Defining equations of algebraic sets

Defining equations of algebraic sets

Matteo Varbaro (University of Genoa, Italy)

December 10th 2015, MPI-Oberseminar, Bonn, Germany

Algebraic sets

K = K field, S = K [x 0 , . . . , x n ] polynomial ring.

Given a set A ⊆ S of homogeneous polynomials, its zero-locus is: Z(A) = {P ∈ P ⁿ (K ) : f (P) = 0 ∀ f ∈ A} ⊆ P ⁿ (K ). Subsets of P ⁿ = P ⁿ (K ) as above are called algebraic sets.

Since Z(A) = Z((A)) (where (A) means the ideal of S generated by A) by the Hilbert’s basis theorem every algebraic set is the zero-locus of finitely many (homogeneous) polynomials. Today, we will discuss the following question:

How many polynomials are needed to define an algebraic set?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Algebraic sets

K = K field,

S = K [x 0 , . . . , x n ] polynomial ring.

Given a set A ⊆ S of homogeneous polynomials, its zero-locus is: Z(A) = {P ∈ P ⁿ (K ) : f (P) = 0 ∀ f ∈ A} ⊆ P ⁿ (K ). Subsets of P ⁿ = P ⁿ (K ) as above are called algebraic sets.

Since Z(A) = Z((A)) (where (A) means the ideal of S generated by A) by the Hilbert’s basis theorem every algebraic set is the zero-locus of finitely many (homogeneous) polynomials. Today, we will discuss the following question:

How many polynomials are needed to define an algebraic set?

Algebraic sets

K = K field, S = K [x 0 , . . . , x n ] polynomial ring.

Given a set A ⊆ S of homogeneous polynomials, its zero-locus is: Z(A) = {P ∈ P ⁿ (K ) : f (P) = 0 ∀ f ∈ A} ⊆ P ⁿ (K ). Subsets of P ⁿ = P ⁿ (K ) as above are called algebraic sets.

Since Z(A) = Z((A)) (where (A) means the ideal of S generated by A) by the Hilbert’s basis theorem every algebraic set is the zero-locus of finitely many (homogeneous) polynomials. Today, we will discuss the following question:

How many polynomials are needed to define an algebraic set?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Algebraic sets

K = K field, S = K [x 0 , . . . , x n ] polynomial ring.

Given a set A ⊆ S of homogeneous polynomials, its zero-locus is:

Z(A) = {P ∈ P ⁿ (K ) : f (P) = 0 ∀ f ∈ A} ⊆ P ⁿ (K ).

Subsets of P ⁿ = P ⁿ (K ) as above are called algebraic sets.

Since Z(A) = Z((A)) (where (A) means the ideal of S generated by A) by the Hilbert’s basis theorem every algebraic set is the zero-locus of finitely many (homogeneous) polynomials. Today, we will discuss the following question:

How many polynomials are needed to define an algebraic set?

Algebraic sets

K = K field, S = K [x 0 , . . . , x n ] polynomial ring.

Given a set A ⊆ S of homogeneous polynomials, its zero-locus is:

Z(A) = {P ∈ P ⁿ (K ) : f (P) = 0 ∀ f ∈ A} ⊆ P ⁿ (K ).

Subsets of P ⁿ = P ⁿ (K ) as above are called algebraic sets.

Since Z(A) = Z((A)) (where (A) means the ideal of S generated by A) by the Hilbert’s basis theorem every algebraic set is the zero-locus of finitely many (homogeneous) polynomials. Today, we will discuss the following question:

How many polynomials are needed to define an algebraic set?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Algebraic sets

K = K field, S = K [x 0 , . . . , x n ] polynomial ring.

Given a set A ⊆ S of homogeneous polynomials, its zero-locus is:

Z(A) = {P ∈ P ⁿ (K ) : f (P) = 0 ∀ f ∈ A} ⊆ P ⁿ (K ).

Subsets of P ⁿ = P ⁿ (K ) as above are called algebraic sets.

Since Z(A) = Z((A)) (where (A) means the ideal of S generated by A) by the Hilbert’s basis theorem every algebraic set is the zero-locus of finitely many (homogeneous) polynomials.

Today, we will discuss the following question:

How many polynomials are needed to define an algebraic set?

Algebraic sets

K = K field, S = K [x 0 , . . . , x n ] polynomial ring.

Given a set A ⊆ S of homogeneous polynomials, its zero-locus is:

Z(A) = {P ∈ P ⁿ (K ) : f (P) = 0 ∀ f ∈ A} ⊆ P ⁿ (K ).

Subsets of P ⁿ = P ⁿ (K ) as above are called algebraic sets.

Since Z(A) = Z((A)) (where (A) means the ideal of S generated by A) by the Hilbert’s basis theorem every algebraic set is the zero-locus of finitely many (homogeneous) polynomials. Today, we will discuss the following question:

How many polynomials are needed to define an algebraic set?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Algebraic sets

Given a subset X ⊆ P ⁿ , the set of polynomials vanishing at X is:

I(X ) = {f ∈ S : f (P) = 0 ∀ P ∈ X }.

As it turns out, I(X ) is a radical ideal of S and, by Hilbert’s Nullstellensatz, for any ideal I ⊆ S we have:

I(Z(I )) = √ I

Further, Z(I(X )) = X whenever X ⊆ P ⁿ is an algebraic set. At a first thought, one could imagine that the optimal number of polynomials defining an algebraic set X = Z(I ) ⊆ P ⁿ is exactly the number of minimal generators of I(X ) ⊆ S .

This, however, is far to be true ...

Algebraic sets

Given a subset X ⊆ P ⁿ , the set of polynomials vanishing at X is:

I(X ) = {f ∈ S : f (P) = 0 ∀ P ∈ X }.

As it turns out, I(X ) is a radical ideal of S and, by Hilbert’s Nullstellensatz, for any ideal I ⊆ S we have:

I(Z(I )) = √ I

Further, Z(I(X )) = X whenever X ⊆ P ⁿ is an algebraic set. At a first thought, one could imagine that the optimal number of polynomials defining an algebraic set X = Z(I ) ⊆ P ⁿ is exactly the number of minimal generators of I(X ) ⊆ S .

This, however, is far to be true ...

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Algebraic sets

Given a subset X ⊆ P ⁿ , the set of polynomials vanishing at X is:

I(X ) = {f ∈ S : f (P) = 0 ∀ P ∈ X }.

As it turns out, I(X ) is a radical ideal of S and, by Hilbert’s Nullstellensatz, for any ideal I ⊆ S we have:

I(Z(I )) = √ I

Further, Z(I(X )) = X whenever X ⊆ P ⁿ is an algebraic set.

At a first thought, one could imagine that the optimal number of polynomials defining an algebraic set X = Z(I ) ⊆ P ⁿ is exactly the number of minimal generators of I(X ) ⊆ S .

This, however, is far to be true ...

Algebraic sets

Given a subset X ⊆ P ⁿ , the set of polynomials vanishing at X is:

I(X ) = {f ∈ S : f (P) = 0 ∀ P ∈ X }.

As it turns out, I(X ) is a radical ideal of S and, by Hilbert’s Nullstellensatz, for any ideal I ⊆ S we have:

I(Z(I )) = √ I

Further, Z(I(X )) = X whenever X ⊆ P ⁿ is an algebraic set.

