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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Problem Set 6}
\centerline{March 10, 2008}
\vskip 10pt
\noindent
{\it Section 10.5}
\vskip 10pt
\noindent
13.  The inclusion $\Q(\sqrt{p} + \sqrt{q}) \subseteq \Q(\sqrt{p},\sqrt{q})$
is automatic.  To show equality, we will use the multiplicativity
of degrees in towers.  From $\Q \subset \Q(\sqrt{p} + \sqrt{q}) \subseteq \Q(\sqrt{p},\sqrt{q})$, 
we see $[\Q(\sqrt{p} + \sqrt{q}): \Q]$ is a divisor of 
$[\Q(\sqrt{p},\sqrt{q}): \Q]$.  By the same reasoning used in the example
on page 349 in the text (and Exercises 3,4,5,6 in Section 10.3), 
$[\Q(\sqrt{p},\sqrt{q}): \Q] = 4$.  Hence $[\Q(\sqrt{p}+\sqrt{q}): \Q]$ is either
$2$ or $4$.  We claim that it cannot be $2$, hence $\Q(\sqrt{p}+\sqrt{q}) = \Q(\sqrt{p},\sqrt{q})$.
Suppose on the other hand that $[\Q(\sqrt{p}+\sqrt{q}): \Q] = 2$.  Then $\sqrt{p} + \sqrt{q}$
has minimal polynomial of the form $x^2 + bx + c \in \Q[x]$ by Theorem 10.7.  
But then 
\[
0 = 2\sqrt{pq} + b\sqrt{p} + b\sqrt{q} + c + p + q = 0.
\]
However, we know from Theorem 10.4 that $\beta = \{1,\sqrt{p},\sqrt{q},\sqrt{pq}\}$ is a 
basis of $\Q(\sqrt{p},\sqrt{q})$ over $\Q$, so we get a contradiction because
the above equation would imply that $\beta$ was linearly dependent over $\Q$.  Hence
$\Q(\sqrt{p} + \sqrt{q}) = \Q(\sqrt{p},\sqrt{q})$.
\vskip 10pt
{\it Note:}  It is also possible to adapt the proof of Theorem 10.18 for this problem.  You 
need to show that taking $c = 1$ in $u = \sqrt{p} + c\sqrt{q}$ satisfies the 
condition on $c$ needed for the portion of that proof showing $F(v,w) = F(u)$
for some $u = v + cw$.  

\vskip 10pt
\noindent
14.  ({\it Note:}  The point of this exercise is to reexamine the proof of Theorem 10.18
to see that only the hypothesis that $w$ is the root of a separable polynomial is
needed to show that $F(v,w)$ is a simple extension of $F$ when $F$ is infinite. )

