statementsabout base loci of line bundles.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 2
/ 25
Asymptotic base loci: definition and basic properties
Let X be a projective scheme over an algebraically closed field k .Later, we will assume that char(k) = p > 0.
If L is a line bundle on X and s ∈ Γ(X , L), then Z (s) denotes thezero-locus of s.
The base-locus of L is
Bs(L) =⋂
s∈Γ(X ,L)
Z (s).
This is interesting, since L defines a map to a projective space onX r Bs(L)red. Of course, it is quite hard to make general statementsabout base loci of line bundles.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 2
/ 25
The stable base locus
First weakening: work asymptotically.Since Z (sr )red = Z (s)red, it follows that Bs(Lmr )red ⊆ Bs(Lm)red.
By Noetherian property, the stable base locus
SB(L) :=⋂m≥1
Bs(Lm)red
is equal to Bs(Lm)red if m is divisible enough.
It follows from this that SB(L) = SB(Lr ) for all r ≥ 1. Using this, we candefine SB(L) when L ∈ Pic(X )Q.
It is particularly interesting to determine when SB(L) is empty, that is, L issemiample. However, this is hard in general, since it is not a numericalproperty: if L ≡ 0, then SB(L) = ∅ if L is torsion, and SB(L) = X ,otherwise.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 3
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The stable base locus
First weakening: work asymptotically.Since Z (sr )red = Z (s)red, it follows that Bs(Lmr )red ⊆ Bs(Lm)red.
By Noetherian property, the stable base locus
SB(L) :=⋂m≥1
Bs(Lm)red
is equal to Bs(Lm)red if m is divisible enough.
It follows from this that SB(L) = SB(Lr ) for all r ≥ 1. Using this, we candefine SB(L) when L ∈ Pic(X )Q.
It is particularly interesting to determine when SB(L) is empty, that is, L issemiample. However, this is hard in general, since it is not a numericalproperty: if L ≡ 0, then SB(L) = ∅ if L is torsion, and SB(L) = X ,otherwise.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 3
/ 25
The stable base locus
First weakening: work asymptotically.Since Z (sr )red = Z (s)red, it follows that Bs(Lmr )red ⊆ Bs(Lm)red.
By Noetherian property, the stable base locus
SB(L) :=⋂m≥1
Bs(Lm)red
is equal to Bs(Lm)red if m is divisible enough.
It follows from this that SB(L) = SB(Lr ) for all r ≥ 1. Using this, we candefine SB(L) when L ∈ Pic(X )Q.
It is particularly interesting to determine when SB(L) is empty, that is, L issemiample. However, this is hard in general, since it is not a numericalproperty: if L ≡ 0, then SB(L) = ∅ if L is torsion, and SB(L) = X ,otherwise.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 3
/ 25
The stable base locus
First weakening: work asymptotically.Since Z (sr )red = Z (s)red, it follows that Bs(Lmr )red ⊆ Bs(Lm)red.
By Noetherian property, the stable base locus
SB(L) :=⋂m≥1
Bs(Lm)red
is equal to Bs(Lm)red if m is divisible enough.
It follows from this that SB(L) = SB(Lr ) for all r ≥ 1. Using this, we candefine SB(L) when L ∈ Pic(X )Q.
It is particularly interesting to determine when SB(L) is empty, that is, L issemiample. However, this is hard in general, since it is not a numericalproperty: if L ≡ 0, then SB(L) = ∅ if L is torsion, and SB(L) = X ,otherwise.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 3
/ 25
The augmented base locus
Second weakening: consider small perturbations. Let L ∈ Pic(X )Q.
The augmented base locus (Nakamaye) is
B+(L) =⋂A
SB(L− A) ⊇ SB(L),
where the intersection is over ample elements A of Pic(X )Q.
Note: this is a Zariski closed subset of X . If X is integral, thenSB(L) 6= X iff L is bigSB(L) = ∅ iff L is ampleConclusion: can consider B+(L) as “locus of points where L is not ample”.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 4
/ 25
The augmented base locus
Second weakening: consider small perturbations. Let L ∈ Pic(X )Q.
The augmented base locus (Nakamaye) is
B+(L) =⋂A
SB(L− A) ⊇ SB(L),
where the intersection is over ample elements A of Pic(X )Q.
Note: this is a Zariski closed subset of X . If X is integral, thenSB(L) 6= X iff L is bigSB(L) = ∅ iff L is ampleConclusion: can consider B+(L) as “locus of points where L is not ample”.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 4
/ 25
The augmented base locus
Second weakening: consider small perturbations. Let L ∈ Pic(X )Q.
The augmented base locus (Nakamaye) is
B+(L) =⋂A
SB(L− A) ⊇ SB(L),
where the intersection is over ample elements A of Pic(X )Q.
Note: this is a Zariski closed subset of X . If X is integral, thenSB(L) 6= X iff L is bigSB(L) = ∅ iff L is ampleConclusion: can consider B+(L) as “locus of points where L is not ample”.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 4
/ 25
The augmented base locus, cont’d
If A− A′ is ample, thenSB(L− A′) = SB ((L− A) + (A′ − A)) ⊆ SB(L− A).
By Noetherian property, it follows that B+(L) = SB(L− A) if A is ample,with image in NS(X )Q is in a small neighborhood of 0.
The augmented base locus is numerically determined: if L ≡ L′, then
B+(L) = B+(L′).
Indeed: A is ample if and only if A− (L′ − L) is ample, and
L′ − (A− (L− L′)) = L− A.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 5
/ 25
The augmented base locus, cont’d
If A− A′ is ample, thenSB(L− A′) = SB ((L− A) + (A′ − A)) ⊆ SB(L− A).
By Noetherian property, it follows that B+(L) = SB(L− A) if A is ample,with image in NS(X )Q is in a small neighborhood of 0.
The augmented base locus is numerically determined: if L ≡ L′, then
B+(L) = B+(L′).
Indeed: A is ample if and only if A− (L′ − L) is ample, and
L′ − (A− (L− L′)) = L− A.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 5
/ 25
The augmented base locus, cont’d
If A− A′ is ample, thenSB(L− A′) = SB ((L− A) + (A′ − A)) ⊆ SB(L− A).
By Noetherian property, it follows that B+(L) = SB(L− A) if A is ample,with image in NS(X )Q is in a small neighborhood of 0.
The augmented base locus is numerically determined: if L ≡ L′, then
B+(L) = B+(L′).
Indeed: A is ample if and only if A− (L′ − L) is ample, and
L′ − (A− (L− L′)) = L− A.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 5
/ 25
The augmented base locus, cont’d
If A− A′ is ample, thenSB(L− A′) = SB ((L− A) + (A′ − A)) ⊆ SB(L− A).
By Noetherian property, it follows that B+(L) = SB(L− A) if A is ample,with image in NS(X )Q is in a small neighborhood of 0.
The augmented base locus is numerically determined: if L ≡ L′, then
B+(L) = B+(L′).
Indeed: A is ample if and only if A− (L′ − L) is ample, and
L′ − (A− (L− L′)) = L− A.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 5
/ 25
Description of B+(L) for L nef
Suppose now that L ∈ Pic(X )Q is nef, that is, (LdimV · V ) ≥ 0 for everyV ⊆ X closed irreducible subvariety.
