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Section 14.6. Today We Discuss Galois Groups of Polynomials. Let Us Briefly Recall What We Know Already

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Section 14.6. Today We Discuss Galois Groups of Polynomials. Let Us Briefly Recall What We Know Already Section 14.6. Today we discuss Galois groups of polynomials. Let us briefly recall what we know already. If L=F is a Galois extension, then L is the splitting field of a separable polynomial f(x) 2 F [x]. If f(x) factors into a product of irreducible polynomials f(x) = f1(x) : : : fk(x) with degrees deg(fj) = nj, the Galois group Gal(L=F ) embeds into the following product of symmetric groups: Gal(L=F ) ⊂ Sn1 × · · · × Snk ⊂ Sn; where n = n1 + ··· + nk. Moreover, Gal(L=F ) acts transitively on the set of roots of each irreducible factor. The following two examples are also well-known to us by now. The Galois 2 2 group of (x − 2)(x − 3) is isomorphic to the Klein 4-group K4 ⊂ S4 and the Galois group 3 of x − 2 is isomorphic to S6. In what follows we shall describe Galois groups of polynomials of degrees less than or equal to 4, and prove that the Galois group of a \general" polynomial of degree n is isomorphic to Sn. Definition 1. Let x1; : : : ; xn be indeterminates. Define k-th elementary symmetric function by X ek = ek(x1; : : : ; xn) = xi1 : : : xin : 16i1<···<ik6n For example, we have e0(x1; x2; x3) = 1; e1(x1; x2; x3) = x1 + x2 + xn; e2(x1; x2; x3) = x1x2 + x1x3 + x2x3; e3(x1; x2; x3) = x1x2x3: Definition 2. Again, let x1; : : : ; xn be indeterminates. Then the general polynomial of degree n is defined as the product (x − x1) ::: (x − xn): It is easy to see that n X k n−k (x − x1) ::: (x − xn) = (−1) ek(x1; : : : ; xn)x k=0 With the above definition, we see that for any field F , the extension F (x1; : : : ; xn) is the splitting field for the general polynomial of degree n. We now make the permutation group Sn act on F (x1; : : : ; xn) by permuting the variables. It is clear that Sn preserves F (e1; : : : ; en), which shows that Sn is a subgroup of the automorphism group of F (x1; : : : ; xn) over F (e1; : : : ; en). On the other hand, the named automorphism group is itself a subgroup of Sn, therefore we have Aut(F (x1; : : : ; xn)=F (e1; : : : ; en)) ' Sn: Now, if the general polynomial was non-separable, it would force the order of the auto- morphism group Aut(F (x1; : : : ; xn)=F (e1; : : : ; en)) to be strictly less than n!. This allows us to conclude that the F (x1; : : : ; xn) is the splitting field of a separable polynomial with coefficients in F (e1; : : : ; en), and hence the extension F (x1; : : : ; xn)=F (e1; : : : ; en) is Galois. 1 2 Corollary 3. Any symmetric rational function f(x1; : : : ; xn) is a rational function in vari- ables e1; : : : ; en. Proof. By the Fundamental Theorem of Galois Theory, F (e1; : : : ; en) is precisely the Sn-fixed subfield of F (x1; : : : ; xn). Sn Remark 4. We have just proved that F (x1; : : : ; xn) = F (e1; : : : ; en). However, a stronger Sn statement is true: F [x1; : : : ; xn] = F [e1; : : : ; en]. Exercises in this chapter outline a boring proof of this statement. A more conceptual proof uses the notion of an integral extension. Check it out if you are curious! Now, let f(x) 2 F [x] be a polynomial of degree d with roots x1; : : : ; xd. We will say that f(x) is generic if its roots are algebraically independent, that is there is no such polynomial g(t1; : : : ; td) that g(x1; : : : ; xd) = 0. We can see that the Galois group of a generic polynomial f(x) is Sn, because there are no algebraic obstructions for realizing a permutation σ 2 Sn as an element of the Galois group. Let us now make the following two remarks. First, the Galois group of a polynomial over a finite field is cyclic (since every finite extension is cyclic), therefore, there are no generic polynomials over finite groups. Second, the condition that roots of f(x) are algebraically dependent is equivalent to the condition that its coefficients are algebraically dependent. We will prove one direction and leave the other as an exercise. Assume that the roots x1; : : : ; xd are dependent, that is there exists a polynomial g(t1; : : : ; td) such that g(x1; : : : ; xd) = 0. Define the polynomial Y g~(t1; : : : ; td) = g(tσ(1); : : : ; tσ(d)): σ2Sd Now one can see that the conditiong ~(x1; : : : ; xd) = 0 provides an algebraic