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\begin{center} {\Large\bf Algebra V63.0349 \\ Assignment 4} \\
Solutions \end{center}
\ben
\item Let $\ah\in C$ be a root of $x^3+x+3$. (This cubic has no special properties.)
Writs $\ah^i$ in the form $a+b\ah+c\ah^2$, $a,b,c\in Q$, for $3\leq i\leq 6$.
Set $\beta=\ah^2$. Find a cubic in $Q[x]$ that has $\beta$ as a root.
\So We have $\ah^3=-\ah-3$ so $\ah^4=-\ah^2-3\ah$ and
\[ \ah^5 =\ah\ah^4 = -\ah^3-3\ah^2 = -3\ah^2+\ah+ 3 \]
and
\[ \ah^6 = \ah\ah^5 = -3\ah^3+\ah^2+3\ah = (3\ah+9) + \ah^2 + 3\ah = \ah^2+6\ah+ 9 \]
(a bit quicker in this case is to write $\ah^6=(\ah^3)^2 = (-\ah-3)^2$.
We use the method of undetermined coefficients. Suppose
\[ \beta^3 + a\beta^2 + b\beta + d = 0 \]
(By dividing we can assume the coefficient of $\beta^3$ is one here.)
So this means
\[ (\ah^2+6\ah+ 9) + a(-\ah^2-3\ah) + b\ah^2+ c = 0 \]
which means that the coefficients of $1,\ah,\ah^2$ all must be zero giving the
equations
\[ 1 - a + b = 0 \]
\[ 6-3a = 0 \]
\[ 9+c = 0 \]
giving the solution (usually the equations are much harder!) $c=-9$, $a=2$, $b=1$ so that
$\beta$ is a root of the cubic $x^3+2x^2+x-9$.
\item Let $p(x)\in Z[x]$ be a {\em monic} polynomial of degree $n$, irreducible over $Q$.
Let $\ah\in C$ be a root of the equation $p(x)=0$. Set
\[ R= \{a_0+a_1\ah+\ldots+a_{n-1}\ah^{n-1}: a_0,\ldots,a_{n-1}\in Z\} \]
Show that $R$ is a ring. (The hard part is closure under multiplication.) Now
suppose further that $p(x)$ has constant term $\pm 1$. Show that $\ah^{-1}\in R$.
\So Write $p(x)=x^n+b_{n-1}x^{n-1}+\ldots+b_0$ with the $b_i\in Z$. We claim that
for any $\gam=a_0+a_1\ah+\ldots+a_{n-1}\ah^{n-1}\in R$ we have $\ah\gam\in R$. All
the terms go down one except $a_{n-1}\ah^{n-1}$ which becomes $a_{n-1}(
(-b_{n-1}\ah^{n-1}+\ldots-b_0)$ so it is still in $R$. Now by induction all $\ah^j\in R$.
and so we get closure under multiplication. Dividing the equation $p(\ah)=0$ by $\ah$ and
isolating the $\ah^{-1}$ gives $b_0\ah^{-1}=b_1+\ldots+b_{n-1}\ah^{n-2}+\ah^{n-1}$. So
that $b_0\ah^{-1}\in R$. When $b_0=1$ we are done and when $b_0=-1$ just take the additive
inverse and we are done.
\item Here we examine the polynomial $p(x)=x^4+1$. Let $\ah,\beta,\gam,\del$ denote
the complex roots of $p(x)=0$.
\ben
\item Find $\ah,\beta,\gam,\del$ both in terms of polar coordinates $\ah=re^{i\theta},\ldots$
(this is actually easier for this particular problem) and in the Cartesian $\ah=a+bi,\ldots$ forms.
and mark them on the complex plane.
\So $-1$ lies on the unit circle and can be written $-1=e^{i\pi}$. So its fourth roots also lie
on the unit circle and so will be of the form $e^{i\theta}$ where $4\theta=\pi$. But the last
equation is as angles (that is, modulo $2\pi$) and so there are four solutions
$\ah=e^{i\pi/4}$,
$\beta=e^{3i\pi/4}$,
$\gam=e^{5\pi/4}=-\ah$,
$\del=e^{7i\pi/4}=-\beta$.
