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!

  • $\begingroup$ What is a secondsuc field? $\endgroup$ – Kenny Lau Jan 2 at 11:59
  • 1
    $\begingroup$ 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

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 $M$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.