Exercise 12.2.4

Question

\def\imp{\Rightarrow}
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\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}}

In the discussion preceding (12.25), we have extensions $F\subset L$, which is a splitting field of $f \in F[x]$, and $F \subset K = F(\beta)$, where $\beta$ is a root of an irreducible polynomial in $F[x]$. Given the many ways in which extension fields can be constructed, these extensions might not have to do with each other. Prove that there is an extension $F\subset M$ that contains subfields $F \subset L_1 \subset M$ and $F \subset K_1 \subset M$ such that $L_1,K_1$ are isomorphic to $L,K$, respectively, where the isomorphisms are the identity on $F$. Thus, by replacing $L,K$ with the isomorphic fields $L_1,K_1$, we can assume that $L,K$ lie in a larger field, as claimed in the text.

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}
Let $g$ be the minimal polynomial of $\beta$ over $F$, and $M$ a splitting field of $fg$ over $F$. Write $\alpha_1,\ldots,\alpha_n$ the roots of $f$ in $M$, and $\beta_1,\ldots,\beta_m$ the roots of $g$ in $M$.

Then $L_1 = F(\alpha_1,\ldots,\alpha_n)$ is a splitting field of $f$ over $F$. Since $L,L_1$ are splitting fields of $f$ over $F$,  there exists by Corollary 5.1.7 an isomorphism $\varphi : L \simeq L_1$ which is the identity on $F$.

Write $K_1 = F(\beta_1)$. Since $g$ is irreducible over $F$, $K_1 \simeq F[x]/\langle g angle \simeq K$, where the isomorphisms are the identity on $F$. Here $K_1, L_1$ are subfields of $M$.

Thus, by replacing $L,K$ with the isomorphic fields $L_1,K_1$, we can assume that $L,K$ lie in a larger field $M$.
\end{proof}

Exercise 12.2.4

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Exercise 12.2.4

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