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\bec {\Large\bf ALGEBRA MIDTERM SOLUTIONS} \ece
{\tt Note:} Other version simila.
\ben
\item (5) How many elements are in $Z_7[x]/(x^2-3)$?
\So $49$. The elements are $a+bx$ with $a,b\in \{0,\ldots,6\}$.
\item (25) Set $\eps=e^{2\pi i/12}$.
\ben
\item (10) Mark $\eps^s$ on the complex plane for $0\leq s < 12$.
\So On the unit circle at a 30 degree angle. (1 o'clock if you feel
like singing.)
\item (5) Find a polynomial $p(x)\in Q[x]$ which has $\eps^8$ as a root
but does {\em not} have $\eps$ as a root.
\So $x^3=1$.
\item (10) Prove that $[Q(\eps):Q]\leq 4$. (For partial credit find
as good an upper bound on $[Q(\eps):Q]$ as you can.)
\So $\eps$ satisfies $x^6+1=0$ whose roots are $\eps^i$ for $i=1,3,5,7,9,11$.
But $\eps^3=i$ and $\eps^9=-i$ satisfy $x^2+1$ which $\eps$ does not satisfy
so they are not roots of the minimal polynomila of $\eps$. That leave four
roots $\eps^i$, $i=1,5,7,11$. We showed in class that minimal polys don't
have multiple roots so the min poly for $\eps$ has degree at most $4$ so
$[Q(\eps):Q]\leq 4$. (Later in the term we will show that it actually is $4$.)
\een
\item (15) Let $T:V\ra W$ be a homomorphism. (Here $V,W$ are vector spaces
over the same field $F$.) Set
\[ S = \{\vec{v}\in V: T(\vec{v})=\vec{0}\} \]
\ben
\item (10) Prove that $S$ is a subspace of $V$.
\So To show subspace we need
\ben
\item $S\neq \emptyset$. True as $T(\vec{0})=\vec{0}$.
\item $S$ closed under sum: Let $\vec{v},\vec{w}\in S$ so $T(\vec{v})=T(\vec{w})=\vec{0}$
so $T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w}) = \vec{0}+\vec{0}=\vec{0}$.
\item $S$ closed under scalar multiple. Let $\vec{v}\in V$ and $\lam\in F$ where $F$ is
the underlying field. Then $T(\lam\vec{v})= \lam T(\vec{v}) = \lam \vec{0} = \vec{0}$.
\een
\item (5) What is the usual term for $S$?
\So Kernel of $T$.
\een
\item (15) Let $\ah$ be a complex number and assume $[Q(\ah):Q]=p$, with
(important!)
$p$ a prime number. Assume $\beta\in Q(\ah)$ and $\beta\not\in Q$.
Prove that there exist $b_0,b_1,\ldots,b_{p-1}\in Q$ with
$\ah=b_0+b_1\beta+\ldots+b_{p-1}\beta^{p-1}$. (Hint: Consider the
relationships between $Q$, $Q(\ah)$, and $Q(\beta)$.)
\So As $\beta\in Q(\ah)$, $Q\subset Q(\beta)\subset Q(\ah)$. As
$\beta\not\in Q$, $Q\neq Q(\beta)$. We showed in class, as
a corollary of the tower theorem, that an
extension of prime dimension has no strictly intermediate fields.
Thus $Q(\beta)=Q(\ah)$. Thus $\ah\in Q(\beta)$ which means $\ah$
can be written as desired above.
\item (10) Let $R=Q[x]/(x^3-2)$. Suppose that in $R$
\[ (1+x+x^2)(a+bx+cx^2) = 2+3x+4x^2 \]
Give (but do {\em not} attempt to solve!!) a system
of three linear equations in three unknowns that $a,b,c$
would satisfy.
\So
\[ (1+x+x^2)(a+bx+cx^2) = a + bx + cx^2+ ax + bx^2 + 2c + ax^2+2b + 2cx \]
where we have used the reductions $x^3=2, x^4=2x$. Identifying the constants
\[ a +2c+2b =2 \]
Identifying the coefficients of $x$ gives
\[ b + a + 2c = 3 \]
and identifying the coefficients of $x^2$ gives
\[ c + b + a = 4 \]
\item (10) Let $\ah,\beta\in C$, both algebraic over $Q$.
