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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!

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  • $\begingroup$ What is a secondsuc field? $\endgroup$ – Kenny Lau Jan 2 at 11:59
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    $\begingroup$ 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 $K(\alpha_j) \cong K(\alpha)$. $\endgroup$ – reuns Jan 2 at 12:10
  • $\begingroup$ 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 $F_1 \to F_j$. $\endgroup$ – reuns Jan 2 at 12:10
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jan 3 at 9:57
  • $\begingroup$ I corrected this part. That symbol was to intend as the field generated by the set union of the fields. $\endgroup$ – Alessandro Pecile Jan 5 at 9:52
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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 $M$.

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