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The Projective Line Over The Integers Ela Celikbas and Christina - - PowerPoint PPT Presentation

The Projective Line Over The Integers Ela Celikbas and Christina Eubanks-Turner Department of Mathematics University of NebraskaLincoln October 2011 AMS Sectional Meeting Lincoln,NE Ela Celikbas and Christina Eubanks-Turner The


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The Projective Line Over The Integers

Ela Celikbas and Christina Eubanks-Turner

Department of Mathematics University of Nebraska–Lincoln

October 2011 AMS Sectional Meeting Lincoln,NE

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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Motivation

For a commutative Noetherian ring R, Spec(R) = {prime ideals of R} , poset under ⊆. Questions.

  • Q1. I. Kaplansky, ’50: What is Spec(R), for a Noetherian ring R?

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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Motivation

For a commutative Noetherian ring R, Spec(R) = {prime ideals of R} , poset under ⊆. Questions.

  • Q1. I. Kaplansky, ’50: What is Spec(R), for a Noetherian ring R?
  • Q2. What is Spec(R) for a two-dimensional Noetherian domain R

related to a polynomial ring?

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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Motivation

For a commutative Noetherian ring R, Spec(R) = {prime ideals of R} , poset under ⊆. Questions.

  • Q1. I. Kaplansky, ’50: What is Spec(R), for a Noetherian ring R?
  • Q2. What is Spec(R) for a two-dimensional Noetherian domain R

related to a polynomial ring?

  • Q3. What is Spec(Z[x])? When is Spec(R) ∼

= Spec(Z[x])?

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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Motivation

For a commutative Noetherian ring R, Spec(R) = {prime ideals of R} , poset under ⊆. Questions.

  • Q1. I. Kaplansky, ’50: What is Spec(R), for a Noetherian ring R?
  • Q2. What is Spec(R) for a two-dimensional Noetherian domain R

related to a polynomial ring?

  • Q3. What is Spec(Z[x])? When is Spec(R) ∼

= Spec(Z[x])?

  • R. Wiegand, ’86: Characterization of Spec(Z[x]), the affine line
  • ver Z.

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5 Axioms for U := Spec(Z[x])

  • Notation. ∀u ∈ poset U, u↑ := {t|t > u}.

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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5 Axioms for U := Spec(Z[x])

  • Notation. ∀u ∈ poset U, u↑ := {t|t > u}.

Theorem 1. [RW] U := Spec(Z[x]) ⇐ ⇒ (1) (Obvious Axioms) • |U| = |Z|,

  • dim(U) = 2,
  • |u↑| = ∞,
  • |(u, v)↑| < ∞ , ∀ u = v, ht(u) = ht(v) = 1.

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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5 Axioms for U := Spec(Z[x])

  • Notation. ∀u ∈ poset U, u↑ := {t|t > u}.

Theorem 1. [RW] U := Spec(Z[x]) ⇐ ⇒ (1) (Obvious Axioms) • |U| = |Z|,

  • dim(U) = 2,
  • |u↑| = ∞,
  • |(u, v)↑| < ∞ , ∀ u = v, ht(u) = ht(v) = 1.

(2) (Subtle Axiom) (RW) For S ⊆ {ht 1’s}, T ⊆ {ht 2’s}, 0 < |S| < ∞, |T| < ∞, ∃ w ∈ {ht 1’s},

  • s∈S

(w, s)↑ ⊆ T ⊂ w↑.

  • Definition. w is a radical element for (S, T) if (RW) holds.

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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The Projective Line Over the Integers

  • Notation. Projective line over Z=Proj (Z)

Proj (Z) := Spec(Z[x]) ∪ Spec(Z 1

x

  • ), with

Spec(Z[x]) ∩ Spec(Z 1

x

  • ) Spec(Z[x, 1

x ]).

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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The Projective Line Over the Integers

  • Notation. Projective line over Z=Proj (Z)

Proj (Z) := Spec(Z[x]) ∪ Spec(Z 1

x

  • ), with

Spec(Z[x]) ∩ Spec(Z 1

x

  • ) Spec(Z[x, 1

x ]).

e.g. (4x3 + 3x2 − 1) (4 + 3 · 1 x − 1 · 1 x3 ) in Spec (Z[x, 1

x ]).

