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Tag 030M

Chapter 9: Fields > Section 9.27: Linearly disjoint extensions

Lemma 9.27.3. Let $E/F$ be a normal algebraic field extension. There exist subextensions $E / E_{sep} /F$ and $E / E_{insep} / F$ such that

  1. $F \subset E_{sep}$ is Galois and $E_{sep} \subset E$ is purely inseparable,
  2. $F \subset E_{insep}$ is purely inseparable and $E_{insep} \subset E$ is Galois,
  3. $E = E_{sep} \otimes_F E_{insep}$.

Proof. We found the subfield $E_{sep}$ in Lemma 9.14.6. We set $E_{insep} = E^{\text{Aut}(E/F)}$. Details omitted. $\square$

    The code snippet corresponding to this tag is a part of the file fields.tex and is located in lines 3424–3435 (see updates for more information).

    \begin{lemma}
    \label{lemma-normal-case}
    Let $E/F$ be a normal algebraic field extension. There exist subextensions
    $E / E_{sep} /F$ and $E / E_{insep} / F$ such that
    \begin{enumerate}
    \item $F \subset E_{sep}$ is Galois and $E_{sep} \subset E$
    is purely inseparable,
    \item $F \subset E_{insep}$ is purely inseparable and $E_{insep} \subset E$
    is Galois,
    \item $E = E_{sep} \otimes_F E_{insep}$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    We found the subfield $E_{sep}$ in Lemma \ref{lemma-separable-first}.
    We set $E_{insep} = E^{\text{Aut}(E/F)}$. Details omitted.
    \end{proof}

    Comments (2)

    Comment #581 by Wei Xu on May 20, 2014 a 4:30 pm UTC

    Line 2511, a typo: "There exist subextensions $E / E_{sep} /F'$" should be "There exist subextensions $E / E_{sep} /F$".

    In the "Waring part", Line 2482 -- Line 2491, the field of rational numbers should all be denoted $\mathbf{Q}$.

    Comment #595 by Johan (site) on May 23, 2014 a 8:10 pm UTC

    Thanks! Fixed here.

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