Let $F \subseteq E$ be an arbitrary algebraic extension. I want to show that there exists $L \supseteq E$ such that $L$ is a normal closure for $E$ over $F$ and this extension is unique up to $E$-isomorphism.

First, I have to give the definition of algebraic closure. $L$ is a normal closure for $E$ over $F$, if $E$ is normal over $F$ and there is no field $K$ ($F\subseteq E \subseteq K \subset L$) such thah $K$ is normal over $F$.

I have a hint: I should work inside an algebraic closure $\bar{E}$ of $E$ for existance and uniqueness.

How can I construct the desired extension in algebraic closure?

  • $\begingroup$ See the comments here. $\endgroup$ – Dietrich Burde Jun 3 '17 at 18:42
  • $\begingroup$ Instead of looking at the algebraic closure, I would show this lemma about uniqueness of splitting field : Let $K/F,L/F$ be two finite extensions such that : $K$ is the splitting field of $f_1,\ldots,f_n$ and $L$ is the splitting field of $g_1,\ldots,g_m$. If each $g_i$ splits in $K$ and each $f_j$ splits in $L$, then $K \cong L$ $\endgroup$ – reuns Jun 3 '17 at 20:49

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