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\bec {\Large\bf HONORS ALGEBRA II\\ FINAL EXAM, MAY 2016 \\ Solutions} \ece
\noi {\bf Part I:} Do any {\em five} of these problems.
\ben
\item (15) Let $\ah\in C$ have minimal polynomial $f(x)\in Q[x]$ and set
$K=Q(\ah_1,\ldots,\ah_n)$ where $\ah=\ah_1,\ldots,\ah_n$ are the
roots of $f(x)$.
Suppose
$[K:Q]=3^w$ for some integer $w$. Prove that $\ah$ is expressible
by radicals.
\So Set $G_0=\Gam[K:Q]$, the Galois Group. Then $|G|=3^w$.
From repeated application of Sylow's Theorem we get a reverse tower
of groups $G_0\supset G_1 \supset \ldots \supset G_{w-1} \supset G_0=\{e\}$
with $|G_i| = 3^{w-i}$. Applying the Galois Correspondence Theorem gives
an increasing tower $Q=K_0 \subset K_1 \subset \ldots \subset K_w=K$ with
$|K_i|=3^i$. Thus $[K_i:K_{i-1}]=3$ for $1\leq i \leq w$. Using Cardano's
Formula we can then express all elements of $K_i$ in terms of radicals involving
$K_{i-1}$ so, by induction, in terms of radicals involving $Q$. That is, all
elements of all $K_i$, including $\ah\in K_w$, are expressible by radicals.
\item (15) Solve the linear recursion $u_n=u_{n-1}+u_{n-2}$ with initial
conditions $u_0=2$, $u_1=1$ in $Z_{11}$.
\So The characteristic equation $x^2=x+1$ has solution $x=4,8$ so the general
solution is $u_n=a4^n+b8^n$. The initial conditions give the equaitons
\[ 2 = u_0 = a+b \]
\[ 1 = u_1 = 4a+8b \]
which have the solution $a=b=1$ so the solution is $u_n=4^n+8^n$.
\item (15) Given $\ah,\beta\in Z[i], \beta\neq 0$, {\em prove} that there
exist $q,r\in Z[i]$ with $\ah=q\beta+r$ and either $r=0$ or
$d(r) < d(\beta)$. Here $d(\beta) = |\beta|^2 = a^2+b^2$ where
$\beta=a+bi$.
\So This is the main part of the proof that $Z[i]$ is a Euclidean Domain, as
given in class.
\item (15) Set $p=701$. (This is a prime.) Let $g$ be a generator
of $Z_p$. In the problems below give a short reason for your
answer and express the solutions in terms of $g$.
\ben
\item (5) Find the number of solutions to $x^5=1$ in $Z_p$.
\So Set $x=g^i$. Then $g^{5i}=1$ so $700|5i$ so $i=140j$ for
$j=0,1,2,3,4$. Five solutions.
\item (5) Find the number of solutions to $x^3=1$ in $Z_p$.
\So Set $x=g^i$. Then $g^{3i}=1$ so $700|3i$ so $i=0$.
One solution.
\item (5) Find the number of solutions to $x^{55}=1$ in $Z_p$.
\So Hmmmm. Set $y=x^5$. Then $y^{11}=1$ so by argument similar to
part two, $y=1$. Thus there are five solutions, the same
solutions as in part one.
\een
\item (15) Let $K=Z_5[x]/(x^2+4x+2)$.
\ben
\item (5) How many elements are in $K$?
\So $25$
\item (5) Find the inverse of $x+1$.
\So $x+3$. (You can do this by finding the $\gcd(x+1,x^2+4x+2)$
but its easier just to ``eyeball" it.
\item (5) Find a $y\in K$ with $y^2=3$.
\So Well, $(x+2)^2 = x^2+4x+4 = 2 = -3$ and $-1 = 2^2$ so
$[2(x+2)]^2 = -(-3) =2$ so the solutions are $\pm(2x+4)$.
Again, maybe you can just eyeball it.
\een
\item (15)
{\em Prove} the Middle Normal Theorem. That is, prove that if $F\subset L \subset K$ and
$K:F$ is normal then $K:L$ is normal.
\So As done in class or in the text.
\item (15) Let $f(x)\in Z_p[x]$ be an irreducible cubic. Set $K=Z_p[x]/(f(x))$.
Let $a\in Z_p$.
Prove there exists $\ah\in K$ with $\ah^3=a$.
\So If $x^3-a$ is reducible in $Z_p[x]$ then it has a root in $Z_p\subset K$.
Otherwise $K=GF(p^3) \cong Z_p[y]/(y^3-a)$ and in that field we can take $\ah=y$.
\noi {\bf Part II:} Do {\em all} of these problems.
\item (20) For the following pairs, tell -- with short reason -- whether or not they are {\em normal} extensions.
\ben
\item (5) $Q(\eps):Q$ with $\eps = e^{2\pi i/12}$. (If you feel like bursting into song \ldots please don't!)
\So Yes. Splitting field of $x^{12}-1$.
