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\begin{center} {\Large\bf Honors Algebra II \\ Assignment 9 Solutions} \\ \end{center}
\ben
\item Let $F\subset L \subset K$ be fields in $C$. Assume that
$K:F$ is normal {\em and} that $L:F$ is normal. Let $\tau\in \Gam[K:F]$
and $\sig\in \Gam[K:L]$. Let $l\in L$
\ben
\item Using a result shown in class argue that $\tau(l)\in L$.
\So We showed that when $L:F$ was normal that any isomorphism $\gam: L\ra L'$
over $F$ had $L'=L$. The restriction of $\tau$ to $L$ is such an isomorphism
so $L'=L$.
\item Show that $(\tau\sig\tau^{-1})(l) = l$.
\So Set $\tau(l)=l'$ so $l'\in L$. So $\sig(l')=l'$. So $\tau^{-1}(l')=l$.
That is $(\tau\sig\tau^{-1})(l) = l$.
\item From the above show that $\Gam[K:L]$ is a {\em normal} subgroup
(get out those Algebra I notes!) of $\Gam[K:F]$. (Assume its already
been shown that it is a subgroup. You only need show the normal part.)
\So We've shown that if $\tau\in \Gam[K:F]$ and $\sig\in \Gam[K:L]$ then
$\tau\sig\tau^{-1}$ fixes all elements of $L$ so $\tau\sig\tau^{-1}\in \Gam[K:L]$.
That is what we need to show that you have a {\em normal} subgroup.
\item In the case $F=Q$, $L=Q(\omega)$, $K=F(\ah,\omega)$ (with
$\ah=2^{1/3}, \omega = e^{2\pi i/3}$ as in our standard example)
give the groups $\Gam[K:L]$ and $\Gam[K:F]$ explicitly in terms of
permutations of $\ah,\beta=\ah\omega, \gam=\ah\omega^2$.
\So $\Gam[K:F]$, as done in class, is all six permutations of $\ah,\beta,\gam$.
Now fix $L$, so $\tau(\omega)=\omega$. There are three cases. Either
$\tau(\ah)=\ah$ so $\tau=e$. Or $\tau(\ah)=\beta=\ah\omega$. Then
$\tau(\beta)=\tau(\ah\omega) = \ah\omega\omega = \gam$ and similarly
$\tau(\gam)=\ah$. So $\tau$ sends $\ah,\beta,\gam$ into $\beta,\gam,\ah$
respectively. Similary if $\tau(\ah)=\gam$, $\tau$ sends $\ah,\beta,\gam$
into $\gam,\ah,\beta$ respectively.
\een
\item Let $F\subset K$ be fields in $C$ with $[K:F]=2$. Prove that
$K:F$ is a normal extension.
\So Let $K:F$ be an extension of degree 2. Then $K = F(\alpha)$ for some $\alpha$,
where the
minimal polynomial of $\alpha$ in $K[x]$ has degree 2. Let $m(x) \in K[x]$ be the minimal
polynomial of $\alpha$, so
\[m(x) = x^2 + bx + c\]
for some $b, c \in K$. (We may assume $m(x)$ is monic to make the calculations easier.) We will
show $K:F$ is normal by showing that $K$ is a splitting field for the polynomial $m(x)$ over $F$.
By the quadratic formula, the roots of $m(x)$ are given by
\[x = \frac{-b \pm \sqrt{b^2 - 4c}}{2}\]
As $m(x)$ is irreducible the roots are distinct. Suppose
\[\alpha = \frac{-b + \sqrt{b^2 - 4c}}{2},\]
(the case where the square root is subtracted is similar)
then
\[\beta = \frac{-b - \sqrt{b^2 - 4c}}{2} = \frac{b - \sqrt{b^2 - 4c} - 2b}{2} = -\alpha - b.\]
Therefore, $\alpha, \beta \in F(\alpha) = K$.
(Another approach: Letting $\ah,\beta$ be the roots, $\ah+\beta=-b$ immediately from the
connection between the roots of a polynomial and its coefficients.)
\item Let $K_1,K_2$ be normal extensions of $Q$. Let $M$ denote the
minimal field containing $K_1\cup K_2$. Prove that $M$ is a normal
extension of $Q$. [One approach: Write $K_1=Q(\ah_1,\ldots,\ah_r)$
where the $\ah_i$ are all the roots of some $p(x)\in Q[x]$ and do
similarly for $K_2$.]
\So Write $K_1$ as above and similarly $K_2=Q(\beta_1,\ldots,\beta_s)$
where the $\beta_j$ are all the roots of some $q(x)\in Q[x]$. Then
$M=Q(\ah_1,\ldots,\ah_r,\beta_1,\ldots,\beta_s)$ which are all
the roots of $p(x)q(x)$ which is in $Q[x]$. ({\tt Reminder:} To
be a splitting field you don't need to have an irreducible
polynomial!)
\item Let $\alpha$ be a root of $f(x) = x^3 + x^2 - 2x - 1 \in Q[x]$.
