Exercise about split closures (Galois Theory) I am studing Galois Theory. I am using the books by Kaplansky, Fields and Rings. I am stuck doing this exercise:
Let $M$ be a split closure of $L$ over $K$ ($M,L,K$ are all fields). Prove that $M=L_1\cup \dots \cup L_r$ (the field generated  by the set union, not the set union itself) where $L_i$ is isomorphic to $L$ over $K$.
The actuall problem is that I do not have fully understood the concept of split closure; in the book it is defined in this way.
Let $K \subset L$ be fields and $[L:K]$ finite.There exists a field $M$ containing $L$ such that $M$ is a splitting field over $K$ and no field othen than $M$ between $M$ and $L$ is a splitting field over $K$. If $M_0$ is a second such field, then there exists an isomorphism of $M$ onto $M_0$ which is the identity on $L$. If $L$ is separable then $M$ is normal over $K$.
We shall call a field having the properties of $M$ a split closure of $L$ over $K$. If $L$ is separable we call $M$ normale closure.
Then problem is that the professor in class did not do man examples, so could you please give me some example of split/normal closure, emphasize their difference and the idea behind the introduction of this concept.
Thank you!
 A: First, I don't think that the union will be finite if the extension is not finite, it will be understood in the following arguments. So I will suppose that $L/K$ is a finite field extension.
Intuitively. As $L/K$ is a finite field extension, we have that $L=K(\alpha_1,\ldots,\alpha_n)$, in such a way that we have a tower of field
$$L\supset K(\alpha_1,\ldots,\alpha_{n-1})\supset\cdots\supset K$$
with non-trivial steps. Let be $f(x)$ the product of the minimal polynomials associated to each step in the previous tower of fields. We have that $L$ is the splitting field of $f(x)$. Construct any tower of fields (the details are a gift for you) this way, using the minimal polynomials (ordered) in the preceding tower of fields and you will get a field isomorphic to $L$. The union, is $K$ attached to all the roots of $f(x)$, so is $M$.
Anyways, you can do the same, using the $\mbox{Aut}\left(M/K\right)$ and the elements of that group acting on $L$ will give you a family of intermediate fields that are isomorphic to $L$ (including $L$), and thinking in the previous idea, you will get that its union is all $M$.
