Homomorphisms and Chinese Remainder Algorithms€¦ · Homomorphisms and Chinese Remainder...

Homomorphisms and Chinese Remainder Algorithms

Homomorphisms and Chinese RemainderAlgorithms

L. Yohanes Stefanus

L. Yohanes Stefanus Slide 06.1


Problem: Intermediate Expression Swell

Most of the time, a computer algebra system computeswith multiprecision integers, not single-precision integerswhich fit into one computer word memory.

Let’s consider a simple problem of solving a system oflinear equations with integer coefficients:

22x + 44y + 74z = 1,

15x + 14y − 10z = −2,

−25x − 28y + 20z = 34.

We can use a variation of Gaussian elimination to solvethis system using only integer arithmetic.



To eliminate the x term from the second equation, wemultiply equation one by 15, equation two by 22 and takethe difference. The system of equations, without the x termin equations two and three, is as follows:

22x +44y +74z = 1,−352y −1330z = −59,

484y +2290z = 773.

Continuing in this manner until we isolate x , y , and z, weobtain the reduced system:

1257315840x = 7543895040,−57150720y = 314328960,

162360z = 243540.



The solutions are

x = 75438950401257315840 = 6,

y = 314328960−57150720 = −11

2 ,

z = 243540162360 = 3

2 .

Notice the growth in the size of the coefficients. But thefinal answer has reasonable size.

Of course we can always solve the system by using rationalarithmetic, but this also involves a hidden cost. Eachrational arithmetic operation requires gcd computationwhich is expensive compared to addition or multiplication.

We will see that the problem of intermediate expressionswell is not unique to solving linear systems.



We are going to study an approach, commonly called themodular or multiple homomorphism approach, to reducingintermediate expression swell. This method involvesbreaking down a problem posed in one algebraic domaininto a similar problem in a number of much simpleralgebraic domains.

For example, rather than solving our system of equationsover Z we solve it over Zp, for a number of primes p.These domains are simpler because they are finite fields.The coefficients that arise during the solution are all offixed size. Thus we avoid the problem of intermediatecoefficient growth.

Of course we still must know how to piece together thesolutions from these simpler domains. This is the subjectof the Chinese remainder algorithm.



Ring Morphisms

Some concepts from universal algebra are needed to studythe mapping of an algebraic system onto a simpler modelof itself.

Definition (5.1)

An algebra (or algebraic system) is a set S together with acollection of operations defined on S. A subset S′ of the set Sis called a subalgebra if S′ is closed under the operationsdefined on S.

Examples of algebras: commutative ring, integral domain,UFD, Euclidean domain, field.



Definition (5.2)

Let R and R′ be two rings. Then a mapping φ : R → R′ is calleda ring morphism if

(i) φ(a + b) = φ(a) + φ(b) for all a, b ∈ R,

(ii) φ(ab) = φ(a)φ(b) for all a, b ∈ R,

(iii) φ(1) = 1.



The general concept of a morphism between two algebrasis that of a mapping which preserves all of the operationsdefined on the algebras.Morphisms are classified according to their properties asfunctions. If φ : R → R′ is a morphism then it is called

a monomorphism if the function φ is injective (one-to-one),an epimorphism if the function φ is surjective (onto),an isomorphism if the function φ is bijective (one-to-one andonto).

The term "homomorphism" is commonly used in place ofepimorphism. In particular, if φ : R → R′ is an epimorphismthen R′ is called a homomorphic image of R.



Example

For any commutative ring R, the polynomial domains R[x , y ],R[y ][x ], and R[x ][y ] are isomorphic. The natural mappingφ : R[x , y ] → R[x ][y ] defined by

φ(m∑

i=0

n∑j=0

aijxiy j) =

n∑j=0

(m∑

i=0

aijxi)y j

is an isomorphism.



Example

Let Z be the set of integers and let Z6 be the set ofintegers modulo 6.

