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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Problem Set 9}
\centerline{April 7, 2008}
\vskip 10pt
\noindent
{\it Section 11.2}
\vskip 10pt 
\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.
\vskip 10pt 
\noindent
9. (a)  By the result of Exercise 8, we know that
${\rm Gal}_E K$ and ${\rm Gal}_F E$ are both 
abelian.  Now, we use some standard facts about 
cyclic groups.  First, every subgroup of a cyclic
group is cyclic (see Theorem 7.16 in Hungerford, or
your text from Algebraic Structures).    Then every
quotient group $G/H$ of a cyclic group is cyclic.  This follows
since if $a$ is a generator of $G$, then every $b \in G$
satisfies $b = a^k$ for some $k \in \Z$.  But then
$Hb = Ha^k = (Ha)^k$, which shows that $G/H$ is cyclic, 
generated by $Ha$.  

(b)  Since we assume $K$ is Galois over $F$, the
Galois correspondence is a one-to-one correspondence
by Theorem 11.11.  In the cyclic group $G = {\rm Gal}_F K \simeq \Z_n$,
there is exactly one subgroup of order $d$
for each divisor $d$ of $n$.  (If $a$ generates $G$, then 
this subgroup is generated by $a^{n/d}$; see Exercise 40 in 
Section 7.3 of Hungerford.)  Hence there is one intermediate
field for each divisor $d$ of $n$, and these are the only 
intermediate fields.  
\vskip 10pt 
\noindent
11.  (a)  The roots of $x^4 - 2 = 0$ are 
\[
x = {\root 4\of 2}, i{\root 4\of 2}, -{\root 4\of 2}, -i{\root 4\of 2}.
\]
Hence the splitting field over $\Q$ is 
\[
K = \Q({\root 4\of 2}, i{\root 4\of 2}, -{\root 4\of 2}, -i{\root 4\of 2}).
\]
We claim that $K = \Q({\root 4\of 2}, i)$.  First, it is clear that
$K \subseteq \Q({\root 4\of 2}, i)$, since all the roots of $x^4 - 2 = 0$
are contained in this field.  On the other hand 
$\Q({\root 4\of 2}, i) \subseteq K$ since ${\root 4\of 2} \in K$ and 
$i = \frac{i{\root 4\of 2}}{{\root 4\of 2}} \in K$.  Therefore the splitting
field is equal to $\Q({\root 4\of 2}, i)$.

(b)  From the tower of field extensions 
\[
\Q \subset \Q({\root 4\of 2}) \subset \Q({\root 4\of 2},i),
\]
we see that 
\[
[\Q({\root 4\of 2},i):\Q] = [\Q({\root 4\of 2})(i): \Q({\root 4\of 2})][\Q({\root 4\of 2}):\Q].
\]
The second factor on the right is $4$ by Theorem 10.7, since the minimal polynomial
of ${\root 4\of 2}$ over $\Q$ is the irreducible polynomial $x^4 - 2\in \Q[x]$. 
The first factor on the right is $2$ by Theorem 10.7, since the 
polynomial $x^2 + 1$ has $i$ as a root and it is still irreducible in 
$\Q({\root 4\of 2})[x]$.  (This follows since $x^2 + 1$ has no roots in $\Q({\root 4\of 2}) \subset \R$.)
Hence $[\Q({\root 4\of 2},i):\Q] = 8$.

(c)  Consider the extension $K/\Q(i)$.  $K$ is still the splitting 
field of $x^4 - 2$ over $\Q(i)$.  The polynomial $x^4 - 2$
is still irreducible in $\Q(i)[x]$ since there are no roots in $\Q(i)$ and
there are no polynomials of degree 2 with coefficients in $\Q(i)$
that divide $x^4 - 2$.  (In other words there are no pairs $r,s$ of 
roots of $x^4 - 2$ such that $(x - r)(x - s) \in \Q(i)[x]$. 
So by Theorem 11.3, there is a $\Q(i)$-automorphism $\sigma$
of $K$ that satisfies $\sigma({\root 4\of 2}) = i{\root 4\of 2}$.
Since $\sigma$ is a $\Q(i)$-automorphism, it satisfies $\sigma(i) = i$.
But $\sigma(a) = a$ for all $a \in \Q$ as well, so $\sigma$ defines
an element of ${\rm Gal}_\Q K$. 

