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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Midterm Exam 2 -- April 16, 2008}
\vskip 10pt
\noindent
{\it Directions:}  Place all work you want to have considered for credit
in the blue exam book.  You have until 9:00pm to work on the exam if
you need that much time.  There are 100 regular and 10 Extra Credit points.  Some partial 
credit will be given for relevant definitions, statements of theorems, etc.  
Even if you cannot see how to complete some part, don't leave it blank.
\vskip 10pt
\noindent
I.  {\it Terminology.}
\vskip 10pt
\begin{enumerate}
\item[(A)] (5) What does it mean to say that a polynomial $f(x) \in F[x]$
is \emph{separable}?

\sol
The polynomial $f(x)$ is separable if it has \emph{distinct} roots in
some (hence all) splitting fields.

\item[(B)] (5) What is an \emph{$F$-automorphism} of a field $K$ containing
a subfield $F$?

\sol
An $F$-automorphism of $K$ is a mapping $\sigma : K \to K$, such that
\begin{itemize}
\item $\sigma$ is one-to-one and onto,
\item for all $a,b\in K$, $\sigma(a+b) = \sigma(a) + \sigma(b)$ and $\sigma(ab) = \sigma(a)\sigma(b)$, and
\item for all $c \in F$, $\sigma(c) = c$.
\end{itemize}

\item[(C)] (5) What does it mean to say that an extension field
$K$ of a field $F$ is \emph{Galois} over $F$?

\sol $K$ is Galois over $F$ if $[K:F]$ is finite and $K$ is normal and separable over $F$.
\end{enumerate}
\vskip 10pt
\noindent
II. 
\begin{enumerate}
\item[(A)] (5)  State a necessary condition for the real number $r$ 
to be {\it constructible} by straightedge and compass,
starting from two given points $(0,0)$ and $(1,0)$.   

\sol
A necessary condition is that $[\Q(r) : \Q] = 2^m$ for some $m \ge 1$.  

\item[B)] (10)  Given a \emph{sphere} of radius 1, is it possible
to construct the radius of a sphere with 5 times the volume of the original
sphere?  Why or why not?

\sol
Not it is not possible.  The volume of a sphere of radius $r$ is $\frac{4\pi r^3}{3}$.  Hence
the radius $r$ of the new sphere would have to satisfy $\frac{4\pi r^3}{3} = 5\cdot \frac{4\pi}{3}$,
or $r^3 - 5 = 0$.  The polynomial $x^3 - 5$ is irreducible in $\Q[x]$ (by Eisenstein with $p = 5$, for
instance).  Therefore if $r$ is any root, $[\Q(r) : \Q] = 3$.  Since this is not
a power of 2, $r = {\root 3\of 5}$ is not constructible. 
\end{enumerate}
\vskip 10pt
\noindent
III.  Let $K$ be a finite field of order $81$.  
\begin{enumerate}
\item[(A)] (5) What is the characteristic of $K$?

\sol
Since $81 = 3^4$, the characteristic of $K$ would be $p = 3$.  

\item[(B)] (10) Show that the mapping $\varphi : K \to K$ given by 
$\varphi(x) = x^3$ is an automorphism of $K$.

\sol
By the ``freshman dream,'' if $a, b\in K$, then
\[
\varphi(a + b) = (a + b)^3 = a^3 + b^3 = \varphi(a) + \varphi(b),
\]
and by general rules for exponents (which hold since the multiplication
in a field is commutative),
\[
\varphi(ab) = (ab)^3 = a^3b^3 = \varphi(a)\varphi(b).
\]
The mapping $\varphi$ is onto because $K$ is a splitting field for $f(x) = x^{81} - x$.
This implies that for all $a\in K$, $a = a^{81} = (a^{27})^3 = \varphi(a^{27})$.  Therefore,
$\varphi$ is onto.  Since $K$ is a finite set, this also implies that $\varphi$ is one-to-one.
Hence $\varphi$ is an automorphism of $K$.

\item[(C)] (5) Does $K$ have a subfield of order $27$?  Why or 
why not? 

\sol
No, $K$ has no subfield of order $27 = 3^3$, since $81 = 3^4$ and $3 \not|\, 4$ (in
the exponents).  
\end{enumerate}
\vskip 10pt
\noindent  
IV.  
\begin{enumerate}
\item[(A)] (15) State the Fundamental Theorem of Galois Theory. 

