Nov. 17, 2006
3.5. splitting fields and normal extensions. We have seen that given a field K, there is a unique (up to isomorphism) algebraic closure, denoted K. Then it is convenient for our further study of roots of polynomial. Even though we do not know the roots explicitly, we know that there are in its algebraic closure. This make the discussion of root of polynomials more concrete.
Let K be a field and f (x) ∈ K[x]. Let {u1, ..., ur} be the roots of f (x) in its algebraic closure K. Then the field K(u1, ..., ur) is called the splitting field of f (x) over K. The splitting field is the smallest field that containing all roots.
Given a set of polynomial S ⊂ K[x], we can similarly define the splitting field of S to be the field generated by all roots of polynomials in S.
In this section, we are going to prove the existence and uniqueness of splitting fields. And we introduce the notion of normal extension.
Proposition 3.5.1. Let K be a field. And S be a set of polynomial in K[x]. Then
(1) Any two splitting field are isomorphic.
(2) If F1, F2 are two splitting fields in a fixed algebraic closure K, then F1 = F2.
Proof. Let F1 and F2 be two splitting fields, one has an K-embedding σ : K → F2 = K. This embedding can be extended to ˜σ : F1 → F2 by the extension theorem. One can prove that image of ˜σ is in F2. Hence one has an injective homomorphism ˜σ : F1 → F2. Similarly there is another one ˜τ : F2 → F1. It’s easy to show that these give the
isomorphism. ¤
Proposition 3.5.2. Let N be an algebraic extension over K contained in K. Then the following are equivalent:
(1) Any K-embedding σ : N → K induces an K-automorphism of N.
(2) N is a splitting field of some S ⊂ K[x] over K.
(3) Every irreducible polynomial in K[x] having a root in N splits in N.
Proof. For (1) ⇒ (2), (3), we prove that for every u ∈ N, with minimal polynomial p(x), then v ∈ N for every root of p(x). To this end, start with an isomorphism σ : K(u) → K(v). By extension theorem, one can extend it to an embedding N → K(v) = K. The embedding is an automorphism by (1). Thus, v = σ(u) ∈ N.
(3) ⇒ (2) is trivial.
For (2) ⇒ (1). Suppose that N is a splitting field of S over K. Let u be a root of f (x) ∈ S. Let σ : N → K be any K-embedding. It’s clear
that σ(u) is a root of f (x), hence σ(u) ∈ N. Thus σ(N) ⊂ N. Since σ is injective and N/K is algebraic, σ is in fact an isomorphism. ¤ The property of being normal is not as well-behaved as begin alge- braic or finite. For example, it’s not preserve after ”extension”
Example 3.5.3. If F/E and E/K are normal, then F/K is not nec- essarily normal. For example, take F = Q(√4
2), E = Q(√
2), K = Q.
It’s easy to see that a degree 2 extension is always normal, however, Q(√4
2) is not normal over Q.
Also let’s consider K ⊂ E ⊂ F . Then F is normal over K implies that F is normal over E. But it doesn’t imply that E is normal over K. For example, take F = Q(√4
2, i), E = Q(√4
2), K = Q Being normal is preserved by ”lifting” and ”compositum”
Proposition 3.5.4. Let E, F be extensions over K and contained in a field L. If E/K is normal then EF/F is normal. Moreover, if both E/K, F/K are normal, then EF/K is normal.
Proof. In order to show that EF is normal over F , we look at F - embedding σ : EF → F . Since σ is identity on F , hence on K. By the extension theorem and the proof of the previous Proposition, one can show that σ|E is an automorphism. Hence σ(E) = E. It follows that
σ(EF ) = σ(E)F = EF.
Thus EF is normal over F .
Suppose now that E/K, F/K are normal. Let σ : EF → K be a K-embedding. We have that σ|E, σ|F are K-embeddings. One sees that σ(E) = E and σ(F ) = F by the normal assumption. If follows that
σ(EF ) = σ(E)σ(F ) = EF.
¤ 3.6. finite dimensional Galois extension. In this section, we are going to prove the fundamental theorem for finite dimensional Galois extension.
Let F/K be an field extension, we define the Galois group of F over K, denoted GalF/K or GF/K or AutK(F ), as
GalF/K := {σ|σ ∈ AutF, σ|K = 1K}.
It’s clear that for σ ∈ GalF/K and u ∈ F algebraic over K with mini- mal polynomial p(x), then σ(u) satisfies the same minimal polynomial.
