Exercise 6.3.2 (Galois group defines a group)

Question

Check that this really does define a group.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
Firstly we have that $Gal(M:K) = Aut(M:K)$.
By definition we have $Aut(M:K)= \{f: M:K ightarrow M:K \ |\ f \ 	ext{is an isomorphism of } M:K\}$
and $\circ: M:K 	imes M:K ightarrow M:K$.
We show that the pair $(Aut(M:K), \circ)$ is a group.\\
Firstly, for $f,g \in Aut(M:K)$ and $\forall a,b \in M:K$, we have that:
\begin{equation*}
    \begin{aligned}
        (g \circ f) (ab) & = g(f(ab))\
        & = g(f(a)f(b))\
        & = g(f(a))g(f(b))\
        & = (g \circ f)(a)(g \circ f)(b)
    \end{aligned}
\end{equation*}
Since $f,g$ are bijective by defintion, $g \circ f$ is bijective, and by above is a homomorphism, hence it is an automporhism.
\\
We now show associativity. $\forall f,g,h \in Aut(M:K)$ and $a \in M:K$ we have:
\begin{equation*}
    \begin{aligned}
        ((h \circ g) \circ f) (a) & = h(g(f(a)))\
        & = h(g \circ f(a))\
        & = h \circ (g \circ f(a))\
        & = (h \circ (g \circ f))(a) \quad \square
    \end{aligned}
\end{equation*}
\\
Next, we check for an identity. $\forall f \in Aut(M:K)$ and $e_{M:K}: (M:K) ightarrow (M:K) \ : a \mapsto a$ we have, 
$f \circ e_{M:K} = e_{M:K} \circ f  = f$. Hence $e_{M:K}$ is the identity element.
\\
Finally, we check for the inverse. $\forall f \in Aut(M:K)$, we have that $f \circ f^{-1} = e_{M:K} = f^{-1} \circ f$.
This follows by definition since $f$ is an isomorphism.

Richard Ganaye · Answer

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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}\,}

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

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

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

\begin{proof} 
By definition, $\Gal(M:K)$ is the group $\mathrm{Aut}_K(M)$ of the automorphisms of $M$ over $K$, that is the set of bijections $f : M 	o M$ which are ring homomorphisms, and such that $f(a) = a$ for all $a \in K$, the law being the composition $\circ$.

It is sufficient to prove that $ \Gal(M:K)$ is a subgroup of the group $S(M)$ of bijections of $M$ (or of the group $\mathrm{Aut}(M)$ of ring automorphisms of $M$), with composition as the group operation.

\begin{itemize}
\item Since $1_M$ is an automorphism of $M$, and $1_M(a) = a$ for all $a \in K$, $1_M \in \Gal(M:K)$, thus $\Gal(M:K) 
e \varnothing$.

\item Let $f, g \in \Gal(M:K)$. We know that $f \circ g$ is an automorphism of $M$, i.e. $f \circ g : M 	o M$ is bijective,  and for all $x,y \in M$, 
\begin{align*}
(f\circ g)(x + y) &= (f\circ g)(x) + (f \circ g)(y),\
 (f\circ g)(x y) &= (f\circ g)(x) (f \circ g)(y),\
  (f\circ g)(1) &= 1.
 \end{align*}

Moreover, for all $a \in K$, $(f \circ g)(a) = f(g(a)) = f(a) =a$.

Therefore $f\circ g \in \Gal(M:K)$.

\item If $f \in \Gal(M : K)$, then $f^{-1}$ is an automorphism of $M$, and for all $a \in K,\ f(a) = a$, thus $a = f^{-1}(a)$. Therefore $f^{-1} \in \Gal(M:K)$.
\end{itemize}
This proves that $\Gal(M:K)$ is a subgroup of the group $S(M)$.
\end{proof}

Exercise 6.3.2 (Galois group defines a group)

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Exercise 6.3.2 (Galois group defines a group)

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