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\begin{center} {\Large\bf Algebra V63.0349 \\ Assignment 9 Solutions}
\end{center}
\ben
\item\label{one} Let $K:Q$ be a normal extension, set $G=\Gam[K:Q]$. Let
$H$ be a subgroup of $G$. Let $a=|H|$, $s=|G|$.
Let $\ah\in K$. Set $\gam = \sum_{\sig\in H} \sig(\ah)$.
\ben
\item Show that $\tau(\gam)=\gam$ for all $\tau\in H$.
\So As $\sig$ ranges over $H$, $\sig\tau$ hits every element
of $H$ exactly once so that
\[ \tau(\gam)= \sum_{\sig\in H} (\sig\tau)(\ah) = \sum_{\mu\in H} \mu(\ah)=\gam
\]
\item Show that $H\subseteq Q(\gam)^*$.
\So The $\tau\in H$ fix $\gam$ and so they fix $Q(\gam)$ and so they belong
to $Q(\gam)^*$.
\item\label{two} Deduce an upper bound on $[Q(\gam):Q]$.
\So As $|Q(\gam)^*|\geq |H|=a$, $[Q(\gam):Q]= |G|/|Q(\gam)^*| \leq \frac{s}{a}$.
\een
\item Let $K:Q$ be a normal extension and assume that $K$ is {\em not} a subset of
the reals.
\ben
\item Show that $K$ is closed under complex conjugation. That is, let $\ah=a+bi\in K$
with $a,b$ real. Show that $a-bi\in K$. (Warning: This is not true for any field
extension, you must use that $K:Q$ is normal.
\So Complex conjugation is an isomorphism $\sig$ from any field $K$ to a field $K'$.
But isomorphisms fix normal fields so $K'=K$. Another approach: $K$ is the splitting
field of some $f(x)\in Q[x]$ so $K=Q(\ah_1,\ldots,\ah_n)$ where $\ah_1,\ldots,\ah_n$
are the roots of $f(x)$.
Let $\ol{\ah}$ denote the complex
conjugate of $\ah$. Any $\beta\in K$ is a polynomial in $\ah_1,\ldots,\ah_n$
with rational coefficients. Then $\ol{\beta}$ is a polynomial in
$\ol{\ah_1},\ldots,\ol{\ah_n}$. But the roots of $f(x)\in Q[x]$ is either real
or in complex conjugate pairs so
$\ol{\ah_1},\ldots,\ol{\ah_n}$ is just a reordering of $\ah_1,\ldots,\ah_n$.
or in complex conjugate pairs so
\item Let $a+bi\in K$ with $a,b$ real. Show that $a\in K$ and $bi\in K$.
\So As $a+bi\in K$, $a-bi\in K$ from above. Thus $a=\frac{1}{2}((a+bi)+(a-bi))\in K$.
Thus $bi = (a+bi)-a \in K$.
\item Let $L$ be the field of real numbers $a\in K$. Prove that $[K:L]=2$.
\So In problem \ref{one} let $H=\{e,\sig\}$ with $\sig$ being complex conjugation.
A complex number is fixed by $\sig$ if and only if it is real so $L=H^{\dagger}$.
By the Galois Correspondence Theorem $[Q:L]=|H| = 2$.
\een
\item Let $G=(Z_4\times Z_6, +)$
\ben
\item Find three subgroups $H\subset G$ with precisely $12$ elements.
\So $H_1=\{(2a,b):a\in Z_4, b\in Z_6\}$
\\ $H_2=\{(a,2b):a\in Z_4, b\in Z_6\}$
\\ $H_3= \{ (x,y)\in Z_4\times Z_6: x+y {\mbox{ even}} \}$
\item (*) Prove there aren't any more $H$.
\So Hmmmmm. Well, since $G/H$ has two cosets (everything is Abelian
so no sweat about normality here) $H$ must have all $x+x$ so it
must have $(0,2)$ and $(2,0)$. If it has $(1,0)$ that forces $H_2$,
if it has $(0,1)$ that forces $H_1$ and if it has $(1,1)$ that forces
$H_3$. If it had none of $(1,0),(0,1),(1,1)$ then $(0,0),(1,0),(0,1),(1,1)$
would all lie in different cosets, which they can't as there are only
two cosets.
\item Let $K:Q$ be normal with $\Gamma(K:Q)$ isomorphic to $G$ above.
Show that $K$ has precisely three distinct nontrivial squareroots.
(We count $\sqrt{a}$ and $\sqrt{q^2a}$ as the same.)
\So We are counting intermediate $L$ with $[L:Q]=2$. By the Galois
Correspondence Theorem this is the same as subgroups $H$ with $|G/H|=2$,
so $|H|=12$.
\een
\item Let $\ah,\beta,\gam$ be the three roots of $x^3+2x+1=0$.
\ben
\item Find $\ah+\beta+\gam$ as an explicit integer.