\sol
Let $K$ be a Galois extension of $F$.
\begin{enumerate}
\item[(1)]  There is a one-to-one correspondence between subgroups $H$
of ${\rm Gal}_F K$ and intermediate fields $F \subseteq E \subseteq K$
given by $H \mapsto E_H$ and $E \mapsto {\rm Gal}_E K$.  In 
this correspondence, for each intermediate field $E$, $[E:F] = [{\rm Gal}_F K: {\rm Gal}_E K]$, 
and $[K:E] = |{\rm Gal}_E K|$.  
\item[(2)]  The intermediate field $E$ is normal over $F$ if and only
if ${\rm Gal}_E K$ is a normal subgroup of ${\rm Gal}_F K$.  If so, 
then ${\rm Gal}_F E \simeq {\rm Gal}_F K/{\rm Gal}_E K$. 
\end{enumerate}

\item[(B)] (10) Show that if $K$ is a finite dimensional extension of $F$,
$H$ is a subgroup of ${\rm Gal}_F K$, and $E$ is the fixed field $E = E_H$, 
then $K$ is a simple, normal, separable extension of $E$.   

\sol
(This is the content of Lemma 11.7 in Hungerford.)  If $u$ is any element 
of $K$, then $u$ is algebraic over $F$, hence over $E$ as well.  If $p(x)$
is the minimal polynomial of $u$ over $E$, then we know $\sigma(u)$ must
be a root of $p(x)$ for all $\sigma$ in $H$.  Let $u = u_1,u_2,\ldots,u_t$
be the distinct images of $u$ under all the elements in $H$, and form
the polynomial
\[
f(x) = (x - u_1)(x - u_2) \cdots (x - u_t).
\]
Since each $\sigma\in H$ permutes the roots of $f$, the coefficients of $f$ must
be fixed under $\sigma$.  Therefore $f(x) \in E[x]$ and is a separable polynomial.  
This shows that every element $u$ in $K$ is the root of a separable polynomial with 
coefficients in $E$, so $K$ is separable over $E$.  By the Primitive 
Element Theorem, we have $K = E(u)$ for some $u \in K$.  But, then
applying the above argument for the polynomial $f(x)$ constructed
from this $u$, we see that $K$ is the splitting field
of $f(x) \in E[x]$.  It follows that $K$ is normal over $E$ (by Theorem 10.15 in 
Hungerford). 
\end{enumerate}
\vskip 10pt
\noindent
V.  Let $K$ be a Galois extension of $F$ with ${\rm Gal}_F K = \langle \sigma\rangle$ 
a cyclic group of order 12.
\begin{enumerate}
\item[(A)] (10) What would the Galois correspondence look like in this case? 
(Construct the parallel diagrams of subgroups and intermediate fields, showing
inclusions, giving the intermediate fields other than $F$ and $K$ ``generic'' names 
like $E_1,E_2,$ etc.)

\sol
Every subgroup of a cyclic group of order $n$ is cyclic, with order some $d | n$.
Moreover, there is exactly one such subgroup for each $d$.  Here $n = 12$, 
so there are subgroups of orders $1,2,3,4,6,12$:
\begin{itemize}
\item order 1:  $\langle {\rm id}\rangle$
\item order 2:  $\langle \sigma^6\rangle$
\item order 3:  $\langle \sigma^4\rangle$
\item order 4:  $\langle \sigma^3\rangle$
\item order 6:  $\langle \sigma^2\rangle$
\item order 12: $\langle \sigma\rangle = {\rm Gal}_F K$.
\end{itemize}  
The following diagram shows the inclusions:
\[
\SelectTips {cm}{}
\xymatrix{
&&\{{\rm id}\} \ar[dl] \ar[dr] & \\
&\langle \sigma^6\rangle \ar[dl]\ar[dr]&& \langle \sigma^4\rangle\ar[dl]\\
\langle \sigma^3\rangle \ar[dr]&& \langle \sigma^2\rangle\ar[dl]& \\
&\langle \sigma \rangle = {\rm Gal}_F K&&
}
\]
The corresponding diagram of intermediate fields is
\[
\SelectTips {cm}{}
\xymatrix{
&&K& \\
&E_1\ar[ur]&& E_2\ar[ul]\\
E_3\ar[ur] && E_4\ar[ul]\ar[ur]& \\
&F\ar[ul]\ar[ur]&&
}
\]
where $E_1 = E_{\langle\sigma^6\rangle}$, $E_2 = E_{\langle\sigma^4\rangle}$,
$E_3 = E_{\langle\sigma^3\rangle}$, and $E_4 = E_{\langle\sigma^2\rangle}$.
\item[(B)] (5) What are the degrees of each of the intermediate fields over $F$? 

