10.5. THE EQUATION xn
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10.5. THE EQUATION xn b D 0
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10.5. The Equation xn b D 0
Consider the polynomial xn b 2 KŒx, where b ¤ 0, and n is relatively prime to the characteristic of K. The polynomial xn b has n distinct roots in a splitting field E, and the ratio of any two roots is an nth root of unity, so again E contains n distinct roots of unity and, therefore, a primitive nth root of unity u.
If a is one root of xn b in E, then all the roots are of the form uj a, and E D K.u; a/. We consider the extension in two stages K K.u/ E D K.u; a/.
A K.u/-automorphism of E is determined by .a/, and .a/ must be of the form .a/ D ui a.
Lemma 10.5.1. The map 7! .a/a 1 is an injective group homomorphism from AutK.u/.E/ into the cyclic group generated by u. In particular, AutK.u/.E/ is a cyclic group whose order divides n.
Proof. Exercise 10.5.1.
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We have proved the following proposition:
Proposition 10.5.2. If K contains a primitive nth root of unity (where n is necessarily relatively prime to the characteristic) and E is a splitting field of xn b 2 KŒx, then AutK.E/ is cyclic. Furthermore, E D K.a/, where a is any root in E of xn b.
Corollary 10.5.3. If K is a field, n is relatively prime to the characteristic of K, and E is a splitting field over K of xn b 2 KŒx, then AutK.E/ has a cyclic normal subgroup N such that AutK.E/=N is abelian.
Proof. As shown previously, E contains a primitive nth root of unity u, and if a is one root of xn b in E, then K K.u/ K.u; a/ D E, and the intermediate field K.u/ is a Galois over K. By Proposition N D AutK.u/.E/ is cyclic, and by the fundamental theorem, N is normal.
The quotient AutK.E/=N Š AutK.K.u// is a subgroup of ˚.n/ and, in particular, abelian, by Proposition 10.4.6.
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10. SOLVABILITY
Exercises 10.5 10.5.1. Prove Lemma 10.5.1.
10.5.2. Find the Galois group of x13 1 over Q. Find all intermediate fields in as explicit a form as possible.
10.5.3. Find the Galois group of x13 2 over Q. Find all intermediate fields in as explicit a form as possible.
10.5.4. Let n D pm1pm2
1
2
pms
s
. Let be a primitive nth root of unity in C and let i be a primitive pmi root of unity. Show that the Galois group
i of xn 1 over Q is isomorphic to the direct product of the Galois groups
mi of xpi 1, (1 i s). Show that each Q.i / is a subfield of Q./, the intersection of the Q.i / is Q, and the composite of all the Q.i / is Q./.
10.6. Solvability by Radicals Definition 10.6.1. A tower of fields K D K0 K1 Kr is called a radical tower if there is a sequence of elements ai 2 Ki such that K1 D K0.a1/; K2 D K1.a2/; : : : ; Kr D Kr 1.ar /, and there is a sequence of natural numbers ni such that ani
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2 Ki 1 for 1 i r.
We call r the length of the radical tower. An extension K L is called a radical extension if there is a radical tower K D K0 K1 Kr with Kr D L.
The idea behind this definition is that the elements of a radical exten sion L can be computed, starting with elements of K, by rational operations— that is, field operations— and by “extracting roots,” that is, by solving equations of the form xn D b.
We am aiming for the following result:
Theorem 10.6.2. (Galois) If K E L are field extensions with E Galois over K and L radical over K, then the Galois group AutK.E/ is a solvable group.
Lemma 10.6.3. If K L is a radical extension of K, then there is a radical extension L N L such that N L is Galois over K.
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10.6. SOLVABILITY BY RADICALS
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Proof. Suppose there is a radical tower K D K0 K1 Kr D L of length r . We prove the statement by induction on r . If r D 0, there is nothing to do. So suppose r 1 and there is a radical extension F Kr 1 such that F is Galois over K. There is an a 2 L and a natural number n such that L D Kr 1.a/, and b D an 2 Kr 1. Let p.x/ 2 KŒx be the minimal polynomial of b, and let f .x/ D p.xn/. Then a is a root of f .x/.