At a first thought, one could imagine that the optimal number of polynomials defining an algebraic set X = Z(I ) ⊆ P ⁿ is exactly the number of minimal generators of I(X ) ⊆ S .

This, however, is far to be true ...

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Algebraic sets

Given a subset X ⊆ P ⁿ , the set of polynomials vanishing at X is:

I(X ) = {f ∈ S : f (P) = 0 ∀ P ∈ X }.

As it turns out, I(X ) is a radical ideal of S and, by Hilbert’s Nullstellensatz, for any ideal I ⊆ S we have:

I(Z(I )) = √ I

Further, Z(I(X )) = X whenever X ⊆ P ⁿ is an algebraic set.

At a first thought, one could imagine that the optimal number of

polynomials defining an algebraic set X = Z(I ) ⊆ P ⁿ is exactly the

number of minimal generators of I(X ) ⊆ S .

Example I

X 1 = {[s ³ , s ² t, st ² , t ³ ] : [s, t] ∈ P ¹ } ⊆ P ³ is an algebraic set and

I(X ₁ ) = (a, b, c) ⊆ S = K [x 0 , x 1 , x 2 , x 3 ],

where a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ . By setting: f = a = x ₀ x ₂ − x ₁ ² and g = x ₃ c − x ₂ b = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ , we claim that X 1 = Z(a, b, c) = Z(f , g ). Obviously Z(a, b, c) is contained in Z(f , g ), for the other inclusion we have to prove that any point P ∈ Z(f , g ) satisfies b(P) = c(P) = 0.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example I

X 1 = {[s ³ , s ² t, st ² , t ³ ] : [s, t] ∈ P ¹ } ⊆ P ³ is an algebraic set and I(X ₁ ) = (a, b, c) ⊆ S = K [x 0 , x 1 , x 2 , x 3 ],

where a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ .

By setting:

f = a = x ₀ x ₂ − x ₁ ² and g = x ₃ c − x ₂ b = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ ,

we claim that X 1 = Z(a, b, c) = Z(f , g ). Obviously Z(a, b, c) is

contained in Z(f , g ), for the other inclusion we have to prove that

any point P ∈ Z(f , g ) satisfies b(P) = c(P) = 0.

Example I

X 1 = {[s ³ , s ² t, st ² , t ³ ] : [s, t] ∈ P ¹ } ⊆ P ³ is an algebraic set and I(X ₁ ) = (a, b, c) ⊆ S = K [x 0 , x 1 , x 2 , x 3 ],

where a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ . By setting:

f = a = x ₀ x ₂ − x ₁ ² and g = x ₃ c − x ₂ b = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ ,

we claim that X 1 = Z(a, b, c) = Z(f , g ). Obviously Z(a, b, c) is contained in Z(f , g ), for the other inclusion we have to prove that any point P ∈ Z(f , g ) satisfies b(P) = c(P) = 0.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example I

X 1 = {[s ³ , s ² t, st ² , t ³ ] : [s, t] ∈ P ¹ } ⊆ P ³ is an algebraic set and I(X ₁ ) = (a, b, c) ⊆ S = K [x 0 , x 1 , x 2 , x 3 ],

where a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ . By setting:

f = a = x ₀ x ₂ − x ₁ ² and g = x ₃ c − x ₂ b = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ , we claim that X 1 = Z(a, b, c) = Z(f , g ).

Obviously Z(a, b, c) is

contained in Z(f , g ), for the other inclusion we have to prove that

any point P ∈ Z(f , g ) satisfies b(P) = c(P) = 0.

Example I

X 1 = {[s ³ , s ² t, st ² , t ³ ] : [s, t] ∈ P ¹ } ⊆ P ³ is an algebraic set and I(X ₁ ) = (a, b, c) ⊆ S = K [x 0 , x 1 , x 2 , x 3 ],

where a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ . By setting:

f = a = x ₀ x ₂ − x ₁ ² and g = x ₃ c − x ₂ b = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ , we claim that X 1 = Z(a, b, c) = Z(f , g ). Obviously Z(a, b, c) is contained in Z(f , g ),

for the other inclusion we have to prove that any point P ∈ Z(f , g ) satisfies b(P) = c(P) = 0.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example I

X 1 = {[s ³ , s ² t, st ² , t ³ ] : [s, t] ∈ P ¹ } ⊆ P ³ is an algebraic set and I(X ₁ ) = (a, b, c) ⊆ S = K [x 0 , x 1 , x 2 , x 3 ],

where a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ . By setting:

f = a = x ₀ x ₂ − x ₁ ² and g = x ₃ c − x ₂ b = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ ,

we claim that X 1 = Z(a, b, c) = Z(f , g ). Obviously Z(a, b, c) is

contained in Z(f , g ), for the other inclusion we have to prove that

any point P ∈ Z(f , g ) satisfies b(P) = c(P) = 0.

Example I

Recall that a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ and f = x ₀ x ₂ − x ₁ ² , g = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ .

Let P = [P 0 , P 1 , P 2 , P 3 ] ∈ Z(f , g ).

P ₂ = 0 = === ^{f (P)=0} =⇒ P ₁ = 0 =====⇒ P ^{g (P)=0} ₀ P ₃ ² = P ₀ P ₃ = 0. So both b and c vanish at P.

P ₂ = 1 = === ^{f (P)=0} =⇒ P ₀ = P ₁ ² =====⇒ (P ^{g (P)=0} ₁ P ₃ ) ² − 2P ₁ P ₃ + 1 = 0. So P ₁ P ₃ = 1 and b(P) = 0. Further

c(P) = P ₀ P ₃ − P ₁ P ₂ = P ₁ ² P ₃ − P ₁ = P ₁ − P ₁ = 0.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example I

Recall that a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ and f = x ₀ x ₂ − x ₁ ² , g = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ .

Let P = [P 0 , P 1 , P 2 , P 3 ] ∈ Z(f , g ).

P ₂ = 0 = === ^{f (P)=0} =⇒ P ₁ = 0 =====⇒ P ^{g (P)=0} ₀ P ₃ ² = P ₀ P ₃ = 0. So both b and c vanish at P.

P ₂ = 1 = === ^{f (P)=0} =⇒ P ₀ = P ₁ ² =====⇒ (P ^{g (P)=0} ₁ P ₃ ) ² − 2P ₁ P ₃ + 1 = 0. So P ₁ P ₃ = 1 and b(P) = 0. Further

c(P) = P ₀ P ₃ − P ₁ P ₂ = P ₁ ² P ₃ − P ₁ = P ₁ − P ₁ = 0.

Example I

Recall that a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ and f = x ₀ x ₂ − x ₁ ² , g = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ .

Let P = [P 0 , P 1 , P 2 , P 3 ] ∈ Z(f , g ).

P ₂ = 0 = === ^{f (P)=0} =⇒ P ₁ = 0 =====⇒ P ^{g (P)=0} ₀ P ₃ ² = P ₀ P ₃ = 0. So both b and c vanish at P.

P ₂ = 1 = === ^{f (P)=0} =⇒ P ₀ = P ₁ ² =====⇒ (P ^{g (P)=0} ₁ P ₃ ) ² − 2P ₁ P ₃ + 1 = 0. So P ₁ P ₃ = 1 and b(P) = 0. Further

c(P) = P ₀ P ₃ − P ₁ P ₂ = P ₁ ² P ₃ − P ₁ = P ₁ − P ₁ = 0.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example I

Recall that a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ and f = x ₀ x ₂ − x ₁ ² , g = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ .

