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\begin{center} {\Large\bf Algebra II \\ Assignment 6\\ Solutions}
\end{center}
\ben
\item Let $\ah=a+bi\in C$. Let $\beta=c+di\in C$ be such that $\beta^2=\ah$.
(That is, $\beta$ is one of the two square roots of $\ah$.) Let $K$ be a field
with $Q\subseteq K$ and $a,b\in K$. Find an explicit tower (you find the $r$!)
\[ K=K_0\subset K_1 \subset \ldots \subset K_r=L \]
with all $[K_{i+1}:K]=2$ and $c,d\in L$. (This is a bit tricky. Its helpful
to think in terms of polar coordinates and first find the ``$r$"-value for
$\beta$.)
\So Set $R=\sqrt{a^2+b^2}$ so that in polar coordinates $\alpha=(R,\theta)$.
and so $\beta=(\sqrt{R},\frac{\theta}{2})$.
Squaring $\beta$ gives the equations
\[ c^2-d^2 = a \]
\[ 2cd = b \]
Look tough. But we also know
\[ \sqrt{c^2+d^2} = \sqrt{R} \]
so that
\[ c^2+d^2 = R \]
Adding and subtracting the first equation gives
\[ c^2 = \frac{R+a}{2} \]
\[ d^2 = \frac{R-a}{2} \]
which give $c,d$. (Of the four possibilities for $c,d$, taking the
two squareroots, exactly two pair work giving $2cd$ correctly.)
So to get the tower of fields we take $K_1=K(R)$ and then
$K_2=K_1(\sqrt{(R+a)/2})$. As $d=b/2c$ we have $d$ for free.
\\ {\tt Another Approach:} Substituting $d=\frac{b}{2c}$ into the
equation $c^2-d^2=a$ gives $c^2-\frac{b^2}{4c^2}=a$ or
$4c^4-b^2=ac^2$. This looks like a quartic in $c$ but because
we only have even powers we can reduce it by setting $e=c^2$,
solving the now quadratic for $e$, and then getting $c=\sqrt{e}$.
\item Let $\eps=e^{2\pi i/5}$. Set $K=Q(\eps)$.
\ben
\item Find $[K:Q]$ and a basis for $K$ over $Q$.
\So As $[(x^5)-1]/(x-1)$ is irreducible, $[K:Q]=4$ with basis $1,\eps,\eps^2,\eps^3$.
\item Set $\gam=\eps+\eps^4$. Show that $[Q(\gam):Q]=2$ by
finding an explicit quadratic equation, with coefficients in
$Q$, satisfied by $\gam$. (Note: Since you have the basis this
is a linear algebra problem: finding a dependence between $1,\gam,\gam^2$.)
\So We have $\gam^2=\eps^2+2\eps^5+\eps^8$ but as $\eps^5=1$ we have
$\gam^2=2+\eps^2+\eps^3$. Also $\gam= -1-\eps^2-\eps^3$ as $1+\eps+\eps^2+
\eps^3+\eps^4=0$. Now we have $1,\gam,\gam^2$ in terms of the four basis
elements. Normally three vectors in $4$-space aren't dependent but these
are. We either eyeball or we use undetermined coefficient: $a + b\gam + c\gam^2=0$.
Then we get four equations: $a-b+2c=0$ (the constant), $0=0$ (as there is no coefficient of $\eps$),
$b-c=0$ (for $\eps^2$) and $b-c=0$ (for $\eps^3$) which means $-a=b=c$. Thus $a=-1, b=c=1$ is a
solution, $-1+\gam+\gam^2=0$.
\item Use the quadratic formula to solve $\gam$ explicitly. Find an
explicit $d\in Z$ with $Q(\gam)=Q(\sqrt{d})$.
\So $\gam = [-1\pm \sqrt{5}]/2$. From the geometry we see $\gam$ is positive
so it is actually $\gam= [-1+ \sqrt{5}]/2$. Thus $Q(\gam)=Q(\sqrt{5})$.
\item Show $[K:Q(\gam)]=2$. (This is immediate if you see it.)
\So The product theorem as $[K:Q]=[K:Q(\gam)]\cdot [Q(\gam):Q]$ and we know
$[K:Q]=4$ and $[Q(\gam):Q]=2$.
