Rings, Modules, Fields

12 Separability

Note that there might be a lot of embeddings \(L/K \to \bar {K}/K\). In fact, there generally are a lot. We say that the separable degree of an extension \(L/K\) (written \([L:K]_s\)) is the number of embeddings of \(L\) into \(\bar {K}\), which is well defined by the previous proposition. A finite extension is separable if its separable degree equals its degree, and a general algebraic extension is separable if every finite subextension is separable, and an element is separable if the corresponding simple extension is. The next two lemmas imply that these notions of separable agree.

Lemma 12.1. For \(M/L/K\), \([M:L]_s[L:K]_s = [M:K]_s\).

Proof. Just note that \(\bar {L} = \bar {K}\), and that extending an embedding from \(K\) to \(L\) and then to \(M\) is the same as extending from \(K\) to \(M\). □

Lemma 12.2. \([M:L]_s|[M:L]\) for a finite extension.

Proof. By multiplicativity, one reduces to the case of a simple extension, where this follows from the fact that if the minimal polynomial of \(\alpha \) factors in the algebraic closure as \(\Pi _i(x-\alpha _i)^b_i\), the \(b_i\) had better agree, since automorphisms of the simple extension can extend to the algebraic closure, and they fix the minimal polynomial. □

This lemma shows that the following notion is useful. A polynomial of degree \(n\) over a field \(F\) is called separable if it has \(n\) roots in the algebraic closure. Additionally, elements with the same minimal polynomial are called conjugate. The proof of the previous lemma gives the following:

Lemma 12.3. Every element of a separable extension has a separable minimal polynomial, and adjoining roots of separable polynomials gives separable extensions.

Corollary 12.4. Given \(M/L/K\), \(M/K\) is separable iff \(M/L\) and \(L/K\) are separable. The compositum of separable extensions is separable.

Proof. The first statement follows from the finite case, which follows from multiplicativity. The second statement follows from the finite case, which follows from the fact that separable polynomials are still separable in extensions. □

Lemma 12.5. TFAE for an irreducible polynomial \(f\) over \(K\) (the first two are equivalent for arbitrary \(f\)):
- 1. \(f\) is separable.
- 2. \(f,f'\) are coprime.
- 3. \(f' \neq 0\).

Proof. \(f\) is separable iff \(f\) has no double roots in the algebraic closure iff \(f,f'\) are coprime. It is clear that it is equivalent for \(f,f'\) to be coprime and \(f' \neq 0\) by degree considerations and the fact that \(f\) is irreducible. □

Note that \(f' = 0\) happens for irreducible \(f\) iff we are in char \(p\) or \(f\) is a polynomial of \(x^p\).

Now by the basic properties of separable extensions, it follows that given an extension \(L/K\), there is a maximal separable subextension, consisting of the elements separable over \(K\). We call this the separable closure of \(K\) in \(L\). The separable closure of \(K\) in \(\bar {K}\) is just called the separable closure, denoted \(K^{\sep }\). It is the maximal separable extension of \(K\), since any separable extension embeds in \(\bar {K}\), but must land in \(K^{\sep }\). Note that in characteristic \(0\), these are identical extensions.

Complementary to separable extensions are purely inseparable extensions, which are algebraic extensions where every finite subextension has separability degree \(1\). We have seen that every extension \(L/K\) splits into \(L/L\cap K^{\sep }/K\). The extension \(L/L\cap K^{\sep }\) is purely inseparable. To see this, note it suffices to show this when \(L/L\cap K^{\sep }\) is simple, and if \(\alpha \) is a primitive element with minimal polynomial \(f(x^{p^k})\) with \(k\) maximal, then \(\alpha ^{p^k}\) is separable so is in \(L\cap K^{\sep }\). Thus \(f = x^{p^k}-\alpha ^{p^k}\), and since there are no nontrivial \(p^k\) roots of unity in characteristic \(p\), there is only one root of \(f\). In fact we get the following equivalent characterization of purely inseparable extensions:

Proposition 12.6. An extension of char \(p\) fields \(L/K\) is purely inseparable iff \(x^{p^k} \in K\) for all \(x \in L\), and some \(k\).

In most cases we want to deal with separable extensions, which are considered nice. Hence the following definition

Definition 12.7. A perfect field is a field where every extension is separable.

A ring of characteristic \(p\) has an endomorphism called the Frobenius endomorphism, \(Fr\), sending \(x \mapsto x^p\). By the binomial theorem, this is indeed a ring homomorphism, and if the ring is reduced, it is injective.

Proposition 12.8. The perfect fields are exactly those where \(Fr\) is surjective and those of characteristic \(0\).

Proof. This follows from our characterization of purely inseparable extensions. □

We can ask the question “when is a finite extension simple?" It is not true in general. For example, consider \(L = \FF _p(t^{\frac {1}{p}},{u}^{\frac {1}{p}})/\FF _p(t,u) = K\), a degree \(p^2\) purely inseparable extension. If this were simple, it would be generated by an element whose \(p^{th}\) power was not in \(K\), but this is true of every element of \(L\).

Lemma 12.9. The group of automorphisms of \(\bar {K}\) fixing \(K\) act transitively on the conjugates of each element.

Proof. This follows from the embedding lemma. □

Corollary 12.10. The conjugates of \(f(\alpha _i)\) over \(K\) where \(f\) is a polynomial in the \(\alpha _i\) are of the form \(f(\beta _i)\), where \(\beta _i\) is a conjugate of \(\alpha _i\).

Proof. This follows from the previous lemma. □

Lemma 12.11. The degree of an automorphism of \(L\) fixing a finite index subfield \(K\) divides \([L:K]_s\)

Proof. Note that an automorphism is determined by its action \(L\), which sends elements to their conjugates. □

Proposition 12.12. If \(L/K\) is a finite separable extension, then \(L\) is simple.

Proof. We may as well assume \(K\) is infinite, as if it is finite we can assume \(K = \FF _p\) and since the order of \(Fr\) divides the degree over \(\FF _p\) by the previous lemma, every element is the root of \(x^{p^k}-x\). Now since there can be at most \(p^l\) roots to \(x^{p^l}-x\) for each \(l\), so the multiplicative group must by cyclic, and any generator is a primitive element.

If \(K\) is infinite, then it suffices by induction to show that \(K(a,b) = K(c)\) if \(a,b\) are separable. Consider elements of the form \(a+kb, k \in K\). We can consider the subset of \(k \in K\) such that the characteristic polynomial of \(a+kb\) is irreducible. Let \(a_i, b_i\) be the conjugates of \(a,b\) respectively, which we can assume are distinct. Now the things of the form \(a_j+kb_i\) are all conjugates, so if they are all distinct, then \(a_j+kb_i\) is primitive. Now setting two of these equal gives the equation \(k = \frac {a_i-a_j}{b_i-b_j}\), so there are finitely many values of \(k\) that are bad. □

This proof secretly uses some Galois theory, and we will see another one later. This one is useful for computations.