L⊥ :=⋃
L|V 6=big
V
Note: the union is over the positive-dimensional closed irreduciblesubvarieties V such that L|V is not big, that is, such that(Ldim(V ) · V ) = 0.
Theorem 1. (Cascini, McKernan, M-) If char(k) > 0 and L ∈ Pic(X )Q isnef, then
B+(L) = L⊥.
If X is smooth and char(k) = 0, this was proved by Nakamaye, using theKawamata-Viehweg vanishing theorem.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 6
/ 25
Description of B+(L) for L nef
Suppose now that L ∈ Pic(X )Q is nef, that is, (LdimV · V ) ≥ 0 for everyV ⊆ X closed irreducible subvariety.
L⊥ :=⋃
L|V 6=big
V
Note: the union is over the positive-dimensional closed irreduciblesubvarieties V such that L|V is not big, that is, such that(Ldim(V ) · V ) = 0.
Theorem 1. (Cascini, McKernan, M-) If char(k) > 0 and L ∈ Pic(X )Q isnef, then
B+(L) = L⊥.
If X is smooth and char(k) = 0, this was proved by Nakamaye, using theKawamata-Viehweg vanishing theorem.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 6
/ 25
Description of B+(L) for L nef
Suppose now that L ∈ Pic(X )Q is nef, that is, (LdimV · V ) ≥ 0 for everyV ⊆ X closed irreducible subvariety.
L⊥ :=⋃
L|V 6=big
V
Note: the union is over the positive-dimensional closed irreduciblesubvarieties V such that L|V is not big, that is, such that(Ldim(V ) · V ) = 0.
Theorem 1. (Cascini, McKernan, M-) If char(k) > 0 and L ∈ Pic(X )Q isnef, then
B+(L) = L⊥.
If X is smooth and char(k) = 0, this was proved by Nakamaye, using theKawamata-Viehweg vanishing theorem.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 6
/ 25
Description of B+(L) for L nef
Suppose now that L ∈ Pic(X )Q is nef, that is, (LdimV · V ) ≥ 0 for everyV ⊆ X closed irreducible subvariety.
L⊥ :=⋃
L|V 6=big
V
Note: the union is over the positive-dimensional closed irreduciblesubvarieties V such that L|V is not big, that is, such that(Ldim(V ) · V ) = 0.
Theorem 1. (Cascini, McKernan, M-) If char(k) > 0 and L ∈ Pic(X )Q isnef, then
B+(L) = L⊥.
If X is smooth and char(k) = 0, this was proved by Nakamaye, using theKawamata-Viehweg vanishing theorem.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 6
/ 25
The non-nef locus
Given L ∈ Pic(X )Q, the non-nef locus of L is
B−(L) :=⋃
A ample
SB(L + A) ⊆ SB(L).
As with B+(L), it is easy to see:1) If L1 ≡ L2, then B−(L1) = B−(L2).2) If Am are ample divisors whose classes in NS(X )Q go to 0, then
B−(L) =⋃m
SB(L + Am).
Hence B−(L) is a countable union of Zariski closed subsets.
Basic question. Is B−(L) Zariski closed?
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 7
/ 25
The non-nef locus
Given L ∈ Pic(X )Q, the non-nef locus of L is
B−(L) :=⋃
A ample
SB(L + A) ⊆ SB(L).
As with B+(L), it is easy to see:1) If L1 ≡ L2, then B−(L1) = B−(L2).
2) If Am are ample divisors whose classes in NS(X )Q go to 0, then
B−(L) =⋃m
SB(L + Am).
Hence B−(L) is a countable union of Zariski closed subsets.
Basic question. Is B−(L) Zariski closed?
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 7
/ 25
The non-nef locus
Given L ∈ Pic(X )Q, the non-nef locus of L is
B−(L) :=⋃
A ample
SB(L + A) ⊆ SB(L).
As with B+(L), it is easy to see:1) If L1 ≡ L2, then B−(L1) = B−(L2).2) If Am are ample divisors whose classes in NS(X )Q go to 0, then
B−(L) =⋃m
SB(L + Am).
Hence B−(L) is a countable union of Zariski closed subsets.
Basic question. Is B−(L) Zariski closed?
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 7
/ 25
The non-nef locus
Given L ∈ Pic(X )Q, the non-nef locus of L is
B−(L) :=⋃
A ample
SB(L + A) ⊆ SB(L).
As with B+(L), it is easy to see:1) If L1 ≡ L2, then B−(L1) = B−(L2).2) If Am are ample divisors whose classes in NS(X )Q go to 0, then
B−(L) =⋃m
SB(L + Am).
Hence B−(L) is a countable union of Zariski closed subsets.
Basic question. Is B−(L) Zariski closed?
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 7
/ 25
The non-nef locus, cont’d
Note: if L ∈ Pic(X ) is such that the sction ring ⊕m≥0Γ(X , Lm) is finitelygenerated, then B−(L) = SB(L).
Example. If X is a smooth surface, L is big, and L = P + N is the Zariskidecomposition, with P nef and N the negative part, then
B−(L) = Supp(N) and B+(L) = B+(P) =⋃
(P·C)=0
C .
B−(L) = ∅ iff L is nef (hence the reason for its name).B−(L) = X if and only if L is not pseudo-effective (assuming k isuncountable).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 8
/ 25
The non-nef locus, cont’d
Note: if L ∈ Pic(X ) is such that the sction ring ⊕m≥0Γ(X , Lm) is finitelygenerated, then B−(L) = SB(L).
Example. If X is a smooth surface, L is big, and L = P + N is the Zariskidecomposition, with P nef and N the negative part, then
B−(L) = Supp(N) and B+(L) = B+(P) =⋃
(P·C)=0
C .
B−(L) = ∅ iff L is nef (hence the reason for its name).B−(L) = X if and only if L is not pseudo-effective (assuming k isuncountable).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 8
/ 25
The non-nef locus, cont’d
Note: if L ∈ Pic(X ) is such that the sction ring ⊕m≥0Γ(X , Lm) is finitelygenerated, then B−(L) = SB(L).
Example. If X is a smooth surface, L is big, and L = P + N is the Zariskidecomposition, with P nef and N the negative part, then
B−(L) = Supp(N) and B+(L) = B+(P) =⋃
(P·C)=0
C .
B−(L) = ∅ iff L is nef (hence the reason for its name).
B−(L) = X if and only if L is not pseudo-effective (assuming k isuncountable).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 8
/ 25
The non-nef locus, cont’d
Note: if L ∈ Pic(X ) is such that the sction ring ⊕m≥0Γ(X , Lm) is finitelygenerated, then B−(L) = SB(L).
Example. If X is a smooth surface, L is big, and L = P + N is the Zariskidecomposition, with P nef and N the negative part, then
B−(L) = Supp(N) and B+(L) = B+(P) =⋃
(P·C)=0
C .
B−(L) = ∅ iff L is nef (hence the reason for its name).B−(L) = X if and only if L is not pseudo-effective (assuming k isuncountable).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 8
/ 25
Description of the non-nef locus
Suppose that L ∈ Pic(X ) is big. For every m ≥ 1, let am denote the idealdefining Bs(Lm); evaluation of sections induces a surjective map
Γ(X , Lm)⊗OX → am ⊗ Lm.
