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\author{Bryden R. Cais}
\title{Math 593 Assignment \# 4 Solutions}
\begin{document}

\maketitle

\begin{enumerate}
    \item \begin{enumerate}
        \item We differentiate two cases:
        \begin{itemize}
            \item For all $m\in \Z\setminus\{0\}$, we have $\phi(m)\neq 0\in R$.  In this case, $\phi$ is an injective
            map (to its image) and hence $\Z$ is isomorphic to $\phi(\Z)$ (since $\phi$ is obviously surjective onto its image).

            \item There exists $m\in \Z\setminus\{0\}$ such that $\phi(m)=0\in R$.  Observe that if such an $m$ is negative then
            we also necessarily have $\phi(-m):=-\phi(m)=0$ so that we can take $m$ to be the {\em least} positive integer
            for which $\phi(m)=0$.  Suppose that $m$ is not prime, say $m=ab$ for positive integers $1<a,b<m$.  Then we have
            $$0=\phi(m)=\phi(ab)=\phi(a)\phi(b).$$  Since $R$ is an integral domain, and $\phi(a),\phi(b)\in R$ have product zero,
            we must have $\phi(a)=0$ or $\phi(b)=0$.  In either case, we have contradicted the minimality of $m$ and hence $m=p$
            is prime.
        \end{itemize}

        \item Take, for example, $R_p=\Z/(p)[X],$ the ring of polynomials in $X$ with coefficients in $\Z/(p)$.  Clearly this is infinite since
        it is a (countably) infinite dimensional vector space over $\Z/(p)$.  Moreover, let $r,s\in R_p$ have degree $d,e$ respectively.
        Then we can write
        \begin{align*}
            r&=aX^d+\cdots\\
            s&=bX^e+\cdots\\
            rs&=abX^{d+e}+\cdots,
        \end{align*}
        so that if $rs=0$ then in particular, $ab=0\in \Z/(p)$.  Since $\Z/(p)$ is an I.D., we must have $a$ or $b$ equal to $0$, which
        is absurd.  Hence $R_p$ is an integral domain.

        \item Let $a\in R$.  Then we have
        $$p\cdot a=p\cdot 1_R\cdot a=\phi(p)\cdot a.$$
        Since $R$ has characteristic $p$, the image of $\Z$ under $\phi$ must be $\Z/(p),$ that is, $\phi(p)=0\in R$
        so we have $p\cdot a=0\cdot a=0,$ as required.

        \item It is obvious that $F$ preserves the multiplicative structure of $R$ and that $F(1)=1,F(0)=F(0)$.  So it suffices
        to check that $F$ is a homomorphism of abelian groups (under the addition of $R$).  Since the binomial theorem
        holds in any commutative ring, we have
        $$F(x+y)=(x+y)^p=\sum_{i=0}^p \binom{p}{i} x^{p-i}y^i.$$
        It is not difficult to see that $\binom{p}{i}$ is divisible by $p$ for $1\leq i\leq p-1$ and therefore, since $R$
        has characteristic $p$, we find that $\binom{p}{i}=0\in R$ for $1\leq i\leq p-1$ and hence that
        $$F(x+y)=x^p+y^p=F(x)+F(y).$$
    \end{enumerate}

    \item Let
    $$f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0\in R[x]$$ and suppose that $a_i\in P$, where $P$ is a prime ideal of $R$,
    but that $a_0\not\in P^2$.  Suppose that $f$ is not irreducible, that is, we have a factorization
    $$f=gh.$$  Reducing modulo $P$ gives a factorization
    $$\bar{f}=\bar{g}\bar{h}$$ where $\bar{f},\bar{g},\bar{h}$ are the images of $f,g,h$ in $R/P[x]$ under the natural
    quotient map.  By hypothesis, we have
    $\bar{f}=x^n$.  Since $R/P$ is an integral domain ($P$ being afterall a prime ideal) it follows that
    $\bar{g}=x^i,\bar{h}=x^j$ with $i+j=n$ and $0<i,j$.  Writing
    \begin{align*}
        g&=\alpha_i x^i+\cdots+\alpha_0\\
        h&=\beta_j x^j+\cdots+\beta_0,
    \end{align*}
    it follows that $\alpha_0,\beta_0\in P$.  By our factorization of $f$, we must have $a_0=\alpha_0\beta_0\in P^2$,
    a contradiction.  Hence $f$ is irreducible.

