# 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!

• What is a secondsuc field? – Kenny Lau Jan 2 at 11:59
• The usual terminology is normal closure. Assume that $L = K(\alpha)$. Then $M = K(\alpha_1,\ldots,\alpha_n) = \prod_{j=1}^n K(\alpha_j)$ (compositum of fields) where $\alpha_1,\ldots,\alpha_n$ are the roots of the minimal polynomial $f \in K[x]$ of $\alpha$ so $K(\alpha_j) \cong K(\alpha)$. – reuns Jan 2 at 12:10
• In general $L$ doesn't have to be generated by a single element but you can use induction : that if $\prod_{j=1}^n F_j$ is normal over $K$ and $F_j \cong F_1$ then the normal closure of $\prod_{j=1}^n F_j(\beta)$ is $\prod_{j=1}^n \prod_{l=1}^m F_j(\beta_{j,l})$ where $\beta_{j,l}$ are the roots of $\sigma_j(h) \in F_j[x]$ and $h \in F_1[x]$ is the minimal polynomial of $\beta$ and $\sigma_j$ is the given isomorphism $F_1 \to F_j$. – reuns Jan 2 at 12:10
• The claim is a bit strange. If the field $K$ is infinite and $L/K$ is not Galois, then $[M:L]>1$ and hence $M$ cannot be written as a finite union of proper subspaces over $K$ let alone subfields. In other words the claim is false in that case. On the other hand, if $K$ is finite, then $L/K$ is Galois, and hence equal to its normal closure, making the claim trivial. – Jyrki Lahtonen Jan 3 at 9:57
• I corrected this part. That symbol was to intend as the field generated by the set union of the fields. – Alessandro Pecile Jan 5 at 9:52

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 $$M$$.