If x ∈ X , let ordx |Lm| := ordx(am) = ordx(E ), for E ∈ |Lm| general.
We put ordx(‖ L ‖) := infm≥1ordx |Lm|
m = limm→∞ordx |Lm|
m .
Theorem 2. If char(k) > 0 and X is smooth, then the following hold:
i) For every x ∈ X , taking L to ordx(L) extends to a continuous,positive-homogeneous function on the big cone in NS(X )R.
ii) If L is big, then x ∈ B−(L) iff ordx(L) > 0. If this is not the case,then there is an ample line bundle A such that Lm ⊗ A is globallygenerated at x for every m ≥ 1.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 9
/ 25
Description of the non-nef locus
Suppose that L ∈ Pic(X ) is big. For every m ≥ 1, let am denote the idealdefining Bs(Lm); evaluation of sections induces a surjective map
Γ(X , Lm)⊗OX → am ⊗ Lm.
If x ∈ X , let ordx |Lm| := ordx(am) = ordx(E ), for E ∈ |Lm| general.
We put ordx(‖ L ‖) := infm≥1ordx |Lm|
m = limm→∞ordx |Lm|
m .
Theorem 2. If char(k) > 0 and X is smooth, then the following hold:
i) For every x ∈ X , taking L to ordx(L) extends to a continuous,positive-homogeneous function on the big cone in NS(X )R.
ii) If L is big, then x ∈ B−(L) iff ordx(L) > 0. If this is not the case,then there is an ample line bundle A such that Lm ⊗ A is globallygenerated at x for every m ≥ 1.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 9
/ 25
Description of the non-nef locus
Suppose that L ∈ Pic(X ) is big. For every m ≥ 1, let am denote the idealdefining Bs(Lm); evaluation of sections induces a surjective map
Γ(X , Lm)⊗OX → am ⊗ Lm.
If x ∈ X , let ordx |Lm| := ordx(am) = ordx(E ), for E ∈ |Lm| general.
We put ordx(‖ L ‖) := infm≥1ordx |Lm|
m = limm→∞ordx |Lm|
m .
Theorem 2. If char(k) > 0 and X is smooth, then the following hold:
i) For every x ∈ X , taking L to ordx(L) extends to a continuous,positive-homogeneous function on the big cone in NS(X )R.
ii) If L is big, then x ∈ B−(L) iff ordx(L) > 0. If this is not the case,then there is an ample line bundle A such that Lm ⊗ A is globallygenerated at x for every m ≥ 1.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 9
/ 25
Description of the non-nef locus
Suppose that L ∈ Pic(X ) is big. For every m ≥ 1, let am denote the idealdefining Bs(Lm); evaluation of sections induces a surjective map
Γ(X , Lm)⊗OX → am ⊗ Lm.
If x ∈ X , let ordx |Lm| := ordx(am) = ordx(E ), for E ∈ |Lm| general.
We put ordx(‖ L ‖) := infm≥1ordx |Lm|
m = limm→∞ordx |Lm|
m .
Theorem 2. If char(k) > 0 and X is smooth, then the following hold:
i) For every x ∈ X , taking L to ordx(L) extends to a continuous,positive-homogeneous function on the big cone in NS(X )R.
ii) If L is big, then x ∈ B−(L) iff ordx(L) > 0. If this is not the case,then there is an ample line bundle A such that Lm ⊗ A is globallygenerated at x for every m ≥ 1.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 9
/ 25
Description of the non-nef locus
Suppose that L ∈ Pic(X ) is big. For every m ≥ 1, let am denote the idealdefining Bs(Lm); evaluation of sections induces a surjective map
Γ(X , Lm)⊗OX → am ⊗ Lm.
If x ∈ X , let ordx |Lm| := ordx(am) = ordx(E ), for E ∈ |Lm| general.
We put ordx(‖ L ‖) := infm≥1ordx |Lm|
m = limm→∞ordx |Lm|
m .
Theorem 2. If char(k) > 0 and X is smooth, then the following hold:
i) For every x ∈ X , taking L to ordx(L) extends to a continuous,positive-homogeneous function on the big cone in NS(X )R.
ii) If L is big, then x ∈ B−(L) iff ordx(L) > 0. If this is not the case,then there is an ample line bundle A such that Lm ⊗ A is globallygenerated at x for every m ≥ 1.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 9
/ 25
Description of the non-nef locus, cont’d
In characteristic zero, the above theorem was proved independently by• Nakayama• Ein, Lazarsfeld, M-, Nakamaye, PopaBoth proofs make use of vanishing theorems. The proof of [ELMNP]makes use of a global generation statement for twists of asymptoticmultiplier ideals. We will see later what replaces this in positivecharacteristic.
There is a weaker version of the above theorem that also holds forpseudo-effective divisors. It is the same as the corresponding result ofNakayama in characteristic zero; its proof reduces easily to the case of bigdivisors by adding a small ample class.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 10
/ 25
Description of the non-nef locus, cont’d
In characteristic zero, the above theorem was proved independently by• Nakayama• Ein, Lazarsfeld, M-, Nakamaye, PopaBoth proofs make use of vanishing theorems. The proof of [ELMNP]makes use of a global generation statement for twists of asymptoticmultiplier ideals. We will see later what replaces this in positivecharacteristic.
There is a weaker version of the above theorem that also holds forpseudo-effective divisors. It is the same as the corresponding result ofNakayama in characteristic zero; its proof reduces easily to the case of bigdivisors by adding a small ample class.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 10
/ 25
Ingredients in positive characteristic
Suppose char(k) = p > 0 and X is a projective scheme over k .
Main character: the (absolute) Frobenius morphism
F : X → X , taking f to f p.
Important features:
1) If L ∈ Pic(X ), then (F e)∗(L) ' Lpe.
2) If Y → X closed subscheme defined by a, then Ye := (F e)−1(Y ) isdefined by a[pe ] = (hp
e | h ∈ a).
3) If s ∈ Γ(Y , L|Y ), then (F e)∗(s) ∈ Γ(Ye , Lpe |Ye ) is a section whose
restriction to Y is s⊗pe.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 11
/ 25
Ingredients in positive characteristic
Suppose char(k) = p > 0 and X is a projective scheme over k .
Main character: the (absolute) Frobenius morphism
F : X → X , taking f to f p.
Important features:
1) If L ∈ Pic(X ), then (F e)∗(L) ' Lpe.
2) If Y → X closed subscheme defined by a, then Ye := (F e)−1(Y ) isdefined by a[pe ] = (hp
e | h ∈ a).
3) If s ∈ Γ(Y , L|Y ), then (F e)∗(s) ∈ Γ(Ye , Lpe |Ye ) is a section whose
restriction to Y is s⊗pe.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 11
/ 25
Ingredients in positive characteristic
Suppose char(k) = p > 0 and X is a projective scheme over k .
Main character: the (absolute) Frobenius morphism
F : X → X , taking f to f p.
Important features:
1) If L ∈ Pic(X ), then (F e)∗(L) ' Lpe.
2) If Y → X closed subscheme defined by a, then Ye := (F e)−1(Y ) isdefined by a[pe ] = (hp
e | h ∈ a).
