§ Separable extensions via derivation
- Let $R$ be a commutative ring, $M$ an $R$-module. A derivation is a map such that $D(a + b) = D(a) + D(b)$ and $D(ab) = aD(b) + D(a)b$ [ie, the calculus chain rule is obeyed ].
- Note that the map does not need to be an $R$-homomorphism (?!)
- The elements of $R$ such that $D(R) = 0$ are said to be the constants of $R$.
- The set of constants under $X$-differentiation for $K[X]$ in char. 0 is $K$, and $K[X^p]$ in char. p
- Let $R$ be an integral domain with field of fractions $K$. Any derivation $D: R \to K$ uniquely extends to $D': K \to K$ given by the quotient rule: $D'(a/b) = (bD(a) - aD(b))/b^2$.
- Any derivation $D: R \to R$ extends to a derivation $(.)^D: R[x] \to R[x]$. For a $f = \sum_i a_i x^i \in R[x]$, the derivation is given by $f^D(x) \equiv \sum_i D(a_i) X^i$. This applies $D$ to $f(x)$ coefficientwise.
- For a derivation $D: R \to R$ with ring of constants $C$, the associated derivation $(.)^D: R[x] \to R[x]$ has ring of constants $C[x]$.
- Key thm: Let $L/K$ be a field extension and let $D: K \to K$ be a derivation. $D$ extends uniquely to $D_L$ iff $L$ is separable over $K$.
§ If $\alpha$ separable, then derivation over $K$ lifts uniquely to $K(\alpha)$
- Let $D: K \to K$ be a derivation.
- Let $\alpha \in L$ be separable over $K$ with minimal polynomial $\pi(X) \in K[X]$.
- So, $\pi(X)$ is irreducible in $K[X]$, $\pi(\alpha) = 0$, and $\pi'(\alpha) \neq 0$.
- Then $D$ has a unique extension $D': K(\alpha) \to K(\alpha)$ given by:
\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 $D$ has an extension, and then showing that it must agree with $D'$. This tells us why it must look this way.
- Then, after doing this, we start with $D'$ 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 $F_p(u)$ as the base field, and let $L = F_p(u)(\alpha)$ where $\alpha$ is a root of $X^p - u \in F_p(u)[x]$. This is inseparable over $K$.
- The $u$ derivative on $F_p(u)$ [which treats $u$ as a polynomial and differentiates it ] cannot be extended to $L$.
- Consider the equation $\alpha^p = u$, which holds in $L$, since $\alpha$ was explicitly a root of $X^p - u$.
- Applying the $u$ derivative gives us $p \alpha^{p-1} D(\alpha) = D(u)$. The LHS is zero since we are in characteristic $p$. The RHS is 1 since $D$ is the $u$ derivative, and so $D(u) = 1$. This is a contradiction, and so $D$ does not exist [any mathematical operation must respect equalities ].
§ Part 2.a: Extension by inseparable element $\alpha$ does not have unique lift of derivation for $K(\alpha)/K$
- Let $\alpha \in L$ be inseparable over $K$. Then $\pi'(X) = 0$ where $\pi(X)$ is the minimal polynomial for $\alpha \in L$.
- In particular, $\pi'(\alpha) = 0$. We will use the vanishing of $\pi'(\alpha)$ to build a nonzero derivation on $K(\alpha)$ which extends the zero derivation on $K$.
- Thus, the zero derivation on $K$ has two lifts to $K(\alpha)$: one as the zero derivation on $K(\alpha)$, and one as our non-vanishing lift.
- Define $Z: K(\alpha) \to K(\alpha)$ given by $Z(f(\alpha)) = f'(\alpha)$ where $f(x) \in K[x]$. By doing this, we are conflating elements $l \in K(\alpha)$with elements of the form $\sum_i k_i \alpha^i = f(\alpha)$. We need to check that this is well defined, that if $f(\alpha) = g(\alpha)$, then $Z(f(\alpha)) = Z(g(\alpha))$.
- So start with $f(\alpha) = g(\alpha)$. This implies that $f(x) \equiv g(x)$ modulo $\pi(x)$.
- So we write $f(x) = g(x) + k(x)\pi(x)$.
- Differentiating both sides wrt $x$, we get $f'(x) = g'(x) + k'(x) \pi(x) + k(x) \pi'(x)$.
- Since $\pi(\alpha) = \pi'(\alpha) = 0$, we get that $f'(\alpha) = g'(\alpha) + 0$ by evaluating previous equation at $\alpha$.
- This shows that $Z: K(\alpha) \to K(\alpha)$ is well defined.
- See that the derivation $Z$ kills $K$ since $K = K \alpha^0$. But we see that $Z(\alpha) = 1$, so $Z$ extends the zero derivation on $K$ 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 $K(\alpha)/K$ where $\alpha$ inseparable, then derivations cannot be uniquely lifted.
- We want to show that if we have $L/K$ inseparable, then derivation cannot be uniquely lifted. But this is not the same!
- $L/K$ inseparable implies that there is some $\alpha \in L$ which is inseparable, NOT that $L = K(\alpha)/K$ is inseparable!
- So we either need to find some element $\alpha$ such that $L = K(\alpha)$ [not always possible ], or find some field $F$ such that $L = F(\alpha)$ and $\alpha$ is inseparable over $F$.
- Reiterating: Given $L/K$ is inseparable, we want to find some $F/K$ such that $L = F(\alpha)$ where $\alpha$ is inseparable over $F$.
- TODO!
§ Part 1 + Part 2: Separable iff unique lift
- Let $L/K$ be separable. By primitive element theorem, $L = K(\alpha)$ for some $\alpha \in L$, $\alpha$ separable over $K$.
- Any derivation of $K$ can be extended to a derivation of $L$ from results above. Thus, separable implies unique lift.
- Suppose $L/K$ is inseparable. Then we can write $L = F(\alpha)/K$ where $\alpha$ is inseparable over $F$, and $K \subseteq F \subseteq L$.
- Then by Part 2.a, we use the $Z$ derivation to non-zero derivation on $L$ that is zero on $F$. Since it is zero on $F$ and $K \subseteq F$, it is zero on $K$.
- This shows that if $L/K$ is inseparable, then there are two ways to lift the zero derivation, violating uniqueness.
§ Lemma: Derivations at intermediate separable extensions
- Let $L/K$ be a finite extension, and let $F/K$ be an intermediate separable extension. So $K \subseteq F \subseteq L$ and $F/K$ is separable.
- Then we claim that every derivation $D: F \to L$ that sends $K$ to $K$ has values in $F$. (ie, it's range is only $F$, not all of $L$).
- Pick $\alpha \in F$, so $\alpha$ is separable over $K$. We know what the unique derivation looks like, and it has range only $F$.
§ Payoff: An extension $L = K(\alpha_1, \dots, \alpha_n)$ is separable over $K$ iff $\alpha_i$ are separable
- Recursively lift the derivations up from $K_0 \equiv K$ to $K_{i+1} \equiv K_i(\alpha_i)$. 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 $L$ is separable.