\So We use the basic relations given by setting
\[ x^3+2x+1 = (x-\ah)(x-\beta)(x-\gam) \]
Equating the $x^2$ coefficient
\[ \ah+\beta+\gam = 0 \]
Equeating the $x$ coefficient
\[ \ah\beta + \ah\gam + \beta\gam = 2 \]
Equating the constants
\[ \ah\beta\gam = -1 \]
\item Find $\ah^2+\beta^2+\gam^2$ as an explicit integer.
\So $(\ah+\beta+\gam)^2 = (\ah+\beta+\gam)^2 - 2(\ah\beta + \ah\gam + \beta\gam) = 0-2(2)= -4$
\item Find $\ah^3+\beta^3+\gam^3$ as an explicit integer.
\So We first express
\[ \ah^3+\beta^3+\gam^3 - (\ah+\beta+\gam)^3 - 3\Delta - 6\ah\beta\gam \]
Here
\[ \Delta = \ah^2\beta + \ah^2\gam + \beta^2\ah + \beta^2\gam + \gam^2\ah + \gam^2\beta \]
We reduce this by setting
\[ \Delta = (\ah\beta+\ah\gam + \beta\gam)(\ah+\beta+\gam) - 6\ah\beta\gam \]
Plugging in $\Delta = 3$. Backing up we now have
\[ \ah^3+\beta^3+\gam^3 - (0)^3 - 3(3) - 6(-1) = -3 \]
\een
\item Recall that $\ah$ is square root constructible if there
exists a sequence $u_1,\ldots,u_s=\ah$ such that
each $u_i$ is either in $Q$ or the sum or product or
additive inverse or multiplicative inverse or -- critically --
a squareroot of previous element(s). Equivalently (we
showed this in class, you can assume it) $\ah$ is square root
constructible if there
exists a tower of fields $Q=F_0\subset F_1\subset\cdots F_r=F$
with all $[F_{i+1}:F_i]=2$ and $\ah\in F$.
Our object here is to
give necessary {\em and} sufficient conditions on $\ah$ being
square root constructible.
\ben
\item Let $\ah$ be square root constructible by the sequence
$u_1,\ldots,u_s=\ah$. Set $K=Q(u_1,\ldots,u_s)$. Let $\sig: K\ra K_1$
be an isomorphism over $Q$. Prove $\sig(\ah)$ is
square root constructible by the sequence
$\sig(u_1),\ldots,\sig(u_s)=\sig(\ah)$.
\So If $u_i=u_j+u_k$ then $\sig(u_i)=\sig(u_j)+\sig(u_k)$, and similarly
with product, additive inverse and multiplicative inverse. For square
roots, if $u_i^2=u_j$ then $\sig(u_i)^2 = \sig(u_j)$.
\item Let $\ah$ have minimal polynomial $p(x)\in Q[x]$ and let
$\beta$ be a root of $p(x)$. Let $\ah$ be square root constructible.
Prove that $\beta$ is square root constructible.
\So Take the isomorphism $\tau: Q(\ah)\ra Q(\beta)$ over $Q$ with
$\tau(\ah)=\beta$ and
extend it (via the Isomorphism Extension Theorem) to an isomorphism
$\sig: K\ra K_1$. Then $\sig(\ah)=\beta$ and apply the previous part.
\item Let $\ah$ have minimal polynomial $p(x)\in Q[x]$, with roots
$\ah=\ah_1,\ldots,\ah_s$. Let $K=Q[\ah_1,\ldots,\ah_s]$ be the
splitting field of $p(x)$ over $Q$. Prove that if $\ah$ is
square root constructible then every $\gam\in K$ is square root
constructible. Prove further that $[K:Q]$ must be a power of two.
\So If $\ah_1$ is square root constructible then all of $\ah_1,\ldots,\ah_s$
are square root constructible so that any polynomial in $\ah_1,\ldots,\ah_s$
(with coefficients in $Q$) is square root constructible, which is all of $K$.
\item Let $\ah$ have minimal polynomial $p(x)\in Q[x]$, with roots
$\ah=\ah_1,\ldots,\ah_s$. Let $K=Q[\ah_1,\ldots,\ah_s]$ be the
splitting field of $p(x)$ over $Q$. Now {\em assume} $[K:Q]=2^w$
for some integer $w$. Prove that all $\gam\in K$ (so that includes
$\ah$ itself!) are square root constructible.
(Hint: Use the Galois Correspondence Theorem
and the Sylow Theorem 2.12.1 on existence of subgroups.)
\So Set $G_0=\Gamma[K:Q]$ so $|G_0|=2^w$. By the Sylow Theorem
there is a subgroup $G_1\subset G_0$ with $|G_1|=2^{w-1}$.
Continuing we get a descending tower of groups $G_0\supset G_1\supset \cdots
G_{w-1} \supset G_w=\{e\}$ with $|G_i|=2^{w-i}$. By the Galois
Correspondence Theorem we then get an ascending tower of fields
$Q=F_0\subset F_1 \subset \cdots F_w=K$ with $[F_{i+1}:F_i]=2$, so
all $\gam\in K$ are square root constructible.
\een
\een
\end{document}
\een
\end{document}