They form an ``X" pattern, all on the unit circle at these angles.
As $\cos(\pi/4)=\sin(\pi/4)=\frac{\sqrt{2}}{2}$ we express the four roots
as
\[ x = \frac{\sqrt{2}}{2}(\pm 1 \pm i) \]
\item Give the factorization of $p(x)$ into irreducibles in $C[x]$.
\So We have the four roots so
\[ p(x)=(x-\ah)(x-\beta)(x-\gam)(x-\del) \]
\item Give the factorization of $p(x)$ into irreducibles in $Re[x]$.
\So We match up roots with their complex conjugates so that
\[ (x-\ah)(x-\gam)= x^2-\sqrt{2}x+ 1 \]
\[ (x-\beta)(x-\del)= x^2+\sqrt{2}x+ 1 \]
are the factors.
\item Give the factorization of $p(x)$ into irreducibles in $(Q(\sqrt{2}))[x]$.
\So The above factorization works over this field as well. (Note that there can't
be any linear factors as none of the roots $\ah,\beta,\gam\del$ are real.)
\item Give the factorization of $p(x)$ into irreducibles in $(Q(i\sqrt{2}))[x]$.
\So Here
\[ (x-\ah)(x-\beta) = x^2 - i\sqrt{2}x - 1 \]
and
\[ (x-\gam)(x-\del) = x^2 + i\sqrt{2}x - 1 \]
form the factors.
\item Show $p(x)$ is irreducible {\em using} the following idea: If, say, $p(x)=f(x)g(x)$,
then, as $p(x)$ factors into four linear factors in the first part above, $f(x)$ and $g(x)$ must be
a product of some (but not all) of those factors. Try all possiblities for products of the linear factors
(there aren't that many) and check that none of them give an $f(x)\in Q[x]$.
\So There are no rational roots (indeed, no real roots) so there are no linear factors and so
$f(x),g(x)$ must be quadratic so one of them (WLOG, $f(x)$) has a factor $(x-\ah)$ so we need
check the three products $(x-\ah)(x-\kappa)$, $\kappa=\beta,\gam,\del$. We have $\gam,\beta$ above
and $(x-\ah)(x-\del)=x^2-\sqrt{2}x+1\not\in Q[x]$.
\item There are many ways to show that a polynomial is irreducible over $Q[x]$. Show $p(x)$ is
irreducible over $Q[x]$ by some other method -- your choice!
\So By Gauss's Lemma we can restrict to factorizations with monic polynomials. A root $a\in Z$
would need $a|1$ and neither $\pm 1$ are roots, so there is no linear monic factor. The constant
terms must multiply to $1$ so they are either both $+1$ or both $-1$.
If we had a factorization
\[ x^4 + 1 = (x^2+cx+1)(x^2+dx+1) \]
then, by the $x^3$ coefficient, $d=-c$, so by the $x^2$ coefficient $-c^2+2=0$ and $c\not\in Z$.
If we had a factorization
\[ x^4 + 1 = (x^2+cx-1)(x^2+dx-1) \]
then, by the $x^3$ coefficient, $d=-c$, so by the $x^2$ coefficient $-c^2+2=0$ and $c\not\in Z$.
\een
\item Let $f(z)=10+(i+1)z^4+6z^5$. (Here $i=\sqrt{-1}$.) Find an angle $\phi$ (following the
proof of the Fundamental Theorem of Algebra) such that
$|f(10^{-100}e^{i\phi})| < 10 - 1.4\cdot 10^{-400}$.
\So Observe $i+1=\sqrt{2}e^{i\pi/4}$. Set $\phi$ so that $4\phi + \frac{\pi}{4}= \pi$. We
may take $\phi = \frac{3\pi}{4}$. (There are three other possibilities.) Now, with
$z=10^{-100}e^{i\phi}$, $(i+1)z^4 = -\sqrt{2}10^{-400}$. (The crucial point is that it is
in the opposite direction of the $+10$ constant.) Thus $|10+(i+1)z^4| = 10 - \sqrt{2}10^{-400}$.