Let P = [P 0 , P 1 , P 2 , P 3 ] ∈ Z(f , g ).

P ₂ = 0 = === ^{f (P)=0} =⇒ P ₁ = 0

g (P)=0

=====⇒ P ₀ P ₃ ² = P ₀ P ₃ = 0. So both b and c vanish at P.

P ₂ = 1 = === ^{f (P)=0} =⇒ P ₀ = P ₁ ² =====⇒ (P ^{g (P)=0} ₁ P ₃ ) ² − 2P ₁ P ₃ + 1 = 0. So P ₁ P ₃ = 1 and b(P) = 0. Further

c(P) = P ₀ P ₃ − P ₁ P ₂ = P ₁ ² P ₃ − P ₁ = P ₁ − P ₁ = 0.

Example I

Recall that a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ and f = x ₀ x ₂ − x ₁ ² , g = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ .

Let P = [P 0 , P 1 , P 2 , P 3 ] ∈ Z(f , g ).

P ₂ = 0 = === ^{f (P)=0} =⇒ P ₁ = 0 =====⇒ P ^{g (P)=0} ₀ P ₃ ² = P ₀ P ₃ = 0.

So both b and c vanish at P.

P ₂ = 1 = === ^{f (P)=0} =⇒ P ₀ = P ₁ ² =====⇒ (P ^{g (P)=0} ₁ P ₃ ) ² − 2P ₁ P ₃ + 1 = 0. So P ₁ P ₃ = 1 and b(P) = 0. Further

c(P) = P ₀ P ₃ − P ₁ P ₂ = P ₁ ² P ₃ − P ₁ = P ₁ − P ₁ = 0.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example I

Recall that a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ and f = x ₀ x ₂ − x ₁ ² , g = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ .

Let P = [P 0 , P 1 , P 2 , P 3 ] ∈ Z(f , g ).

P ₂ = 0 = === ^{f (P)=0} =⇒ P ₁ = 0 =====⇒ P ^{g (P)=0} ₀ P ₃ ² = P ₀ P ₃ = 0. So both b and c vanish at P.

P ₂ = 1 = === ^{f (P)=0} =⇒ P ₀ = P ₁ ² =====⇒ (P ^{g (P)=0} ₁ P ₃ ) ² − 2P ₁ P ₃ + 1 = 0. So P ₁ P ₃ = 1 and b(P) = 0. Further

c(P) = P ₀ P ₃ − P ₁ P ₂ = P ₁ ² P ₃ − P ₁ = P ₁ − P ₁ = 0.

Example I

Recall that a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ and f = x ₀ x ₂ − x ₁ ² , g = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ .

Let P = [P 0 , P 1 , P 2 , P 3 ] ∈ Z(f , g ).

P ₂ = 0 = === ^{f (P)=0} =⇒ P ₁ = 0 =====⇒ P ^{g (P)=0} ₀ P ₃ ² = P ₀ P ₃ = 0. So both b and c vanish at P.

P ₂ = 1 = === ^{f (P)=0} =⇒ P ₀ = P ₁ ²

g (P)=0

=====⇒ (P ₁ P ₃ ) ² − 2P ₁ P ₃ + 1 = 0. So P ₁ P ₃ = 1 and b(P) = 0. Further

c(P) = P ₀ P ₃ − P ₁ P ₂ = P ₁ ² P ₃ − P ₁ = P ₁ − P ₁ = 0.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example I

Recall that a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ and f = x ₀ x ₂ − x ₁ ² , g = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ .

Let P = [P 0 , P 1 , P 2 , P 3 ] ∈ Z(f , g ).

P ₂ = 0 = === ^{f (P)=0} =⇒ P ₁ = 0 =====⇒ P ^{g (P)=0} ₀ P ₃ ² = P ₀ P ₃ = 0. So both b and c vanish at P.

P ₂ = 1 = === ^{f (P)=0} =⇒ P ₀ = P ₁ ² =====⇒ (P ^{g (P)=0} ₁ P ₃ ) ² − 2P ₁ P ₃ + 1 = 0.

So P ₁ P ₃ = 1 and b(P) = 0. Further

c(P) = P ₀ P ₃ − P ₁ P ₂ = P ₁ ² P ₃ − P ₁ = P ₁ − P ₁ = 0.

Example I

Recall that a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ and f = x ₀ x ₂ − x ₁ ² , g = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ .

Let P = [P 0 , P 1 , P 2 , P 3 ] ∈ Z(f , g ).

P ₂ = 0 = === ^{f (P)=0} =⇒ P ₁ = 0 =====⇒ P ^{g (P)=0} ₀ P ₃ ² = P ₀ P ₃ = 0. So both b and c vanish at P.

P ₂ = 1 = === ^{f (P)=0} =⇒ P ₀ = P ₁ ² =====⇒ (P ^{g (P)=0} ₁ P ₃ ) ² − 2P ₁ P ₃ + 1 = 0.

So P ₁ P ₃ = 1 and b(P) = 0.

Further

c(P) = P ₀ P ₃ − P ₁ P ₂ = P ₁ ² P ₃ − P ₁ = P ₁ − P ₁ = 0.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example I

Recall that a = x ₀ x ₂ − x ₁ ² , b = x ₁ x ₃ − x ₂ ² , c = x ₀ x ₃ − x ₁ x ₂ and f = x ₀ x ₂ − x ₁ ² , g = x ₀ x ₃ ² − 2x ₁ x ₂ x ₃ + x ₂ ³ .

Let P = [P 0 , P 1 , P 2 , P 3 ] ∈ Z(f , g ).

P ₂ = 0 = === ^{f (P)=0} =⇒ P ₁ = 0 =====⇒ P ^{g (P)=0} ₀ P ₃ ² = P ₀ P ₃ = 0. So both b and c vanish at P.

P ₂ = 1 = === ^{f (P)=0} =⇒ P ₀ = P ₁ ² =====⇒ (P ^{g (P)=0} ₁ P ₃ ) ² − 2P ₁ P ₃ + 1 = 0.

So P ₁ P ₃ = 1 and b(P) = 0. Further

c(P) = P ₀ P ₃ − P ₁ P ₂ = P ₁ ² P ₃ − P ₁ = P ₁ − P ₁ = 0.

Arithmetical rank

So we found out that X 1 ⊆ P ³ is the zero-locus of 2 polynomials. Can we do better? No, because at least as many polynomials as the codimension are needed. Let us be more precise ...

The arithmetical rank of an algebraic set X ⊆ P ⁿ is:

ara _P

X = min{r | ∃ f ₁ , . . . , f _r ∈ S : X = Z(f ₁ , . . . , f _r )}. By the Krull’s Hauptidealsatz ara _P

X ≥ codim _P

X .

We can therefore conclude that, in Example I: ara _P

X ₁ = 2 = codim _P

X ₁ .

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Arithmetical rank

So we found out that X 1 ⊆ P ³ is the zero-locus of 2 polynomials.

Can we do better? No, because at least as many polynomials as the codimension are needed. Let us be more precise ...

The arithmetical rank of an algebraic set X ⊆ P ⁿ is:

ara _P

X = min{r | ∃ f ₁ , . . . , f _r ∈ S : X = Z(f ₁ , . . . , f _r )}. By the Krull’s Hauptidealsatz ara _P

X ≥ codim _P

X .

