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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Problem Set 8}
\centerline{March 25, 2008}
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\noindent
{\it Chapter 15}
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\noindent
7.  We must have $x^2y = 3$ from the volume constraint and $2x^2 + 4xy = 7$
from the surface area constraint.  Solving for $x$ in the first
equation, $y = \frac{3}{x^2}$. So then in the second equation $2x^2 + \frac{12x}{x^2} = 
7$, or $2x^3 - 7x + 12 = 0$.  By the rational roots test, the possible rational
roots of this polynomial equation are $x = \pm 1,2,3,4,6,12, \frac{1}{2}, \frac{3}{2}$.
But it is easy to check that none of these are roots.  This implies that $f(x) = 2x^3 - 7x + 12$
is irreducible in $\Q[x]$.  Therefore, $[\Q(x) : \Q] = 3$, so $x$ is not constructible
by Theorem 15.9 in the text, or the Corollary of the main theorem on constructible
numbers we stated in class:  $[\Q(x):\Q]$ is not a power of $2$.  
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\noindent
8.  Use Theorem 15.1.  A segment of length $\sqrt{3}$ can be constructed by
this process:
\begin{itemize}
\item construct a semicircle of diameter 4 -- call the end points $A,B$,
\item erecting a perpendicular to the diameter at a point $C$ dividing the 
diameter into lengths $AC = 1, CB = 3$ and
\item find the intersection of $D$ the perpendicular with the semicircle (see
diagram on page 454 or a slightly different diagram in the class notes).  
\end{itemize}
Then given the segment $\overline{CD}$ of length $\sqrt{3}$, construct
a circle with center $C$ and radius $CD = \sqrt{3}$.  This intersects
the line containing the diameter of the semicircle at $E$, and $AE = 1 + \sqrt{3}$.
(There are several other correct method here too.) 
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\noindent
9.  The answer is \emph{no}. 
Call the sides of the triangle $x,y$ (where $x$ is the length of the two 
equal sides).  Then the perimeter is $2x + y = 8$.
Taking the side of length $y$ as the base, and using the 
Pythagorean theorem, the area is $1 = \frac{1}{2}y \sqrt{x^2 - \frac{y^2}{4}}$, 
which implies, after squaring and rearranging,
\[
4x^2y^2 - y^4 = 16.
\]
Now from the perimeter equation, $y = 8 - 2x$, so 
\[
4x^2(8-2x)^2 - (8 - 2x)^4 = 16,
\]
which implies
\[
8x^3-80x^2+256x-257=0.
\]
This cubic is irreducible in $\Q[x]$, since it has no rational
roots.  Therefore, $[\Q(x):\Q] = 3$, so $x$ is not a constructible
number.  
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\noindent
11.  Consider the unit circle centered at $(0,0)$.  An angle of $t$ degrees
is constructible if and only if we can construct a line $\ell$ meeting the 
$x$-axis at an angle of $t$ degrees (measured counterclockwise from the positive $x$-axis).
That line intersects the unit circle at the points $\pm (\cos(t),\sin(t))$.
Hence the angle is constructible if and only if $\cos(t)$ is a constructible
number by the definition (see page 453 in the text, or the class notes).  
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\noindent
17.  As given in the Hint, the case where $C$ is on the line $L$ is done
in the text.  If $C$ is not in $L$, let $D$ be any constructible point on 
$L$ (there must be such points $D$ since it is given that $L$ is constructible).  
Using the compass, construct the circle $\Gamma$ with center at $C$ and radius $\overline{CD}$.  
The circle $\Gamma$ intersects $L$ either in 
\begin{itemize}
\item Two points $D,E$, or
\item In just one point $D$.
\end{itemize}

In the second case, $L$ is tangent to $\Gamma$ at $D$, and we claim that the tangent
$L$ must be perpendicular to the radius $\overline{CD}$ so we are done.
This is a standard geometric fact; it can be proved most easily 
by parametrizing the circle and computing the tangent line from the parametric equations.
(A strictly Euclidean proof without using the idea of reflection across the line $CD$
is more difficult, though still possible!)

(Alternately, we can argue that any line must contain (at least) two distinct constructible
points. Therefore, there must be some point $D$ such that the circle
with radius $\overline{CD}$ is not tangent to $L$.)