\ben
\item Show $f(x)$ is irreducible over $Q$.
\So It is a cubic and by Gauss's Theorem if it factors it would factor into $Z[x]$ so
it would a factor $x-a$ so it would have an integer root $a$. Then $a$ would have to
divide $-1$ so it would be either $+1$ or $-1$, and neither of them work.
\item Find $[Q(\alpha) : Q]$.
\So $f(x)$ is irreducible over $Q$
and $\alpha$ is a root of $f(x)$, so $f(x)$ is the
minimal polynomial for $\alpha$ over $Q$. Hence $[Q(\alpha):Q] = \textrm{deg}(f) = 3$.
\item Set $\beta=-1/(\ah+1)$. Find $\beta$ in the form $a+b\ah+c\ah^2$.
\So First lets find $1/(1+\ah)$:
$(\ah+1)(a+b\ah+c\ah^2) = a+b\ah+c\ah^2+a\ah+b\ah^2+c(-\ah^2+2\ah+1) = 1$ gives the
equations $a+c=1$, $b+a+2c=0$, $c+b-c=0$ so $b=0$, $a=2$, $c=-1$.
Thus $\beta$ is the negative of that: $\beta=\ah^2-2$.
\item Show that $f(\beta)=0$. (Bit of grunt work here!)
\So We need show
\[ \frac{-1}{(1+\ah)^3} + \frac{1}{(1+\ah)^2} -2\frac{-1}{1+\ah} - 1 = 0 \]
Multiplying out by $(1+\ah)^3$ we get that a polynomial which is zero.
Or, we need show
\[ (\ah^2-2)^3 + (\ah^2-2)^2 - 2(\ah^2-2) -1 = 0 \]
One makes a list of $\ah^3,\ah^4,\ah^5,\ah^6$ and everything works out.:wq
\item Find $\gam\in Q(\ah)$, $\gam,\beta,\ah$ distinct, with $f(\gam)=0$.
(Idea: If $f(x)=(x-\ah)(x-\beta)(x-\gam)$ then $\ah+\beta+\gam$ is determined.)
\So $\ah+\beta+\gam = -1$ so $\gam=-1-\ah-\beta$.
\item Deduce that $Q(\ah) : Q$ is normal.
\So It is the splitting field of $f(x)$.
\item List all of the $Q$-automorphisms of $Q(\alpha)$. What familiar group is $\Gamma(Q(\alpha):Q)$ isomorphic to?.
\So When $[Q(\gam):Q]=r$ and $Q(\gam)$ is normal over $Q$ then the Galois group has precisely $r$ elements.
But there is precisely one group on three elements. So $\Gamma(Q(\ah):Q)$ must be isomorphic to $Z_3$.
One is the identity. Then there is a $\sig$ with $\sig(\ah)=\beta$. We can't have $\sig(\gam)=\gam$
as that would make $\sig$ the identity (as $Q(\ah)=Q(\gam)$) so $\sig(gam)=\ah$ and $\sig(beta)=\gam$.
The final automophims sends $\ah,\beta,\gam$ to $\ah,\gam,\beta$ respectively.
\item Let $K$ be the fixed field of $\Gamma(Q(\alpha):Q)$. Prove $K=Q$.
\So This is the Galois Correspondence Theorem -- the entire Galois Group corresponds to the
ground field.
\een
\item As in last week's assignment set
$\ah=2^{1/4}$, $\beta=i\ah$, $\gam=-\ah$, $\del=-i\ah$. Set
$p(x)=x^4-2$. Set $K=Q(\ah,\beta,\gam,\del)$. Set $L=Q(i)$.
Also, set $M=Q(\ah)$ and $N=Q(\sqrt{2})$.
\ben
\item Give the factorization of $p(x)$ into irreducible factors in $Q[x]$.
\So It is irreducible by the Eisenstein Criterion.
\item Give the factorization of $p(x)$ into irreducible factors in $K[x]$.
\So It completely factors $p(x)=(x-\ah)(x-\beta)(x-\gam)(x-\delta)$.
\item Give the factorization of $p(x)$ into irreducible factors in $M[x]$.
\So Well, $(x-\ah)$ is a factor. But as $\gam=-\ah\in M$ so is $(x-\gam)=(x+\ah)$.
Taking those out $p(x)=(x-\ah)(x+\ah)(x^2+\sqrt{2})$. Here $\sqrt{2}=\ah^2\in M$.
Also, $x^2+\sqrt{2}$ is irreducible in $M[x]$ as its two roots, $\gam,\del$,
are not in $M$ as they are complex numbers.
\item Give the factorization of $p(x)$ into irreducible factors in $N[x]$.
\So $p(x)=(x^2-\sqrt{2})(x^2+\sqrt{2})$. As $[Q(\ah):Q]=4$, $\ah\not\in N$.
Thus $(x-\ah)$ is not a factor in $N[x]$. Similarly $x-\beta$, $x-\gam$,
$x-\del$ are not factors. Thus the quadratics above are irreducible in $N[x]$.
\een
\een
\end{document}