2
$\begingroup$

Let L be a finite extension of K and let $ K\subseteq M \subseteq L$. Show that if the extensions $M \subseteq L$ and $K \subseteq M$ are separable, then $K \subseteq L$ is separable.

Here is what we know:

$1. M \subseteq$ L separable: $\forall l \in L$ the minimal polynomial $m_{l,M}(x)$ is separable in $M$.

$2. K \subseteq$ M separable: $\forall n \in L$ the minimal polynomial $m_{n,K}(x)$ is separable in $K$.

What we want:

$ K \subseteq$ L separable: $\forall l \in L$ the minimal polynomial $m_{l,K}(x)$ is separable in $K$.

If $l\in M$, then it works by hypothesis. However, if $l \in L/M,$ then I don't know how to do it.

  • 0
    Have you introduced the seperable degree? If yes, what theorems did you have? – 2017-02-16
  • 0
    I know that for a $f(x)$, then $deg f(x) = deg(f_{\text{sep}}(x)) * deg f_{\text{ins}}(x)$. Furthermore, $deg f_{\text{ins}}(x)=p^k$ for a $k \geq 0$. – 2017-02-16

1 Answers 1

0

All you have to do is take $\theta\in L$ and show that it is separable in $K$. Take the irreducible polynomial of $\theta$ over $M$, $\mathrm{Irr}(\theta,M)(X)\in M[X]$ and denote its coefficients by $a_0,\dots,a_n$. Since $M|K$ is a separable extension, the coefficients are separable over $K$, so we have that the extension $K(a_0,\dots,a_n)|K$ is separable.

On the other hand, the fact that $\theta$ is separable over $K(a_0,\dots,a_n)$ implies that the irreducible polynomial $\mathrm{Irr}(\theta,M)(X)$ does not have roots with multiplicity greater than one. Also, $\mathrm{Irr}(\theta,M)(X)\in K(a_0,\dots,a_n)[X]$.

Hence, $$[K(a_0,\dots,a_n,\theta):K]_s=[K(a_0,\dots,a_n,\theta):K(a_0,\dots,a_n)]_s[K(a_0,\dots,a_n):K]_s=[K(a_0,\dots,a_n,\theta):K(a_0,\dots,a_n)][K(a_0,\dots,a_n):K]=[K(a_0,\dots,a_n,\theta):K].$$ Therefore, $\theta$ is separable over $K$, and from that follows the result.

$\textbf{Note}:$ Remember that in a finite extension of fields $K|k$, the fact that every element of $K$ is separable over $k$ is equivalent to $[K:k]_s=[K:k]$ where $[K:k]_s$ denotes the separability degree of the extension.

  • 0
    I have a few questions concerning your answer. First of all, you used the fact that if $K \subseteq M \subseteq L$, then $[L:K]_{s} = [L:M]_{s} [M:K]_{s} $. How do you know that this is true? Also, how do you know that these extensions are finitely generated? a finite extension only means that $[L:K] < \infty$, not to be confused with a finite type extension.) Finally, how do you know $[K(a_0,…,a_n,θ):K(a_0,…,a_n)]_s $ is separable? I dont understand. – 2017-02-16
  • 0
    For your first question, it is known that given $\textbf{any}$ algebraic extensions $K|k$ and $L|K$, $[L:k]_s=[L:K]_s[K:k]_s$ (you can find this result in a variety of books). On the other hand, you didn't say anything about finite type extensions, you said a "finite extension of $K$"... and the result I proved is actually true for any algebraic separable extension, regardless of its degree. For the last point, $\theta$ is separable over $K(a_0,\dots,a_n)$, therefore $K(a_0,\dots,a_n,\theta)|K(a_0,\dots,a_n)$ is separable. Hope these answers worked for you! – 2017-02-16