computational linear and commutative algebra - …second).pdf · computational linear and...

Computational Linear and Commutative Algebra

Lorenzo Robbiano

University of Genoa, ItalyDepartment of Mathematics

Lorenzo Robbiano (University of Genoa, Italy) Computational Linear and Commutative Algebra Gandhinagar, February 2016 1 / 92

Standard References

Books are like imprisoned soulstill someone takes them

down from a shelf and frees them.(Samuel Butler, 1875–1941)

M. Kreuzer – L. Robbiano: Computational Commutative Algebra 1, Springer (2000)

M. Kreuzer – L. Robbiano: Computational Commutative Algebra 2, Springer (2005)

M. Kreuzer – L. Robbiano: Computational Linear and Commutative AlgebraSpringer, to appear!

Software CoCoA 5.


Topics I

• One Endomorphism• Characteristic and Minimal Polynomial• Big Kernels, Small Images, and Generalized Eigenspaces• The Module Structure• Commendable Endomorphisms• Some Anticipation about Commuting Families

TUTORIAL 1� Introducing CoCoA� Computing Characteristic and Minimal Polynomials� Computing Kernels� Computing Eigenfactors� The Buchberger Möller Algorithm


Topics II

• Commuting Families of Endomorphisms• Dimension• Kernels and Big Kernels of Ideals• Joint Eigenspaces• Splitting Endomorphisms• Cyclic Vector Spaces• Unigenerated Families• Commendable Families• Dual Families

TUTORIAL 2� Computing Dimension� Ker(I) and BigKer(I)� Computing 1-dimensional joint Eigenspaces� Cyclicity Test


Topics III

• Zero-dimensional Affine Algebras•Multiplication Endomorphisms• Separators, Primitive Elements, Curvilinear Rings• The Canonical Module and the Gorenstein property• Primary Decomposition and Separators• Computing Primary Decomposition in Characteristic Zero• Computing Primary Decomposition in Finite Characteristic• Computing Maximal Ideals Directly• Using Radicals• The Separable Subalgebra

TUTORIAL 3�Multiplication Matrices� Chinese Remaindering� Computing Primary Decompositions� Computing Radicals


Topics IV

• Solving Systems of Polynomial Equations•Meaning of Solving• Computing the Rational Zeros• Solving over a Finite Field• Representations of Fq

• Solving over Q

TUTORIAL 4� The Eigenvalue Method� Solving over a Finite Field� Solving via Approximation


Sources and Motivation 1

Figure: Daniel Lazard

Stickelberger’s Eigenvalue Theorem (1920) in Number Theory,rediscovered in the context of commutative algebra by Lazard (1981).

Let P/I be a zero-dimensional ring, let f ∈ P and ϑf the correspondingmultiplication map of P/I.• If there exists a point p ∈ ZK(I) then f (p) ∈ K is an eigenvalue of ϑf .

• If λ ∈ K is an eigenvalue of ϑf then there is p ∈ ZK(I) with λ = f (p).



Figure: Auzinger, Stetter, Möller

Auzinger-Stetter-Möller (1988,...)Rediscover this connection and obtain several results, mostly related to thenon-exact solutions of a system of polynomial equations.



Figure: David Cox

Cox (2005)Solving equations via algebras, in: A. Dickenstein and I. Emiris (eds.),Solving Polynomial Equations, Algorithms and Comp. in Math. 14,Springer, Berlin 2005, pp. 63–124.Mainly in the case of an algebraically closed field.


LECTURE 1One Endomorphism

I would like to understand things better,but I dont want to understand them perfectly.

(Douglas R. Hofstadter 1945–)


Characteristic and Minimal Polynomials

Let K be a field, let V be a finite-dimensional K-vector space, andlet ϕ ∈ EndK(V) be a K-endomorphism of V.

Definition

The polynomial χϕ(z) = det(z idV −ϕ) is called the characteristic polynomial of ϕ.

The computation of χϕ(z) is done using any matrix which represents ϕ.

Since EndK(V) is a finite-dimensional K-vector space, the kernel of the substitutionhomomorphism ε : K[z] −→ K[ϕ] given by f (z) 7→ f (ϕ) is a non-zero ideal.

Definition

The monic generator of the ideal Ker(ε), i.e. the monic polynomial of smallest degreein this ideal is called the minimal polynomial of ϕ, and is denoted by µϕ(z).


An Example

Example

Let M =(

0 1−1 0

)∈ Mat 2(Q), and consider the two block matrices A,B ∈ Mat 6(Q)

defined by

A =

(M 0 I20 M 00 0 M

)and B =

(M I2 00 M I20 0 M

)For the Q-linear map ϕ : Q6 −→ Q6 given by A, we have

µϕ(z) = (z2 + 1)2 and χϕ(z) = (z2 + 1)3

Observe that the map ϕ has only one irreducible factor, namely p(z) = z2 + 1.

Now we consider the Q-linear map ψ : Q6 −→ Q6 given by B. We have

µψ(z) = χψ(z) = (z2 + 1)3

Thus there is again just one irreducible factor p(z) = z2 + 1.


Invariance

Proposition (Invariance of µϕ(z) and χϕ(z))

Let L ⊇ K be a field extension, let VL = V ⊗K L, and let the L-vector spacehomomorphism ϕL = ϕ⊗K L : VL −→ VL be the extension of ϕ.

(a) In L[z] we have χϕL (z) = χϕ(z).(b) In L[z] we have µϕL (z) = µϕ(z).

Definition

Let Ker(ϕ) and Im(ϕ) denote the kernel and the image of the K-linear map ϕ.(a) The K-vector subspace BigKer(ϕ) =

⋃i≥1 Ker(ϕi) of V is called

the big kernel of ϕ.(b) The K-vector subspace SmIm(ϕ) =

⋂i≥1 Im(ϕi) of V is called

the small image of ϕ.


Fitting’s Lemma

Proposition (Fitting’s Lemma)

Consider the chains of K-vector subspacesKer(ϕ) ⊆ Ker(ϕ2) ⊆ · · · and Im(ϕ) ⊇ Im(ϕ2) ⊇ · · ·

(a) There exists a smallest number m ≥ 1 such that Ker(ϕm) = Ker(ϕt) for allt ≥ m. It is equal to the smallest number m ≥ 1 such that Im(ϕm) = Im(ϕt)for all t ≥ m. In particular, we have strict inclusions

Ker(ϕ) ⊂ Ker(ϕ2) ⊂ · · · ⊂ Ker(ϕm)

(b) The number m satisfies BigKer(ϕ) = Ker(ϕm) and SmIm(ϕ) = Im(ϕm).

(c) We have V = BigKer(ϕ)⊕ SmIm(ϕ).


An Example of BigKer

Example

Let V = Q8 and let ϕ : V → V be the endomorphism represented by the matrix

0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 1 0 0 11 0 0 0 0 0 0 00 1 0 0 0 0 0 00 0 1 0 0 0 0 00 0 0 1 0 1 0 00 0 0 0 0 0 1 0

with respect to the canonical basis of V . A Q-basis of Ker(ϕ) is (v1, v2) wherev1 = (0, 0, 0, 1, 0,−1, 0, 0) = e4 − e6, v2 = (0, 0, 0, 0, 1, 0, 0,−1) = e5 − e8.A Q-basis of Ker(ϕ2) is (v1, v2, v3, v4) where v3 = (1, 0,−1, 0, 0, 0, 0, 0) = e1 − e3,v4 = (0, 1, 0, 0, 0, 0,−1, 0) = e2 − e7.Moreover, we have Ker(ϕ2) = Ker(ϕ3) and we conclude that BigKer(ϕ) = Ker(ϕ2).Consequently SmIm(ϕ) = Im(ϕ2) and it turns out that a Q-basis of Im(ϕ2) is thetuple (e3, e6, e7, e8).


Cayley-Hamilton

Cayley (1821–1895) Hamilton (1805–1865)

Theorem (Cayley-Hamilton)

Let ϕ : V −→ V be a K-endomorphism of V.

(a) The minimal polynomial µϕ(z) is a divisor of the characteristicpolynomial χϕ(z).

(b) The polynomials µϕ(z) and χϕ(z) have the same irreducible factors, and hencethe same squarefree part.


Generalized Eigenspaces

Let ϕ ∈ EndK(V). We decompose its minimal polynomial and getµϕ(z) = p1(z)m1 · · · ps(z)ms

Then the characteristic polynomial of ϕ factors in this wayχϕ(z) = p1(z)a1 · · · ps(z)as with ai ≥ mi

Definition

(a) The polynomials p1(z), . . . , ps(z) are called the eigenfactors of ϕ.(b) If an eigenfactor pi(z) of ϕ is of the form pi(z) = z− λ with λ ∈ K

then λ is called an eigenvalue of ϕ.(c) For i = 1, . . . , s, the K-vector subspace Eig(ϕ, pi(z)) = Ker(pi(ϕ)) is

called the eigenspace of ϕ associated to pi(z). Its non-zero elements arecalled pi(z)-eigenvectors, or simply eigenvectors of ϕ.

(d) For i = 1, . . . , s, the K-vector subspace Gen(ϕ, pi(z)) = BigKer(pi(ϕ))is called the generalized eigenspace of ϕ associated to pi(z).


Generalized Eigenspaces Decomposition

Main Theorem (Generalized Eigenspace Decomposition)

Let ϕ ∈ EndK(V) and let µϕ(z) = p1(z)m1 · · · ps(z)ms .

(a) The vector space V is the direct sum of the generalized eigenspaces of ϕ, i.e. wehave

V = Gen(ϕ, p1(z)) ⊕ · · · ⊕ Gen(ϕ, ps(z))

(b) For i = 1, . . . , s, the restriction of the map pi(ϕ) to Gen(ϕ, pi(z)) is nilpotentand its index of nilpotency is mi.

