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Let $L/K$ be a finite separable extension of a field $K$. A Galois Closure $M$ of $L/K$ is defined as a minimal degree extension of $L$ for which $M/K$ is Galois.

I want to show that the Galois Closure of $L/K$ exists. Here is my approach:

Let ${e_1,\ldots,e_n}$ be a (finite!) basis for $L$ as a $K$-vector space. Each $e_i$ has a minimal (and separable) polynomial $P_i(x)$. Define $L_0$ to be $L$ with adjoined all of the roots of the $P_i$. It is easy to check that $L_0$ is a finite extension and that it is the splitting field of $A(x)=\prod P_i(x)$. However, if any of the $P_i$ have common roots, then $A(x)$ will not be separable, despite each $P_i$ being separable.

Any idea how to fix this or I have to think in a different way? Any help appreciated.

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    $\begingroup$ Why do you think the $P_i$ having a root in common changes anything ? $\endgroup$ – reuns Oct 11 '18 at 22:21
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    $\begingroup$ Look at the compositum of splitting fields $\endgroup$ – user10354138 Oct 11 '18 at 22:26
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    $\begingroup$ At first a separable extension is an algebraic extension $F/K$ such that every $a \in F$ is separable over $K$ (there is a non-zero polynomial $h \in K[x]$ with $h(a) = 0$). It is a theorem that $K(\alpha)$ and $K(\alpha_1,\ldots,\alpha_n)$ are separable extensions of $K$ when the $\alpha_j$ are separable over $K$. Those theorems come before your question. $\endgroup$ – reuns Oct 11 '18 at 22:28
  • $\begingroup$ @reuns Ah yes! A polynomial is separable when all of its irreducible factors are separable... and since it is true for all $P_i$ we are done. Thank you! $\endgroup$ – DesmondMiles Oct 11 '18 at 22:33
  • $\begingroup$ The compositum of two normal extensions is normal (recall normality can be characterized as: for any homomorphism $\sigma$ to the algebraic closure does not leave the extension, and we also have $\sigma(L_1L_2)=\sigma(L_1)\sigma(L_2)$), and compositum of separable is separable (the $L_2$-minimal polynomial of a primitive element $\alpha$ of $L_1/K$ divides the $K$-minimal polynomial of $\alpha$). $\endgroup$ – user10354138 Oct 11 '18 at 22:59

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