Exercise 12.2.5

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Suppose that $F\subset L$ is the splitting field of a separable polynomial $f\in F[x]$. Also suppose that we have another finite extension $F\subset K$ such that the compositum $KL$ is defined. Prove that $K\subset KL$ is the splitting field of $f$ over $K$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
By hypothesis, $K,L$ are subfields of a field $M$.

Write $\alpha_1,\ldots,\alpha_n$ the roots of $f$ in $L$, so $L = F(\alpha_1,\ldots,\alpha_n)$.
Since $F\subset K$ is a finite extension, there are $\beta_1,\ldots,\beta_m$ in $K$ such that $K = F(\beta_1,\ldots,\beta_m)$. Then $F(\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_m)$ is the smallest subfield of $M$ containing $K$ and $L$, so is the compositum $KL$:
$$KL = F(\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_m).$$

Therefore $$KL = F(\beta_1,\ldots,\beta_m)(\alpha_1,\ldots,\alpha_n) = K(\alpha_1,\ldots,\alpha_n).$$
Since $\alpha_1,\ldots,\alpha_n$ are the roots of $f$ in $K$, a fortiori in $KL$,  $KL$ is the splitting field of the separable polynomial $f$ over $ K$, so $K\subset KL$ is a Galois extension.
\end{proof}

Exercise 12.2.5

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Exercise 12.2.5

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