(c) For i, j ∈ {1, . . . , s} with i 6= j, the restriction of the map pi(ϕ) to Gen(ϕ, pj(z))is an isomorphism.

(d) For i = 1, . . . , s, we have

Gen(ϕ, pi(z)) = BigKer(pi(ϕ)) = Ker(pi(ϕ)mi)

In particular, if mi = 1 then we haveGen(ϕ, pi(z)) = Ker(pi(ϕ)) = Eig(ϕ, pi(z)).


The Module Structure

Let ϕ ∈ EndK(V) be an endomorphism of the vector space V .

The vector space acquires a module structure over the ring K[ϕ]via ψ · v = ψ(v) for all ψ ∈ K[ϕ] and all v ∈ V .Using the canonical epimorphism K[z] −→ K[ϕ] given by z 7→ ϕ, it followsthat V acquires also a K[z]-module structure.

Main Theorem (Cyclic Module Decomposition)

Let ϕ ∈ EndK(V) be an endomorphism of V. There exists a tuple (v1, v2, . . . , vr) ofvectors in V such that

(a) V =⊕r

i=1〈vi〉K[z],

(b) 〈µϕ(z)〉 = Ann(v1) ⊆ Ann(v2) ⊆ · · · ⊆ Ann(vr).


Commendable Endomorphisms

Definition

The K-linear map ϕ : V −→ V is called commendable (or non-derogatory) if, forevery i ∈ {1, . . . , s}, the eigenfactor pi(z) of ϕ satisfies

dimK(Eig(ϕ, pi(z))) = deg(pi(z))

Equivalently, we require that dimK[z]/〈pi(z)〉(Eig(ϕ, pi(z))) = 1 for i = 1, . . . , s.

Main Theorem (Characterization of Commendable Endomorphisms)

Let ϕ : V −→ V be a K-linear map. Then the following conditions are equivalent.

(a) The endomorphism ϕ is commendable.

(b) We have µϕ(z) = χϕ(z).

(c) The vector space V is a cyclic K[z]-module via ϕ.


Anticipation

Definition

Given a set of commuting endomorphisms S of V , we let F = K[S] be thecommutative K-subalgebra of EndK(V) generated by S and call it the family ofcommuting endomorphisms, or simply the commuting family, generated by S.

Since EndK(V) is a finite-dimensional K-vector space, also F is a finite-dimensionalvector space hence a zero-dimensional K-algebra. The dimension of F as a K-vectorspace will be denoted by dimK(F).

Definition

Let Φ = (ϕ1, . . . , ϕn) be a system of K-algebra generators of F , and let the K-algebrahomomorphism π : P −→ F be defined by letting π(1) = idV and π(xi) = ϕi fori = 1, . . . , n. Then the kernel of π is called the ideal of algebraic relations of Φ anddenoted by RelP(Φ).


LECTURE 2Commuting Families of Endomorphisms

They are strange types of families,with no fathers, no mothers, no children.Their only concern is to be commutative.


Dimension

Schur (1875–1941) Jacobson (1910–1999)

Definition

Let F be a family of commuting endomorphisms of V . The dimension of F as a ringis zero while dimK(F) is the dimension of the family F as a K-vector space.

Example

If ϕ ∈ EndK(V) and F = K[ϕ], we have dimK(F) = deg(µϕ(z)).Therefore, it is ≤ dim(V), with equality if and only if ϕ is commendable.

• The maximal dimension of a commuting family was determined by J. Schur andN. Jacobson a long time ago: it coincides with bd2/4c+ 1 for d = dimK(V).


Dimension II

Example

Let V = K6, and let F be the L-algebra generated by {idV} and the set of all endomorphisms of V whosematrix with respect to the canonical basis of V is of the form

(0 A0 0

)with a matrix A of size 3× 3. Then the

family F is commuting, since we have(0 A

0 0

)(0 B0 0

)=(0 B

0 0

)(0 A0 0

)=(0 0

0 0

)for all matrices A,B of size 3× 3.

Here we have dimK(V) = 6 and dimK(F) = 10 = 62/4 + 1, the maximal possible dimension.

This example will be checked in the SECOND TUTORIAL.

In 1961 Murray Gerstenhaber proved that if thefamily F is generated by two commutingmatrices, the sharp upper bound for thedimension of F is dimK(V).

� A sharp upper bound for the dimension of a family generated by three commutingmatrices is apparently not known.


BMForMat

As already anticipated, if we let Φ = (ϕ1, . . . , ϕn) be a system of K-algebragenerators of F , then we have the ideal of algebraic relations of Φ denotedby RelP(Φ).

It can be computed using a nice algorithm called

Buchberger-Möller Algorithm for Matrices

In the second tutorial you will learn how to use it.


Kernels of Ideals I

Definition

Let I be an ideal in the commuting family F .(a) The K-vector subspace Ker(I) =

⋂ϕ∈I Ker(ϕ) of V is called

the kernel of I.(b) The K-vector subspace BigKer(I) =

⋂ϕ∈I BigKer(ϕ) of V is called

the big kernel of I.

Proposition

Let I be an ideal in the commuting family F .

(a) The vector space Ker(I) has a natural structure of an F/I-module.

(b) For ϕ1, . . . , ϕn ∈ F and the ideal I = 〈ϕ1, . . . , ϕn〉, we haveKer(I) =

⋂nj=1 Ker(ϕj) and BigKer(I) =

⋂nj=1 BigKer(ϕj)

(c) The vector spaces Ker(I) and BigKer(I) are invariant for F .


Kernels of Ideals II

We have some information about the size of the kernel of a maximal ideal in F .Proposition

Let m be a maximal ideal in F .(a) There exists a non-zero map ϕ ∈ F such that m = AnnF (ϕ).

(b) For a map ϕ ∈ F as in (a), and for every v ∈ Im(ϕ) \ {0}, we haveAnnF (v) = m. In particular, we have v ∈ Ker(m) 6= {0}.

(c) We have dimK(Ker(m)) = dimF/m(Ker(m)) · dimK(F/m).In particular, if we have dimK(Ker(m)) = 1 then F/m ∼= K.

Example

Let K = Q, let V = Q3, and let ϕ be the endomorphism of V defined bythe matrix

(0 0 10 0 00 0 0

)with respect to the standard basis. Since µϕ(z) = z2 and

χϕ(z) = z3, the endomorphism ϕ is not commendable. Next, we consider thecommuting family F = K[ϕ] ∼= K[z]/〈z2〉. It has only one maximal ideal m = 〈ϕ〉and satisfies dimK(F/m) = 1, but dimK(Ker(m)) = dimK(Ker(ϕ)) = 2.


An Algorithm

ALGORITHM (The Kernel of an Ideal in a Commuting Family)Let I = 〈ψ1, . . . , ψs〉 be an ideal of F , and for i ∈ {1, . . . , s} let Mi ∈ Matd(V)be the matrix which represents ψi with respect to a fixed K-basis of V. Then thefollowing algorithm computes the kernel of I.

• Form the block column matrix C = Col(M1, . . . ,Ms).• Compute generators of Ker(C) and return the corresponding vectors of V.


Comaximal Ideals

Theorem (Kernels and Big Kernels of Comaximal Ideals)

Let I1, . . . , Is be pairwise comaximal ideals in the family F , and let I = I1 ∩ · · · ∩ Is.

• We have Ker(I) = Ker(I1)⊕ · · · ⊕ Ker(Is).• We have BigKer(I) = BigKer(I1)⊕ · · · ⊕ BigKer(Is).


EigenfactorsDefinition

For an endomorphism ϕ ∈ F and a maximal ideal m of F , the monic generator of thekernel of the K-algebra homomorphism ε : K[z] −→ F/m defined by z 7→ ϕ+ m iscalled the m-eigenfactor of ϕ and is denoted by pm,ϕ(z).

Every eigenfactor is an m-eigenfactor.Proposition

Let Φ = (ϕ1, . . . , ϕn) be a system of K-algebra generators of F , and let m be amaximal ideal of F . Then the following conditions are equivalent.

(a) The maximal ideal m is linear, i.e. dimK(F/m) = 1.

(b) We have m = 〈ϕ1 − λ1 idV , . . . , ϕn − λn idV〉 for some λ1, . . . , λn ∈ K.

(c) For i = 1, . . . , n, we have pm,ϕi(z) = z− λi with λi ∈ K.

Corollary

Let m = 〈ϕ1 − λ1 idV , . . . , ϕn − λn idV〉 be a linear maximal ideal of F , and letψ ∈ F be of the form ψ = f (ϕ1, . . . , ϕn) with f ∈ K[x1, . . . , xn].We have pm,ψ(z) = z− f (λ1, . . . , λn).


Joint Eigenspaces

Definition

For every endomorphism ϕ ∈ F , let pϕ(z) be an eigenfactor of ϕ, and letC = {pϕ(z) | ϕ ∈ F}. The set C is called a coherent family of eigenfactors of F if wehave a non-trivial intersection

⋂ϕ∈F Eig(ϕ, pϕ(z)) 6= {0}.

(a) Given a coherent family of eigenfactors C of F , the corresponding intersection⋂ϕ∈F Eig(ϕ, pϕ(z)) 6= {0} is called a joint eigenspace of F .

Its non-zero elements are called joint eigenvectors of F .(b) If

⋂ϕ∈F Gen(ϕ, pϕ(z)) 6= {0}, it is called a joint generalized eigenspace of F .

Main Theorem (Joint Generalized Eigenspace Decomposition)

Let F be a family of commuting endomorphisms of V, and let m1, . . . ,ms be themaximal ideals of F .

(a) We have V =⊕s

i=1 BigKer(mi).

(b) The joint eigenspaces of F are Ker(m1), . . . ,Ker(ms).

(c) The joint generalized eigenspaces of F are BigKer(m1), . . . ,BigKer(ms).


