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\title{Math 129, Test 1}
\author{Bryden Cais}

\begin{document}
\maketitle

\section{The Prime Divisors of $\ind(K)$}
\begin{enumerate}
	\item Let $q=p^r$ and let $\alpha$ be a root of a degree $r$ monic irreducible polynomial in $\F_p[X]$.
	Then $\F_p[\alpha]$ is a degree $r$ extension of $\F_p$ and is therefore isomorphic to $\F_q$.
	We therefore have a surjective map $\Z[\alpha]\longrightarrow \F_q$, which shows that $F_q$ is
	primitive.

	\item In one direction, suppose $A=\Z[\alpha]$ is primitive and that 
	$A=k_1\times k_2\times\cdots\times k_m$ is a product of finite fields.  Write
	$\alpha=(\alpha_1,\ldots,\alpha_m)$ with $\alpha_j\,\in\,k_j$ for each $j$.  Certainly,
	$\Z[\alpha_j]=k_j$ for each $j$.  Notice that for any
	polynomial $f$ we have $f(\alpha)=(f(\alpha_1),\ldots,f(\alpha_m))$.  Therefore, since 
	$(1,\ldots,1,0,1,\ldots,1)\,\in\,A=\Z[\alpha]$, where the $0$ is in the $i$ th position, we have
	some polynomial $f\,\in\,\F_p[X]$ such that $f(\alpha_j)=\delta_{ij}$.  Thus, 
	$\irr(\alpha_i)|f$ and $\irr(\alpha_j)\not|f$ for $i\not=j$.  Hence, we have distinct monic irreducible
	polynomials $P_i,\ldots,P_m\,\in\,\F_p[X]$ with $\deg(P_i)=f_i=[k_i:\F_p]$ since $k_i=\Z[\alpha_i]$.

	In the other direction, suppose that $P_i,\ldots,P_m\,\in\,\F_p[X]$ with $\deg(P_i)=f_i=[k_i:\F_p]$
	are such polynomials.  Then let $\alpha_i$ be a root of $P_i$ in some splitting field and set
	\begin{eqnarray*}
	    \alpha\ =\ (\alpha_1,\ldots,\alpha_m).
	\end{eqnarray*}
	Let $R_i$ be the polynomial in $\F_p[X]$ given by $P_1\cdots\hat{P_i}\cdots\P_m$, where the hat
	indicates omission.  Define $\beta_i=R_i(\alpha_i)$, and set $B_i=\irr(\beta_i)$.  Clearly,
	$\beta_i\not=0$, so that $B_i(0)\not=0$ since $B_i$ is irreducible and $\deg(B_i)\geq 1$.  Now let
	\begin{eqnarray*}
	    Q_i(X)=\frac{-X}{B_i(0)}(B_i(R_i(X))-B_i(0)). 
	\end{eqnarray*}
	It is not difficult to see that $Q_i(X)\,\in\,\F_p[X]$ and 
	$Q_i(\alpha)=(0,\ldots,0,\alpha_i,0,\ldots,0)$.
	Since $\Z[\alpha_i]=k_i$, we have shown that $A=\Z[\alpha]$ is primitive.

	We now claim that this result holds even if we only know that 
	\begin{eqnarray*}
	    A\ =\ D_1\times D_2\times\cdots\times D_m
	\end{eqnarray*}
	with each $D_j=R/Q_j^{r_j}$ for some prime ideal $Q_j$ and nonzero $r_j$.  To see that this
	is true, notice that we only use that the $k_i$ are fields in our proof when we divide by
	$B_i(0)$, and otherwise we do not use that the $k_i$ are of the same characteristic or even 
	integral domains.  So all that we must show to prove this extension is that $B_i(0)$ is always
	a unit.  Notice, however, that since $R$ is a number ring, $R/Q_j^{r_j}$ is always principal,
	so that Bezout's theorem applies.  That is, with $P_i$ as before, since the $P_i$ are
	distinct irreducible polynomials, there exist polynomials $a,\ b$ such that
	\begin{eqnarray*}
	    aP_i+bP_j\ =\ 1
	\end{eqnarray*}
	for $i\not=j$.  Evaluation at $\alpha_i$ (a root of $P_i$) gives $b(\alpha_i)P_j(\alpha_i)=1,$
	so that $P_j(\alpha_i)$ is a unit for each $j\not=i.$  Therefore, $\beta_i$ is also a unit since
	it is a product of units.  It follows that all galois conjugates of $\beta_i$ are units, and
	therefore also that $N(\beta_i)=B_i(0)$ is a unit, as required.

