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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Problem Set 9}
\centerline{April 7, 2008}
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\noindent
{\it Section 11.2}
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\noindent
8.  First, ${\rm Gal}_E K$ is a subgroup of the abelian
group ${\rm Gal}_F K$, so it is an abelian group.  Second,
since $E$ is normal over $F$, by Theorem 11.11 (the Fundamental
Theorem of Galois Theory), we have 
\[
{\rm Gal}_F E \simeq {\rm Gal}_F K/{\rm Gal}_E K.
\]
Any time $G$ is an abelian group and $H$ is a normal 
subgroup the quotient group $G/H$ is abelian.  Proof:
For all $a,b \in G$,
\begin{eqnarray*}
(Ha)(Hb) &=& H(ab) \quad \text{(definition of coset product)}\\
         &=& H(ba) \quad \text{(since $ab = ba$ in $G$)}\\
         &=& (Hb)(Ha).
\end{eqnarray*}
Therefore,  ${\rm Gal}_F E$ is also abelian.
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\noindent
9.
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\noindent
11.
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\noindent
12.
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\noindent
13.






\[
\SelectTips {cm}{}
\xymatrix{
&\{{\rm id}\} \ar[dl] \ar[d] \ar[dr] &\\
\{{\rm id},\alpha\} \ar[dr] & \{{\rm id},\beta\} \ar[d] & \{{\rm id},\gamma\} \ar[dl]\\
&G&. 
}
\]

Now we show the fixed fields of each of these subgroups in a parallel diagram.
For each subgroup $H$ above, the subfield in the corresponding location is the 
fixed field 
\[
E_H = \{u \in \Q(\sqrt{2},i) : \sigma(u) = u \text{ for all $\sigma$ in $H$ }\}.
\]
For instance, if $H = \{{\rm id},\alpha\}$, then by definition, 
$u = u_0 + u_1\sqrt{2} + u_2 i + u_3 i\sqrt{2}$ ($u_i \in \Q$) is in $E_H$ if and only 
if $\alpha(u) = u_0 - u_1\sqrt{2} + u_2 i - u_3 i\sqrt{2} = u$.  This 
implies $u_1 = u_3 = 0$ and $u_0,u_2$ are arbitrary.  Therefore 
\[
E_H = \{u_0 + u_2 i : u_0,u_2 \in \Q\} = \Q(i).
\]
Similarly, if $H = \{{\rm id},\beta\}$, then by definition, 
$u = u_0 + u_1\sqrt{2} + u_2 i + u_3 i\sqrt{2}$ ($u_i \in \Q$) is in $E_H$ if and only 
if $\beta(u) = u_0 + u_1\sqrt{2} - u_2 i - u_3 i\sqrt{2} = u$.  This 
implies $u_2 = u_3 = 0$ and $u_0,u_1$ are arbitrary.  Therefore 
\[
E_H = \{u_0 + u_1 \sqrt{2} : u_0,u_1 \in \Q\} = \Q(\sqrt{2})
\]
in this case.  Finally
if $H = \{{\rm id},\gamma = \alpha\circ\beta\}$, then by definition, 
$u = u_0 + u_1\sqrt{2} + u_2 i + u_3 i\sqrt{2}$ ($u_i \in \Q$) is in $E_H$ if and only 
if $\gamma(u) = u_0 - u_1\sqrt{2} - u_2 i + u_3 i\sqrt{2} = u$.  This 
implies $u_1 = u_2 = 0$ and $u_0,u_3$ are arbitrary.  Therefore 
\[
E_H = \{u_0 + u_3 i\sqrt{2} : u_0,u_3 \in \Q\} = \Q(i\sqrt{2}).
\]
Every element of $\Q(\sqrt{2},i)$ is fixed by the identity subgroup.  The only 
elements of $\Q(\sqrt{2},i)$ fixed by \emph{all} elements of $G$ are the 
rational numbers.  
Hence we obtain the corresponding diagram of fixed fields and inclusions:

\[
\SelectTips {cm}{}
\xymatrix{
&\Q(\sqrt{2},i) &\\
\Q(i) \ar[ur] & \Q(\sqrt{2})\ar[u] & \Q(i\sqrt{2}) \ar[ul]\\
&\Q \ar[ul] \ar[u] \ar[ur]& 
}
\]
(Note that the arrows are all reversed from the first diagram!)

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