Joint EigenspacesALGORITHM (Computing 1-Dimensional Joint Eigenspaces)Let ϕ1, . . . , ϕn ∈ F be a system of K-algebra generators of F . Consider thefollowing sequence of instructions.

1 Compute all linear eigenfactors p11(z), . . . , p1`(z) of ϕ1 and find K-basesB1, . . . ,B` of the corresponding eigenspaces Eig(ϕ1, p1i(z)) = 〈Bi〉 of ϕ1.If ` = 0, return S = ∅ and stop.

2 Form the tuples S and T whose elements are the Bi of cardinality one,and bigger than one respectively.

3 For i = 2, . . . , n, perform the following steps. Then return S and stop.4 Let L = (B′1, . . . ,B

′m) be the current tuple T. For j = 1, . . . ,m, remove B′j

from T and perform the following steps.5 Compute a matrix representing the restriction ϕij of ϕi to 〈B′j〉.6 Compute the linear eigenfactors pij1(z), . . . , pijk(z) of ϕij and K-bases

B′′1 , . . . ,B′′k of the eigenspaces Eig(ϕij, pijκ(z)) = 〈B′′κ〉 of the map ϕij.

7 Append the 1-element tuples B′′κ to S. Append the tuples in {B′′1 , . . . ,B′′k }which have more than one entry to the tuple T.

This is an algorithm which computes a tuple S of K-bases of the 1-dimensionaljoint eigenspaces of F .


Splitting EndomorphismsCan we find a single endomorphism such that its generalized eigenspaces are the jointgeneralized eigenspaces of the family?Definition

Let m1, . . . ,ms be the maximal ideals of F , and let ϕ ∈ F .(a) If we have the equalities Gen(ϕ, pmi,ϕ(z)) = BigKer(mi) for i = 1, . . . , s then ϕ

is called a splitting endomorphism for F .(b) If we have the equalities Eig(ϕ, pmi,ϕ(z)) = Ker(mi) for i = 1, . . . , s then ϕ is

called a strongly splitting endomorphism for F .

Proposition

Let F be a commuting family, let m1, . . . ,ms be the maximal ideals of F ,and let ϕ ∈ F . The map ϕ is a splitting endomorphism for F if and only ifpmi,ϕ(z) 6= pmj,ϕ(z) for i 6= j. Equivalently, ϕ is a splitting endomorphism ifand only if it has s eigenfactors.

Proposition

If we have card(K) ≥ dimK(F) then there exists a splitting endomorphism for F .


Splitting Endomorphisms IIExample

Let K = Q, V = Q4, and consider the commuting family F = Q[ϕ1, ϕ2] generated bythe endomorphisms represented by the matrices

A1 =

0 2 0 01 0 0 00 0 0 20 0 1 0

and A2 =

0 0 8 00 0 0 81 0 0 00 1 0 0

µϕ1(z) = z2 − 2 and µϕ2(z) = z2 − 8 are irreducible. The endomorphismψ = ϕ1 +ϕ2 ∈ F satisfies µψ(z) = z4−20z2 + 36 = (z2−2)(z2−18) = p1(z) ·p2(z)where p1(z) = z2 − 2 and p2(z) = z2 − 18. In particular, ϕ1 and ϕ2 are not splitting.The corresponding eigenspaces of ψ are Eig(ψ, p1(z)) = Ker(ψ2 − 2 idV) = 〈v1, v2〉,where v1 = (0, 2,−1, 0) and v2 = (4, 0, 0,−1), andEig(ψ, p2(z)) = Ker(ψ2 − 18 idV) = 〈w1,w2〉, where w1 = (0,−2,−1, 0) andw2 = (4, 0, 0, 1). The dimensions of the two eigenspaces of ψ add up to four which isthe dimension of V . Therefore these eigenspaces are also the generalized eigenspacesof ψ. Using Algorithm-Computing 1-Dimensional Joint Eigenspaces, we see thatthere are no 1-dimensional joint eigenspaces of F . Hence the map ψ is a splittingendomorphism for the family F .


F-cyclic Vector Spaces

Let S ⊆ EndK(V) be a set of commuting matrices, and let F = K[S]. Recall that thevector space V has a natural structure as an F-module given by ϕ · v = ϕ(v) for allϕ ∈ F and v ∈ V .

Proposition

Let F be a commuting family of endomorphisms of V.

(a) If V is F-cyclic and v ∈ V is an F-module generator of V then AnnF (v) = 0.(b) For a vector v ∈ V, the following conditions are equivalent.

(1) The vector space V is a cyclic F-module with generator v.(2) The K-linear map η : F −→ V which sends ϕ to ϕ(v) for every ϕ ∈ F is an

isomorphism.

(c) If V is F-cyclic then we have dimK(F) = dimK(V).


Test

ALGORITHM (Cyclicity Test)Let S = {ϕ1, . . . , ϕr} ⊆ EndK(V) be a set of commuting endomorphisms, let F be thefamily generated by the set S, and let Φ = (ϕ1, . . . , ϕr). Then let B = {v1, . . . , vd} bea basis of V, and, for i = 1, . . . , r, let Ai ∈ Matd(K) be the matrix representing ϕi inthe basis B. Consider the following instructions.

(1) Using the BM-Algorithm for Matrices compute a tuple of terms O = (t1, . . . , ts)whose residue classes form a K-basis of K[x1, . . . , xr]/RelP(Φ).

(2) If s 6= d then return "Not cyclic" and stop.

(3) Let z1, . . . , zd be indeterminates. Form the matrix C ∈ Matd(K[z1, . . . , zd])whose columns are ti(A1, . . . ,Ad) · (z1, . . . , zd)tr for i = 1, . . . , d.

(4) Compute the determinant ∆ = det(C). If ∆ 6= 0, find a tuple (c1, . . . , cd) ∈ Kd

such that ∆(c1, . . . , cd) 6= 0, return the vector v = c1v1 + · · ·+ cdvd, and stop.

(5) Return "Not Cyclic" and stop.

This is an algorithm which checks whether V is a cyclic F-module and, in theaffirmative case, computes a generator.


Unigenerated Families

Definition

Let F be a commuting family of endomorphisms of V .(a) Let ϕ,ψ ∈ EndK(V). If there exists a polynomial f (z) ∈ K[z] such that

ψ = f (ϕ), we say that ψ is polynomial in ϕ.(b) If the family F contains an endomorphism ϕ such that all endomorphisms in F

are polynomial in ϕ, it is called unigenerated by ϕ, and ϕ is called a generatorof F .

Theorem

Let F be a family of commuting endomorphisms, and let ϕ ∈ F . Then the followingconditions are equivalent.

(a) The endomorphism ϕ is commendable.

(b) The K-algebra homomorphisms K[z]/〈µϕ(z)〉 −→ F induced by z 7→ ϕ andK[z]/〈µϕ(z)〉 −→ K[z]/〈χϕ(z)〉 induced by z 7→ z are isomorphisms.

(c) The family F is unigenerated by ϕ and the vector space V is F-cyclic.


An Example

Example

Let ϕ1, ϕ2 be two Q-endomorphisms of V = Q7 which are given by the matrices

A1 =

0 0 0 0 0 0 00 0 0 0 0 −12 00 0 0 0 0 0 −121 0 0 0 0 0 00 1 0 0 0 7 00 0 0 1 0 0 00 0 0 0 1 3 0

A2 =

0 0 0 0 0 0 01 0 0 0 0 0 00 1 0 0 0 0 00 0 0 0 0 0 00 0 0 1 0 0 00 0 0 0 0 0 00 0 0 0 0 1 0

respectively. We check that A1A2 = A2A1. Hence {ϕ1, ϕ2} generates a commutingfamily F = Q[ϕ1, ϕ2]. We have χϕ1(z) = µϕ1(z) = z7, while µϕ2(z) = z3, hence ϕ1is commendable and ϕ2 is not.The theorem shows that F is unigenerated by ϕ1, and that ϕ2 is polynomial in ϕ1.Using Buchberger-Möller for Matrices we find

A2 = − 84720736 A6

1 − 851728 A5

1 − 7144 A4

1 − 112 A3

1

Please do not try by hand!


Commendable Families

Definition

Let F be a family of commuting endomorphisms of the K-vector space V .The family F is called commendable if we have the equality

dimK(Ker(m)) = dimK(F/m), equivalently dimF/m(Ker(m)) = 1

for every maximal ideal m of F .If the family F is not commendable, we also say that it is derogatory.


Dual Families

As usual, let K be a field and V a finite dimensional K-vector space of dimension d.Recall that the dual vector space of V is V∗ = HomK(V,K), i.e. the set of all K-linearmaps ` : V −→ K. For a map ϕ ∈ EndK(V), the dual endomorphism of ϕ is theK-linear map ϕ : V∗ −→ V∗ given by ϕ (`) = ` ◦ ϕ.

Definition

Let F be a family of commuting endomorphisms in EndK(V).(a) We call F ˇ = K[ϕ | ϕ ∈ F ] the dual family of F .(b) Given an ideal I of F , we call Iˇ = 〈ϕ | ϕ ∈ I〉 the dual ideal of I.

There is a canonical isomorphism ϑ : V −→ V∗∗ given by v 7→ (` 7→ `(v)) for v ∈ Vand ` ∈ V∗. It is called the canonical isomorphism to the bidual.

Proposition

Let B = (v1, . . . , vd) be a basis of V, let ϕ ∈ EndK(V), and let MB(ϕ) be the matrixwhich represents ϕ with respect to B. Then the dual map ϕ is represented byMB∗(ϕ ) = (MB(ϕ))tr with respect to the dual basis B∗.


Fundamental Results I

Theorem

Let F be a family of commuting endomorphisms of V.

(a) If V is a cyclic F-module, then F ˇ is commendable.

(b) If the family F is commendable, then V∗ is a cyclic F -module.

Corollary

Let F be a family of commuting endomorphisms of V.