On the other hand, if F/K is normal, let u, v be two elements having the same minimal polynomial p(x), then we claim that there is an σ ∈ GalF/K such that σ(u) = v. To see this, we fix an algebraic closure K containing F . There is an K-isomorphism σ0 : K(u) → K(v) which extends to an embedding σ : F → K . Since F is normal over K, one has σ(F ) ⊂ F . And hence σ ∈ AutF .
Example 3.6.1.
Consider the field F := Q(√3
2, ω) which is a splitting field of x3− 2 over Q. Thus it’s normal over Q. One can check that the Galois group GalF/Q is generated by σ, τ that σ(√3
2) = √3
2ω, σ(ω) = ω, and τ (√3
2) = √3
2, τ (ω) = ω2. It’s easy to check that GalF/Q ∼= S3. ¤ Example 3.6.2.
Consider the field F := Q(√3
2) over Q. Then it’s easy to check that
GalF/Q= {1F}. ¤
There is a natural correspondence between subgroups of Galois groups and intermediate fields. To be precise, fix an extension F/K. Let H < G := GalF/K be a subgroup. One can define
H0 := {u ∈ F |σ(u) = u, ∀σ ∈ H}.
It’s clear that this is a field. On the other hand, given and intermediate field L such that K ⊂ L ⊂ F , then one can define
L0 := {σ ∈ GalF/K|σ(u) = u, ∀u ∈ L} = {σ ∈ GalF/K|σ|L= 1L}.
It’s easy to check the following properties:
Proposition 3.6.3. Let F/K be an extension with Galois group G.
Let L be an intermediate field, i.e. K ⊂ L ⊂ F , and H < G is a subgroup.
(1) F0 = {1F}, K0 = G, and {1F}0 = F . (2) For any L, one has L ⊂ L00, L0 = L000. (3) For any H, one has H < H00, H0 = H000.
(4) For any intermediate fields L ⊂ M, one has M0 < L0. (5) For any subgroups J < H, one has H0 ⊂ J0.
Proof. Most of the proof follows directly from the definition. We only sketch the proof for L0 = L000.
By L ⊂ L00 and (4), one has
(L00)0 < L0. On the other hand, by (5), one has
L0 < (L0)00.
We are done. ¤
Proposition 3.6.4. There is a one-to-one correspondence between {L|K ⊂ L ⊂ F, L00 = L} ↔ {H|H < G, H00 = H}.
Proof. The correspondence is given by L 7→ L0 (or H 7→ H0).
To show the injective, one sees that if L01 = L02, then L1 = L001 = L002 = L2.
For any H with H00 = H, we take L = H0, then H = L0. It suffices to check that L00 = L. This follows from the fact that H000 = H0. ¤
In the proposition, one might expect that G0 = K. However, this is not always the case (see e.g. Example 3.6.2). For extension with this property, we call it Galois. It turns out that this naive definition is a very delicate one which leads to some nice properties.
Definition 3.6.5. An extension F/K is said to be Galois if (GalF/K)0 = K.
Example 3.6.6.
Keep the notation as in Example 3.6.1. One can check that G0 = Q, hence a Galois extension.
In fact the group G has the following subgroups: {1}, < τ >, < τ σ >
, < τ σ2 >, < σ >, G of order 1, 2, 2, 2, 3, 6 respectively. Their fixed fields are Q(√3
2, ω), Q(√3
2), Q(√3
2ω), Q(√3
2ω2), Q(ω), Q respectively.
Conversely, for these intermediate subfields, their fixed groups are
exactly those corresponding ones. ¤
In general, we have the following:
Theorem 3.6.7 (Fundamental theorem of finite dimensional Galois extension). Let F/K be a finite dimensional Galois extension with Ga- lois group G, then
(1) There is an one-to-one correspondence between {L|K ⊂ L ⊂ F } ↔ {H|H < G}.
(2) The corresponding degree are equal. That is, if K ⊂ L ⊂ M ⊂ F , then [M : L] = [L0 : M0]. And if J < H < G, then [H : J] = [J0 : H0].
(3) An intermediate field E is Galois over K if and only if E0C G.
And in this case, GalE/K ∼= G/E0. Proof. Step 1. [M : L] ≥ [L0 : M0].
We prove the case that M = L(u) for some u ∈ M and by induction on [M : L], we are done. Suppose now that M = L(u) and let p(x) be the minimal polynomial of u over L. Let S be the set of roots of p(x) in F . Then one has a map
Φ : L0 → S, σ 7→ σ(u).
