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{\Large
\centerline{\bf MAT 535, ALGEBRA II,  SPRING 2001}
\centerline{\bf FINAL EXAM}
}

%\vskip80pt
%\centerline{\bf !!! -10 pts for not writing your name, 
%ssn and section below!!!}

\bigskip
\noindent
NAME :
\hskip 9cm
SSN :

%\bigskip
%\centerline{ SECTION :}

\bigskip
\centerline{\bf THERE ARE SIX PROBLEMS.}
\centerline{THEY DO NOT HAVE EQUAL VALUE. }
\centerline{\bf SHOW YOUR WORK!!!}

\bigskip{ YOU MAY USE ANY RESULT DISCUSSED IN CLASS, IN THE HOMEWORKS
OR ANY THEOREM FROM ARTIN, HUNGERFORD OR THE NOTES I HAVE DISTRIBUTED.
HOWEVER, PLEASE SAY WHICH RESULT(S) YOU ARE USING (FOR EXAMPLE
``BY THE STRUCTURE THEOREM FOR FINITELY GENERATED MODULES OVER A P.I.D.,"
UNLESS IT IS OBVIOUS.}

\bigskip
 WORK ALONE. DO NOT USE OR CONSULT ANY OTHER TEXT
OR PERSON. ANY VIOLATION OF THESE SIMPLE RULES IS CHEATING.

\bigskip
PLEASE:

\noindent
HAND-IN A READABLE SET OF SOLUTIONS!
THE TEST IS POSTED ON THE WEB : YOU CAN MAKE A COPY
IF YOU NEED. USE ONLY THE SPACE GIVEN
(FRONT PAGES ONLY!). 


\bigskip
THIS IS A TAKE-HOME TEST. RETURN IT IN PERSON
TO MY OFFICE, MATH TOWER ROOM 3-115
{\bf ON MONDAY MAY 14, 12 NOON.}
NO EXCEPTIONS.
IF YOU CANNOT RETURN IT TO ME
AT 12 NOON, YOU STILL HAVE TO HAND-IN THE TEST 
BY THAT TIME AND DAY; IN THIS CASE
MAKE AND KEEP A COPY,
 SLIDE THE ORIGINAL  UNDER THE DOOR AND E-MAIL 
({\em mde@math.sunysb.edu}) ME
IMMEDIATELY THAT YOU HAVE DONE SO.



\newpage

\noindent
1. 
Prove that every element of a finite field $E$
can be written
as the sum of two squares of elements in $E$.




\noindent
{\em Hints: There are several ways to solve this problem.
The following hints point to one solution.

 1) Do first the case of $ char \, E=2$. 

2) Study the group
homomorphism
$s : E^*\to E^*$, $s(a)=a^2$; consider $E^*/s(E^*)$ and how $E^*$ gets 
subdivided into cosets; consider what happens to these subdivisions if you
multiply by elements in $E^*$.

3) Consider $f: E \to E$ given by $f(b) = 1-b$.}


\bigskip
Let $p:=char\, E$.

If $p=2$, since $E$ is finite,   Frobenius, $c \to c^2$, is bijective  
and every element is a square, $a=b^2$, and we are done $a=b^2 +0^2$.

Let $p\neq 2$.

Let $s : E^* \to E^*$ be defined as $s(a)=a^2$. It is a group homomorphism
with kernel $\{1, -1 \}$. Let $S:= s(E^*)< E^*$ be the image, i.e. the set of 
squares. $[E^*:S]=2$. Let $T = E^* \setminus S$. We have $|S|=|T|$.

The following can be checked immediately:

1) if $b \in S$, then $bS=S$ and $bT=T$;

2) if $b \in T$, then $bS=T$ and, by cardinality reasons,
$bT=S$.

Let $E^{**}= E \setminus \{0, 1\}.$

Define $S^* = S \setminus \{1 \}$.

$E^{**}=S^* \coprod T$.

Let $f: E^{**} \simeq E^{**}$ be  the bijection $u \to 1-u$.

Since $|S^*| = |T|-1$, $f(T) \cap T \neq \emptyset$.