Let N L be a splitting field of f .x/ over F that contains F .a/. We have the following inclusions:
F
F .a/
N
L
[j [j
K Kr 1 L D Kr 1.a/: If g.x/ 2 KŒx is a polynomial such that F is a splitting field of g.x/ over K, then N L is a splitting field of g.x/f .x/ over K and, hence, N L is Galois over K.
Finally, for any root ˛ of f .x/, ˛n is a root of p.x/, and, therefore, ˛n 2 F . It follows that N L is a radical extension of F and, hence, of K.
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Lemma 10.6.4. Suppose K L is a field extension such that L is radical and Galois over K. Let K D K0 K1 Kr 1 Kr D L be a radical tower, and let ai 2 Ki and ni 2 N satisfy Ki D Ki 1.ai / and
ani
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2 Ki 1 for 1 i r. Let n D n1n2 nr , and suppose that K contains a primitive nth root of unity. Then the Galois group AutK.L/ is solvable.
Proof. We prove this by induction on the length r of the radical tower.
If r D 0, then K D L and the Galois group is feg. So suppose r 1, and suppose that the result holds for radical Galois extensions with radical towers of length less than r . It follows from this inductive hypothesis that the Galois group G1 D AutK .L/ is solvable.
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We have K1 D K.a1/ and an1
1
2 K. Since n1 divides n, K contains a primitive nth 1 root of unity. Then it follows from Proposition 10.5.2 that K1 is Galois over K with cyclic Galois group AutK.K1/.
By the fundamental theorem, G1 is normal in the Galois group G D AutK.L/, and G=G1 Š AutK.K1/. Since G1 is solvable and normal in G and the quotient G=G1 is cyclic, it follows that G is solvable.
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Proof of Theorem 10.6.2. Let K E L be field extensions, where E is Galois over K and L is radical over K. By Lemma 10.6.3, we can i
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10. SOLVABILITY
assume that L is also Galois over K. Since AutK.E/ is a quotient group of AutK.L/, it suffices to prove that AutK.L/ is solvable.
This has been done in Lemma 10.6.4 under the additional assumption that K contains certain roots of unity. The strategy will be to reduce to this case by introducing roots of unity.
Let K D K0 K1 Kr 1 Kr D L be a radical tower, and let ai 2 Ki and ni 2 N satisfy Ki D Ki 1.ai / and ani
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2 Ki 1 for 1 i r. Let n D n1n2 nr , and let be a primitive nth root of unity in an extension field of L.
Consider the following inclusions:
K./ L./ D K./ L [j [j
K K./ \ L L: By Proposition 9.5.6, L./ D K./ L is Galois over K./ and, fur thermore, AutK./.L.// Š AutK./\L.L/. But L./ is obtained from K./ by adjoining the elements ai , so L./ is radical over K./, and, furthermore, Lemma 10.6.4 is applicable to the extension K./ L./.
Hence, AutK./.L.// Š AutK./\L.L/ is solvable.
The extension K K./ is Galois with abelian Galois group, by Proposition 10.4.6. Therefore, every intermediate field is Galois over K with abelian Galois group, by the fundamental theorem. In particular, K./ \ L is Galois over K and AutK.K./ \ L/ is abelian, so solvable.
Write G D AutK.L/ and N D AutK./\L.L/. Since K./ \ L is Galois over K, by the fundamental theorem, N is normal in G and G=N Š AutK.K./ \ L/.
Since both N and G=N are solvable, so is G.
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Definition 10.6.5. Let K be a field and p.x/ 2 KŒx. We say that p(x) is solvable by radicals over K if there is a radical extension of K that contains a splitting field of p.x/ over K.
The idea of this definition is that the roots of p.x/ can be obtained, beginning with elements of K, by rational operations and by extraction of roots.
Corollary 10.6.6. Let p.x/ 2 KŒx, and let E be a splitting field of p.x/ over K. If p.x/ is solvable by radicals, then the Galois group AutK.E/ is solvable.
Corollary 10.6.7. (Abel, Galois) The general equation of degree n over a field K is not solvable by radicals if n 5.
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