sign in sign up
<<
Home , Galois group, Galois theory, Generic polynomial, Separable polynomial

Section 14.6. Today we discuss Galois groups of . Let us briefly recall what we know already. If L/F is a , then L is the splitting field of a separable f(x) ∈ 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 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 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 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 by X ek = ek(x1, . . . , xn) = xi1 . . . xin . 16i1<···

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 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 group of F (x1, . . . , xn) over F (e1, . . . , en). On the other hand, the named 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 with coefficients in F (e1, . . . , en), and hence the extension F (x1, . . . , xn)/F (e1, . . . , en) is Galois. 1 2

Corollary 3. Any symmetric f(x1, . . . , xn) is a rational function in vari- ables e1, . . . , en.

Proof. By the Fundamental Theorem of , 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) ∈ 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 f(x) is Sn, because there are no algebraic obstructions for realizing a permutation σ ∈ 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)).

σ∈Sd

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 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 of x1, . . . , xn by Y 2 D = (xi − xj) . i

(4) Assume that f(x) ∈ F [x] is separable. Reformulating property√ (2) above we see that the Galois group of f(x) is a subgroup of An if and only if D ∈ 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 only√ have two possibilities: G ' A3 ' Z/3Z or G ' S3. The former is realized if and only if D ∈ 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 √ √ In the latter case the splitting field is the extension F ( D1, D2), where D1 and D2 are the of the corresponding quadratics. Then G is either isomorphic to the Klein p 4-group if D1/D2 6∈ 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. Namely, we consider elements

θ1 = (α1 + α2)(α3 + α4), θ1 = (α1 + α3)(α2 + α4), θ1 = (α1 + α4)(α2 + α3). Each of them is fixed by a separate Sylow 2-subgroup, and all 3 of them are fixed by the Klein 4-group, which is precisely the intersection of all Sylow 2-subgroups in S4. It is also clear that S4 permutes elements θj, therefore

h(x) = (x − θ1)(x − θ2)(x − θ3) ∈ F [x]. Moreover, it is rather easy to compute that h(x) = x3 − 2px2 + (p2 − 4r)x + q2. This polynomial is called the resolvent cubic for the quartic g(y). It is also easy to see that the discriminants of g(y) and h(x) coincide, which allows us to compute D = 16p4r − 4p3q2 − 128p2r2 + 144pq2r − 27q4 + 256r3. Let us also notice that the splitting field of the resolvent cubic is a subfield in that of g(y), so the Galois group of the former is a quotient of G. √ We are finally ready to explore the Galois group G. Suppose first that h(x) is irreducible. If D 6∈ F , then G 6⊂ A4, and the Galois group of h(x) is S3, which implies that |G| is divisible √by 6. This leaves us with the only option G ' S4. If the resolvent cubic is irreducible but D ∈ F , then G ⊂ A4 and |G| is divisible by 3, the order of the Galois group of h(x). Therefore G = A4. Now, it is only left to consider cases when the cubic is reducible. If it splits completely, then G fixes all of θj which implies G ⊂ K4 and therefore G = K4. Otherwise, if the cubic splits into a linear and a quadratic, then precisely one of θj lies in F , hence G ⊂ D8 and G 6⊂ K4. This leaves us with 2 possibilities: G ' D8 or G ' Z/4Z. To distinguish between the 2, note that D8 ∩A4 ' K4 while Z/4Z∩A4 ' Z/2Z, the former group is√ transitive on the roots of g(y) and the latter is not. Recall also that the fixed√ field of A4 is F ( D). Therefore, we can determine G completely by factoring g(y) over F ( D). Fundamental theorem of . We finish this lecture by proving the Fundamental Theorem of Algebra.

Theorem 6. Every polynomial f(x) ∈ C[x] of degree n has precisely n roots in C (counted with multiplicities). Equivalently, C is algebraically closed. Proof. We shall use the following two statements: 5

(1) Any polynomial g(x) ∈ R[x] of odd degree has a root in R. Equivalently, there are no non-trivial extensions of R of odd degree. (2) Any polynomial h(x) ∈ C[x] of degree 2 has roots in C. Equivalently, there are no quadratic extensions of C. Now, let f(x) ∈ R[x] be a polynomial of degree n, and K be its splitting field. Then K(i) 2 is the splitting field for f(x)(x + 1), hence is Galois over R. Let G = Gal(K(i)/R), and P P2 ⊂ G be its Sylow 2-subgroup. Then the fixed field K(i) 2 is an odd-degree extension over R, hence is trivial. Therefore, G is a 2-group and so is Gal(K(i)/C). But if the latter group is non-trivial, it would have a subgroup of index 2 which would imply the existence of a quadratic extension of C. Therefore Gal(K(i)/C) is trivial and K(i) = C.