3) If s ∈ Γ(Y , L|Y ), then (F e)∗(s) ∈ Γ(Ye , Lpe |Ye ) is a section whose
restriction to Y is s⊗pe.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 11
/ 25
Ingredients in positive characteristic
Suppose char(k) = p > 0 and X is a projective scheme over k .
Main character: the (absolute) Frobenius morphism
F : X → X , taking f to f p.
Important features:
1) If L ∈ Pic(X ), then (F e)∗(L) ' Lpe.
2) If Y → X closed subscheme defined by a, then Ye := (F e)−1(Y ) isdefined by a[pe ] = (hp
e | h ∈ a).
3) If s ∈ Γ(Y , L|Y ), then (F e)∗(s) ∈ Γ(Ye , Lpe |Ye ) is a section whose
restriction to Y is s⊗pe.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 11
/ 25
Use of Frobenius in the proof of Theorem 1
The following is an observation due to Keel:If Y is a closed subscheme of X and L ∈ Pic(X ), then
SB(L|Y ) = SB(L|Yred).
Upshot: when considering stable base loci in positive characteristic, thenon-reduced structure can be ignored. This is useful when doing inductionon dimension.
Proof. The inclusion “⊇” is easy: if s ∈ Γ(Y , Lm|Y ) and q 6∈ Z (s), thenq 6∈ Z (s ′), where s ′ = s|Yred
.
Suppose now that t ′ ∈ Γ(Yred, Lm|Yred
) is such that q 6∈ Z (t ′). For e 0,
the section t ′pe
lifts to a section t of Lmpe |Y : it is enough to consider
(F e)∗(s) and choose e such that I[pe ]Yred⊆ IY . Clearly, q 6∈ Z (t).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 12
/ 25
Use of Frobenius in the proof of Theorem 1
The following is an observation due to Keel:If Y is a closed subscheme of X and L ∈ Pic(X ), then
SB(L|Y ) = SB(L|Yred).
Upshot: when considering stable base loci in positive characteristic, thenon-reduced structure can be ignored. This is useful when doing inductionon dimension.
Proof. The inclusion “⊇” is easy: if s ∈ Γ(Y , Lm|Y ) and q 6∈ Z (s), thenq 6∈ Z (s ′), where s ′ = s|Yred
.
Suppose now that t ′ ∈ Γ(Yred, Lm|Yred
) is such that q 6∈ Z (t ′). For e 0,
the section t ′pe
lifts to a section t of Lmpe |Y : it is enough to consider
(F e)∗(s) and choose e such that I[pe ]Yred⊆ IY . Clearly, q 6∈ Z (t).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 12
/ 25
Use of Frobenius in the proof of Theorem 1
The following is an observation due to Keel:If Y is a closed subscheme of X and L ∈ Pic(X ), then
SB(L|Y ) = SB(L|Yred).
Upshot: when considering stable base loci in positive characteristic, thenon-reduced structure can be ignored. This is useful when doing inductionon dimension.
Proof. The inclusion “⊇” is easy: if s ∈ Γ(Y , Lm|Y ) and q 6∈ Z (s), thenq 6∈ Z (s ′), where s ′ = s|Yred
.
Suppose now that t ′ ∈ Γ(Yred, Lm|Yred
) is such that q 6∈ Z (t ′). For e 0,
the section t ′pe
lifts to a section t of Lmpe |Y : it is enough to consider
(F e)∗(s) and choose e such that I[pe ]Yred⊆ IY . Clearly, q 6∈ Z (t).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 12
/ 25
Use of Frobenius in the proof of Theorem 1
The following is an observation due to Keel:If Y is a closed subscheme of X and L ∈ Pic(X ), then
SB(L|Y ) = SB(L|Yred).
Upshot: when considering stable base loci in positive characteristic, thenon-reduced structure can be ignored. This is useful when doing inductionon dimension.
Proof. The inclusion “⊇” is easy: if s ∈ Γ(Y , Lm|Y ) and q 6∈ Z (s), thenq 6∈ Z (s ′), where s ′ = s|Yred
.
Suppose now that t ′ ∈ Γ(Yred, Lm|Yred
) is such that q 6∈ Z (t ′). For e 0,
the section t ′pe
lifts to a section t of Lmpe |Y : it is enough to consider
(F e)∗(s) and choose e such that I[pe ]Yred⊆ IY . Clearly, q 6∈ Z (t).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 12
/ 25
Use of Frobenius in the proof of Theorem 1, cont’d
A key ingredient in the proof of Theorem 1 is the followingProposition. If L ∈ Pic(X ) is nef, and D is an effective Cartier divisor onX such that A := L(−D) is ample, then
SB(L) = SB(L|D) and B+(L) = B+(L|D).
The first assertion was due to Keel, with a rather complicated argument.The following easier proof extends easily to the case of the augmentedbase locus.
The proof of Theorem 1 goes by induction on dimension. We reduce tothe case when X is integral. If L is not big, then both loci are equal to X .Otherwise, we can find A and D as in proposition, and L⊥ ⊆ Supp(D),hence L⊥ = (L|D)⊥, and we conclude by induction, using the proposition.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 13
/ 25
Use of Frobenius in the proof of Theorem 1, cont’d
A key ingredient in the proof of Theorem 1 is the followingProposition. If L ∈ Pic(X ) is nef, and D is an effective Cartier divisor onX such that A := L(−D) is ample, then
SB(L) = SB(L|D) and B+(L) = B+(L|D).
The first assertion was due to Keel, with a rather complicated argument.The following easier proof extends easily to the case of the augmentedbase locus.
The proof of Theorem 1 goes by induction on dimension. We reduce tothe case when X is integral. If L is not big, then both loci are equal to X .Otherwise, we can find A and D as in proposition, and L⊥ ⊆ Supp(D),hence L⊥ = (L|D)⊥, and we conclude by induction, using the proposition.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 13
/ 25
Use of Frobenius in the proof of Theorem 1, cont’d
A key ingredient in the proof of Theorem 1 is the followingProposition. If L ∈ Pic(X ) is nef, and D is an effective Cartier divisor onX such that A := L(−D) is ample, then
SB(L) = SB(L|D) and B+(L) = B+(L|D).
The first assertion was due to Keel, with a rather complicated argument.The following easier proof extends easily to the case of the augmentedbase locus.
The proof of Theorem 1 goes by induction on dimension. We reduce tothe case when X is integral. If L is not big, then both loci are equal to X .Otherwise, we can find A and D as in proposition, and L⊥ ⊆ Supp(D),hence L⊥ = (L|D)⊥, and we conclude by induction, using the proposition.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 13
/ 25
Use of Frobenius in the proof of Theorem 1, cont’d
Proof of proposition. We only prove the first assertion. The inclusion“⊇” is again trivial, hence we only need to prove “⊆”.
It is enough to show that given s ∈ Γ(D, Lm|D), there is e > 0 such thats⊗p
eextends to a section in Γ(X , Lmpe ).
We have an exact sequence
0→ Lm(−D)→ Lm → Lm|D → 0.
Lm(−D) = Lm−1 ⊗ A is ample, so H1(X , Lmpe (−peD))) = 0 for e 0.