	\item One direction is trivial: if $R/I=\Z[\alpha]$ is primitive then we have $R/I^e=\Z[\alpha^e]$
	is also primitive.  Now suppose that $R/I^e$ is primitive and that
	\begin{eqnarray*}
	    I\ =\ P_1^{r_1}P_2^{r_2}\cdots P_m^{r_m}
	\end{eqnarray*}
	for distinct prime ideals $P_j$ and integers $r_j>0$.  By the chinese remainder theorem, we then have
	\begin{eqnarray*}
	    R/I^e\ =\ R/P_1^{er_1}\times R/P_2^{er_2}\times\cdots\times R/P_m^{er_m},
	\end{eqnarray*}
	so that by item $2$, there exist distinct monic irreducible polynomials $Q_j\,\in\,R/P_j^{r_j}[x]$
	with $\deg(Q_j)=$

	\item Let $I=I_1I_2\cdots I_m$ with $N(I_i)$ a power of $p_i$, with the $p_j$ distinct primes.
	Since $N(I_i)$ is a power of $p_i$, we have
	\begin{eqnarray*}
	    R/I_i\simeq k_{1i}\times\cdots\times k_{r_ii}
	\end{eqnarray*}
	with $k_{ji}$ finite fields of characteristic $p_i$.
	Suppose that $R/I_i$ is primitive for all $i$.  Then by item $2$, there exist  
	distinct monic irreducible polynomials $Q_{ji}\,\in\,\F_{p_i}$ for $i=1,2,\ldots,m$
	with $\deg(Q_{ji})=[k_{ji}:\F_{p_i}]$.  Now by the extended version of item $2$,
	since the $Q_{ji}$ are all distinct for different values of $i$ (by the fact that
	the primes $p_i$ are all distinct), we see that $R/I_i\times R/I_2\times\cdots\times R/I_m$
	is principal.  By the chinese remainder theorem, again using the fact that the $p_i$ are distinct,
	we see that this is isomorphic to $R/I$, so that $R/I$ is principal.  Conversely, since $I_i\supset I$,
	we have $R/I_i\subset R/I$, so that if $R/I$ is principal, so is $R/I_i$ for every $i$.

	\item We factor $I$ (uniquely) as $I=\prod_{i,j}P_{ij}^{r_j}$, where the $P_{ij}$
	are distinct primes and the primes $P_{ij}$ for a fixed $i$ all have the same norm,
	which is a power of some prime $p_i$.  Notice that \textit{every}
	prime ideal containing $I$ occurrs in this decomposition.  By part $4$, since the $P_j$ are
	distinct, we have that $R/I$ is primitive if and only if $R/p_j^{r_j}$ is primitive for all
	$j$.  Now by part $3$, this is the case if and only if $R/P_j$ is primitive for all $j$.
	In turn, again by item $4$, this holds if and only if $R/P_1\cdots P_m$ is primitive.  

