Dec. 1, 2006
Remark 3.6.8. Some of the result we proved still true in a more gen- eral setting. We list some here:
(1) If F/K is an extension, and an intermediate field E is stable, then E0C GalF/K.
(2) Let F/K be an extension. If N C GalF/K, then H0 is stable.
(3) If F/K is Galois, and E is a stable intermediate field, then E is Galois over K. (finite-dimensional assumption is unnecessary here)
(4) An intermediate field E is algebraic and Galois over K, then E is stable.
We conclude this section with the following theorem concerning the relation between Galois extension, normal extension and splitting fields.
Definition 3.6.9. An irreducible polynomial f (x) ∈ K[x] is said to be separable if its roots are all distinct in K.
Let F be an extension over K and u ∈ F is algebraic over K. Then u is separable over K if its minimal polynomial is separable.
An extesnion F over K is separable if every element of F is separable over K.
Theorem 3.6.10. Let F/K be an extension, then the following are equivalent
(1) F is algebraic and Galois over K.
(2) F is separable over K and F is a splitting field over K of a set S of polynomials.
(3) F is a splitting field of separable polynomials in K[X].
(4) F/K is normal and separable.
Proof. Fix u ∈ F with minimal polynomail p(x) over K. Let {u = u1, ..., ur} be distinct roots of p(x) in F . For any σ, then σ permutes {u = u1, ..., ur}. Thus f (x) := Qr
i=1(x − ui) is invariant under σ.
Hence f (x) ∈ K[x]. It follows that f (x) = p(x). This proved that (1) ⇒ (2), (3), (4).
One notices that (2) ⇔ (4). Thus it remains to show that (2) ⇒ (3), and (3) ⇒ (1).
For (2) ⇒ (3), let f (x) ∈ S and let g(x) be an monic irreducible component of f (x). Since f (x) splits in F , it’s clear that g(x) is an minimal polynomial of some element in F . Moreover, since F/K is separable, g(x) is separable. One sees that F is in fact a splitting field of such g(x)’s.
For (3) ⇒ (1), we first note that F/K is algebraic since F is a split- ting field. We shall prove that (4) ⇒ (1). The implication (3) → (4) follows from a general fact about separable extension that an algebraic extension F/K is separable if F is generated by separable elements.
To this end, pick any u ∈ F − K, with minimal polynomial p(x) of degree ≥ 2 and separable. Hence there is a different root, say v, of p(x) in F . It’s natural to consider the K-isomorphism σ : K(u) → K(v).
Which can be extended to ¯σ : F → K. Since F is normal, ¯σ is an automorphism of F , hence in GalF/K sending u to v 6= u. So F/K is Galois.
¤ 3.7. Galois group of a polynomial. In this section, we are going to study Galois group of a polynomial. We will define this notion in general and study polynomial of degree 3,4 in more detail.
Definition 3.7.1. Let f ∈ K[x] be a polynomial with splitting field F . The Galois group of f (x), denoted Gf is the Galois group of F/K.
The Galois group of a polynomial have some basic properties.
Proposition 3.7.2. Let f (x) be a polynomial of degree n, then Gf ,→
Sn. Thus one can viewed Gf as a subgroup of Sn.
If f (x) is irreducible and separable, then Gf is transitive and |Gf| is divided by n.
Sketch of the proof. Let {u1, ..., ur} be roots of f (x) in F . For σ ∈ Gf, σ(ui) = uj. Hence σ gives a permutation of r elements. It follows that Gf can be viewed as a subgroup of Sr hence Sn.
(r could possibly less than n because there might have multiple roots in general).
Now if f (x) is separable. Then we have distinct roots {u1, ..., un} in F . For any ui, we have K[ui] ∼= K[x]/(f (x)) since f (x) is irreducible.
If follows that there is a K-isomorphism σ : K[ui] → K[x]/(f (x)) → K[uj] for all i, j. sigma gives an K-embedding K[ui] → K[uj] = K and extended to a K-embedding ¯σ : F → K. Since F is normal, ¯σ(F ) = F (cf. Theorem ?). Thus ¯σ ∈ Gf and ¯σ(ui) = σ(ui) = uj. Therefore, Gf is transitive.
Moreover, since K ⊂ K[ui] ⊂ F . So |Gf| = [F : K] = [F : K[ui]]n
is divided by n. ¤
So now, we discuss irreducible separable polynomials of small degree.