Pulling-back by F e the above exact sequence gives an exact sequence
0→ Lmpe (−peD)→ Lmpe → Lmpe |peD → 0.
By above vanishing, (F e)∗(s) lifts to a section in Γ(X , Lmpe ), q.e.d.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 14
/ 25
Use of Frobenius in the proof of Theorem 1, cont’d
Proof of proposition. We only prove the first assertion. The inclusion“⊇” is again trivial, hence we only need to prove “⊆”.
It is enough to show that given s ∈ Γ(D, Lm|D), there is e > 0 such thats⊗p
eextends to a section in Γ(X , Lmpe ).
We have an exact sequence
0→ Lm(−D)→ Lm → Lm|D → 0.
Lm(−D) = Lm−1 ⊗ A is ample, so H1(X , Lmpe (−peD))) = 0 for e 0.
Pulling-back by F e the above exact sequence gives an exact sequence
0→ Lmpe (−peD)→ Lmpe → Lmpe |peD → 0.
By above vanishing, (F e)∗(s) lifts to a section in Γ(X , Lmpe ), q.e.d.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 14
/ 25
Use of Frobenius in the proof of Theorem 1, cont’d
Proof of proposition. We only prove the first assertion. The inclusion“⊇” is again trivial, hence we only need to prove “⊆”.
It is enough to show that given s ∈ Γ(D, Lm|D), there is e > 0 such thats⊗p
eextends to a section in Γ(X , Lmpe ).
We have an exact sequence
0→ Lm(−D)→ Lm → Lm|D → 0.
Lm(−D) = Lm−1 ⊗ A is ample, so H1(X , Lmpe (−peD))) = 0 for e 0.
Pulling-back by F e the above exact sequence gives an exact sequence
0→ Lmpe (−peD)→ Lmpe → Lmpe |peD → 0.
By above vanishing, (F e)∗(s) lifts to a section in Γ(X , Lmpe ), q.e.d.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 14
/ 25
Use of Frobenius in the proof of Theorem 1, cont’d
Proof of proposition. We only prove the first assertion. The inclusion“⊇” is again trivial, hence we only need to prove “⊆”.
It is enough to show that given s ∈ Γ(D, Lm|D), there is e > 0 such thats⊗p
eextends to a section in Γ(X , Lmpe ).
We have an exact sequence
0→ Lm(−D)→ Lm → Lm|D → 0.
Lm(−D) = Lm−1 ⊗ A is ample, so H1(X , Lmpe (−peD))) = 0 for e 0.
Pulling-back by F e the above exact sequence gives an exact sequence
0→ Lmpe (−peD)→ Lmpe → Lmpe |peD → 0.
By above vanishing, (F e)∗(s) lifts to a section in Γ(X , Lmpe ), q.e.d.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 14
/ 25
Use of Frobenius in the proof of Theorem 1, cont’d
Proof of proposition. We only prove the first assertion. The inclusion“⊇” is again trivial, hence we only need to prove “⊆”.
It is enough to show that given s ∈ Γ(D, Lm|D), there is e > 0 such thats⊗p
eextends to a section in Γ(X , Lmpe ).
We have an exact sequence
0→ Lm(−D)→ Lm → Lm|D → 0.
Lm(−D) = Lm−1 ⊗ A is ample, so H1(X , Lmpe (−peD))) = 0 for e 0.
Pulling-back by F e the above exact sequence gives an exact sequence
0→ Lmpe (−peD)→ Lmpe → Lmpe |peD → 0.
By above vanishing, (F e)∗(s) lifts to a section in Γ(X , Lmpe ), q.e.d.Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristic
Chulalongkorn University December 22, 2011 14/ 25
Comments about the proof of Theorem 2
A key ingredient in the proof of the statement of Theorem 2 incharacteristic 0 is the following global generation result:
If L is a big line bundle on the n-dimensional smooth variety X , and H isan ample and globally generated line bundle, then
J (‖ L ‖)⊗ ωX ⊗ L⊗ Hd is globally generated
for all d ≥ n + 1.
In the above, J (‖ L ‖) is the asymptotic multiplier ideal of L (itsdefinition is irrelevant for us).
The above statement is a consequence of• Nadel’s vanishing theorem, and• Castelnuovo-Mumford regularity
We will prove a similar statement in positive characteristic for test ideals.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 15
/ 25
Comments about the proof of Theorem 2
A key ingredient in the proof of the statement of Theorem 2 incharacteristic 0 is the following global generation result:
If L is a big line bundle on the n-dimensional smooth variety X , and H isan ample and globally generated line bundle, then
J (‖ L ‖)⊗ ωX ⊗ L⊗ Hd is globally generated
for all d ≥ n + 1.
In the above, J (‖ L ‖) is the asymptotic multiplier ideal of L (itsdefinition is irrelevant for us).
The above statement is a consequence of• Nadel’s vanishing theorem, and• Castelnuovo-Mumford regularity
We will prove a similar statement in positive characteristic for test ideals.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 15
/ 25
Comments about the proof of Theorem 2
A key ingredient in the proof of the statement of Theorem 2 incharacteristic 0 is the following global generation result:
If L is a big line bundle on the n-dimensional smooth variety X , and H isan ample and globally generated line bundle, then
J (‖ L ‖)⊗ ωX ⊗ L⊗ Hd is globally generated
for all d ≥ n + 1.
In the above, J (‖ L ‖) is the asymptotic multiplier ideal of L (itsdefinition is irrelevant for us).
The above statement is a consequence of• Nadel’s vanishing theorem, and• Castelnuovo-Mumford regularity
We will prove a similar statement in positive characteristic for test ideals.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 15
/ 25
Comments about the proof of Theorem 2
A key ingredient in the proof of the statement of Theorem 2 incharacteristic 0 is the following global generation result:
If L is a big line bundle on the n-dimensional smooth variety X , and H isan ample and globally generated line bundle, then
J (‖ L ‖)⊗ ωX ⊗ L⊗ Hd is globally generated
for all d ≥ n + 1.
In the above, J (‖ L ‖) is the asymptotic multiplier ideal of L (itsdefinition is irrelevant for us).
The above statement is a consequence of• Nadel’s vanishing theorem, and• Castelnuovo-Mumford regularity
We will prove a similar statement in positive characteristic for test ideals.Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristic
Chulalongkorn University December 22, 2011 15/ 25
Test ideals
Let X be a smooth variety over k = k, with char(k) > 0.
Key ingredient: a surjective map Tr : F∗(ωX )→ ωX .
This can be described:• A trace map for relative duality with respect to the Frobenius morphism.
• Via the Cartier isomorphism
Hi (F∗(Ω•X )) ' ΩiX for i = dim(X ).
• Explicit description in local coordinates x1, . . . , xn:
Tr(x i11 · · · xinn dx) = x
i1−p+1p
1 · · · xin−p+1
pn dx ,
where dx = dx1 ∧ . . . ∧ dxn and the above expression is zero if someexponent is not an integer.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 16
/ 25
Test ideals
Let X be a smooth variety over k = k, with char(k) > 0.
Key ingredient: a surjective map Tr : F∗(ωX )→ ωX .
This can be described:• A trace map for relative duality with respect to the Frobenius morphism.