\end{enumerate}

\section{A Nice Ring}
\begin{enumerate}
	\item To show commutativity, notice that as $d$ ranges over all divisors of $n$, so does $n/d$, so that
	we may reindex the sum defining $f*g$ to obtain
	\begin{eqnarray*}
	    (f*g)(n)&=&\sum_{d|n}f(d)g(n/d)\\
	    &=&\sum_{d|n}f(n/d)g(d)\\
	    &=&(g*f)(n).
	\end{eqnarray*}
	Associativity of $*$ is proved similarly.  We have
	\begin{eqnarray*}
	    (f*(g*h))(n)&=&\sum_{d|n}\sum_{d^{\prime}|(n/d)}f(d)g(d^{\prime}h(n/dd^{\prime}))\\
	    &=&\sum_{\delta|n}\sum{d|\delta}f(d)g(\delta/d)h(n/d)\\
	    &=&((f*g)*h)(n)
	\end{eqnarray*}
	upon replacing $dd^{\prime}$ by $\delta$ and reindexing the sum.  Let
	\begin{eqnarray*}
	e(n)\ =\ 
	\begin{cases}
	    1 & \ \text{if}\quad n=1\\
	    0 & \ \text{otherwise}
	\end{cases}.
	\end{eqnarray*}
	Then we obviously have $(f*e)(n)=f(n)$ for all $n$ and any $f\,\in\,\mathcal{L}$,
	so that $e$ is the identity elemant.  We finally have to check some trivial properties, for example
	that 
	\begin{eqnarray*}
	    (f+g)*h&=&f*h+g*h\quad\text{and}\\
	    f*\lambda\cdot g&=&\lambda\cdot f*g,
	\end{eqnarray*}
	which follow from 
	\begin{eqnarray*}
	    \sum_{d|n}(f(d)+g(d))h(n/d)&=&\sum_{d|n} f(d)h(n/d)+\sum_{d|n} g(d)h(n/d),\\
	    \sum_{d|n} f(d)\lambda g(n/d)&=&\lambda\sum_{d|n} f(d)g(n/d).
	\end{eqnarray*},
	Now if $f\,\in\,\mathcal{L}$ and $f(1)=0$ then $(f*g)(1)=f(1)g(1)=0$ no matter what $g$ is, so 
	that $f$ is not invertable.  Conversely, if $f(1)\not=0$, then we may define $f^{-1}(1)=1/f(1)$,
	and we can define $f^{-1}(n)$ for $n>1$ inductively as 
	\begin{eqnarray}
	\label{inv}
	    f^{-1}(n)\ =\ -\frac{1}{f(1)}\sum_{\substack{d|n\\d<n}} f^{-1}(d)f(n/d).
	\end{eqnarray}
	It is then not difficult to see that $f*f^{-1}=e$.
	Finally, suppose that $f*g\equiv 0$.  Suppose that $g(1)\not\equiv0$.  Let $m$ be the least integer
	for which $g(m)\not=0$.  Then since $(f*g)(m)=0$, we have 
	\begin{eqnarray*}
	    0\ =\ \sum_{d|m}f(d)g(m/d)\ =\ f(1)g(m), 
	\end{eqnarray*}
	so that $f(1)=0$.  Now we show by induction that $f\equiv 0$.  Suppose that $f(d)=0$ for all
	$d<k$.  Since $(f*g)(mk)=0$, we have
	\begin{eqnarray*}
	    0&=&\sum_{d|mk}f(d)g(n/d)\\
	    &=&\sum_{\substack{d|mk\\d<k}}f(d)g(mk/d)+\sum_{\substack{d|mk\\d>k}}f(d)g(mk/d)+g(m)f(k)\\
	    &=&g(m)f(k), 
	\end{eqnarray*}
	since $f(d)=0$ for $d<k$ and $g(mk/d)=0$ for $d>k$ because we chose $m$ to be the least integer
	for which $g(m)$ was nonzero.  Thus, we have $f(k)=0$ and by induction, $f\equiv 0$ as claimed.
	