Let $v \in f(T) \cap T$.

Now we can solve the problem.

Let $a \in S$. Then we are done.

Let $a \in T$. Then by 2) above,
$va \in S$ and $(1-v)a \in S$.

Clearly $a = va + (1-v)a$ and we are done.

{\em Remark. In reality the above can be summarized by first realizing that
in $char 2$ the problem is trivial and in the other characteristics 
squaring is a group homomorphism with image of index two and this suggests 
trying to find the solution by some parity argument.}



 \newpage

\noindent
2. 
Let $G$ be a group. Let $H$ and $K$ be subgroups.
We consider right cosets only.

\bigskip
a) Prove that $K$ acts by right translation on $G/H$
and that there is a natural bijection between the
 set of $K$-orbits $(G/H)/K$ and the set of $H$-orbits
$(G/K)/H$.

\bigskip
b) Prove that $G$ acts on the set $G/H \times G/K$ by right translation
and find a bijection
$$
(G/H \times G/K)/G 
\longrightarrow
(G/H)/K
$$
between the set of orbits $G$-orbits in $G/H \times G/K$
and $(G/H)/K$.

\bigskip 
c) Let $H=K$. Determine the fixed points of the $H$-action on $G/H$,
i.e. the right cosets $Hg$ such that $(Hg)h=Hg$, $\forall \, h \in H$.


\bigskip
a) As to the $K$-action, the only real thing to check
is that $(Hg)k:= Hgk$ is well defined. Since
$Hg=Hg'$ iff $g{g'}^{-1}\in H$ iff $ gk (g'k)^{-1} \in H$ 
iff $Hgk = Hgk$.


\noindent
A  bijection is given by $\overline{Hg} \to \overline{Kg^{-1}}$,
where we denote by $\overline{Hg} \in (G/H)/K$ the $K$-orbit of $Hg$.
etc. Well defined: $Kg^{-1}$ and $K (hgk)^{-1}$ are in the same
$H$-orbit for every $h\in H$ and $k \in K$. Surjectivity is obvious.
Inectivity: follows from $g=htk$ for some $h\in H$ and some $k\in K$
iff $g^{-1} = k't^{-1} h'$ for some $k' \in K$ and some $h' \in H$.

\bigskip
b) A bijection is given by
$$
\overline{(Ha, Kb)} \longrightarrow \overline{Hab^{-1}}.
$$
You argue as above.

\bigskip
c)
$(Hg)h=Hg$, $\forall h \in H$ iff $ghg^{-1} \in H$, $\forall
h \in H$, i.e. iff $g\in N$, 
the normalizer of $H$ in $G$. The set of fixed points
is $N/H \subseteq G/H$.



\newpage

\noindent
3. 
Compute the Galois group of 
$$
f= x^3 -x^2 + x +1
$$
a) over ${\Bbb Q}$;



\bigskip
\noindent
b) over $E:={\Bbb Q} (i\sqrt{11})$.


\bigskip
First of all, note that $f$ does not have a rational root:
a root $u=c/d$ would have to be either $1$ or $-1$ by
 Hungerford III, Proposition
6.8 (discussed in class and elementary). Since the degree of $f$ is three,
$f$ is irreducible over ${\Bbb Q}$. Since the characteristic is zero, 
the roots are all distinct.


a) For $g= x^3 + ax^2 +bx +c$, the discriminant is 
$$
D(g) = a^2(b^2-4ac) -4a^3 -27c^2+18abc. 
$$
You can also ``complete the cube" and use the shorter formula.
$D=D(f)=-44$. $D$ is not a perfect square in ${\Bbb Q}$.
By the general theory, the Galois group $G$ of the splitting
field is $G=S_3$.

\bigskip
b) $D=-44 = (2i\sqrt{11})^2$. Note that $\Delta= \Delta (f) = 
2i\sqrt{11}$, up to a sign.
$E= {\Bbb Q}(\Delta)=$ the field fixed by $A_3<S_3$ (as discussed in class).
The Galois group is $A_3$.