(d)  Using Corollary 11.13, the restriction of the complex
conjugation mapping $a + ib \mapsto a - ib$ is also an
element $\tau$ of ${\rm Gal}_\Q K$.  Now by part (1) of Theorem 11.11,
$|{\rm Gal}_\Q K| = [K:\Q] = 8$.  Consider the action of the following
combinations of $\sigma$ and 
$\tau$ on the generators of $K$.  We have
\begin{eqnarray*}
{\rm id} : {\root 4\of 2} &\mapsto& {\root 4\of 2}\\
                  i &\mapsto& i\\
\sigma : {\root 4\of 2} &\mapsto& i{\root 4\of 2}\\
                  i &\mapsto& i\\
\sigma^2 : {\root 4\of 2} &\mapsto& -{\root 4\of 2}\\
                  i &\mapsto& i\\
\sigma^3 : {\root 4\of 2} &\mapsto& -i{\root 4\of 2}\\
                  i &\mapsto& i\\
\tau : {\root 4\of 2} &\mapsto& {\root 4\of 2}\\
                  i &\mapsto& -i\\
\sigma\tau : {\root 4\of 2} &\mapsto& i{\root 4\of 2}\\
                  i &\mapsto& -i\\
\sigma^2\tau : {\root 4\of 2} &\mapsto& -{\root 4\of 2}\\
                  i &\mapsto& -i\\
\sigma^3\tau : {\root 4\of 2} &\mapsto& -i{\root 4\of 2}\\
                  i &\mapsto& -i.
\end{eqnarray*}
These give 8 distinct mappings on $K$ by Theorem 11.4.  Hence
this shows 
\[
{\rm Gal}_\Q K = \{{\rm id},\sigma,\sigma^2,\sigma^3,\tau,\sigma\tau,
\sigma^2\tau,\sigma^3\tau\}.
\]

(e)  We claim ${\rm Gal}_\Q K \simeq D_4$, the dihedral group 
of order 8.  To see this note that the generators $\sigma$ 
and $\tau$ satisfy $\sigma^4 = {\rm id}$, $\tau^2 = {\rm id}$
and 
\begin{eqnarray*}
\tau\sigma : {\root 4\of 2} &\mapsto& -i{\root 4\of 2}\\
                  i &\mapsto& -i.
\end{eqnarray*}
Hence $\tau\sigma = \sigma^3\tau = \sigma^{-1}\tau$. This 
shows that the generators satisfy the same relations as
the generators of $D_4$, and both groups have order $8$.
It follows that ${\rm Gal}_\Q K \simeq D_4$.  (Another 
way to say this is that, viewing $D_4$ as the group of
symmetries of the square in $\R^2$, the mapping 
\begin{eqnarray*}
\varphi : {\rm Gal}_\Q K &\to& D_4\\
               \sigma &\mapsto& r_1 \quad \text{(the $\frac{\pi}{2}$ rotation)} \\
                \tau &\mapsto& v \quad \text{(one of the reflections)}
\end{eqnarray*}
is an isomorphism of groups -- it is a one-to-one and onto
group homomorphism.) 
\vskip 10pt 
\noindent
12.  By Theorem 11.11, ${\rm Gal}_{\Q(i)} K$ is the subgroup of 
${\rm Gal}_\Q K$ fixing all elements of $\Q(i)$.  By examining
the list in 11 (d), we see that this subgroup is $H = \{{\rm id},\sigma,\sigma^2,\sigma^3\}$, which is cyclic of order 4.  The other four elements map $i$ to $-i$
so they are not elements of ${\rm Gal}_{\Q(i)} K$.
Therefore ${\rm Gal}_{\Q(i)} K \simeq \Z_4$.
\vskip 10pt 
\noindent
13.  The Galois correspondence is given by the following
diagrams of subgroups and intermediate fields.   $D_4$ contains
$5$ elements of order $2$, so there are $5$ subgroups of order
$2$.  There are also $3$ subgroups of order $4$, one cyclic (namely, 
$\langle \sigma\rangle$), and 
the other two isomorphic to $\Z_2\times\Z_2$.  Here are the subgroups,
with arrows showing the inclusions:
\[
\SelectTips {cm}{}
\xymatrix{
&&\{{\rm id}\} \ar[dl] \ar[dll] \ar[d] \ar[dr] \ar[drr] &&\\
\langle \tau\rangle \ar[dr] & 
\langle \sigma^2\tau\rangle \ar[d] &
\langle \sigma^2\rangle \ar[d]\ar[dl]\ar[dr] &
\langle \sigma\tau\rangle \ar[d]&
\langle \sigma^3\tau\rangle \ar[dl]\\
&\{{\rm id},\tau,\sigma^2,\sigma^2\tau\} \ar[dr]
&\langle \sigma\rangle \ar[d] &
\{{\rm id},\sigma\tau,\sigma^2,\sigma^3\tau\} \ar[dl]&\\
&&{\rm Gal}_\Q K&&. 
}
\]