(a) The vector space V is a cyclic F-module if and only if F ˇ is commendable.

(b) The vector space V∗ is a cyclic F -module if and only if F is commendable.


Fundamental Results II

Theorem

Let K ⊆ L be a field extension. Then we have the following equivalences.

(a) The F-module V is cyclic if and only if VL is a cyclic FL-module.

(b) The family F is commendable if and only if FL is commendable.

ALGORITHM Let F be a commuting family, let {ϕ1, . . . , ϕr} be aset of K-algebra generators of F , and let B be a K-basis of V. Then the followingalgorithm checks whether the family F is commendable.

(1) For i = 1, . . . , r, represent ϕi by a matrix Ai with respect to B.

(2) Apply the Cyclicity Test to the family F ˇ = K[ϕ1 , . . . , ϕr ] where ϕi isrepresented by the matrix Atr

i with respect to the basis B∗ of V∗.


LECTURE 3Zero-Dimensional Affine Algebras

You haven’t seen nothinguntil you’ve seen zero.

(Jarod Kintz 1982–)


Multiplication EndomorphismsIn the following we let K be a field and R a zero-dimensional affine K-algebra i.e azero-dimensional K algebra of type R = K[x1, . . . , xn]/I.Thus R is a finite dimensional K-vector space, and we let d = dimK(R).Definition

(a) For every element f ∈ R, the multiplication by f yields a K-linearmap ϑf : R −→ R such that ϑf (g) = f · g for all g ∈ R. It is calledthe multiplication endomorphism by f on R.

(b) The family F = K[ϑx1 , . . . , ϑxn ] is called the multiplication family of R.(c) Let B = (t1, . . . , td) be a K-basis of R, and let f ∈ R. Then the matrix

MB(ϑf ) ∈ Matd(K) which represents ϑf with respect to the basis B is calledthe multiplication matrix of f with respect to B.

Proposition

Let F = K[ϑx1 , . . . , ϑxn ] be the multiplication family of endomorphisms of R.

(a) We have F = {ϑf | f ∈ R}.(b) The map ı : R −→ F given by f 7→ ϑf is an isomorphism of K-algebras.

Its inverse is the map η : F −→ R given by ϕ 7→ ϕ(1).Lorenzo Robbiano (University of Genoa, Italy) Computational Linear and Commutative Algebra Gandhinagar, February 2016 44 / 92

Primitive ElementsDefinition

Let K ⊆ L be a field extension. An element f ∈ L is said to be a primitive elementfor K ⊆ L if we have L = K(f ).

Proposition

Let L be a finite field extension of K, let F be the multiplication family of L, andlet f ∈ L. Then the following conditions are equivalent.

(a) The element f ∈ L is a primitive element for K ⊆ L.

(b) The endomorphism ϑf ∈ F is commendable.

Example

Let p be a prime, let a, b, x, y be indeterminates, let K = Fp(a, b), and letL = K[x, y]/m where m = 〈xp − a, yp − b〉 is a maximal ideal. The multiplicationfamily F of L is commendable. Using Frobenius we conclude that every f ∈ Lsatisfies f p ∈ K, and hence dimK(K(f )) = dimK(K[f ]) ≤ p. But dimK(L) = p2,hence there exists no primitive element for K ⊆ L, and so there is no commendableendomorphism in F .


Curvilinear Rings

Definition

Let R be a zero-dimensional K-algebra, and let q1, . . . , qs be the primary componentsof its zero ideal. The ring R is called curvilinear if its maximal ideals are linear and ifthere exist numbers mi such that R/qi ∼= K[z]/〈zmi〉 for i = 1, . . . , s.

Corollary

Let R be a zero-dimensional K-algebra having s linear maximal ideals m1, . . . ,ms,and assume that card(K) ≥ s. The following conditions are equivalent.

(a) The ring R is curvilinear.

(b) There exists a commendable endomorphism in the multiplication family of R.


The Canonical Module

Definition

Let us consider the operation of the ring R on the dual vector space R∗ = HomK(R,K)given by

R× R∗ −→ R∗

(f , `) 7−→ (g 7→ `(fg))

It is easy to check that this definition turns R∗ into an R-module. This module isdenoted by ωR and called the canonical module (or the dualizing module) of R.

Remark

Let B = (t1, . . . , td) be a K-basis of R. For an element f ∈ R, the endomorphismϑf : R −→ R is represented with respect to B by the multiplication matrix MB(ϑf ).The matrix which represents the dual map ϑfˇ : ωR −→ ωR with respect to thebasis B∗ is MB∗(ϑf ) = MB(ϑf )

tr.


Gorenstein Rings

Main Theorem

Let R be a zero-dimensional affine K-algebra. The following conditions areequivalent.

(a) The canonical module ωR is a cyclic R-module.

(b) The K-algebra R is locally Gorenstein.

(c) The multiplication family F is commendable.

ALGORITHM Let R be a zero-dimensional affine K-algebra. The followingalgorithm checks whether R is locally Gorenstein or not.

(1) Compute a basis O of R as a K-vector space.

(2) Compute the multiplication matrices M1 = MO(ϑx1),. . . , Mn = MO(ϑxn).

(3) Apply the Cyclicity Test to (Mtr1 , . . . ,M

trn ).

(4) If the test is positive return Is locally Gorenstein, otherwise return Isnot locally Gorenstein.


The Chinese Remainder Theorem

A key ingredient.

Theorem (Chinese Remainder Theorem)

Let R be a ring, let I1, . . . , It be comaximal ideals in R, and let π denote the canonicalR-linear map π : R/(I1 ∩ · · · ∩ It) −→

∏ti=1 R/Ii.

(a) For i = 1, . . . , s, the ideals Ii and Ji =∏

j6=i Ij are comaximal.

(b) The map π is an isomorphism of rings.

Proof (Hint). Let us prove (b). The map π is clearly injective, so we need to prove itssurjectivity. Claim (a) implies that for every i ∈ {1, . . . , t} there are elements ai ∈ Ii

and bi ∈ Ji such that ai + bi = 1. Now it is easy to check that an element(c1 + I1, . . . , ct + It) of

∏ti=1 R/Ii is the image of the residue class of the element

c1b1 + · · ·+ ctbt = c1(1− a1) + · · ·+ ct(1− at) under π.


Primary DecompositionDefinitionLet S be a ring.(a) An ideal I in S is called primary if fg ∈ I for f , g ∈ S implies that we have f ∈ I

or g ∈ Rad(I).(b) A representation of an ideal I in S as a finite intersection of primary ideals is

called a primary decomposition of I.

Example

Be careful! Let I = 〈x2, xy〉. If fg ∈ I, then one of the two is in 〈x〉. Let f be suchelement. Then f 2 ∈ I hence f ∈ Rad(I). But I = 〈x〉 ∩ 〈x2, y〉 hence I is not primary.

Lemma

Let S be a ring, and let I be an ideal in S.

(a) If the ideal I is primary then Rad(I) is a prime ideal.

(b) If m is a maximal ideal of S and Rad(I) = m then I is m-primary.

If I = q1 ∩ · · · ∩ qs and Rad(I) = m1 ∩ · · · ∩ms the ideals qi are called the primarycomponents of I and the ideals mi are called the maximal components of I.


Separators

Definition

Let R be a zero-dimensional K-algebra, let m1, . . . ,ms be its maximal ideals, letq1, . . . , qs be the primary components of its zero ideal, and let i ∈ {1, . . . , s}.(a) An element f ∈ R is called a separator for mi if we have dimK〈f 〉 = dimK(R/mi)

and f ∈ qj for every j 6= i.

Theorem (Characterization of Separators)

Let R be a zero-dimensional K-algebra, let m1, · · · ,ms be its maximal ideals, and leti ∈ {1, . . . , s}. For an element f ∈ R the following conditions are equivalent.

(a) The element f is a separator for mi.

(b) The image of f is a non-zero element of the socle of the local ring R/qi and, forj 6= i, the image of f is zero in R/qj.

(c) The element f is a joint eigenvector of the multiplication family F of R, and it iscontained in the joint eigenspace Ker(mi), where mi is the image of mi under theisomorphism ı : R −→ F .


Generically Extended Linear Forms

Let us have a look at the first algorithm which computes the primary decomposition.

ALGORITHM [Primary Decomposition I]Let I be a zero-dimensional ideal in P = K[x1, . . . , xn], and let R = P/I. Consider thefollowing sequence of instructions.

(1) Let y1, . . . , yn be new indeterminates, let L = y1x1 + · · ·+ ynxn be thegenerically extended linear form, and let ` = L + I ∈ R⊗K K(y1, . . . , yn).Compute the minimal polynomial µ`(z) ∈ K(y1, . . . , yn)[z].

(2) Factor µ`(z) in the form µ`(z) = p1(z)m1 · · · ps(z)ms , where pi(z) is anirreducible polynomial in K(y1, . . . , yn)[z].

(3) For i = 1, . . . , s, let Qi = I + 〈pi(L)mi〉 ⊂ P⊗K K(y1, . . . , yδ) and compute areduced Gröbner basis Gi of Qi. (Notice that Gi ⊂ P.)

(4) For i = 1, . . . , s, let Qi = 〈Gi〉 ⊂ P. Return the tuple (Q1, . . . ,Qs) and stop.

This is an algorithm which computes the primary components Q1, . . . ,Qs of I.

In practice it does not work!


The Heuristic Algorithm

We start with a discussion about “probability of success”, Monte Carlo and Las VegasAlgorithms.

ALGORITHM [Primary Decomposition II]As above, let P = K[x1, . . . , xn], let I be a zero-dimensional ideal in P, let R = P/I, letL = c1x1 + · · ·+ cnxn be a linear form in P, where c1, . . . , cn ∈ K, and let ` = L + Ibe the residue class of L in R = P/I.

