By Pilgrim K. M.
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Proof. (a) The intersection of all subfields of K is a subfield of K. It is the smallest subfield of K, hence a prime field. (b) Let K be any field and K0 its prime field. Clearly, ޚK Â K0 . Now, in case A above, ޚK is itself already a subfield of K, so K0 D ޚK ' = ޚp ޚ. In case B we have K0 D Frac ޚK ' Frac ޚD ޑ. ˜ Remarks. (a) It is customary to write just n instead of nK , and we will do so. But you should keep an eye open in each case for whether the n represents an integer or an element of K.
Gauss’s Theorem 1. Let ˛ 2 ރbe an algebraic number. ˛/=( ޑChapter 1, F9). ˛/ W ޑequals the degree of the minimal polynomial of ˛ over K (Chapter 2, F3). ˛/ D 0, our task is to determine whether f is irreducible. If so, we have found our desired minimal polynomial — it is f (Chapter 3, F6). If not, we must continue the search by looking for irreducible factors of f . Example. Consider ˛ D e 2 X n i=n , for n > 1 a natural number. ˛/ D 0. Is f irreducible? X 1/ is a divisor of f . But if n is prime we will see in F11 that f is indeed irreducible.
Much of the content of this chapter is probably familiar to you from earlier courses. We nonetheless lay it out here because of its fundamental importance; in connection with the problems pursued up to now, we will be particularly interested in the question of irreducibility of polynomials. 1. Given elements a; b in R, we say that a is a divisor of b (or divides b, or that b is divisible by a) if there exists c in R such that b D ca. In this case we write a j b: (1) The negation of (1) is denoted by a - b: (2) The divisibility relation satisfies some obvious rules: (3) a j a (reflexivity); (4) a j b and b j c ) a j c (transitivity); (5) 1 j a; a j 0; (6) a j b and c j d ) ac j bd .
Canonical Decomposition Theorem by Pilgrim K. M.
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