Rather than now precisely calculating $|f(z)|$ we note that $|6z^5| = 6\cdot 10^{-500}$, much
smaller than $10^{-400}$.
Thus
\[ |f(z)| \leq (10-\sqrt{2}10^{-400}) + 6\cdot 10^{-500} < 10 - 1.4\cdot 10^{-400} \]
(Note: The important thing about $1.4$ was that it was less than $\sqrt{2}$.)
\item Let $\ah\in C$ satisfy $\ah^3+\beta\ah^2+\gam\ah+\del=0$ where $\beta,\gam,\del$
satisfy cubics with coefficients in $Q$. (That is, $\beta^3+a\beta^2+b\beta+c=0$ for some
$a,b,c\in Q$, and similarly for $\gam,\del$.)
(Possibly smaller degree
polynomials are satisfied by these numbers.)
\ben
\item Give an upper bound for the degree of
$\ah$ over $Q$.
\So $[Q(\beta):Q]\leq 3$,
$[Q(\gam):Q]\leq 3$,
and $[Q(\del):Q]\leq 3$,
Set $K=Q(\beta,\gam,\del)$, $[K:Q]\leq 3^3=27$. Also
$[K(\ah):K]\leq 3$ so $[K(\ah):Q]\leq 3^4 = 81$.
\item Show that the minimal polynomial of $\ah$ over $Q$ does {\em not} have degree
precisely five.
\So
We have a tower $Q, K_1=Q[\beta]$,
$K_2=K_1[\gam]$, $K_3=K_2[\del]$, $K_4=K_3[\ah]$. Each has
degree $\leq 3$ so $[K_5:Q]$ is the product of numbers $\leq 3$ and so can't have a
factor of $5$. But $Q(\ah)\subset K_4$ so that $[Q(\ah):Q]|[K_4:Q]$ and so can't be
five.
\een
\item If $\ah,\beta\in K$ are algebraic over $F$ of degrees $m$ and $n$ respectively
and if $m,n$ are relatively prime, prove that $F(\ah,\beta)$ is of degree $mn$ over
$F$.
\So From the tower $F\subset F(\ah)\subset F(\ah,\beta)$ we know that $m|[F(\ah,\beta):F]$.
From the tower $F\subset F(\beta)\subset F(\ah,\beta)$ we know that $n|[F(\ah,\beta):F]$. As
the $m,n$ are relatively prime $mn|[F(\ah,\beta):F]$ but this is our {\em upper} bound for
$[F(\ah,\beta):F]$ and so we must have equality.
\item Let $p$ be an odd prime. Assume as a (true!) fact that when $\ah,\beta\in Z_p^*$
are {\em not} squares then $\ah/\beta$ is a square.
Let $L,K$ be fields, both of size precisely $p^2$.
Using the results on extension of degree two show that $L,K$ are isomorphic.
($L,K$ are isomorphic if there exists a bijection $\psi: L\ra K$
that preserves addition and multiplication.)
\So As $[L:Z_p]=2$ we know there exists $u\in L$, $Z_p(u)=L$, and $u^2=\ah\in Z_p$.
Similarly there exists $v\in K$, $Z_p(v)=K$ and $v^2=\beta\in Z_p$. Then $\ah,\beta$
cannot be squares (otherwise $u,v$ respectively would actually be in $Z_p$) so they
are nonsquares so by the fact $\ah=s^2\beta$ for some $\beta\in Z_p$. Now
the isomorphism $\psi: L\ra K$ is given by setting $\psi(u)=sv$, that is,
setting $\psi(a+bu)=a+b(sv)$ for all $a,b\in Z_p$. To show multiplication
is preserved you basically need $\psi(u^2)=\psi(u)^2$ or $\ah=(sv)^2=s^2\beta$
which we have. If we allow a squareroot notation we are showing $Z_p(\sqrt{\ah})$
is isomorphic to $Z_p(\sqrt{\beta})$ and $\sqrt{\ah}= s\sqrt{\beta}$.
\een
\end{document}