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We now investigate some rather special rings. We start with a very familiar result.
The division algorithm for Z
If a, b ∈ Z with b ≠ 0 then ∃ q, r ∈ Z such that a = bq + r with |r| <|b|.
The element q is called the quotient and r is the remainder.
Proof
Assume a > b > 0. Subtract multiples of b from a until just before things go negative. The quotient q is the number of times you subtract and r = a - bq.
Remarks
The greatest common divisor (or highest common factor) of two integers a, b ∈ Z is the largest integer which divides them both.
Remarks
The Euclidean algorithm
If d is the gcd of a, b there are integers x, y such that d = ax + by.
Proof
Here is an example: Take a = 76, b = 32 :
As a consequence of this algorithm we can now prove:
Theorem
Every ideal of Z is principal.
Proof
Let I be a non-zero ideal. Choose the smallest (non-zero) element a in the ideal I. If b is any other element then the gcd(a, b) is in I and this must be either ±a. So b is a multiple of a.
We now prove similar results for another ring.
The division algorithm for R[x]
If a(x), b(x) ∈ R[x] with b(x) ≠ 0 then ∃ q(x), r(x) ∈ R[x] such that a(x) = b(x)q(x) + r(x) with either r(x) = 0 or deg(r(x)) < deg(b(x)).
Remarks
Proof
We use the (slightly less well-known) process of dividing one polynomial by another (of lower degree). It is just like long division.
Definition
The greatest common divisor of two polynomials a(x), b(x) ∈ R[x] is a polynomial of highest degree which divides them both.
Remarks
One can then prove:
The Euclidean algorithm for polynomials.
If d(x) is the gcd of a(x), b(x) there are polynomials p(x), q(x) such that d = a(x)p(x) + b(x)q(x).
Proof
Just the same as for Z -- except that the divisions are more tedious.
Remarks
In the calculating package Maple the integer gcd is implemented with igcd and the Euclidean algorithm with igcdex. For polynomials use gcd and gcdex.
The above result on principal ideals follows for this ring.
Every ideal of R[x] is principal.
Proof
Remarks
Definition
The ring of Gaussian integers is the subring {a + bi | a, b ∈ Z} of C. It is denoted Z[i].
We'll prove the division and Euclidean algorithms for this ring but first we have to decide when one Gaussian integer is bigger or smaller than another.
Definition
The norm (or length) of the Gaussain integer a + bi is a2 + b2. We will write it as N(a + bi).
Remarks
If u, v ∈ Z[i] with v ≠ 0 then ∃ q, r ∈ Z[i] such that u = vq + r with N(r) < N(v).
Remarks
First divide u by v in C. You get a complex number u /v which lies in one of the "cells of the integer lattice".
For most positions of u/v there will be a choice of 2, 3 or even four Gaussian integers for the quotient.
The greatest common divisor of two Gaussian integers u, v ∈ Z[i] is a Gaussian integer of largest norm which divides them both.
Remarks
The Euclidean algorithm for Gaussian integers.
If w is the gcd of u, v there are Gaussian integers g, h such that w = ug + vh.
and then we can deduce:
Every ideal of Z[i] is principal.
Remark
A ring in which one can define a sensible notion of size which leads to a Euclidean algorithm is called a Euclidean ring.
Definition
We can use the division algorithm to prove
In general, use the procedure: divide (say) a by b to get remainder r1. Note that one can write r1 in terms of a and b.
Then divide b by r1 to get remainder r2. Note that one can write r2 in terms of b and r1 and hence in terms of a and b.
Repeat the process getting smaller and smaller remainders r1 , r2 , ... which must eventually lead to a remainder of 0.
At each stage ri can be expressed in terms of a and b. the last non-zero remainder d divides all the previous ones and hence both a and b. Hence it divides gcd(a, b). Since d can be written in terms of a and b the gcd(a, b) divides d and must therefore be ±d.
Thus I = < a > .
One example will suffice!
Take a(x) = 3x4 + 2x3 + x2 - 4x + 1 and b = x2 + x + 1
Just as in the Z case, an ideal is generated by its smallest (in degree) element.
Now let us prove the same result for a completely different ring.
The division algorithm for Z[i]
Proof
This proof is very different to the other proofs above.
At least one corner of the square is within distance 1 of u/v. This is the quotient q. The remainder is then r = u - vq and since |u/v - q| < 1 we have |u - qv| < |v| and so N(r) < N(v) as required.
Remark
Definition
Then we can calculate just as before and prove:
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