Normality and Separability
9.1 Splitting Fields
The most tractable polynomials are products of linear ones, so we are led to single this property out:
Definition 9.1. If K is a subfield of C and f is a nonzero polynomial over K, then f splitsover K if it can be expressed as a product of linear factors
f(t) = k(t − α1) . . . (t − αn) where k, α1, . . . , αn∈ K.
If this is the case, then the zeros of f in K are precisely α1, . . . , αn. The Funda-mental Theorem of Algebra, Theorem 2.4, implies that f splits over K if and only if all of its zeros in C actually lie in K. Equivalently, K contains the subfield generated by all the zeros of f .
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Examples 9.2. (1) The polynomial f (t) = t3− 1 ∈ Q[t] splits over C, because it can be written as
f(t) = (t − 1)(t − ω)(t − ω2)
where ω = e2πi/3 ∈ C. Similarly, f splits over the subfield Q(i,√
3) since ω ∈ Q(i,
√3), and indeed f splits over Q(ω), the smallest subfield of C with that prop-erty.
(2) The polynomial f (t) = t4− 4t2− 5 splits over Q(i,√
5), because f(t) = (t − i)(t + i)(t −√
5)(t +
√ 5) However, over Q(i) the best we can do is factorise it as
(t − i)(t + i)(t2− 5)
with an irreducible factor t2− 5 of degree greater than 1. (It is easy to show that 5 is not a square in Q(i).)
So over Q(i), the polynomial f does not split. This shows that even if a poly-nomial f (t) has some linear factors in an extension field L, it need not split over L.
If f is a polynomial over K and L is an extension field of K, then f is also a polynomial over L. It therefore makes sense to talk of f splitting over L, meaning that it is a product of linear factors with coefficients in L. We show that given K and f we can always construct an extension Σ of K such that f splits over Σ. It is convenient to require in addition that f does not split over any smaller field, so that Σ is as economical as possible.
Definition 9.3. A subfield Σ of C is a splitting field for the nonzero polynomial f over the subfield K of C if K ⊆ Σ and
(1) f splits over Σ.
(2) If K ⊆ Σ0⊆ Σ and f splits over Σ0then Σ0= Σ.
The second condition is clearly equivalent to:
(20) Σ = K(σ1, . . . , σn) where σ1, . . . , σnare the zeros of f in Σ.
Clearly every polynomial over a subfield K of C has a splitting field:
Theorem 9.4. If K is any subfield of C and f is any nonzero polynomial over K, then there exists a unique splitting field Σ for f over K. Moreover, [Σ : K] is finite.
Proof. We can take Σ = K(σ1, . . . , σn), where the σj are the zeros of f in C. In fact, this is the only possibility, so Σ is unique. The degree [Σ : K] is finite since K(σ1, . . . , σn) is finitely generated and algebraic, so Lemma 6.11 applies.
Isomorphic subfields of C have isomorphic splitting fields, in the following strong sense:
Splitting Fields 131 Lemma 9.5. Suppose that ι : K → K0is an isomorphism of subfields of C. Let f be a nonzero polynomial over K and let Σ ⊇ K be the splitting field for f . Let L be any extension field of K0such that ι( f ) splits over L. Then there exists a monomorphism
j: Σ → L such that j|K= ι.
Proof. We have the following situation:
K → Σ
ι ↓ ↓ j K0 → L
where j has yet to be found. We construct j using induction on ∂ f . As a polynomial over Σ,
f(t) = k(t − σ1) . . . (t − σn)
The minimal polynomial m of σ1 over K is an irreducible factor of f . Now ι(m) divides ι( f ) which splits over L, so that over L
ι (m) = (t − α1) . . . (t − αr)
where α1, . . . , αr∈ L. Since ι(m) is irreducible over K0it must be the minimal poly-nomial of α1over K0. So by Theorem 5.16 there is an isomorphism
j1: K(σ1) → K0(α1)
such that j1|K = ι and j1(σ1) = α1. Now Σ is a splitting field over K(σ1) of the polynomial g = f /(t − σ1). By induction there exists a monomorphism j : Σ → L such that j|K(σ1)= j1. But then j|K= ι and we are finished.
This enables us to prove the uniqueness theorem.
Theorem 9.6. Let ι : K → K0be an isomorphism. Let Σ be the splitting field for f over K, and let Σ0be the splitting field for ι( f ) over K0. Then there is an isomorphism
j: Σ → Σ0 such that j|K = ι. In other words, the extensions Σ : K and Σ0: K0 are isomorphic.