We can therefore conclude that, in Example I:

ara _P

X ₁ = 2 = codim _P

X ₁ .

Arithmetical rank

So we found out that X 1 ⊆ P ³ is the zero-locus of 2 polynomials.

Can we do better?

No, because at least as many polynomials as the codimension are needed. Let us be more precise ...

The arithmetical rank of an algebraic set X ⊆ P ⁿ is:

ara _P

X = min{r | ∃ f ₁ , . . . , f _r ∈ S : X = Z(f ₁ , . . . , f _r )}. By the Krull’s Hauptidealsatz ara _P

X ≥ codim _P

X .

We can therefore conclude that, in Example I: ara _P

X ₁ = 2 = codim _P

X ₁ .

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Arithmetical rank

So we found out that X 1 ⊆ P ³ is the zero-locus of 2 polynomials.

Can we do better? No, because at least as many polynomials as the codimension are needed.

Let us be more precise ... The arithmetical rank of an algebraic set X ⊆ P ⁿ is:

ara _P

X = min{r | ∃ f ₁ , . . . , f _r ∈ S : X = Z(f ₁ , . . . , f _r )}. By the Krull’s Hauptidealsatz ara _P

X ≥ codim _P

X .

We can therefore conclude that, in Example I:

ara _P

X ₁ = 2 = codim _P

X ₁ .

Arithmetical rank

So we found out that X 1 ⊆ P ³ is the zero-locus of 2 polynomials.

Can we do better? No, because at least as many polynomials as the codimension are needed. Let us be more precise ...

The arithmetical rank of an algebraic set X ⊆ P ⁿ is:

ara _P

X = min{r | ∃ f ₁ , . . . , f _r ∈ S : X = Z(f ₁ , . . . , f _r )}. By the Krull’s Hauptidealsatz ara _P

X ≥ codim _P

X .

We can therefore conclude that, in Example I: ara _P

X ₁ = 2 = codim _P

X ₁ .

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Arithmetical rank

So we found out that X 1 ⊆ P ³ is the zero-locus of 2 polynomials.

Can we do better? No, because at least as many polynomials as the codimension are needed. Let us be more precise ...

The arithmetical rank of an algebraic set X ⊆ P ⁿ is:

ara _P

X = min{r | ∃ f ₁ , . . . , f _r ∈ S : X = Z(f ₁ , . . . , f _r )}.

By the Krull’s Hauptidealsatz ara _P

X ≥ codim _P

X . We can therefore conclude that, in Example I:

ara _P

X ₁ = 2 = codim _P

X ₁ .

Arithmetical rank

So we found out that X 1 ⊆ P ³ is the zero-locus of 2 polynomials.

Can we do better? No, because at least as many polynomials as the codimension are needed. Let us be more precise ...

The arithmetical rank of an algebraic set X ⊆ P ⁿ is:

ara _P

X = min{r | ∃ f ₁ , . . . , f _r ∈ S : X = Z(f ₁ , . . . , f _r )}.

By the Krull’s Hauptidealsatz ara _P

X ≥ codim _P

X .

We can therefore conclude that, in Example I: ara _P

X ₁ = 2 = codim _P

X ₁ .

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Arithmetical rank

So we found out that X 1 ⊆ P ³ is the zero-locus of 2 polynomials.

Can we do better? No, because at least as many polynomials as the codimension are needed. Let us be more precise ...

The arithmetical rank of an algebraic set X ⊆ P ⁿ is:

ara _P

X = min{r | ∃ f ₁ , . . . , f _r ∈ S : X = Z(f ₁ , . . . , f _r )}.

By the Krull’s Hauptidealsatz ara _P

X ≥ codim _P

X . We can therefore conclude that, in Example I:

ara _P

X ₁ = 2 = codim _P

X ₁ .

Set-theoretic complete intersections

Wouldn’t happen to be

ara _P

X = codim _P

X for any algebraic set X ⊆ P ⁿ ? No, for example: (Faltings, 1980)

If Y ⊆ P ⁿ is an irreducible d -dimensional algebraic set and X ⊆ P ⁿ is the zero-locus of < d polynomials, then X ∩ Y must be

connected. In particular, ara _P

X < n =⇒ X is connected. It therefore makes sense to name X ⊆ P ⁿ a set-theoretic complete intersection if ara _P

X = codim _P

X .

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Set-theoretic complete intersections

Wouldn’t happen to be

ara _P

X = codim _P

X for any algebraic set X ⊆ P ⁿ ?

No, for example: (Faltings, 1980)

If Y ⊆ P ⁿ is an irreducible d -dimensional algebraic set and X ⊆ P ⁿ is the zero-locus of < d polynomials, then X ∩ Y must be

connected. In particular, ara _P

X < n =⇒ X is connected.

It therefore makes sense to name X ⊆ P ⁿ a set-theoretic

complete intersection if ara _P

X = codim _P

X .

Set-theoretic complete intersections

Wouldn’t happen to be

ara _P

X = codim _P

X for any algebraic set X ⊆ P ⁿ ? No, for example:

(Faltings, 1980)

If Y ⊆ P ⁿ is an irreducible d -dimensional algebraic set and X ⊆ P ⁿ is the zero-locus of < d polynomials, then X ∩ Y must be

connected.

In particular, ara _P

X < n =⇒ X is connected. It therefore makes sense to name X ⊆ P ⁿ a set-theoretic complete intersection if ara _P

X = codim _P

X .

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Set-theoretic complete intersections

Wouldn’t happen to be

ara _P

X = codim _P

X for any algebraic set X ⊆ P ⁿ ? No, for example:

(Faltings, 1980)

If Y ⊆ P ⁿ is an irreducible d -dimensional algebraic set and X ⊆ P ⁿ is the zero-locus of < d polynomials, then X ∩ Y must be

connected. In particular, ara _P

X < n =⇒ X is connected.

It therefore makes sense to name X ⊆ P ⁿ a set-theoretic

complete intersection if ara _P

X = codim _P

X .

Set-theoretic complete intersections

Wouldn’t happen to be

ara _P

X = codim _P

X for any algebraic set X ⊆ P ⁿ ? No, for example:

(Faltings, 1980)

If Y ⊆ P ⁿ is an irreducible d -dimensional algebraic set and X ⊆ P ⁿ is the zero-locus of < d polynomials, then X ∩ Y must be

connected. In particular, ara _P

X < n =⇒ X is connected.

It therefore makes sense to name X ⊆ P ⁿ a set-theoretic complete intersection if ara _P

X = codim _P

X .

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example II

Let X 2 = {[s, t, 0, 0] : [s, t] ∈ P ¹ } ∪ {[0, 0, s, t] : [s, t] ∈ P ¹ } ⊆ P ³ .

For what said before, since X ₂ is not connected we have ara _P

X ₂ ≥ 3 > 2 = codim

P

X ₂ . On the other hand, if S = K [x 0 , x 1 , x 2 , x 3 ], we have

I(X ₂ ) = (x 2 , x 3 ) ∩ (x 0 , x 1 ) = (x 0 x 2 , x 0 x 3 , x 1 x 2 , x 1 x 3 ) ⊆ S . If I = (a, b, c) ⊆ S with a = x ₀ x ₂ + x ₀ x ₃ + x ₁ x ₂ + x ₁ x ₃ , b = x ₀ x ₁ x ₂ + x 0 x 1 x 3 + x 0 x 2 x 3 + x 1 x 2 x 3 and c = x 0 x 1 x 2 x 3 , then √

I = I(X 2 ).

