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\begin{document}
\begin{center} {\Large\bf Algebra V63.0349 \\ Assignment 3} \\ Solutions
\end{center}
\ben
\item Let $F=Q[x]/(x^3-2)$. Find the inverse of $2+x+x^2$ by calling it
$a+bx+cx^2$ and getting, and solving, three equations in three unknowns.
\So We have
\[ (2+x+x^2)(a+bx+cx^2)=(2a+2b+2c)+(a+2b+2c)x+(a+b+2c)x^2 \]
by replacing $x^3$ with $2$ and $x^4$ with $2x$. So the equations are
\[ (2a+2b+2c)=1 \]
\[ (a+2b+2c)=0 \]
\[ (a+b+2c)=0 \]
with solution $a=1,b=0,c=-\frac{1}{2}$ and $1-\frac{1}{2}x^2$ is the inverse.
\item Let $F=Z_7[x]/(x^3-2)$. Find the inverse of $2+x+x^2$ by calling it
$a+bx+cx^2$ and getting, and solving, three equations in three unknowns.
\So Same as above. But we want the final values to be $0,1,\ldots,6$ so
$b=0$ and $c=3$ and $1+3x^2$ is the inverse.
\item For which $n\geq 2$ and which $a\in Z$ can we use Eisenstein's
criterion to show $x^n-a$ is irreducible in $Q[x]$?
\So Any $n\geq 2$ and any $a$ for which {\em some} prime $p$ divides $a$ but
$p^2$ does not divide $a$. So $x^6-12$ is irreducible by taking $p=3$.
But $x^4+4$ and $x^{11}-72$ do not fit Eisenstein's Criterion. That does
mean that they are reducible, it only means that the criterion does not
apply. BTW, the tempting conjecture that $x^n-a$
is irreducible when $a$ is not an $n$-th power is {\em incorrect} as
the example
\[ x^4 + 4 = (x^2+2x+2)(x^2-2x+2) \]
shows.
\item By $C^*$ we mean $C-\{0\}$, the nonzero complex numbers.
Here we examine $C^*,\cdot)$, the group under
multiplication. Let $S$ be the set of solutions to the equation $z^{12}=1$.
\ben
\item Draw a nice picture of $S$ on the complex plane $C$.
\So It should look like the hours of a clock.
\item For each $z\in S$ marked above, give the {\em minimal} positive integer $s$ with $z^s=1$.
\So Set $\eps=e^{2\pi i/12}$, one twelfth
around the circle, so that $\eps^j$ is at ``j o'clock." Then the roots are $z=\eps^j$, $0\leq j < 11$.
$j=0$ is $z=1$ with $s=1$, $j=6$ is $z=-1$ with $s=2$, $j=4,8$ give $z=\omega,\omega^2$ where
$\omega=(-1+i\sqrt{3})/2$ is a cube root of unity. For them, $s=3$. $j=3,9$ give $z=i,-i$ with
$s=4$. $j=2,10$ give $s=6$ (e.g. $(\eps^2)^6 = \eps^{12}=1$. Finally, $j=1,5,7,11$ give $s=12$.
\item Prove that $S$ is a subgroup of $C^*$.
\So One approach is that $1\in S$; $a,b\in S$ imply $(ab)^{12}=a^{12}b^{12}=1$ so $ab\in S$; $a\in S$
implies $(a^{-1})^{12} = (a^{12})^{-1} = 1^{-1}=1$ so $a^{-1}\in S$. Another is to note that $S$
under multiplication is isomorphic to $Z_{12}$ under addition by identifying $\eps^j$ with $j$.
\item Prove that $S$ is the {\em only} subgroup of $C^*$ with (precisely) twelve elements.
\So If $T$ were a subgroup with $12$ elements then by Lagrange's Theorem $w^{12}=1$ for each
$w\in T$, so $T\subset S$. As they both have $12$ elements, $T=S$.
\een
\item 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 If $d(\ah)$ is an integer prime then $\ah$ is a prime.
\So If $\ah=\beta\gam$ then $d(\ah)=d(\beta)d(\gam)$ so $d(\beta)=1$
(or $d(\gam)=1$) so $\beta$ is a unit.
\item 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 $Z[\sqrt{-2}]$.
\So $p=(x+y\sqrt{-2})(x-y\sqrt{-2})$
\item If $p$ is an integer prime and there $p=x^2+2y^2$ has no integer solution then $p$
is a prime in $Z[\sqrt{-2}]$.
\So If $p=\beta\gam$ then $p^2=d(p)=d(\beta)d(\gam)$ so $d(\beta)=p$, but writing
$\beta=x+y\sqrt{-2}$ we would have $p=x^2+2y^2$.
\een
\een
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