Let φ : Z → Z6 be the mapping defined by φ(a) = rem(a, 6)for all a ∈ Z . Then φ is an epimorphism and thus φ is aprojection of Z onto the homomorphic image Z6.

Z6 is a commutative ring as Z is.

Z6 is not an integral domain even though Z is.



Modular and Evaluation Homomorphisms

There are two particular classes of homomorphisms whichhave many practical applications in algorithms for symboliccomputation: modular homomorphism and evaluationhomomorphism.

A modular homomorphism

φm : Z [x1, . . . , xν ] → Zm[x1, . . . , xν ]

is a homomorphism defined for a fixed integer m ∈ Z bythe rules:

φm(xi) = xi , for 1 ≤ i ≤ ν;φm(a) = rem(a, m), for all coefficients a ∈ Z .

Note that φm remains well-defined in the case ν = 0 inwhich case it is simply a projection of Z onto Zm.



Example

In Z [x , y ] let a(x , y) = 14x2y2 − x2y + 7x2 − 3xy + 5. Themodular homomorphism φ7 maps this polynomial onto thefollowing simpler polynomial in Z7[x , y ]:

φ7(a(x , y)) = 6x2y + 4xy + 5.



An evaluation homomorphism

φxi−α : D[x1, . . . , xν ] → D[x1, . . . , xi−1, xi+1, . . . , xν ]

is a homomorphism defined for a particular indeterminatexi and a fixed element α ∈ D such that for any polynomiala(x1, . . . , xν) ∈ D[x1, . . . , xν ],

φxi−α(a(x1, . . . , xν)) = a(x1, . . . , xi−1, α, xi+1, . . . , xν).

In other words, an evaluation homomorphism φxi−α is aprojection of D[x1, . . . , xν ] onto D[x1, . . . , xi−1, xi+1, . . . , xν ]obtained by simply substituting the value α ∈ D for the i-thindeterminate xi .

The notation φxi−α can be read as "substitute α for xi ".



Example

In Z [x , y ] let a(x , y) = 14x2y2 − x2y + 7x2 − 3xy + 5.Then φx−1(a(x , y)) = 14y2 − 4y + 12.

Compositions of modular and evaluation homomorphismswill be used frequently later for projecting the multivariatepolynomial domain Z [x1, . . . , xν ] onto simpler homomorphicimages of itself where very efficient algorithms exist.



Ideals

A ring morphism φ : R → R′ can be conveniently characterizedin terms of its action on particular subsets of R known as ideals.

Definition (5.3)

An ideal I in a commutative ring R is a nonempty subset of Rthat satisfies

(i) a, b ∈ I ⇒ a− b ∈ I;

(ii) a ∈ I, r ∈ R ⇒ ar ∈ I.



Example

In the integral domain Z of integers, the subset

〈m〉 = {rm|r ∈ Z}

for some fixed m ∈ Z is an ideal, called the ideal generatedby m.

Proof :For every km, `m ∈ 〈m〉 and s ∈ R,〈m〉 contains km − `m = (k − `)m and s(rm) = (sr)m.Hence 〈m〉 is an ideal.

In particular, the ideal 〈3〉 is the set

〈3〉 = {0,±3,±6, . . .}.



Example

In the polynomial domain Q[x ], the subset

〈p(x)〉 = {p(x)a(x)|a(x) ∈ Q[x ]}

for some fixed p(x) ∈ Q[x ] is an ideal, called the idealgenerated by p(x).

In particular, the ideal 〈x − α〉 for some fixed α ∈ Q is theset of all polynomials over Q which have x − α as a factor.



Example

In the bivariate polynomial domain Z [x , y ], the subset

〈p(x , y), q(x , y)〉 =

{p(x , y)a(x , y) + q(x , y)b(x , y)|a(x , y), b(x , y) ∈ Z [x , y ]}

for some fixed polynomials p(x , y), q(x , y) ∈ Z [x , y ] is anideal, called the ideal generated by p(x , y) and q(x , y).