\\ Prove that $[Q(\ah,\beta):Q(\ah)]\leq [Q(\beta):Q]$.
\So Say $[Q(\beta):Q]= n$. Then the minimal polynomial
$p(x)\in Q[x]$ for $\beta$ has degree $n$. We can consider
$p(x)$ as a polynomial in $Q(\ah)[x]$. It still has $p(\beta)=0$.
If its minimal, $[Q(\ah,\beta):Q(\ah)]=n$.
If it isn't minimal $\beta$
will satisfy a polynomial in $Q(\ah)[x]$ of degree even less
than $n$ so that $[Q(\ah,\beta):Q(\ah)]< n$. Either way
$[Q(\ah,\beta):Q(\ah)] \leq n$.
\item (20) Let $R$ be a Euclidean Domain. {\em Prove} that
$R$ is a Principle Ideal Domain. (Partial credit for defining
Euclidean Domain and Principle Ideal Domain.)
\So We have a size function $d: R-\{0\} \ra \{0,1,2,3,\ldots \} $ such
that
\ben \item $d(\ah\beta)\geq d(\ah)$
\item For all $a,b\in R$ with $b\neq 0$ there exist $q,r$ with $a=bq+r$
and either $r=0$ or $d(r) < d(b)$.
\een
Let $I\subset R$ be an ideal. To show we have a PID we need show $I=(\kappa)$
for some $\kappa\in R$. First, the trivial case: If $I=\{0\}$ then $I=(0)$.
Otherwise $I$ contains some nonzero $b$. Let $\kappa$ be a nonzero element
of $I$ with {\em minimal} $d(\kappa)$, that is, so $d(\kappa)\leq d(\gam)$ for
all nonzero $\gam\in I$. We claim $I=(\kappa)$. As $\kappa\in I$, $(\kappa)\subset I$.
Now let $a\in I$. There exist $q,r$ with $a=q\kappa+r$ and either $r=0$ or
$d(r) < d(\kappa)$. But $r=a-q\kappa \in I$ as $a,\kappa\in I$ and $I$ is closed
under subtraction and multiplying by any $q\in R$. If $r\neq 0$ then $d(r) < d(\kappa)$
contradicting the minimality of $d(\kappa)$. Thus $r=0$. That is, $a=q\kappa$ so $a\in (\kappa)$.
We've shown $I\subset (\kappa)$. Hence $I=(\kappa)$.
\item (15) Let $Z[i]$ be, as usual, the Gaussian integers.
Let $p$ be an integer prime.
\ben
\item (5) Assume there {\em do} exist integers $a,b$ with $p=a^2+b^2$.
Prove that $p$ is {\em not} a prime in the Gaussian integers.
\So $p= (a+bi)(a-bi)$
\item (10) Assume there do {\em not} exist integers $a,b$ with
$p=a^2+b^2$. Prove that $p$ {\em is} a prime in the Gaussian integers.
\So If $p=\ah\beta$ then $p^2=|\ah|^2\cdot |\beta|^2$. If $|\ah|=1$ then
$\ah$ is a unit, similarly $\beta$. So if it weren't a prime we would
need $|\ah|^2=|beta|^2=p$. Writing $\ah=a+bi$, $p=|\ah|^2= a^2+b^2$.
\\ {\tt Alternate Solution:} Suppose $p=(x+iy)(z+iw)$. Equating
real and complex parts
\[ xz-yw = p \]
\[ xw + yz = 0 \]
Multiply the first equation by $w$ and replace $xw$ with $-yz$ to give
\[ -yz^2-yw^2 = wp \]
and $(z^2+w^2)|wp$. Set $s=\gcd(z^2+w^2,w)$. Then $s|(z^2+w^2)=ww = z^2$.
If $s\neq 1$ then $w,z^2$ have a common factor which implies $w,z$ have
some common factor $t$ but then $t|(w+iz)$ so that $t|p$, not possible.
Hecne $s=1$ so $z^2+w^2|p$, but as $p$ is a prime we must have $z^2+w^2=p$.
(Whew! That was {\em much} harder!)
\een
\een
\begin{quote}
The voyage of discovery lies not in seeking new horizons, but
in seeking with new eyes. \\ -- Proust
\end{quote}
\end{document}