So, I'm studying from Lang's Undergraduate Algebra. In it, he defines a separable extension E/F in terms of the number of extensions from an embedding of F to an embedding of E. The first property he gives to prove is that any element of a separable extension is separable (in the normal sense, that it's minimal polynomial is separable). As I understand it, some define a separable extension in terms of this property, but I'd like to stick with the way Lang does it. Anyway, I'm having trouble proving this.
Let E/F be a separable extension, let a be an element of E, and let p be it's minimal polynomial. Lang has already taken care of the case of characteristic 0, so let's assume charF = q. (First question: This does imply charE = q, right?)
I'm trying to show that p is separable, or that it doesn't have repeated roots. If p has repeated roots, then gcd(p,p') is not constant. But p is irreducible, so gcd(p,p') is either 1 or p. Let's assume for contradiction's sake that p is not separable, so gcd(p,p') = p. Now, p' (the formal derivative of p) is a polynomial with degree less than p. Then we necessarily have p' = 0. This means p is a polynomial in x^q, right? But in characteristic q, x^q = x. In particular, if p(x) = f(x^q), then f(a) = 0. This contradicts p being the minimal polynomial.
Now, I feel like I must have made a mistake because I didn't use the property of E/F being separable. But otherwise, I can't catch a mistake besides the possibility that maybe x^q = x doesn't apply to a. So, is this proof valid or is there something I'm missing?
Closed.
Best Answer
Here's what I ended up doing:
First, note that the number of extensions of an embedding $\sigma$ of $F$ to an embedding of $F(a)$ for some algebraic element $a$ is less than or equal to the degree of $a$; the extension is uniquely determined by the choice of $\sigma a$, which must be a root of the minimal polynomial $f$ of $a$. Furthermore, $a$ is said to be separable when $f$ is separable, and this is the case if and only if there are deg $a$ extensions of $\sigma$, meaning that $a$ is separable iff $F(a)$ is separable.
Now, suppose we have $a,b$ algebraic elements over $F$. An extension of $\sigma$ to $F(a,b)$ is uniquely determined by $\sigma a$ and $\sigma b$. By considering separately the subfields $F(a)$ and $F(b)$, we see that each pair of choices gives a unique extension. Thus the number of extensions to $F(a,b)$ is the product of the number of distinct roots in the minimal polynomials of $a$ and $b$. Put otherwise, the number of extensions is multiplicative over the tower $F \subset F(a) \subset F(a,b)$. Therefore we have that the number of extensions is less than or equal to deg $a$ times deg $b$, or $[F(a,b):F]$.
By induction, we see that for any finite extension $E/F$, the number of extensions is less than or equal to $[E:F]$.
Let $E/F$ be a separable extension. If $a \in E$, then it is a fact of linear algebra (considering E as a vector space over F) that $E = F(a,a_1,...,a_r)$. Considering the tower $F \subset F(a) \subset F(a,a_1) \subset ... \subset E$, and since the number of extensions of an embedding of $F$ is $[E:F]$, it follows in particular that the number of extensions to $F(a)$ is deg $a$, so $a$ is separable.
There may be a simpler or more elegant way to arrive at this fact, but this proof also takes care of most of the work for the other separability results Lang puts in the book. It works for me.