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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The Projective Line Over the Integers

  • Notation. Projective line over Z=Proj (Z)

Proj (Z) := Spec(Z[x]) ∪ Spec(Z 1

x

  • ), with

Spec(Z[x]) ∩ Spec(Z 1

x

  • ) Spec(Z[x, 1

x ]).

e.g. (4x3 + 3x2 − 1) (4 + 3 · 1 x − 1 · 1 x3 ) in Spec (Z[x, 1

x ]).

Characterize Proj (Z)?

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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The Projective Line Over the Integers

  • Notation. Projective line over Z=Proj (Z)

Proj (Z) := Spec(Z[x]) ∪ Spec(Z 1

x

  • ), with

Spec(Z[x]) ∩ Spec(Z 1

x

  • ) Spec(Z[x, 1

x ]).

e.g. (4x3 + 3x2 − 1) (4 + 3 · 1 x − 1 · 1 x3 ) in Spec (Z[x, 1

x ]).

Characterize Proj (Z)? Proj (Z) satisfies the obvious axioms.

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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The Projective Line Over the Integers

  • Notation. Projective line over Z=Proj (Z)

Proj (Z) := Spec(Z[x]) ∪ Spec(Z 1

x

  • ), with

Spec(Z[x]) ∩ Spec(Z 1

x

  • ) Spec(Z[x, 1

x ]).

e.g. (4x3 + 3x2 − 1) (4 + 3 · 1 x − 1 · 1 x3 ) in Spec (Z[x, 1

x ]).

Characterize Proj (Z)? Proj (Z) satisfies the obvious axioms. What about (RW)? ∃ a radical element for finite subsets ∅ = S of height 1’s and T of height 2’s in Proj (Z)?

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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Proj (Z) − "Roughly"

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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Preliminaries, Previous Results

  • Fact. Proj (Z) does not satisfy (RW). [A. Li, S. Wiegand ’97]

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Preliminaries, Previous Results

  • Fact. Proj (Z) does not satisfy (RW). [A. Li, S. Wiegand ’97]
  • Example. For S := {( 1

x ), (2), (5)} and T := {(x, 2), ( 1 x , 2), ( 1 x , 3)},

(S, T) does not have a radical element in Proj (Z).

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Preliminaries, Previous Results

  • Fact. Proj (Z) does not satisfy (RW). [A. Li, S. Wiegand ’97]
  • Example. For S := {( 1

x ), (2), (5)} and T := {(x, 2), ( 1 x , 2), ( 1 x , 3)},

(S, T) does not have a radical element in Proj (Z). Sketch. If w radical for (S, T) in Proj (Z), then w ⊆ ( 1

x , 2) ∩ ( 1 x , 3) =

⇒ w = (p), p ∈ Z prime. Also w = ( 1

x ).

Thus w = (g(x)), for g irreducible, deg(g) > 0. Since (5) ∈ S, (g, 5)↑ = ∅. But (g, 5)↑ T, a contradiction. ∴ ∄ radical element.

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Preliminaries, Previous Results

  • Fact. Proj (Z) does not satisfy (RW). [A. Li, S. Wiegand ’97]
  • Example. For S := {( 1

x ), (2), (5)} and T := {(x, 2), ( 1 x , 2), ( 1 x , 3)},

(S, T) does not have a radical element in Proj (Z). Sketch. If w radical for (S, T) in Proj (Z), then w ⊆ ( 1

x , 2) ∩ ( 1 x , 3) =

⇒ w = (p), p ∈ Z prime. Also w = ( 1

x ).

Thus w = (g(x)), for g irreducible, deg(g) > 0. Since (5) ∈ S, (g, 5)↑ = ∅. But (g, 5)↑ T, a contradiction. ∴ ∄ radical element.

  • Question. When ∃ radical elements?