\item (5) $Q(7^{1/6}):Q(7^{1/3})$
\So Yes. Splitting field of $x^2-7^{1/3}$. Alternately, extension of dimension two, which is always normal.
\item (5) $Q(5^{1/3}):Q$
\So No. $x^5-3$ is irreducible by Eisenstein and $5^{1/3}\omega$ is not in the field $Q(5^{1/3})$ as it contains
only reals.
\item (5) $Q(\sqrt{2},\sqrt{11},\sqrt{13}): Q$
\So Yes. Splitting field of $(x^2-2)(x^2-11)(x^2=13)$.
\een
\item (25) Set $Z[\sqrt{-2}]=\{a+b\sqrt{-2}\}$. This is a Euclidean Domain (a nice exercise but
not requested) with $d(\ah) = |\ah|^2=a^2+2b^2$. Assumming that show:
\ben
\item (5) If $d(\ah)$ is an integer prime then $\ah$ is a prime.
\item (10) If $p$ is an integer prime and there $p=x^2+2y^2$ has an integer solution then $p$
is {\em not} a prime in $Z[\sqrt{-2}]$.
\item (10) If $p$ is an integer prime and there $p=x^2+2y^2$ has no integer solution then $p$
is a prime in $Z[\sqrt{-2}]$.
\een
{\tt Solution:} This problem was given in hw3. Check website.
\noi {\bf PART III:} Do any {\em three} of these problems.
\item (20) Set $\ah=2^{1/3}$, $\omega=e^{2\pi i/3}$, $\beta=\ah\omega$, $\gam=\ah\omega^2$.
\ben
\item (5) Prove $Q(\beta)\neq Q(\gam)$.
\So If $Q(\beta)=Q(\gam)$ then $\omega = \frac{\gam}{\beta}\in Q(\beta)$ and so
$\ah = \frac{\beta}{\omega}\in Q(\beta)$ so $Q(\beta)=Q(\beta,\omega)$ so
$[Q(\beta):Q]=6$ which it doesn't.
\item (5) Find a nonsquare integer $n$ with $\sqrt{n}\in Q(\ah,\beta,\gam)$.
\So $n=-3$ as $\omega\in Q(\ah,\beta,\gam)$ and $2\omega+1=\sqrt{-3}$.
\item (5) Find $\ah^{37}+\beta^{37}+\gam^{37}$.
\So Zero.
\[ \ah^{37}+\beta^{37}+\gam^{37} = 2^{37/3}[1+\omega^{37}+\omega^{74}]=
2^{37/3}[1+\omega+\omega^{2}]= 0 \]
\item (5) Find $\ah^{36}+\beta^{36}+\gam^{36}$.
\So $3\cdot 2^{12}$ as each of them equal $2^{12}$.
\een
\item (20) Let $f(x)\in Q[x]$ be an irreducible quartic with
roots $\ah,\beta,\gam,\del$. Set $K=Q(\ah,\beta,\gam,\del)$
and $G=\Gamma[K:Q]$, the Galois group.
\ben
\item (10) What is the smallest possible size of $G$? Given an $f(x)$
such that $G$ achieves that size. (No proofs requested.)
\So Four. One example is to take $f(x)=\Phi_5(x) = (x^5-1)/(x-1)$.
\item (10) Set $\kappa=(\ah+\beta)^4+(\gamma+\delta)^4$. Give, with argument,
an upper bound on $[Q(\kappa):Q]$
\So Three. There are eight permutations of $\ah,\beta,\gam,\delta$ that preserve
$\kappa$. Thus the index of $Q(\kappa)^*$ in $G$ is at most three and so
$Q(\kappa)$ has degree at most three over $Q$.
\een
\item (20) Let $n\geq 3$. Set $\eps=e^{2\pi i/n}$, $\ah = 2^{1/n}$. Let $L$ be a field with
$\eps\in L$. Set $K=L(\ah)$. Set $\Gam=\Gam[K:L]$, the Galois group
\ben \item (10) Prove that $K:L$ is normal.
\So The equation $x^n-2$ has solutions $\ah,\ah\eps,\ldots,\ah\eps^{n-1}$ which are all
in $K$ so $K$ is the splitting field of $x^n-2$ over $L$.
\item
(10) Prove that $\Gam$ is Abelian.
\So Any $\sig$ is determined by $\sig(\ah)$ and so must be of the form
$\sig_j(\ah)=\ah\eps^j$. As $\sig(\eps)=\eps$,
\[ (\sig_j)(\sig_k)(\ah) = \sig_k(\ah\eps^j) = \sig_k(\ah)\eps^j = \ah\eps^{j+k} =
(\sig_k)(\sig_j)(\ah) \]
so $\sig_j\sig_k = \sig_k\sig_j$.
\een
\item (20) {\em State} clearly the Galois Correspondence Theorem.
Start with a field extension $K:F$ and the Galois Group $G=\Gam[K:F]$.
Include: Definitions of the maps $*$, $\dag$ between fields and groups; conditions
needed on $K:F$;
the relationship between
the sizes of the groups and the dimensions of the field extensions.
\So As done in class and in the text.
\een
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