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 
\[
E_H = \{u \in \Q({\root 4\of 2},i) : \alpha(u) = u \text{ for all $\alpha$ in $H$}\}.
\]
For instance, the fixed field of the subgroup $\langle \sigma\tau\rangle$ would be determined
like this: If $a = a_0 + a_1{\root 4\of 2} + a_2\sqrt{2} + a_3{\root 4\of 8} + 
a_4i + a_5 i{\root 4\of 2} + a_6 i \sqrt{2} + a_7i{\root 4\of 8}$, satisfies 
$\sigma(\tau(a)) = a$, then $a_5 = a_1$, $a_2 = 0$, $a_3 = -a_7$, and $a_4 = 0$.
$a_0,a_6,a_1,a_3$ are arbitrary and the general element of the fixed field looks like
\[
a_0 + a_1(1 + i){\root 4\of 2} + a_6 i\sqrt{2} + a_3(1 - i){\root 4 \of 8}.
\]
We claim this is a general element of the field $\Q((1 + i){\root 4\of 2})$.
To see this note that if we write $\beta = (1 + i){\root 4\of 2}$, then
\begin{eqnarray*}
\beta^2 &=& -2i \sqrt{2}\\
\beta^3 &=& 2(1 - i){\root 4\of 8}\\
\beta^4 &=& -8 \in \Q.
\end{eqnarray*}
This shows that the fixed field of $H = \langle \tau \rangle$ is
$E_H = \Q((1 - i){\root 4\of 2})$ as claimed.  We also see that
$\Q(i\sqrt{2})$ will be contained in $\Q(\beta)$ and 
$\Q(\gamma) = \Q((1 - i){\root 4\of 2})$
from this computation.  (Note that $\beta^2$ and $\gamma^2$ are
rational multiples of $i\sqrt{2}$.)  That is the fixed field of the subgroup 
$H = \{{\rm id},\sigma\tau,\sigma^2,\sigma^3\tau\}$.
\[
\SelectTips {cm}{}
\xymatrix{
&&\Q({\root 4\of 2},i) &&\\
\Q({\root 4\of 2})\ar[urr] &\Q(i{\root 4\of 2}) \ar[ur] & \Q(\sqrt{2},i) \ar[u] & \Q((1+i){\root 4\of 2}) \ar[ul] & \Q((1-i){\root 4\of 2}) \ar[ull]\\
&\Q(\sqrt{2}) \ar[ur]\ar[u]\ar[ul] & \Q(i)\ar[u] & \Q(i\sqrt{2}) \ar[ul]\ar[u]\ar[ur]&\\
&&\Q \ar[ul] \ar[u] \ar[ur]&&
}
\]
(Note that, as always, the arrows are all reversed from the first diagram!)

\end{document}