Proof. Consider the following diagram:
K → Σ
ι ↓ ↓ j K0 → Σ0
We must find j to make the diagram commute, given the rest of the diagram. By Lemma 9.5 there is a monomorphism j : Σ → Σ0such that j|K= ι. But j(Σ) is clearly the splitting field for ι( f ) over K0, and is contained in Σ0. Since Σ0is also the splitting field for ι( f ) over K0, we have j(Σ) = Σ0, so that j is onto. Hence j is an isomorphism, and the theorem follows.
Examples 9.7. (1) Let f (t) = (t2− 3)(t3+ 1) over Q. We can construct a splitting field for f as follows: over C the polynomial f splits into linear factors
f(t) = (t +√
so there exists a splitting field in C, namely
Q a splitting field is afforded by Q(1 +√
3, i) which equals Q(√
3, i). This is the same field as in the previous example, although the two polynomials involved are different.
(3) It is even possible to have two distinct irreducible polynomials with the same splitting field. For example t2− 3 and t2− 2t − 2 are both irreducible over Q, and both have Q(√
3) as their splitting field over Q.
9.2 Normality
The idea of a normal extension was explicitly recognised by Galois (but, as always, in terms of polynomials over C). In the modern treatment it takes the following form:
Definition 9.8. An algebraic field extension L : K is normal if every irreducible polynomial f over K that has at least one zero in L splits in L.
For example, C : R is normal since every polynomial (irreducible or not) splits in C. On the other hand, we can find extensions that are not normal. Let α be the real cube root of 2 and consider Q(α) : Q. The irreducible polynomial t3− 2 has a zero, namely α, in Q(α), but it does not split in Q(α). If it did, then there would be three real cube roots of 2, not all equal. This is absurd.
Compare with the examples of Galois groups given in Chapter 8. The normal extension C : R has a well-behaved Galois group, in the sense that the Galois cor-respondence is a bijection. The same goes for Q(√
2,√ 3,√
5) : Q. In contrast, the non-normal extension Q(α) : Q has a badly behaved Galois group. Although this is not the whole story, it illustrates the importance of normality.
There is a close connection between normal extensions and splitting fields which provides a wide range of normal extensions:
Theorem 9.9. A field extension L : K is normal and finite if and only if L is a splitting field for some polynomial over K.
Separability 133 Proof. Suppose L : K is normal and finite. By Lemma 6.11, L = K(α1, . . . , αs) for certain αjalgebraic over K. Let mjbe the minimal polynomial of αjover K and let f = m1. . . ms. Each mjis irreducible over K and has a zero αj∈ L, so by normality each mjsplits over L. Hence f splits over L. Since L is generated by K and the zeros of f , it is the splitting field for f over K.
To prove the converse, suppose that L is the splitting field for some polynomial g over K. The extension L : K is then obviously finite; we must show it is normal. To do this we must take an irreducible polynomial f over K with a zero in L and show that it splits in L. Let M ⊇ L be a splitting field for f g over K. Suppose that θ1and θ2are zeros of f in M. By irreducibility, f is the minimal polynomial of θ1and θ2
over K.
We claim that
[L(θ1) : L] = [L(θ2) : L]
This is proved by an interesting trick. We look at several subfields of M, namely K, L, K(θ1), L(θ1), K(θ2), L(θ2). There are two towers
K⊆ K(θ1) ⊆ L(θ1) ⊆ M K⊆ K(θ2) ⊆ L(θ2) ⊆ M
The claim will follow from a simple computation of degrees. For j = 1 or 2 [L(θj) : L][L : K] = [L(θj) : K] = [L(θj) : K(θj)][K(θj) : K] (9.1) By Proposition 6.7, [K(θ1) : K] = [K(θ2) : K]. Clearly L(θj) is the splitting field for g over K(θj), and by Corollary 5.13 K(θ1) is isomorphic to K(θ2). Therefore by Theorem 9.6 the extensions L(θj) : K(θj) are isomorphic for j = 1, 2, so they have the same degree. Substituting in (9.1) and cancelling,
[L(θ1) : L] = [L(θ2) : L]
as claimed. From this point on, the rest is easy. If θ1∈ L then [L(θ1) : L] = 1, so [L(θ2) : L] = 1 and θ2∈ L also. Hence L : K is normal.
9.3 Separability
Galois did not explicitly recognise the concept of separability, since he worked only with the complex field, where, as we shall see, separability is automatic. However, the concept is implicit in his work, and must be invoked when studying more general fields.
