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Let $L/K$ be a finite extension of non-archimedian henselian valued fields. Why if $L$ is a compositum of unramified subextensions, then finitely of them suffice ? (see Unramified algebraic extensions of local fields).

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  • $\begingroup$ I'm not sure I understand your question. What do you mean by "finitely of them suffice"? $\endgroup$
    – Mathmo123
    Apr 30, 2020 at 21:14
  • $\begingroup$ @Mathmo123 I mean that if $L$ is the compositum of the family $(L_i)_{i \in I}$, then there exists a finite subset $J$ of $I$ such that $L$ is the compositum of the finite subfamily $(L_j)_{j \in J}$. $\endgroup$
    – Eskil
    Apr 30, 2020 at 21:18
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    $\begingroup$ A finite separable extension of fields $L/K$ has only finitely many intermediate fields, so $L$ could not be a compositum of infinitely manny different field extensions of $K$. $\endgroup$
    – KCd
    Apr 30, 2020 at 21:19
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    $\begingroup$ Without separability : when adding the $L_i$ one by one to the compositum you get an increasing tower of fields, that $[L:K]<\infty$ means this tower ends being constant. $\endgroup$
    – reuns
    Apr 30, 2020 at 22:55

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