• Via the Cartier isomorphism
Hi (F∗(Ω•X )) ' ΩiX for i = dim(X ).
• Explicit description in local coordinates x1, . . . , xn:
Tr(x i11 · · · xinn dx) = x
i1−p+1p
1 · · · xin−p+1
pn dx ,
where dx = dx1 ∧ . . . ∧ dxn and the above expression is zero if someexponent is not an integer.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 16
/ 25
Test ideals
Let X be a smooth variety over k = k, with char(k) > 0.
Key ingredient: a surjective map Tr : F∗(ωX )→ ωX .
This can be described:• A trace map for relative duality with respect to the Frobenius morphism.
• Via the Cartier isomorphism
Hi (F∗(Ω•X )) ' ΩiX for i = dim(X ).
• Explicit description in local coordinates x1, . . . , xn:
Tr(x i11 · · · xinn dx) = x
i1−p+1p
1 · · · xin−p+1
pn dx ,
where dx = dx1 ∧ . . . ∧ dxn and the above expression is zero if someexponent is not an integer.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 16
/ 25
Test ideals
Let X be a smooth variety over k = k, with char(k) > 0.
Key ingredient: a surjective map Tr : F∗(ωX )→ ωX .
This can be described:• A trace map for relative duality with respect to the Frobenius morphism.
• Via the Cartier isomorphism
Hi (F∗(Ω•X )) ' ΩiX for i = dim(X ).
• Explicit description in local coordinates x1, . . . , xn:
Tr(x i11 · · · xinn dx) = x
i1−p+1p
1 · · · xin−p+1
pn dx ,
where dx = dx1 ∧ . . . ∧ dxn and the above expression is zero if someexponent is not an integer.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 16
/ 25
Test ideals
Let X be a smooth variety over k = k, with char(k) > 0.
Key ingredient: a surjective map Tr : F∗(ωX )→ ωX .
This can be described:• A trace map for relative duality with respect to the Frobenius morphism.
• Via the Cartier isomorphism
Hi (F∗(Ω•X )) ' ΩiX for i = dim(X ).
• Explicit description in local coordinates x1, . . . , xn:
Tr(x i11 · · · xinn dx) = x
i1−p+1p
1 · · · xin−p+1
pn dx ,
where dx = dx1 ∧ . . . ∧ dxn and the above expression is zero if someexponent is not an integer.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 16
/ 25
Test ideals, cont’d
Iterating e times Tr, we obtain Tre : F e∗ (ωX )→ ωX .
If b is an ideal on X , we define the ideal b[1/pe ] such that Tre induces asurjective map
F e∗ (b · ωX )→ b[1/pe ] · ωX .
This is like taking the pe-root of an ideal in the following sense: b[1/pe ] isthe unique smallest ideal J such that b ⊆ J [pe ].
Suppose now that a is a nonzero ideal on X , and λ ∈ R≥0.
Easy to see: (adλpee)[1/pe ] ⊆ (adλp
e+1e)[1/pe+1] for every e ≥ 1.
The Noetherian property implies that for e 0, the ideal (adλpee)[1/pe ] is
constant. This is the test ideal τ(aλ).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 17
/ 25
Test ideals, cont’d
Iterating e times Tr, we obtain Tre : F e∗ (ωX )→ ωX .
If b is an ideal on X , we define the ideal b[1/pe ] such that Tre induces asurjective map
F e∗ (b · ωX )→ b[1/pe ] · ωX .
This is like taking the pe-root of an ideal in the following sense: b[1/pe ] isthe unique smallest ideal J such that b ⊆ J [pe ].
Suppose now that a is a nonzero ideal on X , and λ ∈ R≥0.
Easy to see: (adλpee)[1/pe ] ⊆ (adλp
e+1e)[1/pe+1] for every e ≥ 1.
The Noetherian property implies that for e 0, the ideal (adλpee)[1/pe ] is
constant. This is the test ideal τ(aλ).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 17
/ 25
Test ideals, cont’d
Iterating e times Tr, we obtain Tre : F e∗ (ωX )→ ωX .
If b is an ideal on X , we define the ideal b[1/pe ] such that Tre induces asurjective map
F e∗ (b · ωX )→ b[1/pe ] · ωX .
This is like taking the pe-root of an ideal in the following sense: b[1/pe ] isthe unique smallest ideal J such that b ⊆ J [pe ].
Suppose now that a is a nonzero ideal on X , and λ ∈ R≥0.
Easy to see: (adλpee)[1/pe ] ⊆ (adλp
e+1e)[1/pe+1] for every e ≥ 1.
The Noetherian property implies that for e 0, the ideal (adλpee)[1/pe ] is
constant. This is the test ideal τ(aλ).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 17
/ 25
Test ideals, cont’d
Iterating e times Tr, we obtain Tre : F e∗ (ωX )→ ωX .
If b is an ideal on X , we define the ideal b[1/pe ] such that Tre induces asurjective map
F e∗ (b · ωX )→ b[1/pe ] · ωX .
This is like taking the pe-root of an ideal in the following sense: b[1/pe ] isthe unique smallest ideal J such that b ⊆ J [pe ].
Suppose now that a is a nonzero ideal on X , and λ ∈ R≥0.
Easy to see: (adλpee)[1/pe ] ⊆ (adλp
e+1e)[1/pe+1] for every e ≥ 1.
The Noetherian property implies that for e 0, the ideal (adλpee)[1/pe ] is
constant. This is the test ideal τ(aλ).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 17
/ 25
Test ideals, cont’d
Iterating e times Tr, we obtain Tre : F e∗ (ωX )→ ωX .
If b is an ideal on X , we define the ideal b[1/pe ] such that Tre induces asurjective map
F e∗ (b · ωX )→ b[1/pe ] · ωX .
This is like taking the pe-root of an ideal in the following sense: b[1/pe ] isthe unique smallest ideal J such that b ⊆ J [pe ].
Suppose now that a is a nonzero ideal on X , and λ ∈ R≥0.
Easy to see: (adλpee)[1/pe ] ⊆ (adλp
e+1e)[1/pe+1] for every e ≥ 1.
The Noetherian property implies that for e 0, the ideal (adλpee)[1/pe ] is
constant. This is the test ideal τ(aλ).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 17
/ 25
Test ideals, cont’d
These ideals have been introduced by Hara and Yoshida in the context of(generalized) tight closure theory. The above definition is due toBlickle-M-Smith and Schwede.
Test ideals are interesting for two reasons:• They satisfy some analogous properties to those satisfied by multiplierideals.
• They are related to multiplier ideals via reduction to prime characteristic(results of Hara-Yoshida, as well as open questions).
Note: the smallest λ such that τ(aλ) 6= OX is the F -pure threshold fpt(a).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 18
/ 25
Test ideals, cont’d
These ideals have been introduced by Hara and Yoshida in the context of(generalized) tight closure theory. The above definition is due toBlickle-M-Smith and Schwede.
Test ideals are interesting for two reasons:• They satisfy some analogous properties to those satisfied by multiplierideals.
• They are related to multiplier ideals via reduction to prime characteristic(results of Hara-Yoshida, as well as open questions).
