separability and inseparability We first recall something about separable extension.
To start with, let f (x) be an irreducible polynomial in K[x] and f0(x) be its derivative (formally). More precisely, if f (x) = sumni=0aixi, then f0(x) :=Pn
i=1iaixi−1. One has the following equivalence:
(1) f (x) is separable, i.e. no multiple roots in K.
(2) (f (x), f0(x)) = 1 ∈ K[x].
(3) (f (x), f0(x)) = 1 ∈ K[x].
(4) f0(x) = 0.
Therefore, the only possibility to have non-separable polynomial is char(K) = p and f (x) = g(xp).
Given an element u algebraic over K, one can define the separable degree to be the number of distinct roots of minimal polynomial. This notion can be extended to a general setting:
Definition 0.1. Let F/K be an extension. Fix an embedding σ : K → L = L. We define the separable degree of F/K, denoted [F : K]s, to be the cardinality of
Sσ := {τ : F → L|τ|K = σ}.
One can check that [F : K]s is independent of σ and L. Hence the definition is well-defined. Moreover, if F = K(u) for u algebraic over K, then [F : K]s = [K(u) : K]s is the number of distinct roots of the minimal polynomial p(x) of u. (By considering K-embedding τ : K(u) → K, τ (u) must be a root of p(x) and τ is determined by τ (u)).
Proposition 0.2. If K ⊂ E ⊂ F , then [F : K]s = [F : E]s[E : K]s. Moreover, if F/K is finite, then [F : K]s ≤ [F : K].
Then we have the following useful criterion:
Proposition 0.3. If F/K is finite, then F/K is separable if and only if [F : K]s = [F : K].
we can then prove the following:
Theorem 0.4. Suppose that F = K(S) such that each elements of S is separable over K, then F/K is separable.
In particular, let
S := {u ∈ F |u is separable over K}.
Then S is a field extension over K. And we have [F : K]s = [S : K]s = [S : K]¯
¯[F : K].
On the other hand, one have proved the following Proposition which actually gives the definition of inseparable degree
1
Proposition 0.5. If F/K is a finite extension, then [F : K]s¯
¯[F : K].
Moreover, [F : K]/[F : K]s = pn for some n.
Before we move onto the study of inseparability, we would like to prove the famous theorem of primitive element.
Proposition 0.6. If K is a finite field and F/K is an algebraic field extension. The following are equivalent:
(1) F/K is finite.
(2) F = K(α) for some α ∈ f . That is, F/K is a simple extension.
(3) There is only finitely many intermediate fields.
Proof. For (1) ⇒ (2), if F/K is finite, then F is finite. F∗ is a cyclic multiplicative group, say F∗ =< α >. Then it’s clear that F = K(α).
(2) ⇒ (1) is trivial.
(1) ⇒ (3). Suppose that |K| = q, |F | = qn. Let E be an intermediate field, then it’s clear that |E| = qd for some d|n. One can prove that for any d|n, there is exactly one intermediate field with qd elements.
Hence there are only finitely many intermediate fields.
(3) ⇒ (1). Suppose on the other hand that F/K is not finite. Then F/K is not finitely generated. We can easily get (by axiom of choice) a infinite sequence of intermediate fields
K ⊂ K(a1) ⊂ K(a1, a2)...
by adding generators. ¤
Proposition 0.7. Let F/K be a finite extension, then F = K(α) if and only if there is only finitely many intermediate fields.
Proof. If K is finite, then we are done by the previous Proposition. We assume that K is infinite.
Suppose that there is only finitely many intermediate fields. For any α, β ∈ F , we can consider intermediate fields K(α + cβ) as c ranging in K. Since K is infinite. There must exists c1, c2 ∈ K such that K(α + c1β) = K(α + c2β). It’s easy to check that
K(α, β) = K(α + cβ).
By induction on number of generators of F/K, we proved that F/K is a simple extension.
Suppose now that F = K(α). We would like to prove the finiteness by using the following map:
φ : {E|K ⊂ E ⊂ F } → S := {pE(x)},
where pE(x) denotes the minimal polynomial of α over E. Since every pE(x) is a divisor of pK(x) in the algebraic closure (or in the splitting field), it’s clear that S is finite.