One might wondering how do we know a polynomial is separable or not.
We have the following easy criteria:
Proposition 3.7.3. Let f (x) ∈ K[x] be an irreducible polynomial The following are equivalent:
1. f (x) is separable.
2. (f (x), f0(x)) = 1 in K[x]
3. (f (x), f0(x)) = 1 in K[x]
4. f0(x) 6= 0
Recall that when f (x) =P
aixi, then f0(x) is its formal differentia- tion which is f0(x) :=P
iaixi−1.
Proof. If f (x) is separable, then f (x) = Qn
i=1(x − ui) with distinct ui in K[x]. Thus f0(x) = PQn
i=1(x−ui)
x−ui . If (f (x), f0(x)) 6= 1 in K[x], then x − ui|f0(x) for some i. However, f0(ui) = Q
j6=i(uj − ui) 6= 0, a contradiction.
Conversely, if f (x) is not separable, then f (x) =Qr
i=1(x − ui)ai with some ai ≥ 2. Let’s say a1 ≥ 2. Then it’s clear that (x − u1) is a factor of f0(x) as well. Hence (f (x), f0(x)) 6= 1. This proved the equivalence of (1) and (2).
To see the equivalence of (2) and (3). Note that if (f (x), f0(x)) = 1 in K[x], then 1 = f (x)s(x) + f0(x)t(x) for some s(x), t(x) ∈ K[x].
One can view this in K[x] and thus conclude that (f (x), f0(x)) = 1 in K[x]. On the other hand, if (f (x), f0(x)) = d(x) 6= 1 in K[x], then d(x) = f (x)s(x) + f0(x)t(x) for some s(x), t(x) ∈ K[x]. One can view this in K[x] and thus conclude that d(x)|(f (x), f0(x)) in K[x]. In particular, (f (x), f0(x)) 6= 1 in K[x]
Now finally, since f (x) is irreducible, (f (x), f0(x)) could only be 1 or f (x). Since f (x)|f0(x) if and only f0(x) = 0. Thus we are done. ¤ One notice that if charK 6= 0, then an irreducible polynomial is always separable. When charK = p, then an irreducible polynomial f (x) is not separable if and only f (x) = g(xp) for some g(x).
One can go a little bit further. If K is finite field with charK = p.
Let f (x) =P
aixi be an irreducible polynomial. f0(x) = 0 means that p|i for all ai 6= 0. Thus f (x) can be rewrite asP
aixip. Recall that each ai can be written as bpi for some bi because K is finite. Thus f (x) = Pbpixip = (P
bixi)p. This contradicts to f (x) being irreducible. To sum up, an irreducible polynomial over a finite field is always separable.
Let’s now turn back to the discussion of Galois groups. If f (x) is irreducible and separable of degree 2, then Gf ∼= S2 ∼= Z2. If f (x) is irreducible and separable of degree 3, then Gf is a subgroup of S3 of order divided by 3. Thus Gf could be A3 or S3. The question now is how to distinguish these two cases.
Lemma 3.7.4. (charK 6= 2) Let f (x) ∈ K[x] be an irreducible and sep- arable polynomial of degree 3 with splitting field F and roots u1, u2, u3. Then (Gf ∩ A3) = K[∆], where ∆ := (u1 − u2)(u1− u3)(u2− u3)
Note that f (x) is irreducible and separable, then F/K is Galois.
And ∆2 is invariant under Gf. Thus D := ∆2 ∈ K. We call D the discriminant of f (x).
If f (x) is written as x3+ bx2+ cx + d, then s1 := u1+ u2+ u3 = −b, s2 := u1u2 + u1u3 + u2u3 = c, s3 := u1u2u3 = −d. We impose an ordering u1 > u2 > u3. Then leading term of D is u41u22, which is the leading term of s21s22. Then we consider D0 := D − s21s22 with lower leading term, which is −4u31u32. This leading term is the same as the
leading term of −4s32. So we consider D(2) := D0 + 4s32. Inductively, one can write D in terms of s1, s2, s3, hence in terms of b, c, d.
If f (x) is normalized as x3+ px + q, then D = −4p3 − 27q2.