	\item We show that $f*g$ is multiplicitive if $f,\ g$ are.  Let $(m,n)=1$.  Then any $d$ with 
	$d|mn$ may be written \textit{uniquely} $d=d_1d_2$ where $d_1|m$ and $d_2|n$.  We need only
	take $d_1=(m,d)$ and $d_2=(n,d)$.  Thus there is a bijection between divisors $d$ of $mn$ and
	pairs $(d_1,d_2)$ with $d_1|m$, $d_2|n$ and $d_1d_2=d$.  (Since the other direction is clear).
	Therefore, we have
	\begin{eqnarray}
	\label{mult1}
	    \sum_{d|mn}f(d)g(mn/d)\ =\ \sum_{d_1|m}\sum_{d_2|n}f(d_1d_2)g(mn/d_1d_2).
	\end{eqnarray}
	Again, since $(m,n)=1$, we have $(d_1,d_2)=(m/d_1,m/d_2)=1$ for any $d_1|m$ and $d_2|n$.
	Thus, since $f,g$ are multiplicative, we find
	\begin{eqnarray}
	\label{mult2}
	    \sum_{d|mn}f(d)g(mn/d)&=&\sum_{d_1|m}\sum_{d_2|n}f(d_1)f(d_2)g(m/d_1)g(n/d_2)\\
	    &=&\sum_{d_1|m}f(d_1)g(m/d_1)\sum_{d_2|n}f(d_2)g(n/d_2)\\
	    &=&(f*g)(m)(f*g)(n).
	\end{eqnarray}
	Since $(f*g)(1)=f(1)g(1)=1$, $f*g$ is multiplicative.
	Obviously, $e$ is multiplicitive.  We may use our definition \ref{inv} to show by induction
	that $f^{-1}$ is also multiplicitive.  Since $f(1)=1$, we have $f^{-1}(1)=1$.  Now suppose that
	for all $m,n\,<N$, we have $f^{-1}(mn)=f^{-1}(m)f^{-1}(n)$ whenever $(m,n)=1$.  Then
	\begin{eqnarray*}
	    f^{-1}(mN)&=&-\frac{1}{f(1)}\sum_{\substack{d|m\\d<m}}\sum{d^{\prime}|N}f^{-1}(dd^{\prime})
	    f(mN/dd^{\prime})\\
	    &&-\frac{f^{-1}(m)}{f(1)}\sum{\substack{d^{\prime}|N\\d^{\prime}<N}}f^{-1}(d^{\prime})
	    f(N/d^{\prime})\\
	    &=&f^{-1}(m)f^{-1}(N),
	\end{eqnarray*}
	where we have used the same summation reindexing techniques as in \ref{mult1}, and \ref{mult2} and
	the assumption that $(m,N)=1$.  This shows, in fact, that $f^{-1}$ is multiplicitive for all
	$(m,n)\leq N$ with $(m,n)=1$, so by induction, $f^{-1}$ is multiplicitive.  Obviously, if $f,g$
	are multiplicitive, then $f\times g$ is also multiplicitive since we have $f(1)g(1)=1$ and
	\begin{eqnarray*}
	    (f\times g)(mn)&=&f(mn)g(mn)=f(m)f(n)g(m)g(n)\\&=&(f(n)g(n))(f(m)g(m))=(f\times g)(n)(f\times g)(m).
	\end{eqnarray*}
	This shows that the set of multiplicitive functions is a subgroup of $\mathcal{L}^*$ which is
	stable under $\times$.

	\item Define $z$ by $z(n)=1$ for all $n$, and let $\mu$ be the inverse of $z$ under $*$.
	First, we clearly have $\mu(1)\cdot 1=1$ so that $\mu(1)=1$.  Now for any prime $p$,
	we must have $\mu(p)+\mu(1)=0$ so that $\mu(p)=-1$ for all primes.  We also have
	\begin{eqnarray*}
	    \mu(1)+\mu(p)+\mu(p^2)+\cdots+\mu(p^r)=0
	\end{eqnarray*}
	for any $r\geq 2$ so that 
	\begin{eqnarray*}
	    \mu(p^2)+\cdots+\mu(p^r)=0
	\end{eqnarray*}
	for any $r\geq 2$ whence by induction, $\mu(p^r)=0$ for all $r\geq 2$.  Since $z$ is multiplicitive
	and $\mu$ is its inverse, $\mu$ is also multiplicitive.  Thus we have completely determined the 
	behavior of $
	\mu$ and have
	\begin{eqnarray*}
	\mu(n)\ =\ 
	\begin{cases}
	    1 & \ \text{if}\quad n=1\\
	    0 & \ \text{if $n$ has a square factor}\\
	    (-1)^k & \ \text{if} n=p_1p_2\cdots p_k
	\end{cases}
	\end{eqnarray*}
	where all the $p_j$ are distinct.  We note an interesting consequence:  Let 
	\begin{eqnarray*}
	f(n)=(z*g)(n)=\sum_{d|n}g(d).
	\end{eqnarray*}
	Then
	\begin{eqnarray*}
	\label{mobius}
	    (\mu*f)(n)=(\mu*z*g)(n)=(e*g)(n)=g(n).
	\end{eqnarray*}
	The converse is also clear.