\item As $\eps\in K$, find an explicit quadratic equation, with coefficients in
$Q(\gam)$, satisfied by $\eps$.
\So One nice way is to write $\gam=\eps+\eps^4$ and multiply by $\eps$ to
give $\eps\gam = \eps^2+\eps^5= \eps^2+1$ so we have the quadratic
\[ \eps^2-\eps\gam + 1 = 0 \]
\item Use the quadratic formula to solve $\eps$ explicitly. (Some of the
terms will be square roots of non-real numbers, but lets allow that. The
object is to write $\eps$ in terms of usual field expressions and square
roots.)
\So By the quadratic formula
\[ \eps = \frac{\gam \pm \sqrt{\gam^2-4}}{2} \]
As $\gam\sim 0.6\cdots$, the square root is imaginary and can be written
$i\sqrt{1-\gam^2}$.
\item Write $\eps=a+bi$. Find $a,b$ in terms of usual field expressions and square
roots, but not involving complex numbers.
\So From above we find
\[ a=\frac{\gam}{2} = \frac{-1+\sqrt{5}}{4} \]
and
\[ b = \frac{\sqrt{1-\gam^2}}{2} = \frac{\sqrt{\frac{\sqrt{5}-1}{2}}}{2} \]
\een
\item Let $f(x)=x^3+ax^2+cx+d \in Q[x]$ be an irreducible cubic with one real root $\ah$
and two nonreal roots $\beta,\gam$.
\ben
\item Argue that $\gam = \ol{\beta}$. ({\tt Note:} Here, and often, we let $\ol{\kappa}$ denote
the complex conjugate of $\kappa$.)
\So $\gam,\beta$ satisfy a quadratic $f(x)/(x-\ah)$ which has real roots, call it $x^2+ax+b=0$.
We don't have $a^2-4b\geq 0$ as then $\gam,\beta$ would be real so we have $a^2-4b <0$ and the
quadratic formula gives two roots which are complex conjugates.
\item Argue that $\gam\not\in Q(\ah)$.
\So All elements in $Q(\ah)$ are real.
\item Argue that $[Q(\ah,\gam):Q(\ah)]=2$.
\So From above $\gam$ satisfies a quadratic $f(x)/(x-\ah)$ in $Q(\ah)[x]$ but as $\gam\not\in Q(\ah)$
it can't satisfy a linear polynomial, so its minimal polynomial over $Q(\ah)$ is of degree $2$ and
so that is $Q(\ah,\gam):Q(\ah)$.
\item Show that $\beta\in Q(\ah,\gam)$ and that $\ah\in Q(\beta,\gam)$.
\So As $f(x)$ has roots $\ah,\beta,\gam$ we can write $f(x)=(x-\ah)(x-\beta)(x-\gam)$. Equating
coefficients gives
\[ -\ah-\beta-\gam = a \]
\[ \ah\beta + \ah\gam+ \beta\gam = b \]
\[ -\ah\beta\gam = c \]
Taking, say, the first, $\beta=-\ah-\gam-a\in Q(\ah,\gam)$ and similarly
$\ah=-\beta-\gam-a \in Q(\beta,\gam)$.
\item Argue that $[Q(\ah,\beta,\gam):Q]=6$.
\So We have a tower $Q\subset Q(\ah)\subset Q(\ah,\gam)=Q(\ah,\beta,\gam)$ and
$[Q(\ah):Q]=3$ and $[Q(\ah,\gam):Q(\ah)]=2$ so be the tower theorem $[Q(\ah,\beta,\gam):Q]=6$.
\item Argue that $\gam\not\in Q(\beta)$. (Idea: If $\gam\in Q(\beta)$ then
show that $Q(\ah,\beta,\gam)=Q(\beta)$ and get a contradiction.)
\So If it were $Q(\beta,\gam)=Q(\beta)$ so $Q(\ah,\beta,\gam)=Q(\beta,\gam)=Q(\beta)$.
But $[Q(\ah,\beta,\gam):Q]=6$ and $[Q(\beta):Q]=3$.
\een
{\tt Remark:} By the last part, $Q(\beta)$ is a field which is {\em not} closed
under complex conjugation.
\een
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