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\begin{center} {\Large\bf Algebra V63.0349 \\ Assignment 4} \\ Due,
Friday, Februray 19 in Recitation \end{center}
\beq
The universe is not only queerer than we suppose but queerer than we {\em can}
suppose. \\ -- J.B.S. Haldane
\enq
\ben
\item Let $F\subset K$, both fields, and consider $K$ as a vector space over
$F$. Let $\ah\in K$, $\ah\neq 0$.
Prove that the map $T_{\ah}:K\ra K$ given by
$T_{\ah}(\beta)=\ah\beta$ is a homomorphism. That is, show
\ben
\item $T_{\ah}(v_1+v_2)=T_{\ah}(v_1)+T_{\ah}(v_2)$ for all $v_1,v_2\in K$
\item $T_{\ah}(\lam v) = \lam T_{\ah}(v)$ for all $v\in K$, $\lam\in F$.
\item
Prove further that $T_{\ah}$
an isomorphism between $K$ and itself.
\een
\item (Just for Fun) Presidential Trivia:
\begin{enumerate}
\item Which president had a great stamp collection?
\item Which was the fattest president?
\item Which two presidents died on the same day?
\item Which president was divorced?
\end{enumerate}
\item Let $\ah\in C$ be a root of $x^3+x+3$. (This cubic has no special properties.)
Write $\ah^i$ 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.
\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.
\item Give the factorization of $p(x)$ into irreducibles in $C[x]$.
\item Give the factorization of $p(x)$ into irreducibles in $Re[x]$.
\item Give the factorization of $p(x)$ into irreducibles in $(Q(\sqrt{2}))[x]$.
\item Give the factorization of $p(x)$ into irreducibles in $(Q(i\sqrt{2}))[x]$.
\item Show $p(x)$ is irreducible in $Q[x]$ {\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 $f(x)\in Q[x]$.
\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 $x=\pm 2$.)
\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.
\item How many elements are in $F^*$.
\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!)
\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!)
\een
\een
\beq
Nothing is more fruitful - all mathematicians know it - than those
obscure analogies, those disturbing reflections of one theory in
another; those furtive caresses, those inexplicable discords; nothing
also gives more pleasure to the researcher. The day comes when the
illusion dissolves; the yoked theories reveal their common source
before disappearing. As the {\em Gita} teaches, one achieves
knowledge and indifference at the same time.
\\ Andr\'e Weil
\\ (Note: ``indifference" is a controversial translation of the original Sanskrit, ``detachment" is often
used instead)
\enq
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