In particular, the ideal 〈x , y〉 is the set of all bivariatepolynomials over Z with constant term zero.



Two very special ideals in any commutative ring R are thesubsets {0} and R because properties (i) and (ii) ofDefinition 5.3 are clearly satisfied by these two subsets.We call {0} the zero ideal and R the universal ideal.

By a proper ideal we mean any ideal I such that I 6= {0}and I 6= R.

Note that {0} is not a subring of R because it is not closedunder the nullary operation "select 1" defined on R (i.e., itdoes not contain 1, the multiplicative identity of R).

In universal algebra terminology, the selection of theadditive identity 0 can be regarded as the nullary operation"select 0". Likewise, the selection of the multiplicativeidentity 1 can be regarded as the nullary operation "select1".



Theorem (5.1)

Every proper ideal I in a commutative ring R is closed under allof the ring operations defined on R except that I is not closedunder the nullary operation "select 1" (i.e., 1 /∈ I).

Proof.

Property (i) in Definition 5.3 guarantees that I is closed underthe binary operation +, the unary operation - (inverse), and thenullary operation "select 0". Property (ii) in Definition 5.3guarantees that I is closed under multiplication. As for thenullary operation "select 1", if 1 ∈ I then by Property (ii) r ∈ I forall r ∈ R, which means that I = R so that I is not a properideal.



The fact that an ideal I in a commutative ring R is closedunder addition and is closed under multiplication by anyelement of R, implies that if I contains the n elementsa1, . . . , an then it must contain all linear combinations ofthese elements, defined by:

〈a1, . . . , an〉 = {r1a1 + · · ·+ rnan|ri ∈ R}.

On the other hand, for any given elements a1, . . . , an ∈ R,the set 〈a1, . . . , an〉 of all linear combinations of theseelements is an ideal in R. This ideal is called the ideal withbasis a1, . . . , an.



Definition (5.4)

An ideal I in a commutative ring R is called an ideal with finitebasis if I can be expressed as the set 〈a1, . . . , an〉 of all linearcombinations of a finite number n of elements a1, . . . , an ∈ R.

Definition (5.5)

An ideal I in a commutative ring R is called a principal ideal if Ican be expressed as the set 〈a〉 of all multiples of a singleelement a ∈ R.



Domains with Special Ideals

Definition (5.6)

An integral domain D is called a Noetherian integral domain ifevery ideal in D is an ideal with finite basis.

Definition (5.7)

An integral domain D is called a principal ideal domain (PID) ifevery ideal in D is a principal ideal.



Remarks

It can be proved that every Euclidean domain is a PID andtherefore the domains Z and Q[x ] are PIDs.

The polynomial domain Z [x , y ] is an example of an integraldomain that is not a PID. It is not possible to generate theideal 〈x , y〉 by a single element.

However, if D is a Noetherian integral domain then so isD[x ]. By induction this implies that any multivariatepolynomial domain over Z or over a field is a Noetherianintegral domain.

In the hierarchy of domains, the PID lies between the UFDand the Euclidean domain. That is, every Euclideandomain is a PID and every PID is a UFD. But, unlike a PID,a Noetherian integral domain is not simply a UFD whichsatisfies additional axioms.



Definition (5.8)

Let I and J be two ideals in a Noetherian integral domain D andsuppose I = 〈a1, . . . , an〉, J = 〈b1, . . . , bm〉 for elementsai ∈ D(1 ≤ i ≤ n), bj ∈ D(1 ≤ j ≤ m).

(i) The sum of the ideals I and J in D is the ideal defined by〈I, J〉 = 〈a1, . . . , an, b1, . . . , bm〉. Note that the ideal 〈I, J〉consists of all possible sums a + b where a ∈ I and b ∈ J.

(ii) The product I · J of the ideals I and J in D is the idealgenerated by all elements aibj such that ai is a basiselement for I and bj is a basis element for J. Thus theproduct of I and J can be expressed asI · J = 〈a1b1, . . . , a1bm, a2b1, . . . , a2bm, . . . , anb1, . . . , anbm〉.