It’s enough to prove that φ is injective. To this end, let E0 be the extension over K generated by coefficient of pE(x). One sees that
pE(x) ∈ E0[x] is irreducible and hence a minimal polynomial of α.
Hence we have
[K(α) : E] = deg(pE(x)) = [K(α) : E0].
It follows that E = E0. Thus, if φ(E) = φ(E0), then E = E0 = E0.
This proved the injectivity. ¤
Theorem 0.8. If F/K is separable and finite, then F = K(α) for some α ∈ F .
Proof. We may assume that K is infinite. By induction on generators of F/K, we may assume that F = K(α, β). Let n := [F : K]s, and σ1, ..., σn be the distinct embedding of F in K. Let
P (x) :=Y
i6=j
(σiα + σiβx − σiα − σjβx).
Since deg(P (x)) = n(n − 1) and there are infinitely many elements in K, there must be an c ∈ K such that P (c) 6= 0. Thus all σi(α + cβ) are all distinct. This gives n distinct embedding of K(α + cβ). One has
n = [F : K]s≤ [K(α + cβ) : K]s.
Since F/K is separable, so is K(α + cβ). It’s easy to see that F =
K(α + cβ). ¤
Definition 0.9. Let F/K be an extension. An element u ∈ F is purely inseparable over K if its minimal polynomial p(x) ∈ K[x] factors in F [x] as (x − u)m. An extension F/K is purely inseparable over K if every element of F is purely inseparable over K.
It’s easy to see that an element u ∈ F which is both separable and purely inseparable over K if and only if u ∈ K.
Another useful observation is:
Lemma 0.10. Let F/K be an extension with char(K) = 0 6= 0. If u ∈ F is algebraic over K, then upn is separable over K for some n ≥ 0.
Proof. The point is that if u is not separable, then its minimal polyno- mial p(x) is of the form f (xp). Then f (x) is the minimal polynomial of up. By induction on degree of u, we are done. ¤ Being purely inseparable has the following equivalent formulation:
Theorem 0.11. Let F/K be an algebraic extension with char(K) = p 6= 0. The following are equivalent:
(1) F/K is purely inseparable, i.e. every element u ∈ F has mini- mal polynomial of the form (x − u)m.
(2) for all u ∈ F , the minimal polynomial is of the form xpn− a ∈ K[x].
(3) for all u ∈ F , upn ∈ K for some n ≥ 0.
(4) S = K, that is, the only element of F which is separable over K are the elements in K.
(5) F/K is generated by purely inseparable elements.
Proof. Let m = pnr.
(x − u)m = (x − u)pnr = (xpn− upn)r= xm− rupnxpn(r−1)+ ... ∈ K[x].
Therefore, upn ∈ K, this proved (1) ⇒ (3).
Moreover, p0(x) := xpn − upn) ∈ K[x] and p0(x)r is the minimal polynomial of u (hence irreducible). Therefore, r = 1. This proved (1) ⇒ (2).
(2) ⇒ (3) is trivial.
For (3) ⇒ (1), let a = upn ∈ K, then f (x) := xpn − a ∈ K[x] and factors in F [x] as (x − u)pn. Hence the minimal polynomial of u over K is a factor of f (x) and factors into (x − u)m in F [x].
We have seen (1) ⇒ (4) and (5), (4) ⇒ (3) follows from the above Lemma 0.10.
It remains to show that (5) ⇒ (3). To see this, first note that F = K(S) where S consists of elements ui such that upin ∈ K for some n (By the proof of (1) ⇒ (3). For any u ∈ F , say u = f (ug(u1,...,ur)
1,...,ur). Pick N such that upiN ∈ K, ∀i = 1, ..., r. Then upN ∈ K. ¤
As a corollary, one can show that
P := {u ∈ F |u is purely inseparable over K}
is a subfield.
Theorem 0.12. Let F/K be an algebraic extension. Keep the notation as above for S, P .
(1) S/K is separable.
