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\begin{center} {\Large\bf Algebra V63.0349 \\ Assignment 4} \\
Solutions \end{center}
\ben
\item Let $F\subset K$, both fields, and consider $K$ as a vector space over
$F$. Let $\ah\in K-\{0\}$. Prove that the map $T_{\ah}:K\ra K$ given by
$T_{\ah}(\beta)=\ah\beta$ is a homomorphism. Prove further that it is
an isomorphism between $K$ and itself.
\So Setting $T=T_{\ah}$ we have
\[ T(\beta_1+\beta_2)=\ah(\beta_1+\beta_2)=\ah\beta_1+\ah\beta_2 = T(\beta_1)+T(\beta_2) \]
and for any $\kappa\in F,\beta\in K$
\[ T(\kappa\beta) = \ah\kappa\beta = \kappa T(\beta) \]
It is an isomorphism because the map $U(\beta)=\ah^{-1}\beta$ is
the inverse.
\item Presidential Trivia:
\begin{enumerate}
\item Which president had a great stamp collection?
\So Franklin Delano Roosevelt (He would ask the State Department to send him stamps!)
\item Which was the fattest president?
\So William Howard Taft (Yet he was considered an excellent dancer)
\item Which two presidents died on the same day?
\So John Adams and Thomas Jefferson, on the $50$-th anniversary of the signing of the
Declaration of Independence.
\item Which president was divorced?
\So Ronald Reagan (but not as often as Donald Trump)
\end{enumerate}
\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 Here we examine $p(x)=x^4+1$. Let $\ah,\beta,\gam,\delta$ 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 Show $p(x)$ is irreducible in $Q[x]$ {\em using} the following idea: First (the easy part) show $p(x)$ has no root
in $Z$. Now suppose $p(x)=f(x)g(x)$ where $f(x),g(x)\in Z[x]$ are monic quadratics. Then $f(i)|p(i)$ for
$i=0,1$. Writing $f(x)=x^2+ax+b$ the values $f(0),f(1)$ determine $a,b$. Show that in each case that $f(x)$ is
not a divisor of $p(x)$. (Some grunt work here. A quick way to show $f(x)$ does not divide $p(x)$ is to check $x=\pm 2$.)
\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 $c\not\in Z$.
\item Show $p(x)$ is irreducible in $Q[x]$ {\em using} the following idea: First (the easy part) show $p(x)$
has no root
in $Z$. Now suppose $p(x)=f(x)g(x)$ where $f(x),g(x)\in Z[x]$ are monic quadratics. Then $f(i)|p(i)$ for
$i=0,1$. Writing $f(x)=x^2+ax+b$ the values $f(0),f(1)$ determine $a,b$. Show that in each case that $f(x)$
is
not a divisor of $p(x)$. (Some grunt work here. A quick way to show $f(x)$ does not divide $p(x)$ is to
check $x=\pm 2$.)
\So A root would be real but $x^4+1 > 0$ for all real $x$. By results from class, factorization in
$Q[x]$ would imply factorization into two monic quadratics. Suppose $f(x)=x^2+ax+b|(x^4+1)$.
Setting $x=0$, $b|1$ so $b=\pm 1$. Setting $x=1$, $a+b+1|2$ so $a+b+1 = \pm 1, \pm 2$. This
gives eight possibilities. Noting $p(2)=p(-2)=17$, $f(2),f(-2) \in \{-1,1,-2,2\}$ which can
be used to quickly eliminate all these possibilities.
\een
\item Assume as a fact that $x^5+x^2+1$ is irreducible in $Z_2[x]$. (It is!)
Set $F=Z_2[x]/(x^5+x^2+1)$. Set $F^*= F - \{0\}$.
\ben
\item How many elements are in $F$. Describe the elements.
\So $32$. They are $a+bx+cx^2+dx^3+ex^4$ with $a,b,c,d,e\in \{0,1\}$.
\item How many elements are in $F^*$.
\So $31$. All but zero.
\item Considering $F^*$ as a group under multiplication find the order
of $x$, that is, the minimal $n$ such that $x^n=1$. (One can do this
by brute force but there is a quick way!)
\So $F^*$ is a group with $31$ elements and $31$ is prime so all elements
(except the identity $1$) have order $31$.
\item Find the remainder, in $Z_2[x]$, when $x^{31000002}+1$ is divided
by $x^5+x^2+1$. (Again, there is a quick way!)
\So This is asking for $x^{31000002}+1$ in $F$. As $x^{31}=1$,
$x^{31000002}+1=x^2+1$.
\een
\een
\end{document}