§ Separable extensions via derivation

§ If α\alpha separable, then derivation over KK lifts uniquely to K(α)K(\alpha)

\begin{aligned} D'(f(\alpha)) \equiv f^D(\alpha) - f'(\alpha) \frac{\pi^D(\alpha)}{pi'(\alpha)} \end{aligned}
  • To prove this, we start by assuming DD has an extension, and then showing that it must agree with DD'. This tells us why it must look this way.
  • Then, after doing this, we start with DD' and show that it is well defined and obeys the derivation conditions. This tells us why it's well-defined .

§ Non example: derivation that does not extend in inseparable case

  • Consider Fp(u)F_p(u) as the base field, and let L=Fp(u)(α)L = F_p(u)(\alpha) where α\alpha is a root of XpuFp(u)[x]X^p - u \in F_p(u)[x]. This is inseparable over KK.
  • The uu derivative on Fp(u)F_p(u) [which treats uu as a polynomial and differentiates it ] cannot be extended to LL.
  • Consider the equation αp=u\alpha^p = u, which holds in LL, since α\alpha was explicitly a root of XpuX^p - u.
  • Applying the uu derivative gives us pαp1D(α)=D(u)p \alpha^{p-1} D(\alpha) = D(u). The LHS is zero since we are in characteristic pp. The RHS is 1 since DD is the uu derivative, and so D(u)=1D(u) = 1. This is a contradiction, and so DD does not exist [any mathematical operation must respect equalities ].

§ Part 2.a: Extension by inseparable element α\alpha does not have unique lift of derivation for K(α)/KK(\alpha)/K

  • Let αL\alpha \in L be inseparable over KK. Then π(X)=0\pi'(X) = 0 where π(X)\pi(X) is the minimal polynomial for αL\alpha \in L.
  • In particular, π(α)=0\pi'(\alpha) = 0. We will use the vanishing of π(α)\pi'(\alpha) to build a nonzero derivation on K(α)K(\alpha) which extends the zero derivation on KK.
  • Thus, the zero derivation on KK has two lifts to K(α)K(\alpha): one as the zero derivation on K(α)K(\alpha), and one as our non-vanishing lift.
  • Define Z:K(α)K(α)Z: K(\alpha) \to K(\alpha) given by Z(f(α))=f(α)Z(f(\alpha)) = f'(\alpha) where f(x)K[x]f(x) \in K[x]. By doing this, we are conflating elements lK(α)l \in K(\alpha)with elements of the form ikiαi=f(α)\sum_i k_i \alpha^i = f(\alpha). We need to check that this is well defined, that if f(α)=g(α)f(\alpha) = g(\alpha), then Z(f(α))=Z(g(α))Z(f(\alpha)) = Z(g(\alpha)).
  • So start with f(α)=g(α)f(\alpha) = g(\alpha). This implies that f(x)g(x)f(x) \equiv g(x) modulo π(x)\pi(x).
  • So we write f(x)=g(x)+k(x)π(x)f(x) = g(x) + k(x)\pi(x).
  • Differentiating both sides wrt xx, we get f(x)=g(x)+k(x)π(x)+k(x)π(x)f'(x) = g'(x) + k'(x) \pi(x) + k(x) \pi'(x).
  • Since π(α)=π(α)=0\pi(\alpha) = \pi'(\alpha) = 0, we get that f(α)=g(α)+0f'(\alpha) = g'(\alpha) + 0 by evaluating previous equation at α\alpha.
  • This shows that Z:K(α)K(α)Z: K(\alpha) \to K(\alpha) is well defined.
  • See that the derivation ZZ kills KK since K=Kα0K = K \alpha^0. But we see that Z(α)=1Z(\alpha) = 1, so ZZ extends the zero derivation on KK while not being zero itself.
  • We needed separability for the derivation to be well-defined.

§ Part 2.b: Inseparable extension can be written as extension by inseparable element

  • Above, we showed that if we have K(α)/KK(\alpha)/K where α\alpha inseparable, then derivations cannot be uniquely lifted.
  • We want to show that if we have L/KL/K inseparable, then derivation cannot be uniquely lifted. But this is not the same!
  • L/KL/K inseparable implies that there is some αL\alpha \in L which is inseparable, NOT that L=K(α)/KL = K(\alpha)/K is inseparable!
  • So we either need to find some element α\alpha such that L=K(α)L = K(\alpha) [not always possible ], or find some field FF such that L=F(α)L = F(\alpha) and α\alpha is inseparable over FF.
  • Reiterating: Given L/KL/K is inseparable, we want to find some F/KF/K such that L=F(α)L = F(\alpha) where α\alpha is inseparable over FF.
  • TODO!

§ Part 1 + Part 2: Separable iff unique lift

  • Let L/KL/K be separable. By primitive element theorem, L=K(α)L = K(\alpha) for some αL\alpha \in L, α\alpha separable over KK.
  • Any derivation of KK can be extended to a derivation of LL from results above. Thus, separable implies unique lift.
  • Suppose L/KL/K is inseparable. Then we can write L=F(α)/KL = F(\alpha)/K where α\alpha is inseparable over FF, and KFLK \subseteq F \subseteq L.
  • Then by Part 2.a, we use the ZZ derivation to non-zero derivation on LL that is zero on FF. Since it is zero on FF and KFK \subseteq F, it is zero on KK.
  • This shows that if L/KL/K is inseparable, then there are two ways to lift the zero derivation, violating uniqueness.

§ Lemma: Derivations at intermediate separable extensions

  • Let L/KL/K be a finite extension, and let F/KF/K be an intermediate separable extension. So KFLK \subseteq F \subseteq L and F/KF/K is separable.
  • Then we claim that every derivation D:FLD: F \to L that sends KK to KK has values in FF. (ie, it's range is only FF, not all of LL).
  • Pick αF\alpha \in F, so α\alpha is separable over KK. We know what the unique derivation looks like, and it has range only FF.

§ Payoff: An extension L=K(α1,,αn)L = K(\alpha_1, \dots, \alpha_n) is separable over KK iff αi\alpha_i are separable

  • Recursively lift the derivations up from K0KK_0 \equiv K to Ki+1Ki(αi)K_{i+1} \equiv K_i(\alpha_i). If the lifts all succeed, then we have a separable extension. If the unique lifts fail, then the extension is not separable.
  • The lift can only succeed to uniquely lift iff the final extension LL is separable.