Note: the smallest λ such that τ(aλ) 6= OX is the F -pure threshold fpt(a).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 18
/ 25
Test ideals, cont’d
These ideals have been introduced by Hara and Yoshida in the context of(generalized) tight closure theory. The above definition is due toBlickle-M-Smith and Schwede.
Test ideals are interesting for two reasons:• They satisfy some analogous properties to those satisfied by multiplierideals.
• They are related to multiplier ideals via reduction to prime characteristic(results of Hara-Yoshida, as well as open questions).
Note: the smallest λ such that τ(aλ) 6= OX is the F -pure threshold fpt(a).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 18
/ 25
Test ideals, cont’d
These ideals have been introduced by Hara and Yoshida in the context of(generalized) tight closure theory. The above definition is due toBlickle-M-Smith and Schwede.
Test ideals are interesting for two reasons:• They satisfy some analogous properties to those satisfied by multiplierideals.
• They are related to multiplier ideals via reduction to prime characteristic(results of Hara-Yoshida, as well as open questions).
Note: the smallest λ such that τ(aλ) 6= OX is the F -pure threshold fpt(a).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 18
/ 25
The subadditivity property
Test ideals satisfy the following subadditivity property (compare to thecorresponding property for multiplier ideals, due toDemailly-Ein-Lazarsfeld):
τ(aλ1+λ2) ⊆ τ(aλ1) · τ(aλ2) for everyλ1, λ2 ∈ R≥0.
This is a consequence of the following fact: if b1, b2 are two ideals then
(b1 · b2)[1/pe ] ⊆ b[1/pe ]1 · b[1/pe ]
2 . (1)
Indeed, we have
b1 · b2 ⊆(b
[1/pe ]1
)[pe ]·(b
[1/pe ]2
)[pe ]=(b
[1/pe ]1 · b[1/pe ]
2
)[pe ].
This implies (1).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 19
/ 25
The subadditivity property
Test ideals satisfy the following subadditivity property (compare to thecorresponding property for multiplier ideals, due toDemailly-Ein-Lazarsfeld):
τ(aλ1+λ2) ⊆ τ(aλ1) · τ(aλ2) for everyλ1, λ2 ∈ R≥0.
This is a consequence of the following fact: if b1, b2 are two ideals then
(b1 · b2)[1/pe ] ⊆ b[1/pe ]1 · b[1/pe ]
2 . (1)
Indeed, we have
b1 · b2 ⊆(b
[1/pe ]1
)[pe ]·(b
[1/pe ]2
)[pe ]=(b
[1/pe ]1 · b[1/pe ]
2
)[pe ].
This implies (1).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 19
/ 25
The subadditivity property
Test ideals satisfy the following subadditivity property (compare to thecorresponding property for multiplier ideals, due toDemailly-Ein-Lazarsfeld):
τ(aλ1+λ2) ⊆ τ(aλ1) · τ(aλ2) for everyλ1, λ2 ∈ R≥0.
This is a consequence of the following fact: if b1, b2 are two ideals then
(b1 · b2)[1/pe ] ⊆ b[1/pe ]1 · b[1/pe ]
2 . (1)
Indeed, we have
b1 · b2 ⊆(b
[1/pe ]1
)[pe ]·(b
[1/pe ]2
)[pe ]=(b
[1/pe ]1 · b[1/pe ]
2
)[pe ].
This implies (1).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 19
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An asymptotic variant
We follow in our setting the procedure for defining asymptotic multiplierideals, associated to the base-loci of the multiples of a line bundle.
Suppose that X is projective and L ∈ Pic(X ) is such that h0(X , Lm) 6= 0for some m ≥ 1. Recall that am denotes the ideal defining Bs(Lm).Easy to see: (am)r ⊆ amr for every r ≥ 1.
Consequence: for every λ ∈ R≥0, we have
τ(aλ/mm ) = τ((arm)λ/mr ) ⊆ τ(a
λ/mrmr ).
By Noetherian property, if m is divisible enough, then τ(aλ/mm ) is
independent of m. This is the asymptotic test ideal τ(λ· ‖ L ‖).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 20
/ 25
An asymptotic variant
We follow in our setting the procedure for defining asymptotic multiplierideals, associated to the base-loci of the multiples of a line bundle.
Suppose that X is projective and L ∈ Pic(X ) is such that h0(X , Lm) 6= 0for some m ≥ 1. Recall that am denotes the ideal defining Bs(Lm).Easy to see: (am)r ⊆ amr for every r ≥ 1.
Consequence: for every λ ∈ R≥0, we have
τ(aλ/mm ) = τ((arm)λ/mr ) ⊆ τ(a
λ/mrmr ).
By Noetherian property, if m is divisible enough, then τ(aλ/mm ) is
independent of m. This is the asymptotic test ideal τ(λ· ‖ L ‖).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 20
/ 25
An asymptotic variant
We follow in our setting the procedure for defining asymptotic multiplierideals, associated to the base-loci of the multiples of a line bundle.
Suppose that X is projective and L ∈ Pic(X ) is such that h0(X , Lm) 6= 0for some m ≥ 1. Recall that am denotes the ideal defining Bs(Lm).Easy to see: (am)r ⊆ amr for every r ≥ 1.
Consequence: for every λ ∈ R≥0, we have
τ(aλ/mm ) = τ((arm)λ/mr ) ⊆ τ(a
λ/mrmr ).
By Noetherian property, if m is divisible enough, then τ(aλ/mm ) is
independent of m. This is the asymptotic test ideal τ(λ· ‖ L ‖).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 20
/ 25
An asymptotic variant
We follow in our setting the procedure for defining asymptotic multiplierideals, associated to the base-loci of the multiples of a line bundle.
Suppose that X is projective and L ∈ Pic(X ) is such that h0(X , Lm) 6= 0for some m ≥ 1. Recall that am denotes the ideal defining Bs(Lm).Easy to see: (am)r ⊆ amr for every r ≥ 1.
Consequence: for every λ ∈ R≥0, we have
τ(aλ/mm ) = τ((arm)λ/mr ) ⊆ τ(a
λ/mrmr ).
By Noetherian property, if m is divisible enough, then τ(aλ/mm ) is
independent of m. This is the asymptotic test ideal τ(λ· ‖ L ‖).
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 20
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A global generation result
The description of the non-nef locus can be deduced from the followingresult.
Theorem 3. Let X be a smooth projective variety over k = k , withchar(k) = p > 0, and let H be an ample divisor on X , with OX (H)globally generated. If D and E are divisors on X such that OX (D) hasnon-negative Iitaka dimension, and λ ∈ Q≥0 is such that E − λD is nef,then
τ(λ· ‖ D ‖)⊗OXωX (E + dH)
is globally generated for all d ≥ dim(X ) + 1.
Keeler and Hara (unpublished) have used ideas similar to the ones belowto recast results about Fujita’s conjecture for ample and globally generatedline bundles in positive characteristic. Schwede is the first one who noticedthat their arguments give global generation results for twists of test ideals.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 21
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A global generation result
The description of the non-nef locus can be deduced from the followingresult.