	\item Let $\varphi(n)$ be Euler's phi function and let $i(n)=n$ for all $n$.  We know
	that $\varphi$ is multiplicitive, so that $\varphi*z$ is also multiplicitive.  Moreover,
	\begin{eqnarray*}
	    \sum_{d|p^r}\varphi(d)\\&=&\sum_{i=1}^r\varphi(p^i)\\&=&\sum_{i=1}^r \left(p^{i}-p^{i-1}\right)\\
	    &=&p^r.
	\end{eqnarray*}
	Therefore, decomposing any integer as a product of primes and applying the above, we have, by the
	multiplicitivity of $\varphi*z$ that $(\varphi*z)(n)=n$ for any $n$.

	\item We have
	\begin{eqnarray*}
	    z*z&=&\sum_{d|n}1,
	\end{eqnarray*}
	which obviously counts the number of divisors of $n$.  Similarly,
	\begin{eqnarray*}
	    i*z&=&\sum_{d|n}d,
	\end{eqnarray*}
	which clearly gives the sum of the divisors of $n$.

	\item Now suppose that $f$ is completely multiplicitive.  The fact that $\times f$ is
	a $\C$-algebra homomorphism of $(\mathcal{L},+,*,\cdot)$ will follow if we can show that
	it preserves the multiplicitive $*$ structure of $\mathcal{L}$, as the other requirements
	are trivial to check.  We have
	\begin{eqnarray*}
	    (f\times g)*(f\times h)(n)&=&\sum_{d|n} f(d)g(d)f(n/d)h(n/d)\\
	    &=&f(n)\sum_{d|n} g(d)h(n/d)\\
	    &=&f\times(g*h)(n)
	\end{eqnarray*}
	for all $n$.  We also have
	\begin{eqnarray*}
	    f*(\mu\times f)&=&\sum_{d|n} f(d)\mu(n/d)f(n/d)\\
	    &=&f(n)\sum_{d|n}\mu(n/d)\\
	    &=&f(n)\mu*z\\
	    &=&f(n)e(n)\\
	    &=&e(n)
	\end{eqnarray*}
	since $f(1)=1$.  Here we have used the fact that $\mu$ is the inverse under $*$ for $z$.  This
	shows that $\mu\times f$ is the inverse under $*$ of $f$.
\end{enumerate}

\section{Irreducible Polynomials Over $\F_q$}
\begin{enumerate}
	\item This is not difficult.  For irreducible polynomials of degree $1,2,3$ it is enough
	to check for polynomials that do not have a root in $\F_p$.  For degree $4$ polynomials, we
	need only exclude polynomials that have a root in $\F_p$ and polynomials that are the product of
	two irreducible quadratics.  Thus, we find
	\begin{eqnarray*}
	    I_2(1)&=&\{x,x+1\}\\
	    I_2(2)&=&\{x^2+x+1\}\\
	    I_2(3)&=&\{x^3+x^2+1,x^3+x+1\}\\
	    I_2(4)&=&\{x^4+x+1,x^4+x^3+1,x^4+x^3+x^2+x+1\}\\\\
	    I_3(1)&=&\{x,x+1,x+2\}\\
	    I_3(2)&=&\{x^2+1,x^2+x+2,x^2+2x+2\}\\
	    I_3(3)&=&\begin{Bmatrix}x^3+2x+1,x^3+2x+2,x^3+x^2+2,x^3+x^2+x+2,\\x^3+x^2+2x+1,x^3+2x^2+1,x^3+2x^2+x+1,
	    x^3+2x^2+2x+2\end{Bmatrix}\\\\
	    I_5(1)&=&\{x,x+1,x+2,x+3,x+4\}\\
	    I_5(2)&=&\begin{Bmatrix}x^2+x+1,x^2+x+2,x^2+2x+3,x^2+2x+4,\\x^2+3x+3,x^2+3x+4,x^2+4x+1,x^2+4x+2
	    \end{Bmatrix}.
	\end{eqnarray*}

	\item We first prove that 
	\begin{eqnarray*}
	    x^{q^m}-x \quad\text{divides}\quad x^{q^n}-x
	\end{eqnarray*}
	if and only if $m|n$.  First suppose that $m|n$, say $n=ms$.  We employ the identity
	\begin{eqnarray}
	\label{iden1}
	    y^d-1\ =\ (y-1)(y^{d-1}+y^{d-2}+\cdots+1)
	\end{eqnarray}
	with $y=q^m$ and $d=s$ to find that $q^m-1|q^n-1$.  Now set $y=x^{q^m-1}$ and $d=(q^n-1)/(q^m-1)$
	in \ref{iden1} to see that $x^{q^m}-x$ divides $x^{q^n}-x$, as claimed.  Conversely, if 
	$m\not|n$ then $a=q^n$ is not a power of $b=q^m$, so $\F_a$ is not an extension
	of $\F_b$.  But since the elements of $\F_a$ are precisely the roots of $x^a-x$,
	we see that $x^{b}-x$ cannot divide $x^a-x$.