We proceed as in the proof of Theorem 10.18.  Let $p(x)$ be the minimal polynomial
of $v$ over $F$ and let $q(x)$ be the minimal polynomial of $w$ over $F$.  Let $L$
be a splitting field of $p(x)q(x)$ and let $v = v_1,\ldots,v_m$ be the roots of $p(x)$
in $L$, while $w=w_1,\ldots,w_n$ are the roots of $q(x)$ in $L$.  Let $u = v + cw$,
where 
\begin{equation}\label{diff}
c \ne \frac{v_i - v}{w - w_j},
\end{equation}
for all $1 \le i\le m$ and $1 < j \le n$.  
We claim that $F(v,w) = F(u)$.  To see this, let $r(x) = p(u - cx) \in F(u)[x]$.
Since $v = u - cw$, this polynomial has $w = w_1$ as a root.  We claim that 
none of the other roots of $q(x) = 0$ is a root of $r(x)$.  If this were true, then 
$u - cw_j = v_i$, so $v + cw = v_i + cw_j$, and $c = \frac{v_i - v}{w - w_j}$.
But this contracdicts (\ref{diff}), so $w$ is the only root of $q(x)$ that is also
a root of $r(x)$.   Let $s(x)$ be the minimal polynomial of $w$ over $F(u)[x]$.  Then
$s(x)$ divides $q(x)$ and $r(x)$ in $F(u)[x]$.  
Since the roots of $q(x)$ are \emph{distinct} by hypothesis,
this implies that $s(x) = x - w$ must be the gcd of $q(x)$ and $r(x)$ in $F(u)[x]$.
Hence $w \in F(u)$.  From $u = v + cw$, it follows that $v \in F(u)$ as well.  Therefore
$F(v,w) = F(u)$.  
\vfill\eject
\noindent
{\it Section 10.6}
\vskip 10pt 
\noindent
5.  We have that $K$ is an extension field of $F$.  Since $K$ is a finite set, 
the dimension of $K$ over $F$ must also be finite, say $[K:F] = n$.  Then
$|K| = |F|^n = q^n$.  
\vskip 10pt
\noindent
10.  By Lemma 10.24 (the ``freshman's dream''), we have $f(a + b) = (a + b)^p = a^p + b^p = f(a) + f(b)$. 
Moreover, $f(ab) = (ab)^p = a^p b^p = f(a)f(b)$ since multiplication in a field is commutative.
This shows that $f : K \to K$ is a ring homomorphism.  The order of $K$ must be $p^n$
for some $n \ge 1$.  By Theorem 10.25,  $K$ is a splitting field of $x^{p^n} - x$.  Hence if
$a,b \in K$ and $f(a) = f(b)$, then $a^p = b^p$.  But then 
\[
a = a^{p^n} = (a^p)^{p^{n-1}} = (b^p)^{p^{n-1}} = b^{p^n} = b, 
\]
which shows that $f$ is one-to-one.   Theorem 10.25 also
shows that every $a$ in $K$ satisfies $a = \left(a^{p^{n-1}}\right)^p = f(a^{p^{n-1}})$.  
Hence $f$ is onto.  
\vskip 10pt
\noindent
12.  By Fermat's ``little theorem,'' every $a\in \Z_p$ satisfies $a^p = a$.  
If $c$ is a root of $f(x) =a_n x^n + \cdots + a_1 x + a_0 \in \Z_p[x]$, 
then Theorem 10.7 implies the simple
extension $K = \Z_p(c)$ is a finite field also, of order $p^n$ for some $n\le \deg(f)$.
It follows from Exercise 10 that $F : K \to K$ defined by $F(x) = x^p$ is an isomorphism
which satisfies $F(a) = a$ for all $a \in \Z_p$.  Hence if we apply $F$ to both
sides of the equation $0 = a_n c^n + \cdots + a_1 c + a_0$, then 
\begin{eqnarray*}
0 = F(0) &=& F(a_n c^n + \cdots + a_1 c + a_0)\\
         &=& F(a_n) F(c)^n + \cdots + F(a_1) F(c) + F(a_0) \quad \text{since $F$ is an isomorphism}\\
         &=& a_n F(c)^n + \cdots a_1 F(c) + a_0 \quad \text{since $F(a) = a$ for all $a \in \Z_p$.}
\end{eqnarray*}
Therefore, $F(c) = c^p$ is also a root of $f(x)$.  
\vskip 10pt
\noindent
15. (a)  Since $x^3 + x + 1$ has degree three, but has no roots in $\Z_2$, it is 
irreducible in $\Z_2[x]$.  By Theorem 5.10, $K = \Z_2[x]/(x^3+x+1)$ is a field, and 
since $\{1,x,x^2\}$ is a basis of $K$ over $\Z_2$, $K$ has order $8$.  The element $x$
generates the multiplicative group of nonzero elements since:
\begin{eqnarray*}
x^0 &=& 1\\
x^1 &=& x\\
x^2 &=& x^2\\
x^3 &=& x + 1\\
x^4 &=& x^2 + x\\
x^5 &=& x^2 + x + 1\\
x^6 &=& x^2 + 1\\
x^7 &=& 1.
\end{eqnarray*}
If we write the elements of $K$ as $K = \{0,1,x,x^2,\ldots,x^6\}$, then the 
addition and multiplication tables for $K$ are as follows:
\[
\begin{array}{cccccccccc}
+ & \vline & 0 & 1 & x & x^2 & x^3 & x^4 & x^5 & x^6\\ \hline
0 & \vline & 0 & 1 & x & x^2 & x^3 & x^4 & x^5 & x^6 \\
1 & \vline & 1 & 0 & x^3 & x^6 & x & x^5 & x^4 & x^2 \\
x & \vline & x & x^3 & 0 & x^4 & 1 & x^2 & x^6 & x^5 \\
x^2 & \vline & x^2 & x^6 & x^4 & 0 & x^5 & x & x^3 & 1 \\
x^3 & \vline & x^3 & x & 1 & x^5 & 0 & x^6 & x^2 & x^4 \\
x^4 & \vline & x^4 & x^5 & x^2 & x & x^6 & 0 & 1 & x^3\\
x^5 & \vline & x^5 & x^4 & x^6 & x^3 & x^2 & 1 & 0 & x \\
x^6 & \vline & x^6 & x^2 & x^5 & 1 & x^4 & x^3 & x & 0\\
\end{array}
\]
and 
\[
\begin{array}{cccccccccc}
\cdot & \vline & 0 & 1 & x & x^2 & x^3 & x^4 & x^5 & x^6\\ \hline
0 & \vline & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & \vline & 0 & 1 & x & x^2 & x^3 & x^4 & x^5 & x^6 \\
x & \vline & 0 & x & x^2 & x^3 & x^4 & x^5 & x^6 & 1 \\
x^2 & \vline & 0 & x^2 & x^3 & x^4 & x^5 & x^6 & 1 & x \\
x^3 & \vline & 0 & x^3 & x^4 & x^5 & x^6 & 1 & x & x^2 \\
x^4 & \vline & 0 & x^4 & x^5 & x^6 & 1 & x & x^2 & x^3\\
x^5 & \vline & 0 & x^5 & x^6 & 1 & x & x^2 & x^3 & x^4 \\
x^6 & \vline & 0 & x^6 & 1 & x & x^2 & x^3 & x^4 & x^5\\
\end{array}
\]
(Note:  The multiplicative group of nonzero elements is cyclic of
order $7$.)
\vskip 10pt
\noindent
18.  (a)  Following the hint, let $n = qk + r$ in $\Z$ where $0 \le r < k$.
Then we note that
\begin{equation}\label{div}
x^n - 1 = (x^{n-k} + x^{n-2k} + \cdots + x^{n-qk})(x^k - 1) + (x^r - 1),
\end{equation}
as can be seen by multiplying out the right hand side.  Since $r < k$, this
equation is also the result of dividing $x^k - 1$ into $x^n - 1$ using 
polynomial division in $K[x]$ (by the uniqueness of the quotient and remainder
in polynomial division).  

