Composition of separability
Some people asked me whether F >= E >= K separable implies F >= K separable.
(Here >= means greater than or equal to.)
Here's my argument yes.
Lemma 1. A >= B >= C >= D, with A >= D separable, implies B >= C separable.
Proof. A >= D is a condition on individual elements of A, so we can restrict to A=B.
Given b in B, let f(x) be its minimal polynomial in D[x], separable.
This factors in C[x], one factor of which (still separable) is the new min poly of b.
QED.
Lemma 2. If A >= B is separable and purely inseparable, they are equal.
We did this in class.
Okay, now for the theorem.
Let f be the element we want to test, with m(x) in K[x] its minimal polynomial.
If m(x) isn't separable, then (by the derivative and GCD trick from the homework) all its exponents are divisible by pb for some positive power of p; choose b largest. Then m(x) = n(xpb) for a unique n(x), necessarily separable and irreducible. In particular fpb is separable over K.
Now consider the chain F >= E[f] >= E[fpb] >= E.
The middle extension is purely inseparable. By lemma 1, it's separable. Hence by lemma 2, E[f] = E[fpb].
Now we have the chain E[fpb] = E[f] >= K[f] >= K.
Since fpb is separable over K, this big extension is separable;
hence by lemma 1, K[f] is separable over K, which was to be proven.
(Here >= means greater than or equal to.)
Here's my argument yes.
Lemma 1. A >= B >= C >= D, with A >= D separable, implies B >= C separable.
Proof. A >= D is a condition on individual elements of A, so we can restrict to A=B.
Given b in B, let f(x) be its minimal polynomial in D[x], separable.
This factors in C[x], one factor of which (still separable) is the new min poly of b.
QED.
Lemma 2. If A >= B is separable and purely inseparable, they are equal.
We did this in class.
Okay, now for the theorem.
Let f be the element we want to test, with m(x) in K[x] its minimal polynomial.
If m(x) isn't separable, then (by the derivative and GCD trick from the homework) all its exponents are divisible by pb for some positive power of p; choose b largest. Then m(x) = n(xpb) for a unique n(x), necessarily separable and irreducible. In particular fpb is separable over K.
Now consider the chain F >= E[f] >= E[fpb] >= E.
The middle extension is purely inseparable. By lemma 1, it's separable. Hence by lemma 2, E[f] = E[fpb].
Now we have the chain E[fpb] = E[f] >= K[f] >= K.
Since fpb is separable over K, this big extension is separable;
hence by lemma 1, K[f] is separable over K, which was to be proven.
Labels: special request
posted by Allen Knutson @ 5:45 PM
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