So X 2 = Z(I ), and ara _P

X 2 = 3.

Example II

Let X 2 = {[s, t, 0, 0] : [s, t] ∈ P ¹ } ∪ {[0, 0, s, t] : [s, t] ∈ P ¹ } ⊆ P ³ . For what said before, since X ₂ is not connected we have

ara _P

X ₂ ≥ 3 > 2 = codim

P

X ₂ .

On the other hand, if S = K [x 0 , x 1 , x 2 , x 3 ], we have

I(X ₂ ) = (x 2 , x 3 ) ∩ (x 0 , x 1 ) = (x 0 x 2 , x 0 x 3 , x 1 x 2 , x 1 x 3 ) ⊆ S . If I = (a, b, c) ⊆ S with a = x ₀ x ₂ + x ₀ x ₃ + x ₁ x ₂ + x ₁ x ₃ , b = x ₀ x ₁ x ₂ + x 0 x 1 x 3 + x 0 x 2 x 3 + x 1 x 2 x 3 and c = x 0 x 1 x 2 x 3 , then √

I = I(X 2 ). So X 2 = Z(I ), and ara _P

X 2 = 3.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example II

Let X 2 = {[s, t, 0, 0] : [s, t] ∈ P ¹ } ∪ {[0, 0, s, t] : [s, t] ∈ P ¹ } ⊆ P ³ . For what said before, since X ₂ is not connected we have

ara _P

X ₂ ≥ 3 > 2 = codim

P

X ₂ . On the other hand, if S = K [x 0 , x 1 , x 2 , x 3 ], we have

I(X ₂ ) = (x 2 , x 3 ) ∩ (x 0 , x 1 ) = (x 0 x 2 , x 0 x 3 , x 1 x 2 , x 1 x 3 ) ⊆ S .

If I = (a, b, c) ⊆ S with a = x ₀ x ₂ + x ₀ x ₃ + x ₁ x ₂ + x ₁ x ₃ , b = x ₀ x ₁ x ₂ + x 0 x 1 x 3 + x 0 x 2 x 3 + x 1 x 2 x 3 and c = x 0 x 1 x 2 x 3 , then √

I = I(X 2 ).

So X 2 = Z(I ), and ara _P

X 2 = 3.

Example II

Let X 2 = {[s, t, 0, 0] : [s, t] ∈ P ¹ } ∪ {[0, 0, s, t] : [s, t] ∈ P ¹ } ⊆ P ³ . For what said before, since X ₂ is not connected we have

ara _P

X ₂ ≥ 3 > 2 = codim

P

X ₂ . On the other hand, if S = K [x 0 , x 1 , x 2 , x 3 ], we have

I(X ₂ ) = (x 2 , x 3 ) ∩ (x 0 , x 1 ) = (x 0 x 2 , x 0 x 3 , x 1 x 2 , x 1 x 3 ) ⊆ S . If I = (a, b, c) ⊆ S with a = x ₀ x ₂ + x ₀ x ₃ + x ₁ x ₂ + x ₁ x ₃ , b = x ₀ x ₁ x ₂ + x 0 x 1 x 3 + x 0 x 2 x 3 + x 1 x 2 x 3 and c = x 0 x 1 x 2 x 3 , then √

I = I(X 2 ).

So X 2 = Z(I ), and ara _P

X 2 = 3.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example II

Let X 2 = {[s, t, 0, 0] : [s, t] ∈ P ¹ } ∪ {[0, 0, s, t] : [s, t] ∈ P ¹ } ⊆ P ³ . For what said before, since X ₂ is not connected we have

ara _P

X ₂ ≥ 3 > 2 = codim

P

X ₂ . On the other hand, if S = K [x 0 , x 1 , x 2 , x 3 ], we have

I(X ₂ ) = (x 2 , x 3 ) ∩ (x 0 , x 1 ) = (x 0 x 2 , x 0 x 3 , x 1 x 2 , x 1 x 3 ) ⊆ S . If I = (a, b, c) ⊆ S with a = x ₀ x ₂ + x ₀ x ₃ + x ₁ x ₂ + x ₁ x ₃ , b = x ₀ x ₁ x ₂ + x 0 x 1 x 3 + x 0 x 2 x 3 + x 1 x 2 x 3 and c = x 0 x 1 x 2 x 3 , then √

I = I(X 2 ).

So X ₂ = Z(I ), and ara _P

X ₂ = 3.

A general upper bound

What happened before is not a case:

Eisenbud-Evans, 1972

For any nonempty algebraic set X ⊆ P ⁿ : ara _P

X ≤ n. Summarizing, so far we learnt that:

codim _P

X ≤ ara _P

X ≤ n for any algebraic set ∅ 6= X ⊆ P ⁿ ; ara _P

X = n whenever X is not connected.

How to produce other lower bounds?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

A general upper bound

What happened before is not a case:

Eisenbud-Evans, 1972

For any nonempty algebraic set X ⊆ P ⁿ : ara _P

X ≤ n.

Summarizing, so far we learnt that:

codim _P

X ≤ ara _P

X ≤ n for any algebraic set ∅ 6= X ⊆ P ⁿ ; ara _P

X = n whenever X is not connected.

How to produce other lower bounds?

A general upper bound

What happened before is not a case:

Eisenbud-Evans, 1972

For any nonempty algebraic set X ⊆ P ⁿ : ara _P

X ≤ n.

Summarizing, so far we learnt that:

codim _P

X ≤ ara _P

X ≤ n for any algebraic set ∅ 6= X ⊆ P ⁿ ;

ara _P

X = n whenever X is not connected. How to produce other lower bounds?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

A general upper bound

What happened before is not a case:

Eisenbud-Evans, 1972

For any nonempty algebraic set X ⊆ P ⁿ : ara _P

X ≤ n.

Summarizing, so far we learnt that:

codim _P

X ≤ ara _P

X ≤ n for any algebraic set ∅ 6= X ⊆ P ⁿ ; ara _P

X = n whenever X is not connected.

How to produce other lower bounds?

A general upper bound

What happened before is not a case:

Eisenbud-Evans, 1972

For any nonempty algebraic set X ⊆ P ⁿ : ara _P

X ≤ n.

Summarizing, so far we learnt that:

codim _P

X ≤ ara _P

X ≤ n for any algebraic set ∅ 6= X ⊆ P ⁿ ; ara _P

X = n whenever X is not connected.

How to produce other lower bounds?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Cohomological dimension

Let X ⊆ P ⁿ be an algebraic set such that X = Z(f 1 , . . . , f r ).

Notice that U(f _i ) = P ⁿ \ Z(f _i ) is affine, and that: P ⁿ \ X = U(f ₁ ) ∪ . . . ∪ U(f r ).

Therefore, by means of the ˘ Cech cohomology, H ⁱ (P ⁿ \ X , F ) = 0 for any i ≥ r and each quasi-coherent sheaf F on P ⁿ .

The (coherent) cohomological dimension of an open set U ⊆ P ⁿ : cd(U) = sup{s : H ^s (U, F ) 6= 0 for some coherent sheaf on P ⁿ }. For what said above, for any algebraic set X ⊆ P ⁿ we have:

ara _P

X ≥ cd(P ⁿ \ X ) + 1.