(2) P/K is purely inseparable.
(3) F/S is purely inseparable.
(4) F/P is separable if and only F = P S.
(5) P ∩ S = K.
(6) if F/K is normal, then S/K and F/P are Galois. And GalF/K = GalF/P ∼= GalS/K.
Proof. We have seen (1), (2), (5). (3) follows from Lemm 0.10. For (4), look at P ⊂ SP ⊂ F . If F/P is separable, then F/SP is separable.
Look at S ⊂ SP ⊂ F now. We have F/K is purely inseparable, thus so is F/SP . Thus F = SP .
On the other hand, if F = SP = P (S), then clearly F = P (S) is separable over P .
Lastly, we look at G := GalF/K. We claim that G0 = P , hence F/P is Galois with Galois group GalF/P = GalF/K.
To see the claim, if u ∈ P , then it’s clear that σ(u) = u for all σ ∈ G.
Therefore, P ⊂ G0. On the other hand, if u ∈ G0 and v is another root
of p(x), the minimal polynomial of u. There is an σ such that σ(u) = v.
Since F/K is normal, this σ can be extended to G. But u ∈ G0, thus v = u, in other words, p(x) = (x − u)m.
F is Galois over P because P = G0. Hence F/P is separable. By (5), F = P S.
Lastly, we consider GalF/P = GalF/K → GalS/K by restriction. This is well-defined since S is stable. More precisely, for u ∈ S, σ(u) ∈ S for all σ ∈ G because σ(u) has the same minimal polynomial as u does. This is surjective by extension theorem. It remains to show the injectivity. If σ|S = τ |S, then for all u ∈ F we have upn ∈ S. Thus,
σ(u)pn = σ(upn) = τ (upn) = τ (u)pn. It follows that σ(u) = τ (u).
It remains to show that S/K is Galois. To see this, suppose u ∈ S is fixed by all σ ∈ G, then u ∈ G0 = P . Hence u ∈ K. We are done. ¤
If char(K) = p 6= 0, we write Kp = {up|u ∈ K}.
Definition 0.13. K is said to be perfect if Kp = K Example 0.14. Finite fields are perfect.
Zp(x) is not perfect.
Corollary 0.15. Let F/K be an algebraic extension with char(K) = p 6= 0. We have
(1) If F/K is separable, then F = KFpn for each n ≥ 1.
(2) If F/K is finite and F = KFp, then F/K is separable.
(3) In particular, u ∈ F is separable over K if and only if K(up) = K(u).
Note that Fp is not necessarily an extension over K. So is Fpn. But we can take KFpn, which is an extension over K.
Proof. We first suppose that F/K is finite, hence finitely generated.
Write F = K(u1, ..., ur). It’s clear that there is N ≥ 1 such that upN ∈ S. Hence FpN ⊂ S, therefore, KFpN ⊂ S.
We claim that S = KFpN. To see this, one notices that F is purely inseparable over KFpN, so is S purely inseparable over KFpN. And on the other hand, S is separable over K, so is over KFpN. Hence S = KFpN.
For (1), if F/K is separable and finite, then we have F = KFpN. However, in the proof, one can choose N to be arbitrary large. More precisely, one has F = KFpN for all N ≥ N0. By looking at the inclusion
F = KFpN ⊂ KFpN −1 ⊂ ... ⊂ KFp ⊂ F.
One has F = KFpn for all n ≥ 1.
Suppose now that F/K is separable but not necessarily finite. For any u ∈ F , we consider F0 := K(u) which is separable and finite over K. Thus u ∈ F0 = KF0pn ⊂ KFpn for all n ≥ 1. This proves (1).
We now prove (2). If F = KFp, then F = K(KFp)p = KFp2. Inductively, one has F = KFpn for all n ≥ 1. Since we have show that S = KFpN, it follows that F = S.
Apply the statement to a single element. We consider F = K(u).
Fp ⊂ Kp(up) ⊂ K(up) . Indeed, KFp = K(up). By (2), if K(u) = K(up), then u is separable. By (1), if u is separable, then K(u) =
K(up). ¤
