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Baker. Galois Theory, U Glasgow course(88s).pdf

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Example 1.4. The following rings are integral domains. (i) The ring of integers, Z. (ii) If p is a prime, the ring of integers modulo p, Fp = Z/p = Z/(p). (iii) The rings of rational numbers, Q, real numbers, R, and complex numbers, C. (iv) The polynomial rings Z[X ], Q[X ], R[X ] and C[X ]. Definition 1.5. Let I R be a proper ideal in a commutative ring R. • I is a prime ideal if for u, v ∈ R, uv ∈ I =⇒ u ∈ I or v ∈ I . Similarly, I is a maximal ideal R if whenever J R is a proper ideal and I ⊆ J then J = I. • I R is principal if I = (p) = {rp : r ∈ R} for some p ∈ R. Notice that if p, q ∈ R, then (q ) = (p) if and only if q = up for some unit u ∈ R. We also write p | x if x ∈ (p). • p ∈ R is prime if (p) R is a prime ideal; this is equivalent to the requirement that whenever p | xy with x, y ∈ R then p | x or p | y . • R is a principal ideal domain if it is an integral domain and every ideal I R is principal. The following fundamental example should be familiar.
1...

There are no factors of p appearing in (p − 1)!, j ! or (p − j )!, so since this number is an integer it must be divisible by p, i.e., p, (1.1a) p| j...

We need to verify that these operations are well deﬁned. For example, if [a , b ] = [a, b] and [c , d ] = [c, d], then (a d + b c )bd = a d bd + b c bd = ab d d + b bcd
=...

≡ (X − 1)p since by (1.1a), p divides p k when k = 1, . . . , p − 1. Hence Φp (X ) ≡ (X − 1)p−1 Also, p 1 = =...

In this section we work within the complex numbers and take k ⊆ C; in practice we will have k = R or k = C. For monic linear (degree 1) or quadratic (degree 2) polynomials, methods of ﬁnding roots are very familiar. Let us consider the cases of cubic (degree 3) and quartic (degree 4) polynomials....

Clearly ﬁnding the roots of f (X ) is equivalent to ﬁnding those of g (X ), so we may as well assume that we want to ﬁnd the complex roots of f (X ) = X 4 + pX 2 + q X + r ∈ C[X ]. Suppose that x is a root and introduce numbers y , z such that z = x2 + y (we will ﬁx the values of these later). Then z 2 = x4 + 2x2 y + y 2 = −px2 − q x − r + 2xy + y 2 = (2y − p)x2 − q x + y 2 − r. Now choose y to make the last quadratic expression in x a square, (1.15) (1.16) (2y − p)x2 − q x + (y 2 − r) = (Ax + B )2 . q 2 − 4(2y − p)(y 2 − r) = 0. This can be done by requiring the vanishing of the discriminant...

Definition 2.5. Given two extensions L/K and M /L, we say that L/K is a subextension of M /K and sometimes write L/K M/L. Theorem 2.6. Let L/K be a subextension of M /K . (i) If one or both of the dimensions [L : K ] or [M : L] is inﬁnite then so is [M : K ]. (ii) If the dimensions [L : K ] and [M : L] are both ﬁnite then so is [M : K ] and [M : K ] = [M : L] [L : K ].
23...

Let us ﬁnd a non-zero polynomial in ker ε√2+√3 Q[X]. √ √ √ Referring to Example 2.10 or Proposition 2.12 we see that 2 + 3 ∈ Q( 2), hence / √ √ degQ(√2) ( 2 + 3) = 2. √ One polynomial in ker ε√2+√3 Q( 2)[X ] is √ √ √ √ √ (X − ( 2 + 3))(X − ( 2 − 3)) = X 2 − 2 2X − 1. Since this is monic and of degree 2, Similarly, Consider √ minpolyQ(√2),√2+√3 (X ) = X 2 − 2 2X − 1....