(1) Compute the minimal polynomial µ`(z) and factor it in the formµ`(z) = p1(z)m1 · · · ps(z)ms , with pairwise distinct, monic, irreduciblepolynomials p1(z), . . . , ps(z) ∈ K[z].

(2) For i = 1, . . . , s, let Qi = I + 〈pi(L)mi〉 and check whether Qi is a primary ideal.

(3) Return the tuple (Q1, . . . ,Qs). If all the tests in step (2) yielded the answerTRUE, output the string Primary Decomposition. Otherwise, output thestring Partial Decomposition.

This is an algorithm which computes a tuple of ideals (Q1, . . . ,Qs) in P such that wehave I = Q1 ∩ · · · ∩Qs. Moreover, it checks whether this intersection is the primarydecomposition of I or merely a partial decomposition and outputs the correspondingstring.


How do we Check? Idempotents

An idempotent of a ring R is an element f ∈ R such that f 2 = f . Clearly, theelements 0 and 1 are idempotents.

Proposition

Let R = P/I be a zero-dimensional affine K-algebra and let 〈0〉 = q1 ∩ · · · ∩ qs be theprimary decomposition of the zero ideal of R. For i = 1, . . . , s, we let Ji = ∩j 6=iqj.

(a) For i = 1, . . . , s, there exist uniquely determined elements ei ∈ Ji and fi ∈ qi

such that ei + fi = 1.

(b) Under the isomorphism π : R ∼=∏s

i=1 R/qj we haveπ(ei) = (0, . . . , 0, 1, 0, . . . , 0) with 1 in i-th position.In particular, the elements e1, . . . , es are idempotents of R.They are called primitive idempotents.

(c) The ring R is local if and only if its only idempotents are 0 and 1.


Primary Test

ALGORITHM [Primary Ideal Test]As above, let I be a zero-dimensional ideal in P. The following instructions define analgorithm which checks whether I is a primary ideal and outputs the correspondingboolean value.

(1) Compute a tuple B = (t1, . . . , td) of terms in Tn such that t1 = 1 and such thatthe residue classes of the entries of B form a K-basis of P/I.

(2) Let y1, . . . , yd be new indeterminates, let f =∑d

i=1 yi ti, and use the expressionsof the products titj in terms of the basis B to write f 2 =

∑di=1 gi ti + h with

gi ∈ K[y1, . . . , yd] and h ∈ I · P[y1, . . . , yd].(3) In the polynomial ring K[y1, . . . , yd], let J = 〈g1 − y1, . . . , gd − yd〉.

Find the K-rational zeros of J.(4) If the only K-rational zeros of J are (0, 0, . . . , 0) and (1, 0, . . . , 0) then return

TRUE. Otherwise, return FALSE.

Later we shall examine how to find the rational zeros.


The Case of Finite Fields

If K is “small”, the heuristic method can fail completely.For instance if K = F2 there only two constants, hence very few linear forms.

Other methods are available, based on a fundamental tool.

In the following we let K be a finite field. Then the characteristic of K is a primenumber p and the number of its elements is of the form q = pe with e > 0.

Furthermore, it is known that all fields having q elements are isomorphic. In thefollowing we say that a field K with pe elements represents Fq.

Definition

Let p be a prime number, and let R be a ring of characteristic p.(a) The map φp : R −→ R defined by a 7→ ap is a ring endomorphism of R. It is

called the Frobenius endomorphism of R.(b) Suppose that R is an algebra over the field Fq. Then the map φq : R −→ R

defined by a 7→ aq is Fq-linear. It is called the q-Frobenius endomorphism of R.


The Frobenius Space

Proposition

Let R be a zero-dimensional Fq-algebra.

(a) For every a ∈ Fq, we have φq(a) = a.

(b) The polynomial z− 1 is an eigenfactor of φq.

(c) For every f ∈ R, we have φq(f )− f = f q − f =∏

a∈Fq(f − a).

(d) Given a primary ideal q in R and f ∈ R, we have φq(f )− f = f q − f ∈ q if andonly if f − a ∈ q for some a ∈ Fq.

DefinitionThe set

Frobq(R) = Eig(φq, z− 1) = {f ∈ R | f q − f = 0}

i.e., the fixpoint space of R with respect to φq, is called the q-Frobenius space of R.


A Fundamental Theorem

Main Theorem (Properties of the q-Frobenius Space)

Let R be a zero-dimensional affine Fq-algebra, and let s be the number of primarycomponents of the zero ideal of R.

(a) We have Frobq(R) = {∑s

i=1 ciei | c1, . . . cs ∈ Fq} where e1, . . . , es are theprimitive idempotents of R.

(b) We have dim Fq(Frobq(R)) = s.

(c) For f ∈ R, we have f ∈ Frobq(R) if and only if µf (z) splits into distinct linearfactors.

Proof: To prove (a), we first show that Frobq(R) ⊆ {∑s

i=1 ciei | c1, . . . cs ∈ Fq}. Letπ : R ∼=

∏si=1 R/qi be the decomposition of R into local rings. Given f ∈ R, we write

π(f ) = (f1, . . . , fs) and get Frobq(R) = {f ∈ R | f qi − fi = 0 for i = 1, . . . , s}. In the

rings R/qi the zero-ideals are primary. From (c) of the preceding proposition, wededuce that there exist c1, . . . , cs ∈ Fq such that fi = ci for i = 1, . . . , s. Hence, ...Next we prove the other inclusion. Every fundamental idempotent ei satisfies e2

i = ei,and hence eq

i = ei. Moreover, ...Notice that (b) is a direct consequence of (a).The proof of (c) is left as an exercise.


Computing the Number of Primary Components

ALGORITHM Let R be a zero-dimensional affine Fq-algebra, let d = dimK(R),and let B = (b1, . . . , bd) be an Fq-basis of R.Consider the following sequence of instructions.

(1) Compute the matrix MB(φq).(2) Compute an Fq-basis C of Ker(MB(φq)− Id) where Id is the identity

matrix of size d.(3) Return the list of all elements of C and return the cardinality s of C.

This is an algorithm which computes the elements in the q-Frobenius space of Rand the number s of primary components of the zero ideal of R.

Using this algorithm one can construct an algorithm which computes the primarydecomposition of a zero-dimensional ideal in characteristic p.


Computing Maximal Ideals Directly

The idea is to split the minimal polynomials of the indeterminates in a series ofsuccessive extension fields. A key ingredient is the following.

ALGORITHM [Minimal Polynomials over Extension Fields)]Let M be a maximal ideal in P, let L = P/M, let {y1, . . . , ym} be a set of furtherindeterminates, let Q = K[x1, . . . , xn, y1, . . . , ym], and let I be a zero-dimensional idealin Q which contains MQ. Moreover, assume that we are given an element f ∈ Q. Wedenote its residue class in Q/IQ ∼= (P/M)[y1, . . . , ym]/I = L[y1, . . . , ym]/I by f .Consider the following instructions.(1) Choose a term ordering σ′′ on T(x1, . . . , xn). Extend it to a term ordering σ′ on

T(x1, . . . , xn, z) which is an elimination ordering for z. Then extend σ′ to a termordering σ on T(x1, . . . , xn, z, y1, . . . , ym) which is an elimination ordering for(y1, . . . , ym). Compute the reduced σ-Gröbner basis G of the ideal I + 〈z− f 〉.

(2) Let g ∈ P[z] be the unique polynomial in G which is monic in z.(3) Return the residue class g of g in L[z].

This is an algorithm which computes the minimal polynomial µf ,L(z) ∈ L[z] of f .

The next tool is an algorithm which shows how to factorize such minimalpolynomials in polynomial L-algebras.


Computing the Radical

Another possibility to get the primary decomposition is the following.

ALGORITHM [The Radical of a Zero-Dimensional Ideal]Let K be a perfect field, let I be a zero-dimensional ideal in P = K[x1, . . . , xn].

1 Let J = I.2 For i = 1, . . . , n, perform steps 3 – 5.3 Compute gi(xi) = sqfree(µxi, P/J(xi)).4 Replace J with J + 〈gi(xi)〉.5 If deg(gi(xi)) = dimK(P/J), return the ideal J and stop.6 Return the ideal J.

This is an algorithm which computes Rad(I).


Primary from Maximal

ALGORITHM [Primary Components via Generators of Maximal Components]Let R = P/I be a zero-dimensional affine K-algebra, and let M be a maximal idealof P whose residue class ideal in R is the maximal ideal m of R. Let M = 〈p1, . . . , pn〉and let Mi = 〈pi

1, . . . , pin〉 for i ∈ N. Consider the following instructions.

(1) Compute successive Mi until Ma + I = Ma+1 + I.(2) Let Q = Ma + I and return the ideal q of R, where q is the residue class ideal

of Q in R.This is an algorithm which computes the ideal Q in P such that Q is M-primary andthe ideal q in R such that q is m-primary.


Primary from Big Kernels

ALGORITHM [Primary Components via Big Kernels]Let F be the family of multiplication endomorphisms of R. Via the isomorphismı : R −→ F let m the image of m in F . Consider the following instructions.(1) Compute BigKer(m).(2) For i = 1, . . . , n, compute the restriction ϑi of ϑxi to BigKer(m).(3) Let Θ = (ϑ1, . . . , ϑn), compute RelP(Θ) and let q = RelP(Θ) + I.(4) Return q.

This is an algorithm which computes the ideal q in R such that q is m-primary.


The Separable Subalgebra

DefinitionLet K be a perfect field and let R be a zero-dimensional affine K-algebra. Then anelement a ∈ R is said to be separable if its minimal polynomial is square-free,equivalently if it is coprime with its derivative.

Proposition

Let K be a perfect field, let R be a zero-dimensional affine K-algebra, let a ∈ R,let µa(z) be the minimal polynomial of a, and let f (z) = sqfree(µa(z)).

(a) The element f ′(a) ∈ R is invertible.