Proof. σ(∆) = ∆ if and only σ is an even permutation. So ∆ ∈ (Gf ∩ A3)0 clearly. Hence we have K[∆] < (Gf ∩ A3)0. Thus K[∆]0 > (Gf ∩ A3). If σ ∈ K[∆]0, then σ(∆) = ∆, hence σ is even. Thus K[∆]0 <
(Gf ∩ A3). So we have K[∆]0 = (Gf ∩ A3) and K[∆] = (Gf ∩ A3)0. ¤ We thus conclude that Gf = A3 if and only if Df is square in K.
And Gf = S3 if and only if Df is not a square in K Example 3.7.5.
Let f (x) = x3+ x + 1 ∈ Q[x]. It’s irreducible.
Now we consider the case of degree 4 polynomial. One can also define
∆ and discriminant D similarly. However, it turns out that this is not enough to classify all cases. The idea is to consider another normal subgroup V4C S4.
Let’s first list at all possible subgroup in S4. Since Gf is transitive with order divided by 4. We can have following
|Gf| Gf Gf ∩ V4 |Gf|/|Gf ∩ V4|
24 S4 V4 6
12 A4 V4 3
8 ∼= D8 V4 2
4 ∼= Z4 6= V4 2
4 V4 V4 1
Also we have the following
Lemma 3.7.6. Let f (x) be an irreducible separable polynomial of de- gree 4 with splitting field F and roots u1, K, u4. Let α = u1u2 + u3u4 β = u1u3+ u2u4, γ = u1u4 + u2u3. Then K[α, β, γ] = (Gf ∩ V4).
Let g(x) = (x − α)(x − β)(x − γ), then one can check that σ(g(x) = g(x) for all σ ∈ Gf. Thus g(x) ∈ K[x] for F/K is Galois. The cubic g(x) is call the resolvant cubic of f (x). If f (x) = x4+bx3+cx2+dx+e, then its resolvant cubic is g(x) = x3− cx2+ (bd − 4e)x − b2e + 4ce − d2 by computation on symmetric polynomials as we exhibited.
Proof. It clear that K[α, β, γ] < (Gf∩ V4)0. Hence we have (Gf∩ V4) <
K[α, β, γ]0. Now if σ ∈ K[α, β, γ]0 and σ 3 V4. We claim that this would lead to a contradiction. And thus we are done.
The claim can be verified directly by exhausting all cases. For ex- ample, if σ = (1, 3), then σ(α) = α gives u3u2+ u1u4 = u1u2+ u3u4. Thus (u2 − u4)(u1− u3) = 0 contradict to reparability of f (x). The other cases can be computed similarly.
¤
Let m := |Gf|/|Gf∩ V4| = [K[α, β, γ] : K]. By using this correspon- dence, one sees that:
1. m = 1 ⇔ Gf = V4 ⇔ g(x) splits into linear factors in K[x].
2. m = 3 ⇔ Gf = A4 ⇔ g(x) is irreducible in K[x] and Dg is a square in K.
3. m = 6 ⇔ Gf = S4 ⇔ g(x) is irreducible in K[x] and Dg is not a square in K.
The only remaining unclear case is m = 2. This case corresponding to the case that g(x) splits into a linear and a quadratic factors in K[x].
To see the Galois group, we claim that Gf ∼= D8 if and only if f (x) is irreducible in K[α, β, γ][x].
First of all, if f (x) is irreducible in K[α, β, γ][x], then
4 = [K[α, β, γ][u1] : K[α, β, γ]] ≤ [F : K[α, β, γ]] = |Gf ∩ V4|.
So Gf ∼= D8.
On the other hand, F is the splitting field of f (x) over K[α, β, γ] as well. Suppose that f (x) is reducible. If f (x) factors into a linear and a cubic factor in K[α, β, γ], then the Galois group of f (x) over K[α, β, γ], which is Gf∩V4, can only ∼= A3 or S3. This is a contradiction. Running over all cases, one sees that the only possible case is f (x) factors into two linear and one quadratic factors. Thus |Gf ∩ V4| = 2 and hence Gf ∼= Z4.
3.8. finite fields. The Galois theory on finite fields is comparatively easy and basically governed by Frobenius map.
Recall that given a finite field F of q elements, it’s prime field must be of the form Fp for some prime p. Let n = [F : Fp], then |F | = pn. Theorem 3.8.1. F is a finite field with pn elements if and only if F is a splitting field of xpn − x over Fn.
