groebner basis 3

8/8/2019 Groebner Basis 3

1/27

Grobner Bases Introduction 3

Daniel Cabarcas, University of Cincinnati

Cryptography SeminarCincinnati, April 20, 2009

Cabarcas Grobner Bases
http://goforward/http://find/http://goback/


2/27

Construction of Grobner bases

Theorem (characterization of Grobner bases)G = {g1, . . . , gt} I is a Grobner basis of I w.r.t. a monomialorder iffFor all f K[ x] there exist unique r K[ x] and g I s.t.

f = g + r

and no monomial of r is divisible by any of LM(g1), . . . , LM(gt).



3/27

Construction of Grobner bases

Theorem (characterization of Grobner bases)G = {g1, . . . , gt} I is a Grobner basis of I w.r.t. a monomialorder iffFor all f K[ x] there exist unique r K[ x] and g I s.t.

f = g + r

and no monomial of r is divisible by any of LM(g1), . . . , LM(gt).

The crucial idea in Grobner basis theory is the

observation that these infinitely many tests can bereplaced by the consideration of finitely many critical

situations that can be characterized by the so-called

S-polynomials. Bruno Buchberger 1999



4/27

Construction of Grobner basis

LT(g1), . . . , LT(gt) = LT(I)



5/27


LT(g1), . . . , LT(gt) = LT(I)

Let f, g K[ x]



6/27


LT(g1), . . . , LT(gt) = LT(I)

Let f, g K[ x]

let m = lcm(LM(f), LM(g))



7/27


LT(g1), . . . , LT(gt) = LT(I)

Let f, g K[ x]


The s-polynomial of f and g is



8/27


LT(g1), . . . , LT(gt) = LT(I)

Let f, g K[ x]


The s-polynomial of f and g is

S(f, g) :=m

LT(f) fm

LT(g) g



9/27


LT(g1), . . . , LT(gt) = LT(I)

S(f, g) :=m

LT(f)f

m

LT(g)g



10/27


LT(g1

), . . . , LT(gt) = LT(I)

S(f, g) :=m

LT(f)f

m

LT(g)g

e.g. S(x2y 1, xy2 1)



11/27


LT(g1

), . . . , LT(gt) = LT(I)

S(f, g) :=m

LT(f)f

m

LT(g)g

e.g. S(x2y 1, xy2 1)

=x2y2

x2y(x2y 1)

x2y2

xy2(xy2 1)



12/27


LT(g1

), . . . , LT(gt) = LT(I)

S(f, g) :=m

LT(f)f

m

LT(g)g

e.g. S(x2y 1, xy2 1)

=x2y2

x2y(x2y 1)

x2y2

xy2(xy2 1)

=y(x2y 1) x(xy2 1)



13/27


LT(g1

), . . . ,

LT(gt) = LT(I)

S(f, g) :=m

LT(f)f

m

LT(g)g

e.g. S(x2y 1, xy2 1)

=x2y2

x2y(x2y 1)

x2y2

xy2(xy2 1)

=y(x2y 1) x(xy2 1)

=(x2y2 y) (x2y2 x)



14/27


LT

(g1

), . . . , LT

(gt) =

LT(

I)

S(f, g) :=m

LT(f)f

m

LT(g)g

e.g. S(x2y 1, xy2 1)

=x2y2

x2y(x2y 1)

x2y2

xy2(xy2 1)

=y(x2y 1) x(xy2 1)

=(x2y2 y) (x2y2 x)

=x y



15/27


TheoremLet I be an ideal of K[ x]. Then a set of generatorsG = {g1, . . . , gt} for I is a Grobner basis for I iff for all i = j

S(gi, gj)

G

0


G


16/27



S(gi, gj)

G

0

LemmaLet f 1, . . . , fs K[ x] and c1, . . . , cs K. Suppose that forLM(f1) = LM(f2) = = LM(fs) =: m and that

LM(

si=1 ci fi) < m.


C i f G b b i


17/27



S(gi, gj)

G

0


LM(

si=1 ci fi) < m.Then

si=1 ci fi is a linear combination with

coefficients in K of the s-polynomials S(fi, fj) for 1 j < k s.


C i f G b b i


18/27



S(gi, gj)

G

0


LM(

si=1 ci fi) < m.Then

si=1 ci fi is a linear combination with

coefficients in K of the s-polynomials S(fi, fj) for 1 j < k s.Furthermore, LM(S(fi, fj)) < m


C i f G b b i


19/27


TheoremLet I = f1, . . . , fs be a polynomial ideal. Then a Grobner basisfor I can be constructed in a finite number of steps by the

following algorithm:


C t ti f G b b i


20/27




Buchberger algorithm(F = {f1, . . . , fs})


C t ti f G b b i


21/27





G := F




22/27





G := F

REPEAT




23/27





G := F

REPEAT G := G




24/27





G := F

REPEAT G := G FOR EACH pair {f, g} f = g in G




25/27





G := F


Compute S(f,

g)

G r




26/27





G := F


Compute S(f,

g)

G r IF r = 0 THEN G := G {r}




27/27





G := F


Compute S(f,

g)

G r IF r = 0 THEN G := G {r}

UNTIL G = G


groebner basis 3

Documents