(b) There exist elements b, r ∈ R such that we have a = b + r with r ∈ Rad(0) andµb(z) = f (z), hence b ∈ R sep.


Computing the Decomposition R = R sep ⊕ Rad(0)

ALGORITHM Let K be a perfect field, let R be a zero-dimensional affine K-algebra,let a ∈ R, let µa(z) be the minimal polynomial of a, and finally letf (z) = sqfree(µa(z)). Consider the following sequence of instructions.

(1) Let i = 0, bi = a, ri = 0(2) Let i = i + 1, compute f (bi)

f ′(bi), and let bi+1 = bi − f (bi)

f ′(bi), ri+1 = ri + f (bi)

f ′(bi).

(3) Repeat steps (2) until f (bi) = 0.(4) Return b = bi and r = ri.

This is an algorithm which computes an element b with µb(z) = f (z) , and henceb ∈ R sep, and an element r ∈ Rad(0) such that a = b + r.


LECTURE 4Solving Zero-Dimensional Systems

of Polynomial Equations

A tool knows exactlyhow it is meant to be handled,

while the user of the toolcan only have an approximate idea.

(Milan Kundera 1929–)


Meaning of Solving: Examples I

As usual, we let K be a field.1 A linear system

In this case the existence of solutions depends only on rank of matrices.Extending the base field does not change the nature of the problem.

2 x2 + 1 = 0In this case there are no solutions in R, there are in C.

3 x2 − y = 0In this case there are many solutions. An infinite number, if K is infinite,as many as #(K), if K is finite.Moreover, they can be parametrized as (t, t2), for t ∈ K.

4 x2 − 4x + 1 = 0No solutions in Q. The solution in R are 2±

√3.

5 x5 + x− 1 = 0Using CoCoA we get x5 + x− 1 = (x2 − x + 1)(x3 + x2 − 1)

Moreover x2 − x + 1 = 0 has two complex solutions 12 ±

√3

2 i.There exist also a formula for solving x3 + x2 − 1 = 0 using radicals.It is due to Cardano (1545).


Meaning of Solving: Examples II

6 x5 − x− 1 = 0Using CoCoA we check that x5 − x− 1 is irreducible.

Following Galois we know that there is no formula which represents thesolutions via radicals. But there are solutions in C.

So how and where do we represent them?Using CoCoA we get that there is only one real solutionwhich is approximately 1.16730499267578But approximation leads to different kinds of problems.


Meaning of Solving: Examples III

Back to the equation x5 − x− 1 = 0. Let f (x) = x5 − x− 1.

1 K = F2: f (x) = (x2 + x + 1)(x3 + x2 + 1).

2 K = F3: f (x) irreducible.

3 K = F17: f (x) = (x− 8)(x− 6)(x3 − 3 x2 − 5 x + 6).

4 K = F19: f (x) = (x3 + 7 x2 − 6 x− 9)(x + 6)2.

5 K = F101: f (x) = (x + 40)(x4 − 40 x3 − 16 x2 + 34 x− 48).

For instance, where do we find the other three solution in the case of F17?


Polynomial Systems

Let K be a field, let P = K[x1, . . . , xn] be a polynomial ring over K, andlet f1, . . . , fr ∈ P.We want to solve the polynomial system

f1(x1, . . . , xn) = 0...

fr(x1, . . . , xn) = 0

What does it mean “solve this system”?

If K ⊆ L is a field extension then the set of all tuples (a1, . . . , an) ∈ Ln such thatfi(a1, . . . , an) = 0 for i = 1, . . . , r is called the zero set of the ideal I = 〈f1, . . . , fr〉in Ln and is denoted by ZL(I). The elements of ZL(I) are also called the solutionsof the polynomial system in Ln.

Solving makes most sense if the zero set of I is finite.The Finiteness Criterion says that ZK(I) is finite if and only if dimK(P/I) is finite,i.e. if and only if P/I is a zero-dimensional affine algebra over K.

This is our usual setting, and we are ready to apply the techniques developed so far.Lorenzo Robbiano (University of Genoa, Italy) Computational Linear and Commutative Algebra Gandhinagar, February 2016 70 / 92

General Considerations

In general, it may happen that there are no solutions of the polynomial system in Kn.On the other hand, by Hilbert’s Nullstellensatz there exists a finite algebraic extensionfield L of K such that ZL(I) is not empty, if I is a proper ideal.

Therefore, it is true that if we allow a base field extension we have a better chance tofind a solution.It is also true that if K = Q there are approximate solutions.

We know that a rational solution (λ1, . . . , λn) ∈ Kn corresponds to a linear maximalideal 〈x1 − λ1, . . . , xn − λn〉 of P which contains I. On the other hand, a solution withcoordinates in an extension field of K corresponds to a non-linear maximal ideal Mof P which contains I.

Consequently, a first way to represent this solution is simply to write generators of M.

If we are more sophisticated, we can even compute and write generators of thecorresponding primary component Q of I.

Is there a better method to represent this situation, more similar to the description viacoordinates as in rational case?


Rational Zeros

Proposition (Eigenvalues and Rational zeros)

Let f ∈ P, and let ϑf : R −→ R be the corresponding multiplication map.

(a) If there exists a point p ∈ ZK(I) then f (p) ∈ K is an eigenvalue of ϑf .

(b) If λ ∈ K is an eigenvalue of ϑf then there exist an algebraic field extensionL ⊇ K and a point p ∈ ZL(I) such that λ = f (p).

Proof. Let us first prove (a). We let p = (a1, . . . , an) be a point of ZK(I). Then we let(x1 − a1, . . . , xn − an) be the corresponding maximal ideal in P which contains I,let m be its isomorphic image in R, and let ı(m) be its image in F .Then we deduce that f (p) is an eigenvalue of ϑf .Now we prove (b). Let λ ∈ K be an eigenvalue of ϑf . Then z− λ is an eigenfactorof ϑf . There is a maximal ideal m of R such that f − λ ∈ m. Let M be the preimageof m in P. By Hilbert’s Nullstellensatz, there exists a tuple p = (a1, . . . , an) ∈ K n

such that MK[x1, . . . , xn] ⊆ 〈x1 − a1, . . . , xn − an〉. Then the field L = K(a1, . . . , an)is an algebraic extension of K for which we have the inclusionm⊗K L ⊆ 〈x1 − a1, . . . , xn − an〉, and therefore p = (a1, . . . , an) ∈ ZL(I).Finally, we note that f − λ ∈ m implies λ = f (p).


Some Remarks

Example

Let P = Q[x, y], let I = 〈x2, y2 + 1〉, and let R = P/I. For the polynomial x, themultiplication map ϑx : R −→ R is nilpotent and has the eigenvalue λ = 0. However,there is no point p = (a1, a2) ∈ ZQ(I) such that a1 = x(p) = 0. In fact, the zeros of Iin Q 2

are (0, i) and (0, −i).

Corollary

For i ∈ {1, . . . , n}, the i-th coordinates of the points of ZK(I) are the eigenvalues ofthe multiplication map ϑxi .

Remark (The “Bad” Algorithm)

For i = 1, . . . , n, compute the eigenvalues λi1, . . . , λiνi ∈ K of the multiplication mapϑxi : R −→ R. Then check for every tuple (λ1j1 , . . . , λnjn) with 1 ≤ ji ≤ νi whether itsolves the system. The ones who do form the set ZK(I) of K-rational solutions of I.

This method has a major drawback, namely that we have to check for∏n

i=1 νi tupleswhether they solve the system.


The “Good” Algorithm

ALGORITHM (The Eigenvalue Method)Let K be a field, let P = K[x1, . . . , xn], and let I be an ideal in P such that P/I is azero-dimensional K-algebra. Consider the following sequence of instructions.

1 T = {I}.2 For i = 1, . . . , n, perform steps 2 – 7.3 S = ∅.4 For each J ∈ T perform steps 5 – 6.5 Compute the set L of linear eigenfactors of xi ∈ P/J.6 For each ` ∈ L append the ideal J + 〈`〉 to S.7 Replace T with S.8 Return T .

This is an algorithm which computes a tuple S consisting of tuples of generators of alllinear maximal ideals in P which contain I.


The “Good” Algorithm

ALGORITHM (The Eigenvalue Method)Let K be a field, let P = K[x1, . . . , xn], and let I be an ideal in P such that P/I is azero-dimensional K-algebra. Consider the following sequence of instructions.(1) Compute the linear eigenfactors z− a11, . . . , z− a1`1 with aij ∈ K.

If `1 = 0 then return S = ∅ and stop.(2) Form the tuple S = ((x1 − a11), . . . , (x1 − a1`1)).(3) For i = 2, . . . , n, perform the following steps. Then return S and stop.(4) Let S = (L1, . . . ,Lm) be the current tuple S. For j = 1, . . . ,m, remove Lj from S

and perform the following steps.(5) Let Lj = (x1 − a1 j1 , . . . , xi−1 − ai−1 ji−1). Form the ideal Jj = 〈Lj〉+ I. By

substituting xk 7→ ak jk for k = 1, . . . , i− 1, split Jj in the form Jj = 〈Lj〉+ J ′jwith J ′j ⊆ K[xi, . . . , xn].

(6) Compute the minimal polynomial µxi(z) of the residue class xi of xi inK[xi, . . . , xn]/J ′j and its distinct linear factors z− ai1, . . . , z− ai`i .

(7) For each z− aik, append the tuple (x1 − a1 j1 , . . . , xi−1 − ai−1 ji−1 , xi − aik) to S.This is an algorithm which computes a tuple S consisting of tuples of generators of alllinear maximal ideals in P which contain I.


Hint at the Eigenvector MethodTheorem (Characterization of Evaluation Forms)

Let K be a field, let P = K[x1, . . . , xn], let I be a zero-dimensional ideal in P, letR = P/I, let F be the multiplication family of R, and let p ∈ ZK(I).