Cohomological dimension

Let X ⊆ P ⁿ be an algebraic set such that X = Z(f 1 , . . . , f r ).

Notice that U(f _i ) = P ⁿ \ Z(f _i ) is affine,

and that: P ⁿ \ X = U(f ₁ ) ∪ . . . ∪ U(f r ).

Therefore, by means of the ˘ Cech cohomology, H ⁱ (P ⁿ \ X , F ) = 0 for any i ≥ r and each quasi-coherent sheaf F on P ⁿ .

The (coherent) cohomological dimension of an open set U ⊆ P ⁿ : cd(U) = sup{s : H ^s (U, F ) 6= 0 for some coherent sheaf on P ⁿ }. For what said above, for any algebraic set X ⊆ P ⁿ we have:

ara _P

X ≥ cd(P ⁿ \ X ) + 1.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Cohomological dimension

Let X ⊆ P ⁿ be an algebraic set such that X = Z(f 1 , . . . , f r ).

Notice that U(f _i ) = P ⁿ \ Z(f _i ) is affine, and that:

P ⁿ \ X = U(f ₁ ) ∪ . . . ∪ U(f r ).

Therefore, by means of the ˘ Cech cohomology, H ⁱ (P ⁿ \ X , F ) = 0 for any i ≥ r and each quasi-coherent sheaf F on P ⁿ .

The (coherent) cohomological dimension of an open set U ⊆ P ⁿ : cd(U) = sup{s : H ^s (U, F ) 6= 0 for some coherent sheaf on P ⁿ }. For what said above, for any algebraic set X ⊆ P ⁿ we have:

ara _P

X ≥ cd(P ⁿ \ X ) + 1.

Cohomological dimension

Let X ⊆ P ⁿ be an algebraic set such that X = Z(f 1 , . . . , f r ).

Notice that U(f _i ) = P ⁿ \ Z(f _i ) is affine, and that:

P ⁿ \ X = U(f ₁ ) ∪ . . . ∪ U(f r ).

Therefore, by means of the ˘ Cech cohomology, H ⁱ (P ⁿ \ X , F ) = 0 for any i ≥ r and each quasi-coherent sheaf F on P ⁿ .

The (coherent) cohomological dimension of an open set U ⊆ P ⁿ : cd(U) = sup{s : H ^s (U, F ) 6= 0 for some coherent sheaf on P ⁿ }. For what said above, for any algebraic set X ⊆ P ⁿ we have:

ara _P

X ≥ cd(P ⁿ \ X ) + 1.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Cohomological dimension

Let X ⊆ P ⁿ be an algebraic set such that X = Z(f 1 , . . . , f r ).

Notice that U(f _i ) = P ⁿ \ Z(f _i ) is affine, and that:

P ⁿ \ X = U(f ₁ ) ∪ . . . ∪ U(f r ).

Therefore, by means of the ˘ Cech cohomology, H ⁱ (P ⁿ \ X , F ) = 0 for any i ≥ r and each quasi-coherent sheaf F on P ⁿ .

The (coherent) cohomological dimension of an open set U ⊆ P ⁿ : cd(U) = sup{s : H ^s (U, F ) 6= 0 for some coherent sheaf on P ⁿ }.

For what said above, for any algebraic set X ⊆ P ⁿ we have:

ara _P

X ≥ cd(P ⁿ \ X ) + 1.

Cohomological dimension

Let X ⊆ P ⁿ be an algebraic set such that X = Z(f 1 , . . . , f r ).

Notice that U(f _i ) = P ⁿ \ Z(f _i ) is affine, and that:

P ⁿ \ X = U(f ₁ ) ∪ . . . ∪ U(f r ).

Therefore, by means of the ˘ Cech cohomology, H ⁱ (P ⁿ \ X , F ) = 0 for any i ≥ r and each quasi-coherent sheaf F on P ⁿ .

The (coherent) cohomological dimension of an open set U ⊆ P ⁿ : cd(U) = sup{s : H ^s (U, F ) 6= 0 for some coherent sheaf on P ⁿ }.

For what said above, for any algebraic set X ⊆ P ⁿ we have:

ara _P

X ≥ cd(P ⁿ \ X ) + 1.

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example III

Let X 3 ⊆ P ⁷ be the set of 2 × 4 matrices of rank at most 1.

So S = K [Z ] where

Z = z ₁₁ z ₁₂ z ₁₃ z ₁₄ z ₂₁ z ₂₂ z ₂₃ z ₂₄



and I(X 3 ) = I 2 (Z ) is minimally generated by the six 2-minors of Z , [i , j ] with 1 ≤ i < j ≤ 4. One can show that five polynomials are enough to generate I ₂ (Z ) up to radical, namely:

[1, 2], [1, 3], [1, 4] + [2, 3], [2, 4], [3, 4].

In fact [1, 4] ² = [1, 4]([1, 4] + [2, 3])+[1, 2][3, 4] − [1, 3][2, 4]. We

want to show that ara _P

X 3 = 5, however codim _P

X 3 = 3. In

characteristic 0, we can prove that cd(P ⁷ \ X ₃ ) = 4 ...

Example III

Let X 3 ⊆ P ⁷ be the set of 2 × 4 matrices of rank at most 1. So S = K [Z ] where

Z = z ₁₁ z ₁₂ z ₁₃ z ₁₄ z ₂₁ z ₂₂ z ₂₃ z ₂₄



and I(X 3 ) = I 2 (Z ) is minimally generated by the six 2-minors of Z , [i , j ] with 1 ≤ i < j ≤ 4. One can show that five polynomials are enough to generate I ₂ (Z ) up to radical, namely:

[1, 2], [1, 3], [1, 4] + [2, 3], [2, 4], [3, 4].

In fact [1, 4] ² = [1, 4]([1, 4] + [2, 3])+[1, 2][3, 4] − [1, 3][2, 4]. We want to show that ara _P

X 3 = 5, however codim _P

X 3 = 3. In characteristic 0, we can prove that cd(P ⁷ \ X ₃ ) = 4 ...

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example III

Let X 3 ⊆ P ⁷ be the set of 2 × 4 matrices of rank at most 1. So S = K [Z ] where

Z = z ₁₁ z ₁₂ z ₁₃ z ₁₄ z ₂₁ z ₂₂ z ₂₃ z ₂₄



and I(X 3 ) = I 2 (Z ) is minimally generated by the six 2-minors of Z , [i , j ] with 1 ≤ i < j ≤ 4.

One can show that five polynomials are enough to generate I ₂ (Z ) up to radical, namely:

[1, 2], [1, 3], [1, 4] + [2, 3], [2, 4], [3, 4].

In fact [1, 4] ² = [1, 4]([1, 4] + [2, 3])+[1, 2][3, 4] − [1, 3][2, 4]. We

want to show that ara _P

X 3 = 5, however codim _P

X 3 = 3. In

characteristic 0, we can prove that cd(P ⁷ \ X ₃ ) = 4 ...

Example III

Let X 3 ⊆ P ⁷ be the set of 2 × 4 matrices of rank at most 1. So S = K [Z ] where

Z = z ₁₁ z ₁₂ z ₁₃ z ₁₄ z ₂₁ z ₂₂ z ₂₃ z ₂₄



and I(X 3 ) = I 2 (Z ) is minimally generated by the six 2-minors of Z , [i , j ] with 1 ≤ i < j ≤ 4. One can show that five polynomials are enough to generate I ₂ (Z ) up to radical, namely:

[1, 2], [1, 3], [1, 4] + [2, 3], [2, 4], [3, 4].