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Coefficient Subset

  • Definition. For U poset, dim(U) = 2, C ⊆ {ht 1’s},

C = coefficient subset of U if ∀ p ∈ C, |p↑| = ∞; ∀ p = q ∈ C, (p, q)↑ = ∅;

  • p∈C p↑ = {ht 2’s};

p ∈ C, u ∈ {ht 1’s} \ C ⇒ (p, u)↑ = ∅, p↑ =

(v∈{ht 1’s}\C)(p, v)↑ ;

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Coefficient Subset

  • Definition. For U poset, dim(U) = 2, C ⊆ {ht 1’s},

C = coefficient subset of U if ∀ p ∈ C, |p↑| = ∞; ∀ p = q ∈ C, (p, q)↑ = ∅;

  • p∈C p↑ = {ht 2’s};

p ∈ C, u ∈ {ht 1’s} \ C ⇒ (p, u)↑ = ∅, p↑ =

(v∈{ht 1’s}\C)(p, v)↑ ;

Proposition 1. [Arnavut] C0 = {pZ[x] | p ∈ Spec(Z)} = the unique coefficient subset of Proj (Z).

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Conjecture

Proposition 2. [Arnavut] Let S ⊆ {ht 1’s}, T ⊆ {ht 2’s}, 0 < |S| < ∞, |T| < ∞. Suppose: S ∩ C0 ⊆

m∈T(m↓ ∩ C0), and

Either

  • s∈S(s, 1

x )↑ ⊆ T

  • r
  • s∈S(s, x)↑ ⊆ T.

Then (S, T) has ∞ radical elements in Proj (Z).

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Conjecture

Proposition 2. [Arnavut] Let S ⊆ {ht 1’s}, T ⊆ {ht 2’s}, 0 < |S| < ∞, |T| < ∞. Suppose: S ∩ C0 ⊆

m∈T(m↓ ∩ C0), and

Either

  • s∈S(s, 1

x )↑ ⊆ T

  • r
  • s∈S(s, x)↑ ⊆ T.

Then (S, T) has ∞ radical elements in Proj (Z).

  • Conjecture. [Arnavut] Same hypothesis. Suppose:

S ∩ C0 ⊆

m∈T(m↓ ∩ C0), and

(x) ∈ S and ( 1

x ) ∈ S.

Then ∃ ∞ radical elements for (S, T).

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Conjecture when T = ∅

Proposition 3. [–, E-T] Let S ⊆ {ht 1’s}, 0 < |S| < ∞ and T = ∅. Suppose

  • S ∩ C0 = ∅;
  • (x) ∈ S

& ( 1

x ) ∈ S.

Then ∃ ∞ radicals for (S, ∅).

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Conjecture when T = ∅

Proposition 3. [–, E-T] Let S ⊆ {ht 1’s}, 0 < |S| < ∞ and T = ∅. Suppose

  • S ∩ C0 = ∅;
  • (x) ∈ S

& ( 1

x ) ∈ S.

Then ∃ ∞ radicals for (S, ∅). Proposition 4. [–, E-T] (S, T) as usual. If T ⊆ ( 1

x )↑,

S ∩ C0 ⊆ {(p) |( 1

x , p) ∈ T}, p ∈ Spec(Z) ,

(x) ∈ S, ( 1

x ) ∈ S,

S − C0 = {(x), ( 1

x )} ∪ {(x + αi), αi ∈ Z}m i=1,

= ⇒ ∃ ∞ radicals for (S, T).

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More Results and an Example

Proposition 5. [–, E-T] S = {(x), ( 1

x ), (p1), ..., (pn)}, and

T = {(x, p1), ..., (x, pn), ( 1

x , p1), ..., ( 1 x , pn)} =

⇒ ∃ radical elt for (S, T).

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers

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More Results and an Example

Proposition 5. [–, E-T] S = {(x), ( 1

x ), (p1), ..., (pn)}, and

T = {(x, p1), ..., (x, pn), ( 1

x , p1), ..., ( 1 x , pn)} =

⇒ ∃ radical elt for (S, T). Example of Prop 5. Let S = {(x), ( 1

x ), (2), (3), (5)} and

T = {(x, 2), (x, 3), (x, 5), ( 1

x , 2), ( 1 x , 3), ( 1 x , 5)}.

