Galois Theory, Fourth Edition
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Since 1973, Galois Theory has been educating undergraduate students on Galois groups and classical Galois theory. In Galois Theory, Fourth Edition, mathematician and popular science author Ian Stewart updates this well-established textbook for today’s algebra students.
New to the Fourth Edition
- The replacement of the topological proof of the fundamental theorem of algebra with a simple and plausible result from point-set topology and estimates that will be familiar to anyone who has taken a first course in analysis
- Revised chapter on ruler-and-compass constructions that results in a more elegant theory and simpler proofs
- A section on constructions using an angle-trisector since it is an intriguing and direct application of the methods developed
- A new chapter that takes a retrospective look at what Galois actually did compared to what many assume he did
- Updated references
This bestseller continues to deliver a rigorous yet engaging treatment of the subject while keeping pace with current educational requirements. More than 200 exercises and a wealth of historical notes augment the proofs, formulas, and theorems.
defining a special kind of automorphism. DEFINITION 8.1 Let L:K be a field extension, so that K is a subfield of the subfield L of A K-automorphism of L is an automorphism α of L such that 94 Galois Theory (8.1) We say that α fixes if (8.1) holds. Effectively, condition (8.1) makes α an automorphism of the extension L:K, rather than an automorphism of the large field L alone. The idea of considering automorphisms of a mathematical object relative to a subobject is a useful general method; it
contrary to Lemma 1. Therefore, m≥n. 2. Next, suppose for a contradiction that Then there exists a set of n+1 elements of K that are linearly independent over K0; let such a set be By Lemma 10.3 there exist not all zero, such that for j=1,…, n (10.7) We shall subject this equation to a combinatorial attack, similar to that used in proving Lemma 10.1. Choose y1,…, yn+1 so that as few as possible are nonzero, and renumber so that Equation (10.7) now becomes (10.8) Let and operate on (10.8) with
such and since we may divide by k, which shows that the xi are linearly dependent over K0. This is a contradiction. Therefore, [K:K0] is not greater than n, so by the ﬁrst part of the proof, [K:K0]=n=|G| as required. COROLLARY 10.6 If G is the Galois group of the ﬁnite extension L:K, and H is a ﬁnite subgroup of G, then PROOF By the tower law, But this equals [L:K]/|H| by Theorem 10.5. so Example 10.7 We illustrate Theorem 10.5 by two examples, one simple, the other more intricate. 1. Let G
are the conjugacy classes of G. 154 Galois Theory If the conjugacy classes of G are C1,…, Cr, then one of them, say C1, contains only the identity element of G. Therefore, |C1|=1. Since the conjugacy classes form a partition of G we have (14.2) which is the class equation for G. then the centralizer CG(x) of x in G is the DEFINITION 14.10 If G is a group and set of all for which xg=gx. It is always a subgroup of G. There is a useful connection between centralizers and conjugacy classes. LEMMA
divide m. Let b be in A but not in M, and let B be the cyclic subgroup generated by b. Then MB is a subgroup of A, larger than M, so by maximality A=MB. From the First Isomorphism Theorem so that p divides the order r of B. Since B is cyclic, the element br/p has order p. The lemma follows. From this result we can derive a more general theorem of Cauchy. Solubility and Simplicity 155 THEOREM 14.15 (Cauchy’s Theorem) If a prime p divides the order of a ﬁnite group G, then G has an element of