In fact [1, 4] ² = [1, 4]([1, 4] + [2, 3])+[1, 2][3, 4] − [1, 3][2, 4]. We want to show that ara _P

X 3 = 5, however codim _P

X 3 = 3. In characteristic 0, we can prove that cd(P ⁷ \ X ₃ ) = 4 ...

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example III

Let X 3 ⊆ P ⁷ be the set of 2 × 4 matrices of rank at most 1. So S = K [Z ] where

Z = z ₁₁ z ₁₂ z ₁₃ z ₁₄ z ₂₁ z ₂₂ z ₂₃ z ₂₄



and I(X 3 ) = I 2 (Z ) is minimally generated by the six 2-minors of Z , [i , j ] with 1 ≤ i < j ≤ 4. One can show that five polynomials are enough to generate I ₂ (Z ) up to radical, namely:

[1, 2], [1, 3], [1, 4] + [2, 3], [2, 4], [3, 4].

In fact [1, 4] ² = [1, 4]([1, 4] + [2, 3])+[1, 2][3, 4] − [1, 3][2, 4].

We

want to show that ara _P

X 3 = 5, however codim _P

X 3 = 3. In

characteristic 0, we can prove that cd(P ⁷ \ X ₃ ) = 4 ...

Example III

Let X 3 ⊆ P ⁷ be the set of 2 × 4 matrices of rank at most 1. So S = K [Z ] where

Z = z ₁₁ z ₁₂ z ₁₃ z ₁₄ z ₂₁ z ₂₂ z ₂₃ z ₂₄



and I(X 3 ) = I 2 (Z ) is minimally generated by the six 2-minors of Z , [i , j ] with 1 ≤ i < j ≤ 4. One can show that five polynomials are enough to generate I ₂ (Z ) up to radical, namely:

[1, 2], [1, 3], [1, 4] + [2, 3], [2, 4], [3, 4].

In fact [1, 4] ² = [1, 4]([1, 4] + [2, 3])+[1, 2][3, 4] − [1, 3][2, 4]. We want to show that ara _P

X 3 = 5, however codim _P

X 3 = 3.

In characteristic 0, we can prove that cd(P ⁷ \ X ₃ ) = 4 ...

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example III

Let X 3 ⊆ P ⁷ be the set of 2 × 4 matrices of rank at most 1. So S = K [Z ] where

Z = z ₁₁ z ₁₂ z ₁₃ z ₁₄ z ₂₁ z ₂₂ z ₂₃ z ₂₄



and I(X 3 ) = I 2 (Z ) is minimally generated by the six 2-minors of Z , [i , j ] with 1 ≤ i < j ≤ 4. One can show that five polynomials are enough to generate I ₂ (Z ) up to radical, namely:

[1, 2], [1, 3], [1, 4] + [2, 3], [2, 4], [3, 4].

In fact [1, 4] ² = [1, 4]([1, 4] + [2, 3])+[1, 2][3, 4] − [1, 3][2, 4]. We

Example III

Notice that H ⁴ (P ⁷ \ X ₃ , O(k)) ∼ = H _I(X ⁵

3

) (S ) _k for any k ∈ Z.

The K -algebra R = K [[i , j ] : 1 ≤ i < j ≤ 4] ⊆ S is the coordinate ring of the Grassmannian of lines in P ³ . In particular dim(R) = 5. Moreover in characteristic zero R is a direct summand of S . So

H _I(X ⁵

3

) (S ) = H _(I(X ⁵

3

)∩R)S (S ) ∼ = H _I(X ⁵

_)∩R (S ) ←- H _I(X ⁵

3

)∩R (R). Thus H ⁴ (P ⁷ \ X ₃ , O(k)) 6= 0 ∀ k  0, so that in characteristic 0:

5 ≤ cd(P ⁷ \ X ₃ ) + 1 ≤ ara _P

X ₃ ≤ 5. How can we do in positive characteristic?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example III

Notice that H ⁴ (P ⁷ \ X ₃ , O(k)) ∼ = H _I(X ⁵

3

) (S ) _k for any k ∈ Z. The K -algebra R = K [[i , j ] : 1 ≤ i < j ≤ 4] ⊆ S is the coordinate ring of the Grassmannian of lines in P ³ .

In particular dim(R) = 5. Moreover in characteristic zero R is a direct summand of S . So

H _I(X ⁵

3

) (S ) = H _(I(X ⁵

3

)∩R)S (S ) ∼ = H _I(X ⁵

_)∩R (S ) ←- H _I(X ⁵

3

)∩R (R). Thus H ⁴ (P ⁷ \ X ₃ , O(k)) 6= 0 ∀ k  0, so that in characteristic 0:

5 ≤ cd(P ⁷ \ X ₃ ) + 1 ≤ ara _P

X ₃ ≤ 5.

How can we do in positive characteristic?

Example III

Notice that H ⁴ (P ⁷ \ X ₃ , O(k)) ∼ = H _I(X ⁵

3

) (S ) _k for any k ∈ Z. The K -algebra R = K [[i , j ] : 1 ≤ i < j ≤ 4] ⊆ S is the coordinate ring of the Grassmannian of lines in P ³ . In particular dim(R) = 5.

Moreover in characteristic zero R is a direct summand of S . So H _I(X ⁵

3

) (S ) = H _(I(X ⁵

3

)∩R)S (S ) ∼ = H _I(X ⁵

_)∩R (S ) ←- H _I(X ⁵

3

)∩R (R). Thus H ⁴ (P ⁷ \ X ₃ , O(k)) 6= 0 ∀ k  0, so that in characteristic 0:

5 ≤ cd(P ⁷ \ X ₃ ) + 1 ≤ ara _P

X ₃ ≤ 5. How can we do in positive characteristic?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example III

Notice that H ⁴ (P ⁷ \ X ₃ , O(k)) ∼ = H _I(X ⁵

3

) (S ) _k for any k ∈ Z. The K -algebra R = K [[i , j ] : 1 ≤ i < j ≤ 4] ⊆ S is the coordinate ring of the Grassmannian of lines in P ³ . In particular dim(R) = 5.

Moreover in characteristic zero R is a direct summand of S .

So H _I(X ⁵

3

) (S ) = H _(I(X ⁵

3

)∩R)S (S ) ∼ = H _I(X ⁵

_)∩R (S ) ←- H _I(X ⁵

3

)∩R (R). Thus H ⁴ (P ⁷ \ X ₃ , O(k)) 6= 0 ∀ k  0, so that in characteristic 0:

5 ≤ cd(P ⁷ \ X ₃ ) + 1 ≤ ara _P

X ₃ ≤ 5.

How can we do in positive characteristic?

Example III

Notice that H ⁴ (P ⁷ \ X ₃ , O(k)) ∼ = H _I(X ⁵

3

) (S ) _k for any k ∈ Z. The K -algebra R = K [[i , j ] : 1 ≤ i < j ≤ 4] ⊆ S is the coordinate ring of the Grassmannian of lines in P ³ . In particular dim(R) = 5.

Moreover in characteristic zero R is a direct summand of S . So H _I(X ⁵

3

) (S ) = H _(I(X ⁵

3

)∩R)S (S ) ∼ = H _I(X ⁵

_)∩R (S ) ←- H _I(X ⁵

3

)∩R (R).