Theorem 3. Let X be a smooth projective variety over k = k , withchar(k) = p > 0, and let H be an ample divisor on X , with OX (H)globally generated. If D and E are divisors on X such that OX (D) hasnon-negative Iitaka dimension, and λ ∈ Q≥0 is such that E − λD is nef,then
τ(λ· ‖ D ‖)⊗OXωX (E + dH)
is globally generated for all d ≥ dim(X ) + 1.
Keeler and Hara (unpublished) have used ideas similar to the ones belowto recast results about Fujita’s conjecture for ample and globally generatedline bundles in positive characteristic. Schwede is the first one who noticedthat their arguments give global generation results for twists of test ideals.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 21
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Proof of the global generation result
We first fix m such that τ(λ· ‖ D ‖) = τ(aλ/mm ), where am is such that we
have a surjective map
Γ(X ,OX (mD))⊗OX → am ⊗OX (mD). (2)
This implies that for every i ≥ 1, we can find a linear space V = Vi suchthat we have a surjective map
V ⊗k L−mi → aim. (3)
We now consider e large enough (to be fixed later) such that
τ(aλ/mm ) =
(adλpe/mem
)[1/pe ].
By definition of inverse Frobenius power, we have a surjective map
(F e)∗(adλpe/mem ⊗ ωX )→ τ(λ· ‖ D ‖)⊗ ωX . (4)
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Proof of the global generation result
We first fix m such that τ(λ· ‖ D ‖) = τ(aλ/mm ), where am is such that we
have a surjective map
Γ(X ,OX (mD))⊗OX → am ⊗OX (mD). (2)
This implies that for every i ≥ 1, we can find a linear space V = Vi suchthat we have a surjective map
V ⊗k L−mi → aim. (3)
We now consider e large enough (to be fixed later) such that
τ(aλ/mm ) =
(adλpe/mem
)[1/pe ].
By definition of inverse Frobenius power, we have a surjective map
(F e)∗(adλpe/mem ⊗ ωX )→ τ(λ· ‖ D ‖)⊗ ωX . (4)
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 22
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Proof of the global generation result
We first fix m such that τ(λ· ‖ D ‖) = τ(aλ/mm ), where am is such that we
have a surjective map
Γ(X ,OX (mD))⊗OX → am ⊗OX (mD). (2)
This implies that for every i ≥ 1, we can find a linear space V = Vi suchthat we have a surjective map
V ⊗k L−mi → aim. (3)
We now consider e large enough (to be fixed later) such that
τ(aλ/mm ) =
(adλpe/mem
)[1/pe ].
By definition of inverse Frobenius power, we have a surjective map
(F e)∗(adλpe/mem ⊗ ωX )→ τ(λ· ‖ D ‖)⊗ ωX . (4)
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 22
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Proof of the global generation result
We first fix m such that τ(λ· ‖ D ‖) = τ(aλ/mm ), where am is such that we
have a surjective map
Γ(X ,OX (mD))⊗OX → am ⊗OX (mD). (2)
This implies that for every i ≥ 1, we can find a linear space V = Vi suchthat we have a surjective map
V ⊗k L−mi → aim. (3)
We now consider e large enough (to be fixed later) such that
τ(aλ/mm ) =
(adλpe/mem
)[1/pe ].
By definition of inverse Frobenius power, we have a surjective map
(F e)∗(adλpe/mem ⊗ ωX )→ τ(λ· ‖ D ‖)⊗ ωX . (4)
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 22
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Proof of the global generation result, cont’d
Tensoring the previous map with OX (E + dH) and using the projectionformula, we get a surjective map
(F e)∗(adλpe/mem ⊗ ωX (pe(E + dH))
)→ τ(λ· ‖ D ‖)⊗ ωX (E + dH). (5)
On the other hand, we have seen that we have a surjective map from
(F e)∗(V ⊗k ωX (pe(E + dH)−mdλpe/meD))
to the left-hand side of (5).By combining this with (5), we deduce that it is enough to show that
E := (F e)∗(ωX (pe(E + dH)−mdλpe/meD))
is globally generated.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 23
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Proof of the global generation result, cont’d
Tensoring the previous map with OX (E + dH) and using the projectionformula, we get a surjective map
(F e)∗(adλpe/mem ⊗ ωX (pe(E + dH))
)→ τ(λ· ‖ D ‖)⊗ ωX (E + dH). (5)
On the other hand, we have seen that we have a surjective map from
(F e)∗(V ⊗k ωX (pe(E + dH)−mdλpe/meD))
to the left-hand side of (5).By combining this with (5), we deduce that it is enough to show that
E := (F e)∗(ωX (pe(E + dH)−mdλpe/meD))
is globally generated.
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 23
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Proof of the global generation result, cont’d
For this, it is enough to show that E is 0-regular, in the sense ofCastelnuovo-Mumford regularity, with respect to H, that is,
H i (X , E(−iH)) = 0 for 1 ≤ i ≤ n = dim(X ).
Using the fact that F is affine and the projection formula, we see that it isenough to show that
H i (X , ωX (pe(E + (d − i)H)−mdλpe/meD)) = 0 for 1 ≤ i ≤ n. (6)
It is easy to see that since λ ∈ Q, there are finitely many divisorsT1, . . . ,Tr such that for every e we can write
peE −mdλpe/meD − Ti = non-negative multiple of (E − λD),
hence it is nef. The vanishing in (6) follows for e 0 from the following
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 24
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Proof of the global generation result, cont’d
For this, it is enough to show that E is 0-regular, in the sense ofCastelnuovo-Mumford regularity, with respect to H, that is,
H i (X , E(−iH)) = 0 for 1 ≤ i ≤ n = dim(X ).
Using the fact that F is affine and the projection formula, we see that it isenough to show that
H i (X , ωX (pe(E + (d − i)H)−mdλpe/meD)) = 0 for 1 ≤ i ≤ n. (6)
It is easy to see that since λ ∈ Q, there are finitely many divisorsT1, . . . ,Tr such that for every e we can write
peE −mdλpe/meD − Ti = non-negative multiple of (E − λD),
hence it is nef. The vanishing in (6) follows for e 0 from the following
Mircea Mustata (University of Michigan) Asymptotic base loci in positive characteristicChulalongkorn University December 22, 2011 24
/ 25
Proof of the global generation result, cont’d
For this, it is enough to show that E is 0-regular, in the sense ofCastelnuovo-Mumford regularity, with respect to H, that is,
H i (X , E(−iH)) = 0 for 1 ≤ i ≤ n = dim(X ).
Using the fact that F is affine and the projection formula, we see that it isenough to show that
H i (X , ωX (pe(E + (d − i)H)−mdλpe/meD)) = 0 for 1 ≤ i ≤ n. (6)
It is easy to see that since λ ∈ Q, there are finitely many divisorsT1, . . . ,Tr such that for every e we can write
peE −mdλpe/meD − Ti = non-negative multiple of (E − λD),
hence it is nef. The vanishing in (6) follows for e 0 from the followingMircea Mustata (University of Michigan) Asymptotic base loci in positive characteristic
Chulalongkorn University December 22, 2011 24/ 25
Theorem (Fujita) If F is a coherent sheaf on a projective variety X , andH is an ample divisor on X , then there is m0 such that
H i (X ,F ⊗OX (mH + F )) = 0
for every i ≥ 1, every m ≥ m0, and every nef divisor F on X .
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