Exercise 14.2.14

Question

\def\imp{\Rightarrow}
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ewcommand{\ssi} {si et seulement si }

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Here are some examples to illustrate Galois's definition of imprimitive. We will use the notation of Exercise 8. Let $F$ be a field of characteristic different from 2 or 3.
\be
\item[(a)] Let $f = x^6 + bx^4+cx^2 +d \in F[x]$ be irreducible with splitting field $F \subset L$. Show that the splitting field of $x^3+bx^2+cx+d$ gives an intermediate field $F\subset K \subset L$ such that $F \subset K$ is Galois and $f=f_1f_2f_3$, where $f_i \in K[x]$ has degree 2 for $i=1,2,3$. Also explain how $K$ relates to the field $K$ constructed in Exercise 8.
\item[(b)] Work out the analogous theory when $f = x^6 + bx^3+c \in F[x]$ is irreducible.
\ee

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
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\def\dcro{\mbox{]\hspace{-.15em}]}}

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\begin{proof}

\item[(a)] By hypothesis, $f = x^6 + bx^4+cx^2 +d$ is irreducible. Since the characteristic of $F$ is different from 2 or 3, $f' = 6x^5+\cdots 
e 0$, thus $\gcd(f,f') =1$, and $f$ is separable.

If $\alpha$ is a root of $f$, then $-\alpha$ is a root of $f$. Moreover $\alpha 
e 0$, otherwise $d= 0$ and $f$ would not be irreducible, therefore $\alpha 
e -\alpha$. Since $f$ is separable, the roots of $f$ can be partitioned into three blocks
$$R_1 = \{\alpha,-\alpha\},\qquad R_2 = \{\beta,-\beta\},\qquad R_3 = \{\gamma, -\gamma\}.$$
If $\sigma \in \Gal(L/F)$ and if $\lambda \in R_i$ is a root of $f$, then $\sigma(\lambda) \in R_j$ for some index $j$, and $\sigma(-\lambda) = - \sigma(\lambda) \in R_j$, thus $\sigma(R_i) = R_j$. Therefore $f$ is imprimitive, with blocks $R_1,R_2,R_3$. By Corollary 14.2.10, $\Gal(L/F)$ is isomorphic to a subgroup of $S_3 \wr S_2$.

Since $\alpha,-\alpha, \beta,-\beta,\gamma,-\gamma$ are the distinct roots of $f$, then $\alpha^2,\beta^2,\gamma^2$ are distinct and they are the roots of $g = x^3 +bx^2+cx+d$. Therefore
$$g =  x^3 +bx^2+cx+d = (x-\alpha^2)(x-\beta^2)(x-\gamma^2),$$
so that a splitting field $K$ of $g$ over $F$ is 
$$K = F(\alpha^2,\beta^2,\gamma^2),\qquad F \subset K \subset L.$$
Since $K$ is the splitting field of the separable polynomial $g$, $F \subset K$ is a Galois extension. Note that $g$ is irreducible over $F$, otherwise any non trivial factorization of $g$ over $F$ gives a factorisation of $f$ over $F$. Therefore $K 
e F$.

Moreover,
$$f = g(x^2) = (x^2-\alpha^2)(x^2-\beta^2)(x^2-\gamma^2) = f_1f_2f_3,$$
where $f_1 = x^2 - \alpha^2, f_2 = x^2 - \beta^2,f_3 = x^2 -\gamma^2 \in K[x]$. This proves the first assertion of part (a).

\bigskip

It remains to prove that $K$ is the fixed field of the subgroup $G$ of $\Gal(L/F)$ defined in Exercise 8:
$$G = \{\sigma \in \Gal(L/F)\, | \, \forall i \in \gcro 1, 3 \dcro, \ \sigma(R_i) = R_i \}.$$
To give an explicit description of $G$, note that
\begin{align*}
\sigma \in G &\iff \sigma(\alpha) = \pm \alpha,\ \sigma(\beta) = \pm \beta, \ \sigma(\gamma) = \pm \gamma\
&\iff \sigma(\alpha^2) = \alpha^2,\ \sigma(\beta^2) = \beta^2, \sigma(\gamma^2) = \gamma^2\
&\iff \forall \lambda \in K,\ \sigma(\lambda) = \lambda,
\end{align*}
where the last equivalence is explained by the fact that every $\lambda \in K = F(\alpha^2,\beta^2,\gamma^2)$ is a polynomial in $\alpha^2,\beta^2,\gamma^2$.

This proves that every element of $K$ is fixed by every $\sigma \in G$, thus $K \subset L_G$, where $L_G$ is the fixed field of $G$.

Since the Galois correspondence is order reversing, $K \subset L_G$ implies $\Gal(L/K) \supset G$. To prove the inverse inclusion, take $\sigma \in \Gal(L/K)$. Then $\sigma(\lambda) = \lambda$ for all $\lambda \in K$, and the preceding equivalence shows that $\sigma \in G$. Thus $\Gal(L/K) = G$. Applying the Galois correspondence once more, the fixed fields of $\Gal(L/K)$ and $G$ are equal, that is
$$K = L_G.$$
$K$ is the fixed field of $G =  \{\sigma \in \Gal(L/F)\, | \, \sigma(R_1) = R_1,\sigma(R_2) = R_2, \sigma(R_3) = R_3 \}$.

\item[(b)] We have proved in Exercise 13 that $f$ is imprimitive, with blocks
$$R_1 = \{\alpha, \beta, \gamma\},\qquad R_2 = \{\alpha',\beta',\gamma'\}.$$
With the same notations as in Exercise 13 and 8, we have 
$$G = \{\sigma \in \Gal(L/F) \,|\, \sigma(R_1) = R_1,\ \sigma(R_2) = R_2\}.$$
Since $\alpha,\beta,\gamma$ are the roots of $x^3 -\lambda$, and $\alpha',\beta',\gamma'$ the roots of $x^3 -\mu$, where $\{\lambda, \mu\} = \{\sigma(\lambda), \sigma(\mu)\}$, then, for all $\sigma \in \Gal(L/F)$,
\begin{align*}
\sigma \in G &\iff \sigma(\{\alpha,\beta, \gamma\} = \{\alpha,\beta,\gamma\},\  \sigma(\{\alpha',\beta', \gamma'\}) = \{\alpha',\beta',\gamma'\}\
&\iff \sigma(\lambda) = \lambda,\ \sigma(\mu) = \mu\
&\iff \forall \xi \in K,\ \sigma(\xi) \in K.
\end{align*}
This proves as in part (a) that $K = L_G$.
\end{proof}

Exercise 14.2.14

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Exercise 14.2.14

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