Definition 9.10. An irreducible polynomial f over a subfield K of C is separable over K if it has simple zeros in C, or equivalently, simple zeros in its splitting field.
This means that over its splitting field, or over C, f takes the form f(t) = k(t − σ1) . . . (t − σn)
where the σjare all different.
Example 9.11. The polynomial t4+t3+t2+t + 1 is separable over Q, since its zeros in C are e2πi/5, e4πi/5, e6πi/5, e8πi/5, which are all different.
For polynomials over R there is a standard method for detecting multiple zeros by differentiation. To obtain maximum generality later, we redefine the derivative in a purely formal manner.
Definition 9.12. Suppose that K is a subfield of C, and let f(t) = a0+ a1t+ · · · + antn∈ K[t]
Then the formal derivative of f is the polynomial
D f= a1+ 2a2t+ · · · + nantn−1∈ K[t]
For K = R (and indeed for K = C) this is the usual derivative. Several useful properties of the derivative carry over to D. In particular, simple computations (Ex-ercise 9.3) show that for all polynomials f and g over K,
D( f + g) = D f + Dg D( f g) = (D f )g + f (Dg) Also, if λ ∈ K then D(λ ) = 0, so
D(λ f ) = λ (D f )
These properties of D let us state a criterion for the existence of multiple zeros without knowing what the zeros are.
Lemma 9.13. Let f 6= 0 be a polynomial over a subfield K of C, and let Σ be its splitting field. Then f has a multiple zero (in C or Σ) if and only if f and D f have a common factor of degree≥ 1 in K[t].
Proof. Suppose f has a repeated zero in Σ, so that over Σ f(t) = (t − α)2g(t) where α ∈ Σ. Then
D f = (t − α)[(t − α)Dg + 2g]
so f and D f have a common factor (t − α) in Σ[t]. Hence f and D f have a common factor in K[t], namely the minimal polynomial of α over K.
Now suppose that f has no repeated zeros. Suppose that f and D f have a common factor, and let α be a zero of that factor. Then f = (t − α)g and D f = (t − α)h.
Differentiate the former to get (t − α)h = D f = g + (t − α)Dg, so (t − α) divides g, hence (t − α)2divides f .
Exercises 135 We now prove that separability of an irreducible polynomial is automatic over subfields of C.
Proposition 9.14. If K is a subfield of C then every irreducible polynomial over K is separable.
Proof. An irreducible polynomial f over K is inseparable if and only if f and D f have a common factor of degree ≥ 1. If so, then since f is irreducible the common factor must be f , but D f has smaller degree than f , and the only multiple of f having smaller degree is 0, so D f = 0. Thus if
f(t) = a0+ · · · + amtm
then this is equivalent to nan= 0 for all integers n > 0. For subfields of C, this is equivalent to an= 0 for all n > 0.
EXERCISES
9.1 Determine splitting fields over Q for the polynomials t3− 1,t4+ 5t2+ 6,t6− 8, in the form Q(α1, . . . , αk) for explicit αj.
9.2 Find the degrees of these fields as extensions of Q.
9.3 Prove that the formal derivative D has the following properties:
(a) D( f + g) = D f + Dg (b) D( f g) = (D f )g + f (Dg) (c) If f (t) = tn, then D f (t) = ntn−1
9.4 Show that we can extend the definition of the formal derivative to K(t) by defining
D( f /g) = (D f · g − f · Dg)/g2 when g 6= 0. Verify the relevant properties of D.
9.5 Which of the following extensions are normal?
(a) Q(t) : Q (b) Q(√
−5) : Q
(c) Q(α) : Q where α is the real seventh root of 5 (d) Q(√
5, α) : Q(α), where α is as in (c) (e) R(√
−7) : R
9.6 Show that every extension in C, of degree 2, is normal. Is this true if the degree is greater than 2?
9.7 If Σ is the splitting field for f over K and K ⊆ L ⊆ Σ, show that Σ is the splitting field for f over L.
9.8* Let f be a polynomial of degree n over K, and let Σ be the splitting field for f over K. Show that [Σ : K] divides n! (Hint: Use induction on n. Consider separately the cases when f is reducible or irreducible. Note that a!b! divides (a + b)! (why?).)
9.9 Mark the following true or false.
(a) Every polynomial over Q splits over some subfield of C.
(b) Splitting fields in C are unique.
(c) Every finite extension is normal.
(d) Q(√
19) : Q is a normal extension.
(e) Q(√4
19) : Q is a normal extension.
(f) Q(√4
19) : Q(√
19) is a normal extension.
(g) A normal extension of a normal extension is a normal extension.