Thus H ⁴ (P ⁷ \ X ₃ , O(k)) 6= 0 ∀ k  0, so that in characteristic 0: 5 ≤ cd(P ⁷ \ X ₃ ) + 1 ≤ ara _P

X ₃ ≤ 5.

How can we do in positive characteristic?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Example III

Notice that H ⁴ (P ⁷ \ X ₃ , O(k)) ∼ = H _I(X ⁵

3

) (S ) _k for any k ∈ Z. The K -algebra R = K [[i , j ] : 1 ≤ i < j ≤ 4] ⊆ S is the coordinate ring of the Grassmannian of lines in P ³ . In particular dim(R) = 5.

Moreover in characteristic zero R is a direct summand of S . So H _I(X ⁵

3

) (S ) = H _(I(X ⁵

3

)∩R)S (S ) ∼ = H _I(X ⁵

_)∩R (S ) ←- H _I(X ⁵

3

)∩R (R).

Thus H ⁴ (P ⁷ \ X ₃ , O(k)) 6= 0 ∀ k  0, so that in characteristic 0:

5 ≤ cd(P ⁷ \ X ₃ ) + 1 ≤ ara _P

X ₃ ≤ 5.

How can we do in positive characteristic?

Example III

Notice that H ⁴ (P ⁷ \ X ₃ , O(k)) ∼ = H _I(X ⁵

3

) (S ) _k for any k ∈ Z. The K -algebra R = K [[i , j ] : 1 ≤ i < j ≤ 4] ⊆ S is the coordinate ring of the Grassmannian of lines in P ³ . In particular dim(R) = 5.

Moreover in characteristic zero R is a direct summand of S . So H _I(X ⁵

3

) (S ) = H _(I(X ⁵

3

)∩R)S (S ) ∼ = H _I(X ⁵

_)∩R (S ) ←- H _I(X ⁵

3

)∩R (R).

Thus H ⁴ (P ⁷ \ X ₃ , O(k)) 6= 0 ∀ k  0, so that in characteristic 0:

5 ≤ cd(P ⁷ \ X ₃ ) + 1 ≤ ara _P

X ₃ ≤ 5.

How can we do in positive characteristic?

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Positive characteristic

Peskine-Szpiro, 1973

If char(K ) > 0, then for any algebraic set X = Z(I ) ⊆ P ⁿ : cd(P ⁿ \ X ) ≤ projdim(I ).

In the previous example projdim(I ₂ (Z )) = 2, therefore we have

cd(P ⁷ \ X ₃ ) =

( 2 if char(K ) > 0 4 if char(K ) = 0

However, in 1990 Bruns and Schw¨ anzl managed to prove that

ara _P

X 3 = 5 also in positive characteristic ...

Positive characteristic

Peskine-Szpiro, 1973

If char(K ) > 0, then for any algebraic set X = Z(I ) ⊆ P ⁿ : cd(P ⁿ \ X ) ≤ projdim(I ).

In the previous example projdim(I ₂ (Z )) = 2, therefore we have

cd(P ⁷ \ X ₃ ) =

( 2 if char(K ) > 0 4 if char(K ) = 0

However, in 1990 Bruns and Schw¨ anzl managed to prove that ara _P

X 3 = 5 also in positive characteristic ...

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Positive characteristic

Peskine-Szpiro, 1973

If char(K ) > 0, then for any algebraic set X = Z(I ) ⊆ P ⁿ : cd(P ⁿ \ X ) ≤ projdim(I ).

In the previous example projdim(I ₂ (Z )) = 2,

therefore we have

cd(P ⁷ \ X ₃ ) =

( 2 if char(K ) > 0 4 if char(K ) = 0

However, in 1990 Bruns and Schw¨ anzl managed to prove that

ara _P

X 3 = 5 also in positive characteristic ...

Positive characteristic

Peskine-Szpiro, 1973

If char(K ) > 0, then for any algebraic set X = Z(I ) ⊆ P ⁿ : cd(P ⁿ \ X ) ≤ projdim(I ).

In the previous example projdim(I ₂ (Z )) = 2, therefore we have

cd(P ⁷ \ X ₃ ) =

( 2 if char(K ) > 0 4 if char(K ) = 0

However, in 1990 Bruns and Schw¨ anzl managed to prove that ara _P

X 3 = 5 also in positive characteristic ...

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Positive characteristic

Peskine-Szpiro, 1973

If char(K ) > 0, then for any algebraic set X = Z(I ) ⊆ P ⁿ : cd(P ⁿ \ X ) ≤ projdim(I ).

In the previous example projdim(I ₂ (Z )) = 2, therefore we have

cd(P ⁷ \ X ₃ ) =

( 2 if char(K ) > 0 4 if char(K ) = 0

However, in 1990 Bruns and Schw¨ anzl managed to prove that

Etale cohomology ´

So far we considered the Zariski topology on P ⁿ , which has as closed sets the algebraic sets.

Another useful topology is the ´ etale topology. For a Zariski-open subset U ⊆ P ⁿ , denote by U _{´ et} the set U equipped with the ´ etale topology and define its ´ etale cohomological dimension as:

´ ecd <sub>`</sub> (U) = sup{s : H <sup>i</sup> (U <sub>´ et</sub> , F ) 6= 0 for some `-torsion sheaf on P ⁿ }

´ ecd(U) = max{´ ecd <sub>`</sub> (U) : GCD(`, char(K )) = 1} Because ´ ecd(U) ≤ n whenever U is affine, Mayer-Vietoris yields:

ara _P

X ≥ ´ ecd(U) − n + 1 for any algebraic set X ⊆ P ⁿ .

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets

Etale cohomology ´

So far we considered the Zariski topology on P ⁿ , which has as closed sets the algebraic sets. Another useful topology is the ´ etale topology.

For a Zariski-open subset U ⊆ P ⁿ , denote by U _{´ et} the set U equipped with the ´ etale topology and define its ´ etale cohomological dimension as:

´ ecd <sub>`</sub> (U) = sup{s : H <sup>i</sup> (U <sub>´ et</sub> , F ) 6= 0 for some `-torsion sheaf on P ⁿ }

´ ecd(U) = max{´ ecd <sub>`</sub> (U) : GCD(`, char(K )) = 1} Because ´ ecd(U) ≤ n whenever U is affine, Mayer-Vietoris yields:

ara _P

X ≥ ´ ecd(U) − n + 1 for any algebraic set X ⊆ P ⁿ .

Etale cohomology ´

So far we considered the Zariski topology on P ⁿ , which has as closed sets the algebraic sets. Another useful topology is the ´ etale topology. For a Zariski-open subset U ⊆ P ⁿ , denote by U _{´ et} the set U equipped with the ´ etale topology

and define its ´ etale cohomological dimension as:

´ ecd <sub>`</sub> (U) = sup{s : H <sup>i</sup> (U <sub>´ et</sub> , F ) 6= 0 for some `-torsion sheaf on P ⁿ }

´ ecd(U) = max{´ ecd <sub>`</sub> (U) : GCD(`, char(K )) = 1} Because ´ ecd(U) ≤ n whenever U is affine, Mayer-Vietoris yields:

ara _P

X ≥ ´ ecd(U) − n + 1 for any algebraic set X ⊆ P ⁿ .

Matteo Varbaro (University of Genoa, Italy) Defining equations of algebraic sets