Sx := {(x), (2), (3), (5)}, Tx := {(x, 2), (x, 3), (x, 5)} S 1

x := {( 1

x ), (2), (3), (5)}, T 1

x := {( 1

x , 2), ( 1 x , 3), ( 1 x , 5)}

∴ {Sx, Tx} ⊆ Spec(Z[x]), {S 1

x , T 1 x } ⊆ Spec(Z[ 1

x ]).

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More Results and an Example

Proposition 5. [–, E-T] S = {(x), ( 1

x ), (p1), ..., (pn)}, and

T = {(x, p1), ..., (x, pn), ( 1

x , p1), ..., ( 1 x , pn)} =

⇒ ∃ radical elt for (S, T). Example of Prop 5. Let S = {(x), ( 1

x ), (2), (3), (5)} and

T = {(x, 2), (x, 3), (x, 5), ( 1

x , 2), ( 1 x , 3), ( 1 x , 5)}.

Sx := {(x), (2), (3), (5)}, Tx := {(x, 2), (x, 3), (x, 5)} S 1

x := {( 1

x ), (2), (3), (5)}, T 1

x := {( 1

x , 2), ( 1 x , 3), ( 1 x , 5)}

∴ {Sx, Tx} ⊆ Spec(Z[x]), {S 1

x , T 1 x } ⊆ Spec(Z[ 1

x ]).

= ⇒ ∃ rad. elt. for (Sx, Tx) : vx = (15x + 2) ∈ Spec(Z[x]), = ⇒ ∃ rad. elt. for (S 1

x , T 1 x ) : v 1 x = (2 + 15

x ) ∈ Spec(Z[ 1 x ]).

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More Results and an Example

Proposition 5. [–, E-T] S = {(x), ( 1

x ), (p1), ..., (pn)}, and

T = {(x, p1), ..., (x, pn), ( 1

x , p1), ..., ( 1 x , pn)} =

⇒ ∃ radical elt for (S, T). Example of Prop 5. Let S = {(x), ( 1

x ), (2), (3), (5)} and

T = {(x, 2), (x, 3), (x, 5), ( 1

x , 2), ( 1 x , 3), ( 1 x , 5)}.

Sx := {(x), (2), (3), (5)}, Tx := {(x, 2), (x, 3), (x, 5)} S 1

x := {( 1

x ), (2), (3), (5)}, T 1

x := {( 1

x , 2), ( 1 x , 3), ( 1 x , 5)}

∴ {Sx, Tx} ⊆ Spec(Z[x]), {S 1

x , T 1 x } ⊆ Spec(Z[ 1

x ]).

= ⇒ ∃ rad. elt. for (Sx, Tx) : vx = (15x + 2) ∈ Spec(Z[x]), = ⇒ ∃ rad. elt. for (S 1

x , T 1 x ) : v 1 x = (2 + 15

x ) ∈ Spec(Z[ 1 x ]).

Now (2 + 15

x ) (2x + 15).

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Example Cont’d

Let y be another variable. gcd((15x + 2)(2x + 15), 30x) = 1 = ⇒ ((15x + 2)(2x + 15) + y(30x)) ∈ Spec(Z[x, y]). [Kaplansky ex]

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Example Cont’d

Let y be another variable. gcd((15x + 2)(2x + 15), 30x) = 1 = ⇒ ((15x + 2)(2x + 15) + y(30x)) ∈ Spec(Z[x, y]). [Kaplansky ex] Hilbert’s Irreducibility Theorem ⇒ ∃ a q ∈ Z, q prime, r(x) = (15x + 2)(2x + 15) + q(30x) irreducible in Z[x].

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Example Cont’d

Let y be another variable. gcd((15x + 2)(2x + 15), 30x) = 1 = ⇒ ((15x + 2)(2x + 15) + y(30x)) ∈ Spec(Z[x, y]). [Kaplansky ex] Hilbert’s Irreducibility Theorem ⇒ ∃ a q ∈ Z, q prime, r(x) = (15x + 2)(2x + 15) + q(30x) irreducible in Z[x]. ∴ w = (r(x)) = (30x2 + (229 + 30q)x + 30) a rad. elt. for (S, T)

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Question

  • Question. ∃ radical element for

S := {(p1), ..., (pn), (x), (1 x ), (x + a), (x + b)}, T := {(x, p1), ..., (x, pℓ), (1 x , pℓ+1), ..., (1 x , pn)}, if 0 ≤ ℓ ≤ n, gcd(ab, p1...pℓ) = 1, pi, a, b distinct primes ∈ Z ?

