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\begin{document}

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\begin{center}
{{\Large\bf MODERN ALGEBRA I}\\
\bigskip

{\Large\bf HOMEWORK ASSIGNMENT VIII}\\
\bigskip
{\bf (Turn in underlined problems.)}}
\end{center}

\vspace{.2in}

\noindent
\underline{\large\bf PROBLEMS}
\vspace{.1in}

{\bf 

\underline{1.}
Explain why a subring of a finite field is a finite field.

\medskip

\underline{2.}
Show that if $\BF$ is a finite field with an odd
number of elements, then $1\neq -1$ in $\BF$.
(Hint: First show that if $1=-1$, then $\BF$ can be
partitioned into $2$-element sets of the form
$\{x, x+1\}$.)

\medskip

\underline{3.}
Use Problem 2 to show that if $\BF$ is a finite field 
with an odd number of elements, then there
is some element $a\in {\BF}$ which is not
a perfect square in $\BF$. (Hint: Show that the
squaring function ${\sf Sq}(x) = x^2$ is not injective.
Then, because $\BF$ is finite, $\sf Sq$ is not surjective.
Choose $a$ to be any element which
is not in the range of ${\sf Sq}$.)

\medskip

4. 
Assume that some element $a\in {\BF}$ is not a perfect square
in $\BF$. Show that the set 
${\BF}[\sqrt{a}] = \{m + n\sqrt{a}\;|\;m, n \in {\BF}\}$
can be made into a field by defining
\[\begin{array}{rl}
(m+n\sqrt{a}) + (m'+n'\sqrt{a}) &= (m+m')+(n+n')\sqrt{a},\\
(m+n\sqrt{a})(m'+n'\sqrt{a}) &= (mm'+nn'a)+(mn'+m'n)\sqrt{a},\\
-(m+n\sqrt{a}) &= (-m)+(-n)\sqrt{a},\\
0 = 0+0\sqrt{a}, &\hphantom{=} 1 = 1 + 0\sqrt{a}.
\end{array}\]

\medskip

\underline{5.} Let $p$ be an odd prime. Use Problems 2, 3 \& 4 to
explain why there exist fields of size $p, p^2, p^4, p^8, \ldots$.
(Hint: Show that if $\BF$ has $q$ elements, then 
${\BF}[\sqrt{a}]$ has $q^2$ elements.)

\medskip

6. Can you show that there is a field of size $p^3$ for
some prime $p$? 

\medskip

\underline{7.} Give a new proof of Fermat's Little
Theorem using the following outline:
Choose $a\not\equiv 0 \;({\rm mod}\;p)$.
Show that the function $f(x) = ax$ is an
injective function from ${\BZ}_p$ to ${\BZ}_p$
which maps zero to zero.
Deduce that the set
$\{1, 2, 3, \cdots, p-1\}$ 
of elements of ${\BZ}_p$ is the same
as the set $\{a, 2a, 3a, \ldots, (p-1)a\}$.
Argue that the product of the elements
in the first set equals the product in the second set:
\[1\cdot 2\cdot 3\cdots (p-1) \equiv a\cdot (2a)\cdot (3a)\cdots
((p-1)a)\;({\rm mod}\;p).\]
Finish the proof.

}
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