In the first case, construct the circles $\Gamma_1$ with center $D$ and radius $DC$, and 
$\Gamma_2$ with center $E$ and radius $EC$.  The circles 
$\Gamma_1$ and $\Gamma_2$ meet at the constructible points $C$
and $Q$.  We claim that $\overline{CQ} \perp \overline{DE}$, so extending the line 
$\overline{CQ}$ gives the line
we are looking for.  First, we have $CD = CE$ since both are radii of the same
circle.  Similarly, $CD = DQ$ and $CE = EQ$.
This implies that $\Delta CDE$ is isosceles, so $\angle CDE = \angle CED$.
Similarly $\angle QDE = \angle QED$.  Hence 
\[
\angle QDE + \angle EDC = \angle QDC = \angle QEC = \angle QED + \angle DEC.
\]
From this, we see that $\Delta CDQ \cong \Delta CEQ$ by the SAS congruence theorem.
Call $S$ the intersection point of the lines $\overline{DE}$ and $\overline{CQ}$.  
By the triangle congruence $\Delta CDQ \cong \Delta CEQ$ above, $\angle DCS = \angle ECS$.  This 
implies $\Delta DCS \cong \Delta ECS$ (by SAS again, since $CD = CE$ and $CS$ is 
in both triangles).  Hence $\angle DSC$ and $\angle ESC$ are
right angles since they are equal and add up to a straight angle (i.e. $180^\circ$).   
Thus $\overline{CQ} \perp \overline{DE}$, which is what we wanted to show.
\vskip 10pt
\noindent
19.  Let $A$ be a constructible point not on the constructible line $L$.
By Exercise 17, we can construct a line $M$ through $A$ perpendicular to $L$.
Then using the example from the text on page 451, we can construct a 
line $N$ passing through $A$ that is perpendicular to $M$.  Since the $L$
and $N$ are both perpendicular to $M$, but they meet $M$ at different
points (one at $A$, one at a point different from $A$), $L$ and $M$ must
be parallel.   
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\noindent
{\it Section 11.1}
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\noindent
2.  Yes this is true.  If $\{v_1,\ldots,v_n\}$ is a basis for $K$ over $F$, then 
each $a \in K$ is uniquely $a = c_1 v_1 + \cdots + c_n v_n$ for $c_1,\ldots,c_n \in F$.
Suppose $\phi,\psi$ are two $F$-automorphisms of $K$ such that $\phi(v_i) = \psi(v_i)$
for all $i = 1,\ldots, n$.  Then for each $a \in K$, 
\begin{eqnarray*}
\phi(a) &=& \phi(c_1 v_1 + \cdots + c_n v_n) \\
&=& c_1\phi(v_1) + \cdots + c_n \phi(v_n) \quad \text{since $\phi$ is an $F$-automorphism of $K$}\\
&=& c_1\psi(v_1) + \cdots + c_n \psi(v_n) \quad \text{by assumption}\\
&=& \psi(c_1 v_1 + \cdots + c_n v_n) \quad \text{since $\psi$ is an $F$-automorphism of $K$}\\
&=& \psi(a).
\end{eqnarray*}
In other words, $F$-automorphisms are completely determined by their action
on a basis of $K$ over $F$.  
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\noindent
3.  We are given that $\sigma \in {\rm Gal}_F(K)$, so $\sigma$ is an $F$-automorphism
of $K$.  If $u \in K$ satisfies $\sigma(u) = u$, then we claim that $\sigma(a) = a$
for all $a \in F(u)$.  This will suffice to show that $\sigma \in {\rm Gal}_{F(u)} (K)$ 
since by definition, elements of ${\rm Gal}_{F(u)} (K)$ are automorphisms of $K$
fixing all the elements of $F(u)$.  Now we are given that $[K:F]$ is finite. since 
$F \subseteq F(u) \subseteq K$, by Theorem 10.4, 
\[
[K:F] = [K:F(u)][F(u):F]
\]
and hence $[F(u):F]$ is finite as well.  Theorem 10.7 implies that if
$[F(u):F] = n$, then $1,u,\ldots,u^{n-1}$ is a basis for $F(u)$ over $F$.
Hence if $a \in F(u)$, then $a = c_0 + c_1 u + \cdots + c_{n-1} u^{n-1}$
for some $c_i \in F$.  But then as in the previous exercise,
\begin{eqnarray*}
\sigma(a) &=& \sigma(c_0 + c_1 u + \cdots + c_{n-1} u^{n-1}) \\
&=& c_0  +  c_1 \sigma(u) + \cdots + c_{n-1} (\sigma(u))^{n-1} \quad \text{since $\sigma$ is an $F$-automorphism of $K$}\\
&=& c_0  + c_1 u + \cdots + c_{n-1} u^{n-1} \quad \text{since $\sigma(u) = u$}\\
&=& a.