\sol
By part (1) of the FTGT, we have $[E:F] = [{\rm Gal}_F K: {\rm Gal}_E K]$ for
each intermediate field $E$.  Here 
\begin{itemize}
\item $[E_1 : F] = [\langle\sigma\rangle : \langle\sigma^6\rangle] = 12/2 = 6$,
\item $[E_2 : F] = [\langle\sigma\rangle : \langle\sigma^4\rangle] = 12/3 = 4$,
\item $[E_3 : F] = [\langle\sigma\rangle : \langle\sigma^3\rangle] = 12/4 = 3$,
\item $[E_4 : F] = [\langle\sigma\rangle : \langle\sigma^2\rangle] = 12/6 = 2$.
\end{itemize}
\item[(C)] (10)  Let $E = E_H$ for $H = \langle \sigma^3\rangle$.  Is $E$ a 
normal extension of $F$?  Why or why not?  What is ${\rm Gal}_F E$?

\sol
By part (2) of the FTGT, $E$ is normal over $F$ if and only if ${\rm Gal}_E K$
is a normal subgroup of ${\rm Gal}_F K$.  The Galois group ${\rm Gal}_F K$
is abelian here so all its subgroups are normal, and the answer to the first
part of the question is \emph{yes}.  Then 
\[
{\rm Gal}_F E \simeq {\rm Gal}_F K/{\rm Gal}_E K = \langle\sigma\rangle/\langle\sigma^3\rangle \simeq \Z_3.
\]
\end{enumerate}
\vskip 10pt
\noindent
{\it Extra Credit}  (10) Let $K$ be the splitting field of the polynomial
$f(x) = x^3 - 2 \in \Q[x]$.  By Discussion 2, we know that $G = {\rm Gal}_\Q K \simeq S_3$.
What is the fixed field $E_H$ if $H$ is the cyclic subgroup of $G$ of order $3$?  (In $S_3$,
this is the subgroup generated by the permutation $(123)$.)  
Compute the fixed field using a basis for $K$ over $\Q$.

\sol
The splitting field of $f(x)$ is $\Q(\alpha,\omega)$, where $\alpha = {\root 3\of 2}$, and 
$\omega = \frac{-1 + i\sqrt{3}}{2}$ is a primitive cube root of 1, which satisfies
$\omega^2 + \omega + 1 = 0$.  By Theorem 10.4,
a basis for $K$ over $\Q$ is
\[
\{1,\alpha,\alpha^2, \omega, \omega\alpha, \omega\alpha^2\}.
\]
The cyclic subgroup of $G$ of order 3 consists of the identity mapping, $\sigma$ and $\sigma^2$
where 
\begin{eqnarray*}
\sigma : \alpha &\mapsto& \omega\alpha\\
         \omega &\mapsto& \omega.
\end{eqnarray*}
Hence 
\[
a = a_0 + a_1 \alpha + a_2 \alpha^2 + a_3 \omega + a_4 \omega\alpha + a_5 \omega\alpha^2
\]
is in the fixed field if and only if 
\begin{eqnarray*}
a = \sigma(a) &=& a_0 + a_1\omega\alpha + a_2 \omega^2\alpha^2 + a_3 \omega + a_4\omega^2\alpha + a_5\alpha^2\\
&=& a_0 + a_1\omega\alpha + a_2 (-1-\omega)\alpha^2 + a_3 \omega + a_4(-1-\omega)\alpha + a_5\alpha^2\\
&=& a_0 - a_4\alpha + (a_5 - a_2) \alpha^2 + a_3 \omega + (a_1  - a_4)\omega\alpha - a_2 \omega\alpha^2
\end{eqnarray*}
Equating coefficients gives
\begin{eqnarray*}
-a_4 &=& a_1\\
a_5 - a_2 &=& a_2\\
a_1 - a_4 &=& a_4\\
a_5 &=& -a_2.
\end{eqnarray*}
From the second and fourth equations, we get $a_2 = a_5 = 0$.  Then the other two equations
imply $a_1 = a_4 = 0$ as well. The coefficients $a_0,a_3$ are arbitrary.  Hence
the fixed field is $\Q(\omega)$.  

\end{document}
