rings lect 3

8/6/2019 Rings Lect 3

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3 The Chinese Remainder Theorem

3.1 Introduction

In the first year course we learned about the Division Algorithm and Euclids Algorithm for

both the ring of integers Z and polynomial rings over fields, K. We are going to slightlyextend these results, proving one of the most versatile theorems of algebra, the ChineseRemainder Theorem.

3.2 Abstract version

Although one of the most important things about the CRT is its efficiency and practical use,we start quite abstractly. This deals with rather dull technicalities once and for all, so thatwhen we come to concrete versions we can concentrate on what is interesting.

Lemma 3.2.1. Let R, S1, S2 be rings and let f1 : R S1 and f2 : R S2 be homomor-phisms. Then f : R S1 S2 by f : a (f1(a), f2(b)) is a homomorphism whose kernel isker f1 ker f2.

Proof. As the operations on S1 S2 are defined coordinatewise it is trivial to see that f isa homomorphism: for example,

f(a + a) =

f1(a + a), f2(a + a

)

=

f1(a) + f1(a), f2(a) + f2(a

)

= (f1(a), f2(a)) +

f1(a), f2(a

)

= f(a) + f(a).

The kernel of f is

ker f = {a | f(a) = 0} = {a | (f1(a), f2(a)) = 0 = (0, 0)} = {a | f1(a) = 0, f2(a) = 0} = ker f1ker f2.

Lemma 3.2.2. Let R be a ring, and let I1, I2 R be ideals such that 1 I1 + I2. Then themap : R R/I1 R/I2 given by : a (a + I1, a + I2) is an onto homomorphism.

Proof. By the previous lemma applied to the natural homomorphisms R R/I1 and R R/I2 we have that is a homomorphism. We must prove it is onto, so let (a1 + I1, a2 + I2)

be an arbitrary member of the codomain. By hypotheses we know that for some i1 I1 andi2 I2 we have that 1 = i1 + i2. Consider (key step) the element x = a2i1 + a1i2. Then wehave that

x + I1 = a2i1 + a1i2 + I1 = a1i2 + I1 = a1(1 i1) + I1 = a1 + I1

as a2i1 and a1i1 lie in the ideal I1. A similar argument deals with the coset modulo I2, andwe get that x has the required image.

We now have:

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Theorem 3.2.3 (Abstract CRT). Let R be a ring, and let I1, I2 R be ideals such thatI1 + I2 = R. Then

R/(I1 I2) = R/I1 R/I2

where the isomorphism is the natural x + (I1 I2) (x + I1, x + I2).

Proof. We consider the map : R R/I1R/I2 given by a (a + I1, a + I2). By the firstlemma this is a homomorphism with kernel I1 I2. As R = I1 + I2 we have that 1 I1 + I2so we can use the second lemma to get that is onto. Now apply the Isomorphism Theoremto and get the result.

3.3 The CRT for Z

Originally the Chinese Remainder Theorem was the proposition that one could, if a, b arecoprime integers, and r, s are any integers, find a solution to the simultaneous congruences

x r (mod a), x s (mod b).

That is only part of what we are now able to prove.

Theorem 3.3.1 (CRT for integers). Let a, b Z have highest common factor 1. Then

Z/abZ = Z/aZ Z/bZ,

the isomorphism being the natural x + abZ (x + aZ, x + bZ).

Proof. By Euclids Algorithm we can find R, S such that 1 = Ra + Sb, so that Z = aZ+ bZ.Also, we have that lcm(a, b) = ab/ hcf(a, b) = ab, so that aZ bZ = abZ. The theorem nowfollows from the abstract version.

3.4 Applications

3.4.1 Simultaneous congruences

The result about simultaneous congruences, that if a, b are coprime integers, and r, s are anyintegers, then we can find a solution to the simultaneous congruences

x r (mod a), x s (mod b).

is an easy corollary of our CRT. For consider the pair of cosets (r + aZ, s + bZ); by the CRTthere exists a unique coset x + abZ such that (x + aZ, x + bZ) = (r + aZ, s + bZ).

Although fine in theory, this is not yet of practical use. How do we find the solutionx? We have answered this implicitly. When we proved the ontoness of the map the keyconsideration was that we could express 1 = i1 + i2 with i1 I1 and i2 I2; then we got xas si1 + ri2.

So in practice, we use the (extended) Euclid Algorithma very efficient processtocalculate the integers R and S such that 1 = aR + bS; and the solution we then seek is, asper the proof of our abstract theorem, x = aRs + bSr.

3.4.2 Eulers function

See problems sheet for an application.

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3.4.3 Speeding up Arithmetic

Suppose again that a, b Z have highest common factor 1. By the CRT we have that

Z/abZ = Z/aZ Z/bZ.

We can then carry out arithmetic calculations in Z/abZ in the following way:

Step 1 For each input x := x + abZ calculate x := x + aZ and x := x + bZ.

Step 2 Carry out the required calculations on the x in Za to get the answer y; and on thex in Zb to get y.

Step 3 Using the inverse of the isomorphism, calculate the value of y which maps to thepair (x, x).

Can this possibly be a good idea? The answer is a resounding Yes! Essentially we have

some setup costs we must pay once and for all: the calculation of integers R and S such thatRa + Sb = 1 which allow us to compute the inverse isomorphism. This is not very expensive:Euclids algorithm is very fast, requiring O(log a) steps. The reductions of step 1 are notexpensive, although an extra overhead. But the savings come in step 2: although we haveto carry out the calculations twice we do so in much smaller systems. If we make a carefulanalysis well see that this really works.

3.5 The CRT for K[X]

If we look at our proof of the CRT for Z we will see that all we needed to use about the ringZ and the elements a, b with hcf(a, b) = 1 were these facts, both consequences of EuclidsAlgorithm:

(i) 1 = Ra + Sb for some R, S;

(ii) aZ bZ = abZ.

We have the Division Algorithm, highest common factors, and Euclids Algorithm in thering K[X], where K is any field; we saw this in the first-year course. Therefore we have, withno more work to be done:

Theorem 3.5.1 (CRT for K[X] ). LetK be a field and let f(X), g(X) K[X] have highestcommon factor 1. Then

K[X]/f(X)g(X)K[X] = K[X]/f(X)K[X] K[X]/g(X)K[X],

the isomorphism being the natural t + f(X)g(X)K[X] (t + f(X)K[X], t + f(X)K[X]).

3.6 An application

There would be no point in this, of course, if there were not important applications.

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3.6.1 Interpolation

Suppose we have k + 1 distinct members ai of a field K; and k + 1 arbitrary members bi ofK.

We can apply the CRT first to the polynomials (X a1) and i>1(X ai), and theninductively and get

K[X]/

(X ai)K[X] =

K[X]/(X ai)K[X].

Let us look for a moment at K[X]/(X a)K[X]. What does a coset b + (X a)K[X]represent? It is, as the Remainder Theorem tells us, the set of all polynomials which takethe value b at the point a.

So what the CRT tells us in part is this: there is a unique coset of polynomials

t(X) +

(X ai)K[X]

consisting of those polynomials which take the values bi at the points ai. By the DivisionAlgorithm we can then find in the coset a unique t(X) of degree at most k with this property.Once again, note that the CRT is actually constructive: Euclids Algorithm lets us

compute the inverse of the isomorphism efficiently, and find t(X) from the data (ai, bi),i = 1, . . . , k + 1.

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rings lect 3

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