By George R. Kempf (auth.)

The legislation of composition contain addition and multiplication of numbers or func tions. those are the elemental operations of algebra. it is easy to generalize those operations to teams the place there's only one legislation. the idea of this ebook was once begun in 1800 by way of Gauss, while he solved the 2000 year-old Greek challenge approximately developing general n-gons via ruler and compass. the idea used to be additional built through Abel and Galois. After years of improvement the idea used to be installed the current shape by means of E. Noether and E. Artin in 1930. at the moment it was once known as smooth algebra and focused on the summary exposition of the idea. these days there are too many examples to enter their information. i believe the scholar may still examine the proofs of the theorems and never spend time trying to find strategies to difficult workouts. The workouts are designed to elucidate the idea. In algebra there are 4 easy buildings; teams, earrings, fields and modules. We current the idea of those uncomplicated constructions. expectantly this can provide an exceptional introduc tion to fashionable algebra. i've got assumed as heritage that the reader has realized linear algebra over the true numbers yet this isn't necessary.

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Let G be a splitting field of fg over F containing E. Then f = 7l"i(X - ai) and 9 = 7l"k(X - (3k) where these (3k are distinct elements of G and (31 = (3 and a1 = a. If k =f 1, the equation ai + x(3k = a + x(3 has a unique solution. As F is infinite we may find r in F such that ai + r(3k =f a + r(3 for all k =f 1. Let 8 = a + r(3. I claim this works. Now (3 satisfies g((3) = 0 and f(8 - r(3) = 0 but f(8 - r(3k) =f 0 for k =f 1. Thus x - (3 is the greatest common divisor of g(x) and f(8 - rx) in G[X].

So it = a I1O"EO(X - 0') = af where a E F. , the irreducible equation of e over F. 1. 0 The proof of this lemma contains interesting information not contained in the statement! Proof of Theorem We can now prove the only if part of the theorem. Let E = F(el, ... ,en). Let f be the product of the irreducible equations of the ei's over F. Then E is the splitting field of f by the calculation in the last proof. For the converse, assume that E is the splitting field of a separable polynomial in F[X].

We will do an induction on n. If n = 0 there is nothing to prove as FO = {a}. Therefore by induction on n we may assume that n ~ 1 and dimp Fn-l = n - 1. Now let el,"" em be a spanning subset of Fn. Now em = (A1, ... , An) where An =I- 0 for some j. Say by reordering, j = n. Consider the linear transformation L : F n -+ F n- l given by L(XI"'" xn) = (YI, ... , Yn-l) where Yi = Xi - Ai . \~-lxn' Then L is surjective as L(YI"'" Yn-l, 0) = (YI, ... , Yn-d. Therefore L(el)"" , L(em) span F n- l but L(e m ) = O.