\newpage

\noindent
4.
Let $f: {\Bbb Z}^3 \to {\Bbb Z}^3$ be the ${\Bbb Z}$-linear map defined by 
the matrix $A$:
$$
\begin{bmatrix}
2 &3& -4 \\
2& 6 & 0\\
2&6&-4
\end{bmatrix}
$$
i.e.  view  elements ${\Bbb Z}^3$ 
as  column vectors, $v$, of integers  and set $f(v):= Av$.
Let $K:= Ker f$ and $C=Coker \; f$.

{\em
Note, but do not prove,
 that $K$ and $C$ are finitely generated abelian groups, so that
they are both isomorphic to a direct sum
of finitely many cyclic groups.}

\bigskip
\noindent
a)
Compute $K$ and $C$, i.e. determine
the cyclic groups appearing in their direct sum decomposition.

\bigskip
\noindent
b) Let $B$ be any $n\times n$ matrix of integers and use it to define
a ${\Bbb Z}$-linear map $g: {\Bbb Z}^n \to {\Bbb Z}^n$ as above.
Prove that if $g$ has non-zero eterminant, then
$Coker \; g$ is a finite abelian group.


\noindent
{\em Hint: Consider the operation $-\otimes_{\Bbb Z} {\Bbb Q}$.}




\bigskip
a) The matrix has non zero determinant. Viewed as a linear map
${\Bbb Q}^3 \to {\Bbb Q}^3$ the map has no kernel so that it has no kernel:
$K=\{0\}$ the trivial cyclic group.

Viewing $C$ as cosets, the elements of $C$ can be described
using a row vector of integers modulo elements in the image of $f$.

We can therefore perform elementary row operations on $A$ like switching
rows and adding or subtracting integer multiples of rows to a given row.

By doing so, one obtains the matrix $B$:
$$
\begin{bmatrix}
2 &0& 0 \\
0& 3 & 0\\
0&0&4
\end{bmatrix}
$$
We can therefore describe the elements of $C$ using three vectors $v_1$,
$v_2$ and $v_3$ subject to $2v_1=0$, $3v_2=0$ and $4v_3=0$.

$C\simeq {\Bbb Z}_2 \oplus {\Bbb Z}_3 \oplus {\Bbb Z}_4$.


\bigskip
b) Consider the exact sequence: 
$$
{\Bbb Z}^n \stackrel{f}\to {\Bbb Z}^n \stackrel{p}\to C \to 0,
$$
where $p$ is the natural quotient map.
Note that $p \circ f =0$. By tensoring with
${\Bbb Q}$ we get the sequence
$$
{\Bbb Q}^n \stackrel{g}\to {\Bbb Q}^n 
\stackrel{p':= p \otimes Id_{Bbb Q}}\longrightarrow
 C\otimes_{\Bbb Z} {\Bbb Q} \to 0,
$$
with $p'\circ g =0$ and $p'$ surjective (note that, as discussed in class,
tensoring does not destroy surjectivity, only, possibly, injectivity).
Since $g$ is an isomorphism by assumption, $p'=0$.
Since $p'$ is surjective and trivial,
$C\otimes_{\Bbb Z} {\Bbb Q}=0$. It follows that $C$, which is 
a finitely 
generated abelian group (it is a quotient of ${\Bbb Z}^n$),
has no free summand and is a finite direct sum of finite cyclic groups.

NB: The converse is also true.


\newpage

\noindent
5.

\bigskip
\noindent
a) Let $R$ be a principal ideal ring, i.e. every ideal $I \subseteq R$
is principal. Let $f: R \to S$ be a surjective ring homomorphism.
Prove that $S$  is a principal ideal ring.

\bigskip 
\noindent
b)
Let $N$ be a positive integer and
$$
f_N (x) := \prod_{n=0}^N{ (x-n) } \in {\Bbb Q}[x].
$$
Determine all the ideals of the quotient ring
$$
{\Bbb Q}[x]/(f_N (x) ).
$$


\bigskip
a) Let $J$ be an ideal in $S$. Since $f$ is surjective, $I:=f^{-1}(J)$
is an ideal and it is principal, say $I= (r)$. It is routine to check
that $J= (f(r))$.