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Question

  • Question. ∃ radical element for

S := {(p1), ..., (pn), (x), (1 x ), (x + a), (x + b)}, T := {(x, p1), ..., (x, pℓ), (1 x , pℓ+1), ..., (1 x , pn)}, if 0 ≤ ℓ ≤ n, gcd(ab, p1...pℓ) = 1, pi, a, b distinct primes ∈ Z ?

  • Answer. Yes, |radical elements| = ∞.

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Theorem 2

Theorem 2. [–, E-T] Assume a and b are relatively prime integers and S := {(p1), . . . , (pn), (x), ( 1

x ), (x − a), (x − b)},

T := {(x, p1), . . . , (x, pℓ), ( 1

x , pℓ+1), . . . , ( 1 x , pn)},

0 ≤ ℓ ≤ n, gcd(ab, p1...pℓ) = 1, the pi are distinct prime integers. Suppose pq divides (1 − pt)(b2 + ab + a2) + qa3b3 and (1 − pt + qb2a2)(b + a), p = p1 . . . pℓ, q = pℓ+1 . . . pn, t = lcm(φ(a2), φ(b2)), and φ is the Euler phi function. Then ∃ ∞ radical elements for (S, T).

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An Example of Theorem 2

For S = {(2), (3), (x), ( 1

x ), (x − 5), (x − 7)} and T = {(x, 2), ( 1 x , 3)},

g(x; u, v, w) := 3x4 + 6ux3 + 6vx2 + 6wx + 2420 ∈ Z[x] generates a radical element for (S, T) for w = 0, u = (1 − 2420 + 3 · 52 · 72)(5 + 7) 2 · 3 · 52 · 72 ∈ Z , and v = (1 − 2420)(52 + 35 + 72) + 3 · 53 · 73 2 · 3 · 52 · 72 ∈ Z .

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An Example of Theorem 2

For S = {(2), (3), (x), ( 1

x ), (x − 5), (x − 7)} and T = {(x, 2), ( 1 x , 3)},

g(x; u, v, w) := 3x4 + 6ux3 + 6vx2 + 6wx + 2420 ∈ Z[x] generates a radical element for (S, T) for w = 0, u = (1 − 2420 + 3 · 52 · 72)(5 + 7) 2 · 3 · 52 · 72 ∈ Z , and v = (1 − 2420)(52 + 35 + 72) + 3 · 53 · 73 2 · 3 · 52 · 72 ∈ Z . u & v ∈ Z since 52 · 72 divides 1 − 2420, and also 2 · 3 = 6 divides (1 − 2420 + 3 · 52 · 72)(5 + 7) and (1 − 2420)(52 + 35 + 72) + 3 · 53 · 73.

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An Example of Theorem 2

For S = {(2), (3), (x), ( 1

x ), (x − 5), (x − 7)} and T = {(x, 2), ( 1 x , 3)},

g(x; u, v, w) := 3x4 + 6ux3 + 6vx2 + 6wx + 2420 ∈ Z[x] generates a radical element for (S, T) for w = 0, u = (1 − 2420 + 3 · 52 · 72)(5 + 7) 2 · 3 · 52 · 72 ∈ Z , and v = (1 − 2420)(52 + 35 + 72) + 3 · 53 · 73 2 · 3 · 52 · 72 ∈ Z . u & v ∈ Z since 52 · 72 divides 1 − 2420, and also 2 · 3 = 6 divides (1 − 2420 + 3 · 52 · 72)(5 + 7) and (1 − 2420)(52 + 35 + 72) + 3 · 53 · 73. For w = 5 · 7 · k = 35k, k ∈ Z, we get different integers u and v. ∴ ∃ ∞ radical elements for (S, T).

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THANKS!

Ela Celikbas and Christina Eubanks-Turner The Projective Line Over The Integers