\noindent
$\Rightarrow$: Suppose first that $x^k - 1$ divides $x^n - 1$
in $F[x]$.  Then the remainder of the polynomial division must be zero.
This is true only if $r = 0$, so $k$ divides $n$ in $\Z$.  

\noindent
$\Leftarrow$:  Conversely, if $k | n$, then $r = 0$, so $x^k - 1$ divides
$x^n - 1$ in $K[x]$.  

\noindent
(b)  Consider equation (\ref{div}) but replace $x$ by $p$ (a prime $\ge 2$ in $\Z$):
\[
p^n - 1 = (p^{n-k} + p^{n-2k} + \cdots + p^{n-qk})(p^k - 1) + (p^r - 1),
\]
This is also the result of dividing $p^r - 1$ into $p^n - 1$ in $\Z$ (since $0 \le p^r - 1 < p^n - 1$,
by the uniqueness of quotient and remainder in integer division).  
\[
(p^k - 1)|(p^n - 1) \Leftrightarrow p^r - 1 = 0 \Leftrightarrow r = 0 \Leftrightarrow k | n.
\] 
\vskip 10pt
\noindent
19. (a)  Let $F$ be a subfield of the finite field $K$ of order $p^n$.  Since $p\cdot x = 0$
for all $x \in K$, $F$ also has characteristic $p$, and hence $|F| = p^d$ for some 
$d\le n$.  However, not all such $d$ are possible orders of $F$.  The 
the multiplicative group of nonzero elements in $F$ is a subgroup of the multiplicative
group of nonzero elements in $K$.  As a result, by Lagrange's Theorem, $p^d - 1$ must
divide $p^n - 1$.  By part (b) of Exercise 18, this implies $d|n$. 

\noindent
(b)  If $d | n$, consider the polynomial $x^{p^d} - x$.  We know that the splitting
field of this polynomial is a field of order $p^d$.  If we factor out the common
factor $x$, then the nonzero elements of the field are the roots of $x^{p^d - 1} - 1$.
By part (b) of Exercise 18, since $d | n$, $p^d - 1$ divides $p^n - 1$.  But
then part (a) of Exercise 18 shows that $x^{p^d - 1} - 1$ divides $x^{p^n - 1} - 1$,
so  $x^{p^d} - x$ divides $x^{p^n} - x$ as well.  Hence if $K$ is a field of
order $p^n$ and $d|n$, then by Theorem 10.25, $K$ contains a subfield of order $p^d$.  

By Corollary 10.27, if there are two such subfields, then they are isomorphic
as fields (in particular the multiplicative orders of all the corresponding 
nonzero elements must be the same).  However, this implies that such a subfield 
is unique, since the subgroup the multiplicative group of nonzero elements in $K$
consisting of elements of orders dividing $p^d - 1$ is unique.   
\vskip 10pt
\noindent
20.  If $f(x)$ is not irreducible, then since it is a quadratic polynomial it must 
have a root $\alpha$ in $K$.  It follows that $[\Z_p(\alpha) : \Z_p] = 2$ by Theorem 10.7,
so $\Z_p(\alpha)$ has order $p^2$.  
This would contradict the result of Exercise 19, though:  $\Z_p(\alpha)$ would be a field
of order $p^2$ contained in $K$ which has order $p^3$.  Since $2 \not|\ 3$, this is 
not possible.  Therefore $f(x)$ must be irreducible in $K[x]$.  
\end{document}