(a) The evaluation form evp ∈ ωR is a joint eigenvector of the dual family F ˇ andgenerates a 1-dimensional joint eigenspace.

(b) Every 1-dimensional joint eigenspace of F ˇ is generated by an evaluation form.

(c) For every f ∈ R, the eigenvalue of ϑfˇ corresponding to evp is f (p).

(d) Given a K-basis B = (t1, . . . , td) of R, we have evp = t1(p) t∗1 + · · ·+ td(p) t∗din ωR.

ALGORITHM (The Second Eigenvector Method)In the setting of the theorem, let B = (t1, . . . , td) be a K-basis of R such that t1 = 1and ti+1 = xi for i = 1, . . . , n. Consider the following instructions.(1) Compute a set S of bases of all 1-dimensional joint eigenspaces of F .

Let S = {(`1), . . . , (`s)} with `i ∈ ωR.(2) For i = 1, . . . , s, write `i = ai1t∗1 + · · ·+ aidt∗d with aij ∈ K. Return the set of

tuples {(ai2/ai1, . . . , ai n+1/ai1) | i = 1, . . . , s}.This is an algorithm which computes the set ZK(I) of all K-rational zeros of I.


Solving over Fq

Definition

Let K be a field, let f (x) ∈ K[x], and let K ⊆ L be a field extension. Iff (x) =

∏ri=1(x− ci) with ci ∈ L, then we say that f (x) splits in L. Moreover, if

L = K(c1, . . . , cr) then L is called a splitting field of f (x).

Let p be a prime number and let q = pe. It is known that there exists a unique, up toisomorphism, finite field with q elements which is denoted by Fq.

The uniqueness depends on the fact that a field with q elements is a splitting fieldof zq − z, and the fact that two splitting fields of the same polynomial are isomorphic.

We say that a field K with pe elements represents Fq.


Splitting a Polynomial

Proposition (Splitting a Polynomial over a Finite Field)

Let p be a prime number, let K = Fp, let f (x) ∈ K[x] be a monic irreduciblepolynomial of degree e. Then let t be a new indeterminate and let L = K[t]/〈f (t)〉.(a) The field L has q = pe elements, hence it represents Fq.

(b) In the ring L[x] we have f (x) = (x− t)(x− t p) · · · (x− t pe−1), hence L is a

splitting field of f (x).

(c) Let I = 〈f (t), f (x)〉 ⊂ K[t, x]. Then Mi = 〈f (t), x− tpi〉 is a maximal ideal forevery i = 0, . . . , e− 1, and I = ∩e−1

i=0 Mi is the primary decomposition of I.

(d) In claims (b) and (c) we can substitute tpiwith the normal form NF(tpi

, 〈f (t)〉)for every i = 0, . . . , e− 1.

(e) Let V be a finite-dimensional K-vector space, let ϕ ∈ EndK(V) be such thatµϕ(z) is irreducible in K[z], and let ϕL be the extension of ϕ to VL = V ⊗K L.Then ϕL is diagonalizable.

There is a nice way to compute the “big powers” of steps (c) and (d).


Solving over a Finite FieldMain Theorem (Solutions of Polynomial Systems over Finite Fields)

Let p be a prime number, let K = Fp, let Mt be a maximal ideal of K[t], and let Mx be amaximal ideal of K[x]. Assume that dimK(K[x]/Mx) = dimK(K[t]/Mt), call e this number, andlet L = K[t]/Mt.

(a) The field L has q = pe elements, hence it represents Fq.

(b) The ideal Mt + Mx of K[t, x] is radical.

(c) For every i = 1, . . . , e there exists a tuple(gi1(t), . . . , gin(t)

)∈ (K[t])n such that, if we let

Mi = Mt + 〈x1 − gi1(t), . . . , xn − gin(t)〉 then M1 ∩ · · · ∩Me is the primarydecomposition of Mt + Mx.

(d) For i = 1, . . . , e and j = 1, . . . , n, let gij(t) be elements of K[t] as described in (c), and letgij(t) be their residue classes in L. Then we have the primary decompositionMxL[x] = ∩e

i=1〈x1 − gi1(t), . . . , xn − gin(t)〉(e) The polynomial system represented by the maximal ideal Mx of K[x] has e solutions in L.

They are(g11(t), . . . , g1n(t)

), . . . ,

(ge1(t), . . . , gen(t)

).

(f) Let V be a finite-dimensional K-vector space, let Φ = (ϕ1, . . . , ϕn) be endomorphisms ofV, and let F = K[ϕ1, . . . , ϕn] be such that RelP(Φ) is a maximal ideal inP = K[x1, . . . , xn]. Moreover, let ΦL = ((ϕ1)L, . . . , (ϕn)L) where (ϕi)L is the extension ofϕi to VL = V ⊗K L for i = 1, . . . , n, and let FL be the family generated by ΦL. Then thefamily FL is simultaneously diagonalizable.


The Algorithm

ALGORITHM (Computing Solutions over Finite Fields)In the setting of the theorem, consider the following instructions.(1) Compute the primary decomposition M1 ∩ · · · ∩Me of Mt + Mx in K[t, x].(2) Choose a term ordering σ on T(t1, . . . , tm, x1, . . . , xn) which is an elimination

ordering for (x1, . . . , xn) and compute the reduced σ-Gröbner bases G1, . . . ,Ge

of the maximal ideals M1, . . . ,Me respectively.(3) For i = 1, . . . , e, let x1 − gi1(t), . . . , xn − gin(t) be the polynomials in Gi which

are linear in the indeterminates x1, . . . , xn, and returns the e tuples(g11(t), . . . , g1n(t)

), . . . ,

(ge1(t), . . . , gen(t)

).

This is an algorithm which computes all the tuples in Ln which are solutions of thepolynomial system associated to the maximal ideal Mx.Proof. From claim (c) of the theorem we deduce that the primary components of Mt + Mx are the emaximal ideals ideals Mi = Mt + 〈x1 − gi1(t), . . . , xn − gin(t)〉. Let σ be a term ordering onT(t1, . . . , tm, x1, . . . , xn) which is an elimination ordering for (x1, . . . , xn), let σ′ be its restriction toT(t1, . . . , tm), let Gi be the reduced σ-Gröbner basis of Mi, and let G′

i be the reduced σ′-Gröbner basisof Mt , so that G′

i ⊂ Gi. We observe that NFσ(xj,Mi) = xj, since otherwise Gi would not contain anypolynomial in xj, contradicting the fact that Mi is zero-dimensional. If we let NFσ(gij(t),Mi) = hij(t)and let G′

i = (g1(t), . . . , gs(t)), it follows that Gi = (g1(t), . . . , gs(t), x1 − hi1(t), . . . , xn − hin(t)).The correctedness of the algorithm is then guaranteed by claim (e) of the theorem, and the fact thathij(t) = gij(t) in K[t]/Mt .


Special Isomorphisms

Theorem (Isomorphisms of K-Algebras)

Let K be a field, let T = K[t1, . . . , tm], let P = K[x1, . . . , xn] and letQ = K[t1, . . . , tm, x1, . . . , xn]. Let f1, . . . , fm be elements of P, let Φ : T → P be definedby Φ(ti) = fi, and let A = 〈t1 − f1, . . . , tm − fm〉. Then let g1, . . . , gn ∈ T, letΨ : P→ T be defined by Ψ(xi) = gi, and let B = 〈x1− g1, . . . , xn− gn〉. Finally, let Jbe and ideal of T and I an ideal in P. Then the following conditions are equivalent.

(a) We have the equality JQ + B = IQ + A.

(b) The two K-algebra homomorphisms Φ and Ψ induce K-algebra isomorphismsϕ : T/J → P/I and ψ : P/I → T/J which are inverse to each other.


Representations of Fq

Theorem (Isomorphisms of Representatives of Fq)

Let p be a prime number, let K = Fp, let Mt be a maximal ideal of K[t], let Mx be amaximal ideal of K[x], let dimK(K[x]/Mx) = dimK(K[t]/Mt), and call e this number.Then let

(gi1(t), . . . , gin(t)

)∈ (K[t])n be tuples as described in the above theorem.

(a) There exist e tuples(fi1(x), . . . , fim(x)

)∈ (K[x])m for i = 1, . . . , e such that

Mt + 〈x1 − gi1(t), . . . , xn − gin(t)〉 = Mx + 〈t1 − fi1(x), . . . , tm − fim(x)〉(b) Let Φi : K[t]→ K[x] be defined by Φi(tj) = gij(x), an let Ψi : K[x]→ K[t] be

defined by Ψi(xj) = fij(t). Then Φi and Ψi induce ϕi : K[t]/Mt → K[x]/Mx andψi : K[x]/Mx → K[t]/Mt which are K-algebra isomorphisms inverse to eachother.

(c) The isomorphisms in (b) are all the K-algebra homomorphisms between thefields K[t]/Mt and K[x]/Mx.


An Algorithm

ALGORITHM (Computing Isomorphisms of Representatives of Fq)Consider the following instructions.(1) Compute the e tuples

(g11(t), . . . , g1n(t)

), . . . ,

(ge1(t), . . . , gen(t)

)using the

solutions over a finite field.(2) Let σ be an ordering on T(t1, . . . , tm, x1, . . . , xn) of elimination of (t1, . . . , tm),

and compute the reduced σ-Gröbner bases G1, . . . ,Ge of the idealsMx + 〈x1 − gi1(t), . . . , xn − gin(t)〉 for i = 1, . . . , e.

(3) For i = 1, . . . , e, let t1 − fi1(x), . . . , tm − fim(x) be the polynomials in Gi whichare linear in the indeterminates t1, . . . , tm, and return the e tuples(f11(t), . . . , g1n(t)

), . . . ,

(ge1(t), . . . , gen(t)

)and the e tuples(

f11(x), . . . , f1m(x)), . . . ,

(fe1(x), . . . , fem(x)

)This is an algorithm which computes the tuples which define the isomorphisms ϕi

and ψi such that ϕi and ψi are inverse to each other for i = 1, . . . , e.


