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    6 The Euclidean Algorithm

    In this section we revisit the Euclidean Algorithm we learned in the first year, and discusshow far it applies in general Euclidean Domains. We will see that the main part of the

    algorithm is still in place.

    6.1 Algorithms

    Sometimes we dont want simple assertions about the existence of a mathematical object, notmatter how carefully theyve been proved: we want a way to find it in practice. In the firstyear courses we met several algorithms which allowed us to find or construct things: explicitprocesses which can be carried out in a finite number of steps, and which are guaranteed todeliver the object we seek.

    For examplebut there are many others if you look closelywe have: the GaussianElimination process provides algorithms for finding all solutions of a set of linear equations(and hence inverting matrices where possible); the Division Algorithm for integers, and

    for polynomials with rational coefficients, allows us to calculate a particular q, r such thata = bq + r; and the Extended Euclidean Algorithm lets us calculate hcf(x, y) and express itas Xx + Y y.

    Some interesting things, though, are beyond our reach: theres no effective way to calcu-late the decimal expansion of a/b where a, b R are given by their decimal expansions.

    6.2 The Division Algorithm for K[X]

    Given f, g K[X] with g = 0 we can find algorithmically q, r K[X] such that f = gq + rwhere r = 0 or deg r < deg g: provided we can somehow carry out the arithmeticoperations in K. There is no way to do this in general, so in this subsection we assume

    that we have an Oracle which will on demand deliver to us 0, x, x + y, 1, xy, and (ify = 0)x/y.

    Provided with this Oracle all we need to do is mimic the steps of the Division Algorithmwe learned for Q[X].

    It is clear what to do if deg f < deg g; just take q = 0 and r = f.Otherwise suppose that f = aXM+lower degree terms and g = bXN+lower degree terms;

    here a, b are nonzero and M N 0. Consider f(X) (a/b)XNMg(X); the oracle helpsus with the coefficients of this polynomial. This is a polynomial of degree less than deg f weuse recursion, and find q, r such that

    f(X) (a/b)XNMg(X) = g(X)q(X) + r(X) where r(X) = 0 or deg r < deg g.

    Now put r(X) := r(X) and q(X) := g(X) + (a/b)XNMg(X), where again the Oracle helpsus with the latter calculation.

    It is easy to prove by induction that this process works.

    6.3 Euclids Algorithm in Euclidean Domains

    Let R equipped with d : R \ {0} N \ {0} be a Euclidean Domain.Note that for a, b R this merely asserts the existence of a q, r such that a = bq + r

    with r = 0 or d(r) < d(b. Weve seen above that sometimes there are algorithms to calculatethese, although we may need an Oracle or Giant Look-up Table to help. Sometimes there

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    may not be. So throughout this subsection assume that there is an Oracle which will deliveron demand a suitable q and r.

    In that case, the Extended Euclidean Algorithm we learned in the first-year course willcalculate hcf(x, y) and express it as Xx + Y y.

    Proposition 6.3.1 (Extended Euclidean Algorithm). Let R equipped with d be a EuclideanDomain, and suppose that for any x, y = 0 we are given a definite q(x, y) and r(x, y) suchthat x = yq(x, y) + r(x, y) where r(x, y) = 0 or d(r(x, y)) < d(y).

    For every pair of elements a, b R, not both 0, proceed as follows:

    1. (a) leta0 := a, m0 := 1, and n0 := 0;

    (b) let a1 := b, m1 := 0, and n1 := 1;

    (c) let i := 1;

    2. while ai = 0 repeat the following steps:

    (a) let qi+1 := q(ai1, ai);(b) i. let ai+1 := ai1 qi+1ai;

    ii. let mi+1 := mi1 qi+1mi;

    iii. let ni+1 := ni1 qi+1ni;

    (c) increase i to i + 1;

    3. (a) letd = ai1;

    (b) let m = mi1;

    (c) let n = ni1.

    Then the following are true:

    1. the process stops after a finite number of steps;

    2. every divisor of both a and b also divides d;

    3. d divides both a and b;

    4. d = ma + nb.

    Proof. The proof is identical to the one for integers.

    6.4 Chinese Remainder Theorem

    A final remark is in order. The Extended Euclidean Algorithm is exactly what we need tocalculate the inverse of the isomorphism R/xyR = R/xR R/yR of the Chinese RemainderTheorem.

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