\end{eqnarray*}
Thus $\sigma(a) = a$ for all $a \in F(u)$.  
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\noindent
8.  The mapping $\sigma : \Q({\root 4 \of 2}) \to \Q({\root 4 \of 2})$
defined by $\sigma({\root 4\of 2}) = -{\root 4\of 2}$ (that is 
\[
\sigma(a_0 + a_1 {\root 4 \of 2} + a_2 \sqrt{2} + a_3 ({\root 4 \of 2})^3) = 
a_0 - a_1 {\root 4 \of 2} + a_2 \sqrt{2} - a_3 ({\root 4 \of 2})^3
\]
is also an element of ${\rm Gal}_{\Q} (\Q({\root 4 \of 2}))$.  This is 
true since $\sigma$ fixes all $a_0 \in \Q$, and if 
$a = a_0 + a_1 {\root 4 \of 2} + a_2 \sqrt{2} + a_3 ({\root 4 \of 2})^3$
and $b = b_0 + b_1 {\root 4 \of 2} + b_2 \sqrt{2} + b_3 ({\root 4 \of 2})^3$, 
then
\begin{eqnarray*}
\sigma(a + b) &=&\sigma((a_0 + b_0) + (a_1 + b_1) {\root 4 \of 2} + (a_2 + b_2) \sqrt{2} + (a_3 + b_3) ({\root 4 \of 2}^3)\\
&=& (a_0 + b_0) - (a_1 + b_1) {\root 4 \of 2} + (a_2 + b_2) \sqrt{2} - (a_3 + b_3) ({\root 4 \of 2}^3\\
&=& (a_0 - a_1 {\root 4 \of 2} + a_2 \sqrt{2} - a_3 ({\root 4 \of 2})^3) + 
(b_0 - b_1 {\root 4 \of 2} + b_2 \sqrt{2} + b_3 ({\root 4 \of 2})^3)\\
&=& \sigma(a) + \sigma(b).
\end{eqnarray*}
Moreover, 
\begin{eqnarray*}
\sigma(a b) &=&\sigma\left((a_0 b_0 + 2a_1b_3 + 2a_2b_2 + 2a_3b_1) + (a_0b_1 + a_1b_0 + 2a_2b_3 + 2a_3b_2){\root 4 \of 2}\right.\\
&&\quad + \left.(a_0b_2 + a_2b_0 + a_1b_1 + 2a_3b_3) \sqrt{2} + (a_0b_3 + a_3b_0 + a_1b_2 + a_2b_1) ({\root 4 \of 2}^3)\right)\\
&=& (a_0 b_0 + 2a_1b_3 + 2a_2b_2 + 2a_3b_1) - (a_0b_1 + a_1b_0 + 2a_2b_3 + 2a_3b_2){\root 4 \of 2}\\
&&\quad + (a_0b_2 + a_2b_0 + a_1b_1 + 2a_3b_3) \sqrt{2} - (a_0b_3 + a_3b_0 + a_1b_2 + a_2b_1) ({\root 4 \of 2}^3)\\
&=& (a_0 - a_1 {\root 4 \of 2} + a_2 \sqrt{2} - a_3 ({\root 4 \of 2})^3) \cdot
(b_0 - b_1 {\root 4 \of 2} + b_2 \sqrt{2} + b_3 ({\root 4 \of 2})^3)\\
&=& \sigma(a) \sigma(b).
\end{eqnarray*}
Therefore $\sigma$ is a $\Q$-automorphism of $\Q({\root 4 \of 2})$.
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\noindent
9. (a) Since $x^3 - 1 = (x - 1)(x^2 + x + 1)$ in $\Q[x]$, the minimal polynomial
of $\omega = \frac{-1 + i\sqrt{3}}{2}$ is $x^2 + x + 1$.  By the quadratic formula,
the roots of this are
\[
x = \frac{-1 \pm i\sqrt{3}}{2}.
\]
Taking the $+$ sign gives $\omega$, the other root $\frac{-1 - i\sqrt{3}}{2}$ equals 
$\omega^2$,

(b)  The Galois group ${\rm Gal}_\Q (\Q(\omega)) = \{id, \sigma\}$
where $\sigma(\omega) = \omega^2$.  This is cyclic of order 2, isomorphic to $\Z_2$.
Note that by Theorem 11.3, every $\Q$-automorphism of $\Q(\omega)$ takes 
$\omega$ to either $\omega$ or to $\omega^2$.  
\end{document}