(iii) The i-th power of the ideal I in D (for i a positive integer) isdefined recursively in terms of products of ideals asfollows:

I1 = I;I i = I · I i−1 for i ≥ 2.



Example

Consider two principal ideals 〈a〉 and 〈b〉 in D. We have

〈a〉 · 〈b〉 = 〈ab〉

〈a〉i = 〈ai〉 for i ≥ 1

If D is a PID then the sum 〈〈a〉, 〈b〉〉 = 〈a, b〉 must be aprincipal ideal:

〈a, b〉 = 〈gcd(a, b)〉



Definition (5.9)

Let R and R′ be commutative rings and let φ : R → R′ be amorphism. The kernel K of the morphism φ is the set definedby:

K = φ−1(0) = {a|a ∈ R and φ(a) = 0}.

Theorem (5.2)

Let R and R′ be commutative rings. The kernal K of amorphism φ : R → R′ is an ideal in R.



Proof.

The set K is not empty since φ(0) = 0.

If a, b ∈ K then φ(a− b) = φ(a)− φ(b) = 0− 0 = 0 so thata− b ∈ K . This proves property (i) of Definition 5.3.

Property (ii) of Definition 5.3 also holds because if a ∈ Kand r ∈ R then φ(ar) = φ(a)φ(r) = 0 · φ(r) = 0 so thatar ∈ K .



Theorem (5.3: Characterization Theorem)

Let R be a commutative ring and let K be an ideal in R. Ifφ1 : R → R′ and φ2 : R → R” are two morphisms both havingkernel K then the correspondence between the twohomomorphic images φ1(R) and φ2(R) defined byφ1(a) ↔ φ2(a) is an isomorphism.

This theorem means that we can specify a homomorphic imageof a commutative ring R by simply specifying the ideal ofelements which is mapped onto zero.



Quotient Rings

Given a commutative ring R and any ideal I in R, we wantto construct a homomorphic image φ(R) such that I is thekernel of the morphism φ.

If φ : R → R′ is to be a morphism with kernel I then wemust have

φ(a) = φ(b) iff a− b ∈ I.

Therefore we can define the following congruence relationon R:

a ≡ b iff a− b ∈ I. (1)

This is an equivalence relation on R and hence it partitionR into equivalence classes, called residue classes.

For any element a ∈ R, the residue class containing a isthe set

a + I = {a + c|c ∈ I}.



The set of all residue classes of the congruence relation ≡defined by (1) is called a quotient set, denoted by

R/I = {a + I : a ∈ R}

(read as "R modulo the ideal I").

The operations of addition and multiplication on thequotient set R/I are defined in terms of the operationsdefined on R as follows:

(a + I) + (b + I) = (a + b) + I (2)

(a + I)(b + I) = (ab) + I (3)

The following theorem expresses that the quotient set R/Iwith the operations (2) and (3) is a commutative ring, andR/I is called the quotient ring of R modulo the ideal I.



Theorem (5.4)

Let R be a commutative ring and let I be an ideal in R. Thequotient set R/I is a commutative ring under the operationsdefined by (2) and (3), and the mapping φ : R → R/I defined by

φ(a) = a + I for all a ∈ R

is an epimorphism with kernel I.



Proof

The residue classes 0 + I and 1 + I are respectively thezero and identity in R/I because from (2) and (3) we have:(a + I) + (0 + I) = a + I for any a + I ∈ R/I;(a + I)(1 + I) = a + I for any a + I ∈ R/I.

It follows from (2) and (3) that for any a, b ∈ R,φ(a + b) = (a + b) + I = (a + I) + (b + I) = φ(a) + φ(b)andφ(ab) = (ab) + I = (a + I)(b + I) = φ(a)φ(b).Furthermore, by definition of φ, φ(1) = 1 + I.Thus, φ is a morphism according to Definition 5.2.By the definition of R/I, φ is surjective and hence anepimorphism.Since R/I is a homomorphic image of R, R/I is acommutative ring.