Solving over Q

ALGORITHM (Computing the Splitting Field)Let K = Q, let P = K[x], and let f (x) be a non linear irreducible polynomial in P.Consider the following sequence of instructions.(1) Let i = 1, let ti be a new indeterminate, and let Li = K[ti]/〈f (ti)〉.(2) Factorize f (x) in Li[x], let Lini(f (x)) be the set of linear factors of f (x) in Li[x].

If Lini(f (x)) yields a complete factorization of f (x), then return the field L = Li,the set Lin(f (x)) = Lini(f (x)), and stop.

(3) Repeat the following steps until Lini(f (x)) yields a complete factorization off (x).

(4) Increase i by 1, let fi(t1, . . . , ti−1, x) be a non linear irreducible factor of f (x) inthe polynomial ring Li−1[x], and then letLi = K[t1, . . . , ti]/〈f (t1), f2(t1, t2), . . . , fi(t1, . . . , ti)〉. Then perform Step 2 withf (x) = fi(t1, . . . , ti−1, x).

(5) Return the field L, the set Lin(f (x)), and stop.This is an algorithm which computes the splitting field L of f (x) and the set Lin(f (x))of the linear factors of f (x) in L[x].


An Example

Example

Let P = Q[x]. The discriminant of f (x) = x3 − 3x + 1 is 81 = 92 which implies thatthe Galois group of f (x) is F3. Then we let Mt = 〈f (t)〉 and Mx = 〈f (x)〉 in Q[t, x].The primary decomposition of the ideal Mt + Mx is given byMt + Mx = M1 ∩M2 ∩M3 where

M1 = Mt + 〈x− t〉M2 = Mt + 〈x− t2 + 2〉M3 = Mt + 〈x + t2 + t − 2〉

Therefore L = Q[t]/Mt is the splitting field of f (x) and dimQ(L) = 3.


An Example (continued)

Example

The discriminant of the polynomial g(x) = x3 − 4x + 1 is 229. This is not a squarewhich implies that the Galois group of g(x) is the symmetric group S3. The differenceof the Galois groups implies a different size of the splitting fields of f (x) and g(x).To check this claim, we let Mt1 = 〈g(t1)〉 and Mx = 〈g(x)〉 in Q[t1, x]. The primarydecomposition of Mt1 + Mx is given by Mt1 + Mx = M1 ∩M2 where

M1 = Mt1 + 〈x− t1〉M2 = Mt1 + 〈x2 + t1x + t21 − 4〉

Then we let M(t1, t2) = Mt1 + 〈t22 + t1t2 + t21 − 4〉. The primary decomposition ofM(t1, t2) + 〈x2 + t1x + t21 − 4〉 is given by M(t1, t2) + 〈x2 + t1x + t21 − 4〉 = M21 ∩M22 where

M21 = M(t1, t2) + 〈x− t2〉M22 = M(t1, t2) + 〈x + t1 + t2〉

We conclude that M(t1, t2) + Mx = M1 ∩M21 ∩M22 where

M1 = M(t1, t2) + 〈x− t1〉M21 = M(t1, t2) + 〈x− t2〉M22 = M(t1, t2) + 〈x + t1 + t2〉

and that L = Q[t1, t2]/M(t1, t2) is the splitting field of g(x). We have dimQ(L) = 6.


Another Example

Example

Let P = Q[x, y] and let I = 〈x2 + x− 1, y3 − x− 1〉. Using the ordering Lex withy > x we get the reduced Gröbner basis (y6 − y3 − 1, x− y3 + 1). From the fact thatthe polynomial y6− y3− 1 is irreducible we deduce that I is a maximal ideal. Then weuse an algorithm described before to find the appropriate extension of Q where I splitsinto linear maximal ideals. Let M1 = 〈t2

1 + t1 − 1〉 and Ix = 〈x2 + x− 1〉. In Q[t1, x]we get the primary decomposition M1 + Ix =

(M1 + 〈x− t1〉

)∩(M1 + 〈x + t1 + 1〉

).

We let L1 = Q[t1]/M1. In L1[x, y] we get the decomposition M = N1 ∩ N2 where

N1 = 〈x− t1, y3 − t1 − 1〉 N2 = 〈x + t1 + 1, y3 − t1 − 1〉

We have L1[x, y]/N1 ∼= L1[y]/〈y3 − t1 − 1〉.The next step is to split the polynomial y3 − t1 − 1 ∈ L1[y].Let M2 = M1 + 〈t3

2 − t1 − 1〉 and Iy = 〈y3 − t1 − 1〉. The ideal M2 is maximal inQ[t1, t2]. Moreover, in the polynomial ring Q[t1, t2, y] we get the primarydecomposition M2 + Iy =

(M2 + 〈y− t2〉

)∩(M2 + 〈y2 + t2y + t2

2〉). Let

M3 = M2 + 〈t23 + t2t3 + t22 − 1〉 = 〈t21 + t1 − 1, t32 − t2 − t1 − 1, t23 + t2t3 + t22〉


Example continued

ExampleThe ideal M3 is maximal in Q[t1, t2, t3], and in Q[t1, t2, t3, y] we get the primary decomposition

M3 + Iy =(M3 + 〈y− t1〉

)∩(M3 + 〈y− t2〉

)∩(M3 + 〈y + t2 + t3〉

)At this point we let M = M3, we let L = Q[t1, t2, t3]/M, and we have three solutions in L of thepolynomial system associated to M. They are the residue classes of

(t1, t2), (t1, t3), (t1,−t2 − t3)

Next, we have L1[x, y]/N2 ∼= L1[y]/〈y3 + t1〉. Let us see how much the polynomial y3 + t1 splits in L. Wecompute the primary decomposition of M + 〈y3 + t1〉 and get

M + 〈y3 + t1〉 =(M + 〈y− t1t2t3〉

)∩(M + 〈y− t1t2t3 − t1t22〉

)∩(M + 〈y + t1t22〉

)We conclude that all the solutions can be found in L. They are

(t1, t2) (t1, t3) (t1, −t2 − t3)(−t1 − 1, t1t2t3) (−t1 − 1, t1t2t3 + t1t22) (−t1 − 1, −t1t22)

Finally we observe that degQ(L) = 12.


Approximation: an Example

Example

Over the field K = Q we consider the polynomial system

f1 = 3x2 − y2 + 2yz− z2 − 8x− 8y + 5z− 5 = 0f2 = x3 − 6x2 − 6xy− 4y2 + z2 + 3x + 7y− 7z + 15 = 0f3 = z3 + 4x2 + 2xy− 3z2 − 13x− 5y + 6z + 5 = 0

In the ring P = Q[x, y, z] we form the ideal I = 〈f1, f2, f3〉. To find the zeros of I weuse the Eigenvalue Method. The minimal polynomial of x evaluated at x factors asf (x)(x− 2)3 where

f (x) = x15 − 102 x14 + 4254 x13 − 91420 x12 + 1108030 x11 − 7617460 x10

+26368737 x9 − 16709678 x8 − 163401402 x7 + 408143116 x6

−102334132 x5 − 561462068 x4 + 119765981 x3 + 1225188788 x2

−1326547350 x + 524273780We add 〈x− 2〉3 to I and getQ1 = 〈x− y− 3, z2 − 2z + 1, yz− y + z− 1, y2 + 2y + 1

2 z + 12 〉. The radical of Q1

is M1 = 〈x− 2, y + 1, z− 1〉. We conclude that I has the unique rational zero(2, −1, 1).


Example continued

Example

Next we add 〈f (x)〉 to I and get the ideal M2 = 〈f1, f2, f3, f4〉 where

f1 = x2 − 13 y2 + 2

3 yz− 13 z2 − 8

3 x− 83 y + 5

3 z− 53

f2 = z3 + 2 xy + 43 y2 − 8

3 yz− 53 z2 − 7

3 x + 173 y− 2

3 z + 353

f3 = xy2 − 2 xyz + xz2 − 10 xy− 463 y2 − 5 xz + 20

3 yz− 13 z2 − 38

3 x− 173 y− 13

3 z + 853

f4 = y4 − 10 y2z2 − 823 y3 − 282 xyz + 50 y2z + 103 xz2 + 4 yz2 − 2917

3 xy− 130699 y2

− 14383 xz + 5033

9 yz + 8459 z2 − 13034

9 x− 107519 y− 3754

9 z + 202159

To check that M2 is a maximal we compute a Lex-Gröbner basis of M2. We get(x− g1(z), y− g2(z), g(z)) where

g(z) = z15 − 15 z14 + 453 z13 − 1723 z12 + 1653 z11 + 39878 z10 − 143949 z9

+44358 z8 + 1493331 z7 − 3545819 z6 + 1822878 z5 + 12465182 z4

−23275915 z3 + 35305982 z2 − 40170300 z + 15978680and g1, g2 are polynomials of degree 14 whose coefficients are fractions wherenumerators and denominators are in the order of 1040. The verification that g(z) isirreducible confirms that M2 is a maximal ideal.


Example continued

Example

If we do not resort to approximation, apparently to best represent the roots associatedto M2 one has simply to write M2 = 〈f1, f2, f3, f4〉. If we leave the world of exactnessand enter the realm of approximation, we can deduce that of the 15 solutions, three ofthem have real coordinates. They are (approximately)

( −2.375085, 0.866703, 2.514900 )( 9.492307, −17.531086, 0.723763 )( 2.029906, −0.970152, 0.998203 )

2 is not equal to 3,not even for very large values of 2.

(Gabriel’s Law)


Next?

The Endor

approximately The End?


computational linear and commutative algebra - …second).pdf · computational linear and...

Documents