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\bec {\Large\bf OLD ALGEBRA MIDTERM SOLUTIONS} \ece
\ben
\item (15) Let $F=Z_7[x]/(x^2+x+4)$. {\em Oooops}: On the
actual exam we had $x^2+x+2$ which does not give a field.
\ben \item (5) How many elements are in $F$?
\So $49$, $a+bx$ for $a,b\in Z_7$.
\item (10) Find
an explicit $\ah\in F$ such that $\ah\not\in Z_7$ and $\ah^2\in Z_7$.
Give $\ah^2$ explicitly.
\So As $x^2+x+4=0$ we complete the square, noting that $\frac{1}{2}=4$,
so that $(x+4)^2= x^2+x+16=12 = 5$ so $\ah=x+4$ and $\ah^2=5$. (There
are other solutions.)
\een
\item (20) Let $p$ be an integer prime of the form $p=4k+1$. (Note:
You can assume earlier parts when working on later parts of this
problem.)
\ben
\item \label{a} (5) Show that there exists $a\in Z_p^*$ with
$a^2=-1$ in $Z_p$. (Hint: Let $g$ be a generator.)
\So $g^{4k}=1$. Set $y=g^{2k}$. Then $y^2=1$ but $y\neq 1$
(as $g$ is a generator) so $y=-1$. Take $a=g^k$.
\item \label{b} (5) Let $a$ be a positive integer. Show that $a^2+1$
is {\em not} a prime in the Gaussian Integers.
\So $(a^2+1)=(a+i)(a-i)$. As $a > 0$ neither of the factors are units.
\item \label{c} (5) Using (\ref{a},\ref{b}) show that $p$ is not a prime
in the Gaussian Integers.
\So Let $a$ be from (\ref{a}). Then $p|(a^2+1)$ but $p$ divides neither of
$a+i,a-i$. But the Gaussian Integers are a Euclidean Domain in which
primes $\kappa$ must satisfy $\kappa|\ah\beta \Rightarrow \kappa|\ah {\mbox or}
\kappa|\beta$.
\item \label{d} (5) Using (\ref{c}) show that there exist integers
$x,y$ with $p=x^2+y^2$. {\tt Note:} This result is called
the Fermat Two Squares Theorem.
\So As $p$ is not prime some $(x+iy)|p$. The size function $d(\ah)=|\ah|^2$
is multiplicative so $x^2+y^2|d(p)=p^2$. But $d(x+iy)$ is neither one nor $p^2$
since the factorization is nontrivial so it must be $p$.
\een
\item (25)
Let $R=\{a+b\sqrt{-2}:a,b\in Z\}$. You may assume, without
proof, that $R$ is a ring. Set $d(\ah)=|\ah|^2$ where,
as usual, $|x+iy| = \sqrt{x^2+y^2}$.
\ben
\item (5) Draw a picture of the points of $R$ on the complex plane.
\So It will be a rectangular lattice but in the $Y$-axis is it every
$\sqrt{2}\sim 1.4$.
\item (5) Find all units of $R$ and prove they are the only units.
\So The only units are $+1,-1$. As $d(\ah)$ is multiplicative and
a positive ($\ah\neq 0$) integer a unit must have $d(\ah)=1$ so
$x^2+2y^2=1$ which has only the solutions $y=0, x=\pm 1$.
\item (5) State the conditions for $R$ to be a Euclidean Domain
with size function $d$.
\So First $d(\ah)$ must always be a nonnegative integer. Then
\ben \item $d(\ah)\leq d(\ah\beta)$
\item For all $\ah,\beta$ with $\beta\neq 0$ there exist $q,r$ with
$\ah=q\beta+r$ and either $r=0$ or $d(r) < d(\beta)$.
\een
\item (10) Prove that $d$ does indeed satisfy those conditions.
\So Clearly $d(\ah)\geq 1$ for all $\ah$. (Note $d$ is not
defined on zero.) As $d$ is multiplicative
$d(\ah\beta) = d(\ah)d(\beta) \geq d(\ah)$.
For the second set $\ah/\beta = x + y\sqrt{-2}$. Round off
$x,y$ to the nearest integers $x_0,y_0$ and set
$q=x_0+y_0\sqrt{-2}$. Then
\[ r = \ah - q\beta = \beta((x-x_0)+ (y-y_0)\sqrt{-2}) \]
and
\[ d(r) = |r|^2 = d(\beta)[(x-x_0)^2+2(y-y_0)^2] < d(\beta) \]
as the bracketed term is at most $3/4$. (One can also argue
geometrically.)
\een
\item (10) Let $F=Q[x]/(x^2-x-1)$ Write each of $x^2,x^3,x^4,x^5$ in
the form $a+bx$.
\So $x^2=x+1$,
$x^3=xx^2 = x^2+x= 2x+1$,
$x^4=xx^3 = 2x^2+x= 3x+2$,
$x^5=xx^4 = 3x^2+2x= 5x+3$. (Fibonacci!)
\item (10) Let $\ah$ be a complex number and assume $[Q(\ah):Q]=p$, with
$p$ 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
$\ah=b_0+b_1\beta+\ldots+b_{p-1}\beta^{p-1}$.
\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 (5) Use Eisenstein's criteria to prove that $f(x)=x^{11}+6x^8 + 12$
is irreducible in $Z[x]$.
\So $3$ divides the coefficients $12,6$ (and the zeroes) but does not divide
the lead coefficient $1$, while $9=3^2$ does not divide the constant $12$.
(Note $p=2$ does not work!)
\item (10) Let $D$ be a P.I.D. Assume $\pi\in D$ is irreducible. Let
$a\in D$. Prove that either $\pi|a$ or that there exist $x,y\in D$
with $x\pi+ya=1$.
\So Set $\ah=\gcd(a,\pi)$. As $\ah|\pi$, either $\ah=1$ or $\ah=p$.
If $\ah=p$ then $p=\ah|a$. If $\ah=1$ then, by the $\gcd$ property,
we can write $\gcd(a,\pi)=xa+y\pi$ for some $x,y$, so
$xa+y\pi = 1$.
\item (5) Let $R=Q[x]/(x^3-x-1)$. Suppose that in $R$
\[ (1+x)(a+bx+cx^2) = 1 \]
Give (but do {\em not} attempt to solve!!) a system
of three linear equations in three unknowns that $a,b,c$
would satisfy.
\So We multiply out
\[ 1=(1+x)(a+bx+cx^2) = a + (a+b)x+ (b+c)x^2+ cx^3 =
a + (a+b)x+ (b+c)x^2+ c(x+1) \]
Equating the constant term gives $a+c=1$. The coefficients
of $x,x^2$ must be zero so $a+b+c=0$ and $b+c=0$.
\item (10) Let $Q=K_0\subset K_1 \subset K_s$ be a tower of fields,
all inside $C$, such that for each $1\leq i\leq s$ either
$K_i=K_{i-1}(\ah_i)$ for some $\ah_i$ with $\ah_i^2\in K_{i-1}$ or
$K_i=K_{i-1}(\ah_i)$ for some $\ah_i$ with $\ah_i^3\in K_{i-1}$.
Prove $2^{1/5}\not\in K_s$.
\So All $[K_i:K_{i-1}]\leq 3$ so $[K_s:K_0]$ is the product of numbers
at most $3$ and so does not have a factor of $5$. But $2^{1/5}$
satisfies $x^5-2=0$ which is irreducible by Eisenstein so if
$2^{1/5}\in K_s$ then $Q\subset Q(2^{1/5})\subset K_s$ so by
the tower theorem $[Q(2^{1/5}):Q]=5|[K_s:Q]$, a contradiction.
\een
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