Can anyone help me with this problem? I can't get through very well. It says:
Let $L/k$ an extension. Show that the set
$\{\alpha \in L : $ $\alpha$ is separable over $k\}$ is a subfield of $L$ that contains $k$
Well I tried the following. We know by hypothesis that k is in L, the K is an intermediate field, and mi set is defined as {α∈L: α is separable over k}, the i try to use tower law... But I got confused
Any hint or sketch of proof is wellcome.
Let's call that set $k_{sep}$. As the polynomial $x - a \in k[x]$ is separable for all $a \in k$, it's clear that $k \subseteq k_{sep}$. Now to show that it's a subfield of $L$, we have to show it's closed under the field arithmetic inherited from $L$. In other words, for $\alpha, \beta \in k_{sep}$ that $\alpha + \beta, \alpha \beta \in k_{sep}$. This will show that's it a subring of $L$. To show that it's a subfield, you have to show that if $\alpha \neq 0$ that $\alpha^{-1} \in k_{sep}$. As you can imagine, doing this manually would be obscenely difficult. After all, what's the relationship between the minimal polynomial of $\alpha$ and $\alpha^{-1}$? However, you mentioned that you have the tower law so here's a hint: use the fact that $\alpha + \beta, \alpha \beta, \alpha^{-1} \in k(\alpha, \beta)$.