Matrices [Exercises] by Marc Sage

By Marc Sage

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2. Let D = Z, the integers, p = 13. Then D mod p has 13 elements, which we may denote by 0, I, ... , 12. Then for example Let us find the inverse of 6. We apply Euclid's algorithm to 6 and 13 to find a linear combination of 6 and 13 equal to 1. We find that 6 . 11 - 13 . 5 = 1. , (6)- = 11. 3. Take D = Z again, but now let p be an arbitrary prime. The resulting important finite field has exactly p elements {O, I, ... , p - l}j it is commonly denoted by either of the two symbols Fp or GF(p}. This construction yields infinitely many finite fields, since there are infinitely many • primes.

6 we know that this polynomial will factor uniquely into a product of irreducible monic polynomials over k. The next theorem tells us something more about this factorization. 1. xqn - X = II Vd(X), din where Vd(X) is the product of all monic irreducible polynomials in k[x] of degree d. Proof: Let d be a divisor of n and let f(x) be a monic irreducible polynomial of degree dover k. Form the field F = k[x] (modf(x))j then F has qd elements. , a = x. 10, a qd = a, and this is equivalent to the statement d f(x) I (x q - x).

Let a k = ak+t be the first repeat in the sequence. Then clearly k = OJ otherwise a k- l = ak+t-l would be an earlier repeat. Thus (1, a, ... , at-I) are all distinct, but at = 1. The integer t ~ 1 is called the order of a. This number will in general be different for different values of a; and given an element a, it may be difficult to calculate t. However, it turns out that we can say exactly how many elements of each order t ~ 1 are contained in F. Our first step in this direction is a special case of a famous theorem of Lagrange.

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