10.7. RADICAL EXTENSIONS
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Proof. According to Theorem 9.7.1, the Galois group of the general equation of degree n is Sn, and it follows from Theorem 10.3.4 that Sn is not solvable when n 5.
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This result implies that there can be no analogue of the quadratic (and cubic and quartic) formula for the general polynomial equation of degree 5 or higher. A general method for solving a degree n equation over K is a method of solving the general equation of degree n, of finding the roots ui in terms of the variables ti . Such a method can be specialized to any particular equation with coefficients in K. The procedures for solving equations of degrees 2, 3, and 4 are general methods of obtaining the roots in terms of rational operations and extractions of radicals. If such a method existed for equations of degree n 5, then the general equation of degree n would have to be solvable by radicals over the field K.t1; : : : ; tn/, which is not so.
10.7. Radical Extensions
This section can be omitted without loss of continuity.
In this section, we will obtain a partial converse to the results of the previous section: If K E is a Galois extension with a solvable Galois group, then E is contained in a radical extension. I will prove this only in the case that the ground field has characteristic 0. This restriction is not essential but is made to avoid technicalities. All the fields in this section are assumed to have characteristic 0.
We begin with a converse to Proposition 10.5.2.
Proposition 10.7.1. Suppose the field K contains a primitive nth root of unity. Let E be a Galois extension of K such that AutK.E/ is cyclic of order n. Then E is the splitting field of an irreducible polynomial xn b 2 KŒx, and E D K.a/, where a is any root of xn b in E.
The proof of this is subtle and requires some preliminary apparatus.
Let K E be a Galois extension. Recall that E denotes the set of nonzero elements of E.
Definition 10.7.2. A function f W AutK.E/ ! E is called a multiplicative 1-cocycle if it satisfies f . / D f ./.f .//.
The basic example of a multiplicative 1-cocycle is the following: If a is any nonzero element of E, then the function g. / D .a/a 1 is a multiplicative 1-cocycle (exercise).
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Proposition 10.7.3. If K E is a Galois extension, and f W AutK.E/ !
E is a multiplicative 1-cocycle, then there is an element a 2 E such that f . / D .a/a 1.
Proof. The elements of the Galois group are linearly independent, by Proposition 9.5.7, so there is an element b 2 E such that
X
f . / .b/ ¤ 0: 2AutK .E/
Call this nonzero element a 1. In the following computation, the 1-cocycle relation is used in the form .f . // D f ./ 1f ./. Now, for any 2 AutK.E/,
0
1
X .a 1/ D @ f . / .b/A
2AutK .E/
X
D
.f . // .b/
2AutK .E/
X
D
f . / 1f . / .b/
2AutK .E/
D f ./ 1
X
f . / .b/
2AutK .E/
D f ./ 1a 1:
This gives f . / D .a/a 1.
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Definition 10.7.4. Let K E be a Galois extension. For a 2 E, the norm of a is defined by
Y N.a/ D N E K .a/ D .a/:
2AutK .E/ Note that N.a/ is fixed by all 2 AutK.E/, so N.a/ 2 K (because E is a Galois extension). For a 2 K, N.a/ D adimK.E/.
Proposition 10.7.5. (D. Hilbert’s theorem 90). Let K E be a Galois extension with cyclic Galois group. Let be a generator of AutK.E/.
If b 2 E satisfies N.b/ D 1, then there is an a 2 E such that b D .a/a 1.
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Proof. Let n be the order of the cyclic group AutK.E/. Define the map f W AutK.E/ ! E by f .id/ D 1, f ./ D b, and f . i / D i 1.b/ .b/b; for i 1.
We can check that N.b/ D 1 implies that f .iCn/ D f .i /, so f is well–defined on AutK.E/ and, furthermore, that f is a multiplicative 1-cocycle (Exercise 10.7.2).
Because f is a multiplicative 1-cocycle, it follows from the previous proposition that there is an a 2 E such that b D f ./ D .a/a 1.
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Proof of Proposition 10.7.1.
Suppose that AutK.E/ is cyclic of order n, that is a generator of the Galois group, and that 2 K is a primitive nth root of unity. Because 2 K, its norm satisfies N./ D n D 1. Hence, by Proposition 10.7.5, there is an element a 2 E such that D .a/a 1, or .a/ D a. Let b D an; we have .b/ D .a/n D .a/n D an D b, so b is fixed by AutK.E/, and, therefore, b 2 K, since E is Galois over K.
The elements i .a/ D i a, 0 i n 1 are distinct roots of xn b in E, and AutK.E/ acts transitively on these roots, so xn b D Qn 1
i D0 .x i a/ is irreducible in KŒx. (In fact, if f is an irreducible factor of xn b in KŒx, and i a is one root of f , then for all j , we have that j i .i a/ D j a is also a root of f ; hence, deg f n and f .x/ D xn b.) Since dimK.K.a// D n D dimK.E/, we have K.a/ D E.
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Theorem 10.7.6. Suppose K E is a Galois field extension. If the Galois group AutK.E/ is solvable, then there is a radical extension L of K such that K E L.
Proof. Let n D dimK.E/, and let be a primitive nth root of unity in a field extension of E. Consider the extensions:
K./ E./ D K./ E [j [j
K K./ \ E E: By Proposition 9.5.6, E./ is Galois over K./ with Galois group AutK./.E.// Š AutK./\E .E/ AutK.E/: Therefore, AutK./.E.// is solvable.
Let G D G0 D AutK./.E.//, and let G0 G1 Gr D feg be a composition series of G with cyclic quotients. Define Ki D Fix.Gi / for 0 i r; thus K./ D K0 K1 Kr D E./:
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10. SOLVABILITY
By the fundamental theorem, each extension Ki 1 Ki is Galois with Galois group AutK
.K
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1
i / Š Gi 1=Gi , which is cyclic. Since Ki 1 contains a primitive d th root of unity, where d D dimK
.K
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1
i /, it fol lows from Proposition 10.5.2 that Ki D Ki 1.ai /, where ai satisfies an irreducible polynomial xd bi 2 Ki 1Œx.
Therefore, E./ is a radical extension of K./. Since also K./ is a radical extension of K, E./ is a radical extension of K containing E, as required.
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Corollary 10.7.7. If E is the splitting field of a polynomial p.x/ 2 KŒx, and the Galois group AutK.E/ is solvable, then p.x/ is solvable by radicals.
Exercises 10.7 10.7.1. If a 2 E, then the function g./ D .a/a 1 is a multiplicative 1-cocycle.
10.7.2. With notation as in the proof of 10.7.5, check that if N.b/ D 1, then f is well–defined on AutK.E/ and f is a multiplicative 1-cocycle.
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