Exercise 14.2.8

Question

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The goal of this exercise is to relate Definition 14.2.2 to Galois's definition of not primitive. Let $f \in F[x]$ be monic, separable, and irreducible with splitting field $F \subset L$. Also assume that $f$ is imprimitive with blocks of roots given by $R_1,\ldots,R_m$, where each block has $n$ elements (thus $\deg(f) = mn$). Let $f_i$ be the monic polynomial whose roots are the elements of $R_i$, and let $K \subset L$ be the fixed field of $$\{\sigma \in \Gal(L/F)\, | \, \sigma(R_i) = R_i 	ext{ for all } i\}.$$
\be
\item[(a)] Show that $f = \prod_{i=1}^m f_i$ and that $f_i \in K[x]$ for all $i$.
\item[(b)] In Galois' definition, $K$ is obtained by adjoining the roots of a separable polynomial of degree $m$. In modern terms, Galois wants $F \subset K$ to be Galois extension such that $\Gal(K/F)$ (*) is isomorphic to a subgroup of $S_m$. Prove that the field $K$ defined in part (a) has these properties.
See Exercise 14 for some examples.

[(*) misprint in Cox.]
\ee

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 }

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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}\,}

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

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\begin{proof}
\be
\item[(a)] By Definition 14.2.2, $R = R_1 \cup \cdots \cup R_m$ (disjoint union) is the set of roots of $f$. Since $f$ is separable, by definition of $f_i$,
$$f = \prod_{\alpha \in R} (x - \alpha) = \prod_{i=1}^m \prod_{\alpha \in R_i} (x - \alpha) = \prod_{i=1}^m f_i.$$
Let 
$$G = \{\sigma \in \Gal(L/F)\, | \, \forall i \in \gcro 1, m \dcro, \ \sigma(R_i) = R_i \}.$$
If $G_i = \{\sigma \in \Gal(L/F)\, | \, \sigma(R_i) = R_i \}$ for $i = 1,\ldots,n$, then $G = \bigcap\limits_{i=1}^m G_i$.

Each $G_i$ is a subgroup of $\Gal(L/F)$: $e(R_i) = R_i$, and if $\sigma, 	au \in R_i$, then  $(\sigma 	au)(R_i) = \sigma(R_i) = R_i$ and $R_i = \sigma^{-1}(R_i)$. Therefore $G = \bigcap\limits_{i=1}^m G_i$ is a subgroup of $\Gal(L/F)$.

Let $K = L_G$ be the fixed field of $G$. By the Galois correspondence, $G = \Gal(L/K)$.

If $\sigma \in G_i$, since $\sigma(R_i) = R_i$, where le restriction of $\sigma$ to $R_i$ is bijective, then 
$$\sigma \cdot f_i = \prod_{\alpha \in R_i} (x - \sigma(\alpha)) = \prod_{\beta \in R_i} (x - \beta) = f_i, \qquad (\beta = \sigma(\alpha)).$$
Therefore, if $\sigma \in G = \bigcap\limits_{i=1}^m G_i$, then for all $i \in \{1,\ldots,m\}$, $\sigma\cdot f_i = f_i$. The coefficients of $f_i$ are in the fixed field $K$ of $G$, so that
$$f_i\in K[x],\ i=1,\ldots,m.$$
To give a first example, $f = x^4 - 2$ is imprimitive with blocks $$R_1 =\{\sqrt[4]{2},-\sqrt[4]{2}\},R_2 = \{i \sqrt[4]{2},-i\sqrt[4]{2}\}.$$

If $	au,\sigma$ are defined by $	au(\sqrt[4]{2}) = \sqrt[4]{2}, 	au(i) = -i$, and $\sigma(\sqrt[4]{2}) = i \sqrt[4]{2}, \sigma(i) = i$, then
$$\Gal(L/F) = \{e,\sigma,\sigma^2,\sigma^3, 	au, \sigma 	au, \sigma^2 	au, \sigma^3	au\} \simeq D_8.$$
Here $G = G_1 = G_2 = \{e,\sigma^2,	au,\sigma^2 	au\}$, and $K = L_G = \Q(\sqrt{2})$ (see Ex. 6.3.2 and Ex. 7.3.3).

We verify $f_1(x) = x^2 -\sqrt{2}, f_2(x) = x^2 + \sqrt{2} \in K[x]$.

\item[(b)] We prove that $G$ is a normal subgroup of $\Gal(L/F)$.

Let $\lambda \in \Gal(L/F)$, and $\sigma \in G$. Since $f$ is imprimitive, $\lambda \in \Gal(L/F)$ permutes the blocks $R_i$: there exists $	au \in S_m$ such that
$$\lambda(R_i) = R_{	au(i)},\ i=1,\ldots,m.$$
Let $j$ be any fixed index in $\{1,\ldots,m\}$, and $i$ such that $	au(i) = j$. Since $\sigma \in G \supset G_i$, $\sigma(R_i) = R_i$, thus
$$(\lambda \sigma \lambda^{-1})(R_j) = (\lambda \sigma)(R_i) = \lambda(R_i) = R_j.$$
Since this is true for all $j \in \{1,\ldots,m\}$, $\lambda \sigma \lambda^{-1} \in G$. This proves that $G$ is a normal subgroup of $\Gal(L/F)$. Therefore $F \subset K$ is a Galois extension (Theorem 7.2.5).

Now we prove that $\Gal(K/F)$ is isomorphic to a subgroup of $S_m$.

Since $f$ is imprimitive with blocks of roots given by $R_1,\ldots,R_m$, for each $\sigma \in \Gal(L/F)$, there exists $	au \in S_m$ such that $\sigma(R_i) = R_{	au(i)},\ i=1,\ldots,m$.
Consider the map $\varphi$ sending $\sigma$ to $	au$:
$$
\varphi
\left\{
\begin{array}{ccl}
\Gal(L/F) & 	o & S_m\
\sigma & \mapsto &	au : \forall i \in \gcro 1,m \dcro,\ \sigma(R_i) = R_{	au(i)}.
\end{array}
ight.
$$
Then , for all $\sigma \in \Gal(L/F)$,
$$\varphi(\sigma) = ()  \iff \forall i \in \gcro 1,m \dcro,\ \sigma(R_i) = R_i \iff \sigma \in G.$$
Therefore, $\ker(\varphi) = G$, and by the Galois correspondence (see part (a)) $G = \Gal(L/K)$, so that
$$\ker(\varphi) = G = \Gal(L/K).$$
Then, by Theorem 7.3.2, since $K$ is Galois over $F$,
$$ S_m \supset \mathrm{im}(\varphi) \simeq \Gal(L/F)/ \ker(\varphi) = \Gal(L/F)/\Gal(L/K) \simeq \Gal(K/F).$$
Therefore $\Gal(K/F)$ is isomorphic to the subgroup $\mathrm{im}(\varphi)$ of $S_m$.
\ee
\end{proof}

Exercise 14.2.8

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Exercise 14.2.8

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