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\author{Bryden R. Cais}
\title{Math 593 Assignment \# 2 Solutions}
\begin{document}

\maketitle

\begin{enumerate}
    \item

    \item   Let $A$ be a commutative ring with unit and let $I,J\subset A$ be ideals with $I+J=A$.
    \begin{enumerate}
        \item Since $I+J=A$ and $1\in A$, there exist $i\in I$ and $j\in J$ such that $i+j=1$.  Now let $a,b\in A$ and consider
        $$x=aj+bi\in A.$$
        We have $x\equiv bi\equiv b\mod J$ since $i\equiv 1\mod J$.  Similarly, $x\equiv a\mod I$.

        \item By part (a), we have a surjective map
        $$A\longrightarrow A/I\oplus A/J$$ given by
        $$a\mapsto (a\mod I,a\mod J).$$  This is obviously a homomorphism.  An element $a$ of $A$ is in the
        kernel of this map if and only if $a\equiv 0\mod I$ and
        $a\equiv 0\mod J$, that is, $a\in I$ and $a\in J$.  In other words, we have an exact sequence
        $$0\longrightarrow I\cap J\longrightarrow A\longrightarrow A/I\oplus A/J\longrightarrow 0.$$  Therefore,
        $$A/(I\cap J)\simeq A/I\oplus A/J,$$ and if $I\cap J=0,$ then $A\simeq A/I\oplus A/J.$

        \begin{rem}
        This is just a special case of the \textbf{Chinese Remainder Theorem}.
        \end{rem}
    \end{enumerate}

    \item \begin{enumerate}
    \item From class, we know that $\Z$ is a PID, so that every ideal is of the form $n\Z$.  If $0\neq n=xy$ is composite
    (with $1<x,y<n$) then $x,y\not\in (n)$ since $n\not|x,y$ but $xy=n\in (n)$.  Therefore $(n)$ is not prime.  Conversely,
    if $p\in\Z$ is a prime and $xy\in (p)$ then $p|xy$ so that $p|x$ or $p|y$; that is, $x\in(p)$ or $y\in(p)$.  In short, $(p)$
    is prime.  Since $\Z$ is an integral domain, by part $d$, $(0)$ is prime.

    \item Let $M\subset R$ be a maximal ideal and suppose that $ab\in M$ and $a\not\in M$.  Then $M+(a)$ is an ideal of
    $R$ properly containing $M$.  It follows that we can write
    $$1=m+ax$$ for some $m\in M$ and $x\in R$.  Therefore, $b=mb+abx\in M$ since $ab\in M$.  It follows that $M$ is prime.
    \begin{rem}
    Of course, we could just observe that $R/M$ is a field and hence an integral domain, and then apply part (e).
    \end{rem}

    \item The ideal $(x)$ is prime since $x$ is irreducible.  Moreover, $(x)\subsetneq (x,y)\subsetneq \Q[x,y]$
    so that $(x)$ is not maximal.
    \begin{rem}
        If you want to use overkill, observe that $(x)$ is the kernel of the homomorphism
        $$\varphi:\Q[x,y]\longrightarrow \Q[y]$$ given by
        $$p(x,y)\mapsto p(0,y).$$  We therefore have an exact sequence
        $$0\longrightarrow (x)\longrightarrow \Q[x,y]\longrightarrow \Q[y]\longrightarrow 0,$$
        and since $\Q[y]$ is an integral domain, it follows that $\Q[x,y]/(x)\simeq \Q[y]$ is an integral domain
        and hence, by part (e), that $(x)$ is prime.
    \end{rem}

    \item This follows from part (e) since $R/(0)\simeq R.$

    \item We have that $R/P$ is an integral domain if and only if for any $x,y\in R/P$, $xy=0$ implies $x=0$ or $y=0$.
    Let $\tilde{x},\tilde{y}$ be any lifts of $x,y$ to $R$.  Then $R/P$ is an integral domain if and only if
    $\tilde{x}\tilde{y}\in P$ implies $\tilde{x}\in P$ or $\tilde{y}\in P$, i.e. if and only if $P$ is prime.
    \end{enumerate}

    \item Since parts (a) and (b) follow from (c), we prove only (c).  Let $I\subset R$ be an ideal and let $\sqrt{I}$ be
    the radical of $I$.  Let $x,y\in \sqrt{I}.$  Then there exist $m,n$ such that $x^m,y^n\in I$.  Set $N=m+n$ and consider
    $$(x+y)^N=\sum_{i=0}^{N}\binom{N}{i}x^{N-i}y^i,$$
    (by the binomial theorem, which holds in any commutative ring).  Observe that for any $0\leq i\leq N$, either
    $N-i\geq m$ or $i\geq n$.  Therefore, every summand is in $I$ (since $I$ is a two sided ideal of $R$) and hence
    $(x+y)^N\in I$.  It follows that $x+y\in \sqrt{I}.$

    \begin{rem}
    Notice that we don't need $R$ to be commutative for this to hold.  All we need is that $xy=cyx$ for come $c$ in the
    center of $R$ when considered as a group under multiplication.
    \end{rem}

    Now let $r\in R,x\in \sqrt{I}.$  Then for some $n$, we have $x^n\in I$ and hence, since $R$ is commutative,
    $$(rx)^n=r^nx^n\in I.$$  It follows that $rx\in\sqrt{I}$ and hence that $\sqrt{I}$ is an ideal.

    Let $f\in R/\sqrt{I}$ and suppose that $f^n=0$.  Let $\tilde{f}$ be any lift of $f$ to $R$.  Then
    $\tilde{f}^n\in \sqrt{I},$ and hence there exists some $m$ such that $\tilde{f}^{nm}\in I$.  It follows
    that $\tilde{f}\in\sqrt{I}$ and hence that $f=0$.  Therefore $R/\sqrt{I}$ is reduced.
\end{enumerate}

\end{document}