    \item \begin{enumerate}
    \item Let $p$ be prime and set
        $$\Phi_p(x)=x^{p-1}+x^{p-2}+\cdots +x+ 1\in \Z[x].$$
    Observe that $$\Phi_p(x)=\frac{x^p-1}{x-1}.$$  Now $\Phi_p(x)$ is irreducible
    if and only if $\Phi_p(x+1)$ is irreducible.  But we have
    $$\Phi_p(x+1)=\frac{(x+1)^p-1}{x}\equiv x^{p-1}\mod p$$ by part d) of question $1$.
    On the other hand, the constant term in $\Phi_p(x+1)$ is clearly
    $$\binom{p}{p-1}=p$$
    so that $\Phi_p(x+1)$ is irreducible since it is $p$-Eisenstein.

    \item It is false that for any natural number the polynomial
    $$x^{m-1}+\cdots x^{m-2}+\cdots +x+1$$
    is irreducible.  A simple counterexample is $m=4$ when we have
    $$x^3+x^2+x+1=(x+1)(x^2+1).$$
    More generally, define
    \begin{align*}
    \Phi_m(x)=\begin{cases}x-1 & \text{if $m=1$}\\ \cfrac{x^m-1}{\prod_{\substack{d|m\\d<m}} \Phi_d(x)} & \text{otherwise}\end{cases}.
    \end{align*}
    Then it can be shown that $\Phi_m(x)\in \Z[x]$ is irreducible for every $m>0$.  The polynomial $\Phi_m(x)$
    is called the $m^{\text{th}}$ {\em cyclotomic} polynomial.
    \end{enumerate}

    \item This is a fairly simple counting argument.  If a degree two monic polynomial $p(x)$ is reducible, then it splits as the product
    of two monic linear factors.  Moreover, these monic linear factors are uniquely determined by $p(x)$.  To see this, observe that
    a monic linear polynomial is completely determined by its constant term, and that $a,b$ are uniquely determined by the requirement
    $$(x+a)(x+b)=x^2+c x+d$$ in $F[x],$ since the system of equations $a+b=c,ab=d$ has a unique (up to transposition
    of $a,b$) solution (if any) modulo $p$.
    It follows that the number of degree 2 reducible polynomials is the same as the number of pairs (where order
    doesn't matter) of monic linear polynomials.  There are
    $$\frac{p(p-1)}{2}$$ such pairs if the factors are different, and
    $$p$$ pairs if they are the same, for a total of
    $$\frac{p(p+1)}{2}$$ reducible degree two monic polynomials.  Therefore, there are
    $$p^2-\frac{p(p+1)}{2}=\frac{p(p-1)}{2}$$ monic irreducible degree 2 polynomials in $F[x]$.

    \item \begin{enumerate}
        \item Suppose that $d<0$ is squarefree and that $a+b\sqrt{d}$ is a unit.  Then there exist $c,d$ such that
        $$(a+b\sqrt{d})(c+e\sqrt{d})=1.$$  Observe that if one of $b,e$ is $0$, so is the other,
        giving the units $\pm 1$.  Otherwise, both $b,e\neq 0$.  There is a unique, nontrivial
        automorphism of $\Q[\sqrt{d}]$ fixing $\Q$ pointwise---namely, complex conjugation, which sends
        $\sqrt{d}$ to $-\sqrt{d}$.  Applying this automorphism gives
        $$(a+b\sqrt{d})(a-b\sqrt{d})(c+e\sqrt{d})(c-e\sqrt{d})=(a^2-db^2)(c^2-de^2)=1.$$
        Now since $a^2-db^2,c^2-de^2\in \Z$ we have $a^2-db^2=\pm 1$.  If $d< -1$, this requires
        $b=0,a=\pm 1$.  If $d=-1$ we can have $a=\pm 1,b=0$ or $b=\pm 1, a=0$.  The same goes for $c,e$.  In any
        case, we have determined that the group of units in $\Z[\sqrt{d}]$ is precisely
        \begin{align*}
        & \left\{\pm 1\right\} & \text{if $d\neq -1$}\\
            & \left\{\pm 1,\pm i\right\} & \text{if $d= -1$}
        \end{align*}
        \item Observe that $(\sqrt{2}+1)(\sqrt{2}-1)=1$ so that $\sqrt{2}-1$ is a unit in $\Z[\sqrt{2}]$.
        It follows that $(\sqrt{2}+1)^n$ is a unit {\em for every} $n\in \Z$.  Moreover, this gives an infinitude
        of units as $|\sqrt{2}+1|>1$ so that $|(\sqrt{2}+1)^n|\neq |(\sqrt{2}+1)^m|$ unless $n=m$.
        In fact, \textit{every} unit in $\Z{\sqrt{2}}$ has the form $\pm (1+\sqrt{2})^n,$ for some $n\in \Z$
        (in other words, the unit group is isomorphic to $\Z\oplus\Z/(2)$).
    \end{enumerate}



\end{enumerate}

\end{document}