	Now let $P\,\in\,I_q(d)$.  Let $\alpha$ be a root of $P$ in some extension of $\F_q$.  Then
	the field $K=\F_q(\alpha)$ has degree $d$ over $\F_q$ and its elements are the roots
	of $x^{q^d}-x$.  Since $\alpha$ is thus a root of $x^{q^d}-x$ and $P$ is irreducible,
	$P|x^{q^d}-x$, and hence by the above, $P|x^{q^r}-x$ for all $r$ divisible by $d$.  Conversely,
	if $P|x^{q^n}-x$ for some $n$, then the field generated by a root of $P$ must be a subfield of
	$\F_{q^n}$.  As above, this implies that $q^n$ is some power of $q^{d}$, that is,
	$d|n$.

	\item By the previous item, we see that the irreducible polynomials of degree $d$ in
	$\F_q[x]$ are precisely the degree $d$ irreducible factors of $X^{q^n}-x$ when $d|n$,
	and hence we may write
	\begin{eqnarray*}
	    x^{q^n}-x\ =\ \prod_{d|n}\prod_{P\in I_q(d)}P.
	\end{eqnarray*}
	Counting degrees now gives
	\begin{eqnarray}
	\label{rq}
	    \sum_{d|n}dr_q(d)\ =\ q^n.
	\end{eqnarray}

	\item Notice that we may rewrite \ref{rq} in the form 
	\begin{eqnarray*}
	    (i\times r_q)*z\ =\ q^n.
	\end{eqnarray*}
	We may apply \ref{mobius} to the above identity to obtain
	\begin{eqnarray*}
	    (i\times r_q)*z*\mu&=&\mu*q^n\\
	    &=&(i\times r_q)*e\\
	    &=&(i\times r_q),
	\end{eqnarray*}
	or what is the same thing,
	\begin{eqnarray*}
	    r_q(n)\ =\ \frac{1}{n}\sum_{d|n}\mu(n/d)q^d.
	\end{eqnarray*}

	\item From our formula for the m{\"o}bius function, we have
	\begin{eqnarray*}
	    nr_q(n)\ \geq\ q^n-(q+q^2+\cdots+q^{n-1})\ =\ q^n-\frac{q^n-q}{q-1}.
	\end{eqnarray*}
	Now using the fact that, for $n\geq 1$ we always have
	\begin{eqnarray*}
	    q^n\geq q,
	\end{eqnarray*}
	we see that 
	\begin{eqnarray*}
	    q^n(q-2)\geq q(q-2),
	\end{eqnarray*}
	with equality only if $q=2$ or $n=1$.  We expand this inequality and rearrange terms to obtain
	\begin{eqnarray*}
	    \frac{q^{n+1}-2q^n+q}{q-1}\geq q,
	\end{eqnarray*}
	that is,
	\begin{eqnarray*}
	    q^n-\frac{q^n-q}{q-1}\geq q,
	\end{eqnarray*}
	so that
	\begin{eqnarray*}
	    nr_q(n)\ \geq\ q,
	\end{eqnarray*}
	with equality only possible when $q=2$ or $n=1$.  It is clear that equality always holds when $n=1$
	since the ploynomial $x+r$ is irreducible for every $r\,\in\,\F_q$.  The possibility of equality 
	when $q=2$ may be ruled out as we are only summing over the divisors of a particular integer $m$ and
	not over all $k\leq m$.  (Moreover, the m{\"o}bius function is not always negative one).  In any case,
	it is easy to see that for $n\not=1$ and $q=2$, we have grossly underestimated things and equality
	above holds if and only if $n=1$.
\end{enumerate}

\section{A Bound for $p|\ind(K)$}
\begin{enumerate}
	\item 

\end{enumerate}

\end{document}