\bigskip
b)  Define a ring homomorphism
$$
u: {\Bbb Q}[x]/(f_N (x) ) \to \prod_{n=0}^N{ {\Bbb Q}[x]/(x-n )}
$$
by 
$$
u ( [f] ) \longrightarrow
( [f], [f] , \ldots, [f] ). 
$$
Note that each ${\Bbb Q}[x]/(x-n )$ is isomorphic to
${\Bbb Q}$: just send an $[f]$ to $f(n)$. 
It is injective since if $u([f])=0$, then (any representative)
$f$ is divisible by $x-n$, $\forall n =0, \ldots, N$. By unique factorization,
$f$ is divisible by $f_N(x)$ so that $[f]=0 \in
{\Bbb Q}[x]/(f_N (x) ) $.  
 To prove surjectivity it is enough to find a polynomial
with prescribed rational values at $x=0, \ldots, N$.
This can be done easily.

By part a) or because the ${\Bbb Q}[x]/(x-n )$
are fields, $\prod_{n=0}^N{ {\Bbb Q}[x]/(x-n )}$ is a principal ideal ring.
Each factor in the product is a field and has no proper ideals.
The ideals are all of the form $\prod_{n=0}^N I_n$
where $I_n \subseteq {\Bbb Q}[x]/(x-n )$ is either the zero ideal
or the whole field. 
\newpage


\noindent
6.
Let $a \in {\Bbb R}$ be such $a^4=5$.

\bigskip
a) Show that ${\Bbb Q}(ia^2)$ is normal over ${\Bbb Q}$.


\bigskip
b) Show that ${\Bbb Q}(a +ia)$ is normal over ${\Bbb Q} (ia^2)$.

\bigskip
c) Prove that  ${\Bbb Q}(a +ia)$ is not  normal over
${\Bbb Q}$$\,$.


\bigskip
a) $ia^2$ is a root of $x^2+5=0$ which is  of degree
two and irreducible over ${\Bbb Q}$.
It follows that $x^2+5$ splits over
${\Bbb Q}(ia^2)$ which is then a splitting field 
(no smaller one does the job).
It follows that ${\Bbb Q}(ia^2)$ is normal over ${\Bbb Q}$. 


\bigskip
b) Same proof as above, just note that 
$a+ia$ is a root of $x^2 + 2ia^2=0$ which is irreducible
over ${\Bbb Q}(ia^2)$. We show this below, but even if it were reducible, 
we would then have the equality of fields
${\Bbb Q}(ia^2) = {\Bbb Q} (a +ia)$ and the trivial extension is normal.

\bigskip
c) It is not normal. 

First note that part a) and b) 
imply that  $[ {\Bbb Q}(a+ia): {\Bbb Q}]=2\cdot 2 =4$.

Note also that $[ {\Bbb Q}(a): {\Bbb Q}]= \deg{(x^4-5)}=4$.

We prove it by contradiction. Assume it is normal.

$a+ia$ is a root of $x^4+20=0$ which is irreducible over ${\Bbb Q}$.
By normality all four roots
$$
a+ia, \quad i(a+ia), \quad -(a+ia) , \quad -i(a+ia)
$$
belong to ${\Bbb Q}(a+ia)$.

It follows that $a= (1/2)[ (a+ia) -i(a+ia)] \in {\Bbb Q}(a+ia)$.

It follows that 
${\Bbb Q}(a)= {\Bbb Q}(a+ia)$: they have the same dimension, $4$, over
${\Bbb Q}$ and one is contained in the other.

It follows that $i \in {\Bbb Q}(a)$, but $a$ is real, contradiction.


\end{document}











\vskip 3cm
\bigskip
\begin{center}
\begin{tabular}
{|r|l|l|}
\hline
1&40&{\phantom{abcdef}}\\
\hline
2&75&\\
\hline
3&60&\\
\hline
4&50&\\
\hline
5&40&\\
\hline
6&40&\\
\hline
7&45&\\
\hline
8&50&\\
\hline
Total&400&\\
\hline
\end{tabular}
\end{center}