One can check that Φ induces an injective map L0/M0 → S. Hence one has
[L0 : M0] = |L0/M0| ≤ |S| ≤ deg(p(x)) = [M : L].
Step 2. [H : J] ≥ [J0 : H0].
Let n = [H : J]. Suppose on the contrary that there are n + 1 elements u1, ..., un+1 ∈ J0 linearly independent over H0.
We consider the equation Pn+1
i=1 uixi = 0 in F Consider now a set of representative of H/J, denoted {e = σ1, ..., σn}. By applying σi to the above equation. Then one has a system of linear equations in F .
(∗)
σ1(u1)x1+ σ1(u2)x2+ ... + σ1(un+1)xn+1 = 0 σ2(u1)x1+ σ2(u2)x2+ ... + σ2(un+1)xn+1 = 0 ...
σn(u1)x1+ σn(u2)x2 + ... + σn(un+1)xn+1 = 0
Pick a solution in F with smallest number of non-zero ai’s, may assume it’s (a1, ..., as, 0..., 0) and a1 = 1.
If there is an τ ∈ H such that τ (a2) 6= a2, then by applying τ to the system (∗), one get the same system of equations with a solution (τ (a1), τ (a2), ..., τ (as), 0, ..., 0 . Hence
(a1, ..., as, 0..., 0) − (τ (a1), τ (a2), ..., τ (as), 0, ..., 0) = (0, a2− τ (a2), ..., 0) is a non-zero solution of smaller length. This is the required contradic- tion.
To find τ . We look at u1a1 + ... + usas = 0. Since {u1, ..., us} is independent over H0, not all a1 is in H0. We may assume that a2 6∈ H0. Hence there is a τ ∈ H such that τ (a2) 6= a2. We are done.
Step 3. We show that every intermediate field L, L00 = L. And every subgroup H < G, H00 = H.
By Step 1, one has
[L00 : K] = [L00 : K00] ≤ [K0 : L0] ≤ [L : K],
however, one has L ⊂ L00. Thus one has L = L00. Similarly, one can prove that H00= H by considering [H00: {1F}].
Step 4. [M : L] = [L0 : M0] and [H : J] = [J0 : H0].
This follows from [M : L] = [M : K]/[L : K] = [K0 : M0]/[K0 : L0] = [L0 : M0]. And the other one is similar.
Step 5. F/K is normal and separable.
Given u ∈ F , with minimal polynomial p(x) over K. As in the proof of Step 1. One has [K(u)0 : K0] ≤ |S| ≤ deg(p(x)) = [K(u) : K]. By Step 4, they are equalities. In particular, every root of p(x) is in F and there is no multiple roots. Thus F is normal and separable over K.
Step 6. If N C G, then N0 is stable. That is, for all σ ∈ G, σ(N0) ⊂ N0 (indeed = N0).
Since N C G, for all σ ∈ G and for all τ ∈ N, one has σ−1τ σ ∈ N.
Thus, σ−1τ σ(N0) = N0. It follows that τ σ(N0) = σ(N0), for all τ ∈ N.
Hence σ(N0) is fixed by all N and thus σ(N0) ⊂ N0.
Step 7. If E is a stable intermediate subfield. Then the restriction map GalF/K → GalE/K is well-defined and surjective.
Since E is stable, then σ|E ∈ GalE/K for any σ ∈ GalF/K. Moreover, let τ ∈ GalE/K, by the extension theorem, there is an extension τ : F → K. Since F is normal over K, τ is in fact an automorphism of F .
Step 8. If an intermediate field E is stable, then E/K is Galois.
To see this, it suffices to show that for any u ∈ E −K, there is an σ ∈ GalE/K such that σ(u) 6= u. Fix any F 3 v 6= u with the same minimal polynomial as u. There is an K-isomorphism σ0 : K(u) → K(v) such that σ(u) = v. σ can be extended to an embedding σ : F → K, which gives an automorphism of F . The restriction σ = σ|E gives an automorphism of E that σ(u) 6= u.
Step 9. If E/K is Galois, then E is stable.
One first notices that E/K is normal. For every σ ∈ GalF/K, σ gives an embedding σ|E : E → K. Since E/K is normal, σ|E is an automorphism of E. And hence E is stable under the Galois group GalF/K action.
Step 10. If E is stable, then E0 is normal.
This can be checked directly. For all σ ∈ G and τ ∈ E0 and for all u ∈ E,
σ−1τ σ(u) = σ−1τ (σ(u)) = σ−1σ(u) = u,
since σ(u) ∈ E. Therefore, σ−1τ σ ∈ E0. ¤ 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.
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