Proof (cont.)

Finally, we prove that the kernel of φ is precisely I asfollows:a ∈ I ⇒ φ(a) = a + I = 0 + Iandφ(a) = 0 + I ⇒ a + I = 0 + I ⇒ a− 0 ∈ I ⇒ a ∈ I.



Example

In the integral domain Z , 〈m〉 is an ideal, for some fixed m ∈ Z .Thus the quotient ring Z/〈m〉 is a homomorphic image of Z and〈m〉 is the kernel of the natural homomorphism φ : Z → Z/〈m〉.Assuming that m is positive, the elements of Z/〈m〉 are givenby:

Z/〈m〉 = {0 + 〈m〉, 1 + 〈m〉, . . . , (m − 1) + 〈m〉}.

Z/〈m〉 is usually denoted by Zm (the ring of integers modulo m)and its elements is denoted simply by {0, 1, . . . , m − 1}. Thenatural homomorphism is precisely the modularhomomorphism φm : Z → Zm defined previously.



Example

In the polynomial domain Q[x ], 〈p(x)〉 is an ideal, for somefixed polynomial p(x) ∈ Q[x ]. Thus the quotient ringQ[x ]/〈p(x)〉 is a homomorphic image of Q[x ] and 〈p(x)〉 isthe kernel of the natural homomorphismφ : Q[x ] → Q[x ]/〈p(x)〉. Two polynomials a(x), b(x) ∈ Q[x ]are in the same residue class if they have the sameremainder after division by p(x).

In particular, if p(x) = x − α for some constant α ∈ Q then

Q[x ]/〈x − α〉 = {r + 〈x − α〉|r ∈ Q}.

Q[x ]/〈x − α〉 can be identified with Q. In this case, thenatural homomorphism is precisely the evaluationhomomorphism φx−α : Q[x ] → Q defined previously.



In practical applications, the most commonly usedhomomorphism will be of the form

φ〈I,p〉 : Z [x1, . . . , xν ] → Zp[x1]

where I = 〈x2 − α2, . . . , xν − αν〉 with αi ∈ Zp and p is aprime integer.

The composite homomorphism φ〈I,p〉 is the composition ofa modular homomorphism

φp : Z [x1, . . . , xν ] → Zp[x1, . . . , xν ]

followed by a multivariate evaluation homomorphism

φI : Zp[x1, . . . , xν ] → Zp[x1, . . . , xν ]/I.



Congruence Arithmetic

Recall that if I is an ideal in a commutative ring R then theresidue classes which form the quotient ring R/I aredetermined by the congruence relation ≡ defined on R by

a ≡ b iff a− b ∈ I.

This relation is read as "a is congruent to b modulo I" anddenoted as a ≡ b (mod I).In particular, when I is a principal ideal 〈q〉 for some fixedelement q ∈ R, we write (mod q) rather than (mod 〈q〉).For any a, b, c, d ∈ R, if a ≡ b (mod I) and c ≡ d (mod I)then

a + c ≡ b + d (mod I) (4)

a− c ≡ b − d (mod I) (5)

ac ≡ bd (mod I) (6)



Theorem (5.6)

Let 〈q〉 be an ideal in a Euclidean domain D and let a ∈ D berelatively prime to q (i.e. gcd(a, q) = 1). Then there exists anelement a−1 ∈ D such that

a a−1 ≡ 1 (mod q).

Proof.

Since D is a Euclidean domain we can apply the extendedEuclidean algorithm (Algorithm 2.2) to a, q ∈ D yieldingelements s, t ∈ D such that sa + tq = 1, where we have usedthe fact that gcd(a, q) = 1. Then sa− 1 ∈ 〈q〉, or sa ≡ 1(mod q). Thus a−1 = s is the desired inverse.


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