dependence of the coefficients of f(x) which are nothing else but the elementary symmetric functions of the roots x1; : : : ; xd. Therefore, we can conclude that the general polynomial is separable over F (e1; : : : ; en). Recall that [Sn : An] = 2, and for n > 5 An is the unique normal subgroup of Sn. Therefore, for n > 5 there is a unique field K such that F (e1; : : : en) ⊂ K ⊂ F (x1; : : : xn) and K=F (e1; : : : en) is Galois. Moreover, K is an extension of degree 2. Definition 5. Define the discriminant of x1; : : : ; xn by Y 2 D = (xi − xj) : i<j The discriminant of a polynomial is defined to be the discriminant of its roots. Note the following properties of the discriminant which will be handy in what follows: (1) D 2 Fp[e1; : : : ; en], and if D is viewed as a discriminant of the polynomial f(x) 2 F (e1; : : : en)[x], then D = Q (x − x ) is contained in the splitting field of f. i<j i pj (2) A permutation σ fixes Dpif and only if σ 2 An. Therefore, given a field F with char(F ) 6= 2, the element D generates the fixed field of An, which is a quadratic extension of F (e1; : : : ; en). (3) Let D be the discriminant of a polynomial f(x) 2 F (e1; : : : en)[x]. Then f is separable if and only if D 6= 0. Therefore, over a perfect field f(x) is reducible if and only if D = 0. 3 (4) Assume that f(x) 2 F [x] is separable. Reformulating propertyp (2) above we see that the Galois group of f(x) is a subgroup of An if and only if D 2 F . Now, let us finally discuss Galois groups of (the splitting fields of) polynomials of degrees 2, 3, and 4. Polynomials of degree 2. Consider f(x) = x2 + ax + b with roots α and β. Then 2 2 2 2 2 D = (α − β) = (α + β) − 4αβ = e1 − 4e2 = (−a) − 4b = a − 4b: Clearly, f(x) is separable if and only if D 6= 0. Moreover, the Galois group is isomorphic to Z=2Z if D 6= 0 and is trivial otherwise. Polynomials of degree 3. Consider f(x) = x3 + ax2 + bx + c. Making a substitution x = y − a=3, the polynomial becomes g(y) = y3 + py + q where 3b − a2 2a3 − 9ab + 27c p = and q = : 3 27 The above substitution does not change the splitting field or the discriminant, so we shall only work with g(y) from now on. Let α; β; γ be its roots, then D = (α − β)(α − γ)(β − γ)2: As before, we can write out D in terms of the roots in a straightforward manner, and express the result in terms of coefficients of g(y). However, we are going to use the following trick, which will simplify the work a little: observe that D = −g0(α)g0(β)g0(γ): Using the fact that g0(y) = 3y2 + p; we can derive D = −4p3 − 27q2: Now, if g(y) is reducible over F , it either splits into three linear factors, in which case the Galois group is trivial, or into a linear and an irreducible quadratic factor. In the latter case, the Galois group is isomorphic to Z=2Z. If g(y) is irreducible then a root of g(y) generates an extension of degree 3, and therefore the degree of the splitting field is divisible by 3. Since the Galois group G is a subgroup of S3 we onlyp have two possibilities: G ' A3 ' Z=3Z or G ' S3. The former is realized if and only if D 2 F . Polynomials of degree 4. As before, we consider a polynomial f(x) = x4+ax3+bx2+cx+d, and make a substitution x = y − a=4. Then f(x) turns into g(y) = y4 + py2 + qy + r where −3a2 + 8b a3 − 4ab + 8c −3a4 + 16a2b − 64ac + 256d p = ; q = ; and r = : 8 8 256 Let the roots of g(y) be α1; : : : ; α4, and let G be the Galois group of the splitting field of g(y). If g(y) is reducible it either splits into a linear and a cubic, in which case G coincides with the Galois group of the cubic which we already considered, or into a pair of quadratics. 4 p p In the latter case the splitting field is the extension F ( D1; D2), where D1 and D2 are the discriminants of the corresponding quadratics. Then G is either isomorphic to the Klein p 4-group if D1=D2 62 F , or to Z=2Z otherwise. Now, we only need to consider the case when g(y) is irreducible. In that case, the Galois group G ⊂ S4 acts transitively on the roots. Inspecting the subgroups of S4 we conclude that the only subgroups with transitive action on the set of 4 elements are (1) S4 (2) A4 (3) Any of the 3 Sylow 2-subgroups, each isomorphic to D8 (4) K4, the Klein 4-group (5) any of the cyclic groups Z=4Z generated by the cycle of length 4 In order to express the discriminant in terms of the coefficients of g(y) we once again use a trick.
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