3.5. MULTIPLICITY OF ROOTS AND SEPARABILITY

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Corollary 3.56. If f (X ) is irreducible in K [X ] then a root u is a multiple root if and only if f (X ) = 0. In particular, this can only happen if char K > 0. Corollary 3.57. If char K = 0 and f (X ) is irreducible in K [X ], then every root of f (X ) is simple. Example 3.58. For n E 1, show that each of the roots of f (X ) = X n − 1 in C is simple. f (ζ ) = nζ n−1 = 0. Solution. We have f (X ) = ∂ (X n − 1) = nX n−1 , so for any root ζ of f (X ), xample 3.59. Show that 2i is a multiple root of f (X ) = X 4 + 8X 2 + 16. Solution. We have f (X ) = 4X 3 + 16X . Using Long Division and the Euclidean Algorithm we ﬁnd that gcd(f (X ), f (X )) = X 2 + 4, where 2i is also a root of X 2 + 4. Hence 2i is a multiple rootE f f (X ). In fact, X 4 + 8X 2 + 16 = (X 2 + 4)2 , so this is obvious. o xample 3.60. Let p > 0 be a prime and suppose that L/Fp is an extension. Show each of that the roots of f (X ) = X p − 1 in L is multiple. f
(ζD= )

Solution. We have f (X ) = ∂ (X p − 1) = pX p−1 = 0, so if ζ is any root of f (X ) then 0. Later we will see that 1 is the only root of X p − 1...

ext we summarize the properties of Galois groups that can be deduced from what we have established so far. Recollection 4.6. Recall that an action of a group G on a set X is transitive if for every pair of elements x, y ∈ X , there is an element g ∈ G such that y = g x (so there is only one orbit); the action is faithful or eﬀective if for every non-identity element h ∈ G, there is a element z ∈ X such that hz = z . For an extension F /K and a polynomial f (X ) ∈ K [X ], recall that Roots(f , F ) denotes the set of roots of f (X ) in F . Theorem 4.7. Let E /K be a ﬁnite Galois extension. Suppose that E is the splitting ﬁeld of a separable irreducible polynomial f (X ) ∈ K [X ] of degree n. Then the fol lowing are true. (i) Gal(E /K ) acts transitively and faithful ly on Roots(f , E ). (ii) Gal(E /K ) can be identiﬁed with a subgroup of the group of permutations of Roots(f , E ). If we order the roots u1 , . . . , un then Gal(E /K ) can be identiﬁed with a subgroup of Sn . (iii) | Gal(E /K )| divides n! and is divisible by n. As we have seen in Examples 4.4 and 4.5, in practise it is often easier to use a not necessarily irreducible polynomial to determine and work with a Galois group. Example 4.8. The Galois extension Q(ζ8 )/Q has degree [Q(ζ8 ) : Q] = 4 and it has the following automorphisms apart from the identity:
3 α : ζ8 −→ ζ8 , 5 β : ζ8 −→ ζ8 , 7 γ : ζ8 −→ ζ8 ....

Notice that {id, (1 2 3), (1 3 2)} S3 and so Q(ζ3 )/Q is a normal extension. Of course Q(ζ3 ) is the splitting ﬁeld of X 3 − 1 over Q. 4.6. Galois extensions inside the complex numb ers and complex conjugation When working with Galois extensions contained in the complex numbers it is often useful to make use of complex conjugation as an element of a Galois group. Let E /Q be a ﬁnite Galois extension with E /Q C/Q. Setting ER = R ∩ E , we have Q ER E. Proposition 4.20. Complex conjugation ( ) : C −→ C restricts to an automorphism of E over Q, ( )E /Q : E −→ E . (i) ( )E /Q which agrees with the identity function if and only if ER = E . (ii) If ER = E then ( = )E /Q {id, ( )E /Q } ∼ Z/2, = ( )E/Q and [E : E ] = 2. hence E = E...