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Let $K/F$ be a finite extension. Prove that there is an extension $E/K$ so that $E/F$ is a splitting field of some polynomial in $F[x]$.


If this proof correct? My notation is all over the place -- is there a neater way to do this?

Proof. Since $K/F$ is finite, it is algebraic. Then $K=F(a_1,a_2 \dots,a_n)$ for some finite natural number, $n$. Consider the minimal polynomial, $Mp_i(x)$ of $a_i$, for each $0 \le i \le n$. Then, by construction, $K$ contains a root of $Mp_i(x)$, for each $0 \le i \le n$.

If $F$ contains all the roots of each $Mp_i(x)$, then $Mp_i(x)$ splits in $K$ and we are done. Suppose $K$ does not contain all the roots of each $Mp_i(x)$. Then for some $i$, $\exists b_1,b_2, \dots,b_m \notin K$ s.t. $b_1,b_2, \dots,b_m$ are roots of $Mp_i(x)$. Let $E$ be the splitting field of this $Mp_i(x)$. Then by $K=F(a_1,a_2 \dots,a_n)\subseteq F(a_1,a_2,\dots,a_n,b_1,b_2, \dots,b_m)=K(b_1,b_2, \dots,b_m)=E \implies K \subseteq E$.

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  • $\begingroup$ LaTex would certainly make the proof more readable; do you know how to use it on this site? $\endgroup$ – Zach Effman Apr 7 '15 at 15:44
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    $\begingroup$ Do I take it right that $D = E$? $\endgroup$ – Andreas Caranti Apr 7 '15 at 16:18
  • $\begingroup$ I have taken some time to edit your question into proper LaTeX (except that I believe $D = E$, but I leave this change to you). Next time, please, please, please, write it up in LaTeX yourself. Just look at the source code of what I wrote, to see what you should have done. If you don't do that, your questions will be hard to read, you may get downvotes, close votes, and other plagues. $\endgroup$ – Andreas Caranti Apr 7 '15 at 16:23
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    $\begingroup$ Re "finite natural number": are you afraid of the non-standard analysis crowd? $\endgroup$ – Marc van Leeuwen Apr 7 '15 at 16:34
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You're nearly there. Hint: consider the splitting field of the product of all $Mp_{i}(x)$.

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