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\begin{center}
  {\large Mathematics 101}\\Fall 2014\\Homework 4
\end{center}


\begin{enumerate}

\item (\#27, 10.3) You will demonstrate that over a noncommutative
  ring, the notion of rank of a module is not well-defined.

  Let $M =\prod_{n=1}^\infty \Z$ be the direct product of a countably
  infinite number of copies of $\Z$, and let $R = \End_\Z(M)$.  Define
  $\varphi_1, \varphi_2 \in R$ by
  \begin{align*}
    \varphi_1(a_1, a_2, \dots) &= (a_1, a_3, a_5, \dots),\\
    \varphi_2(a_1, a_2, \dots) &= (a_2, a_4, a_6, \dots).
  \end{align*}
  \begin{enumerate}
  \item Prove that $\{\varphi_1, \varphi_2\}$ is a basis of the left
    $R$-module $R$.  \textit{D\&F hint:} Define maps $\psi_1,\psi_2
    \in R$ by $\psi_1(a_1, a_2, \dots) = (a_1, 0, a_2, 0, \dots)$ and
    $\psi_2(a_1, a_2, \dots) = (0, a_1, 0, a_2, \dots)$.  Verify that
    $\varphi_i\psi_i = 1$, $\varphi_1\psi_2 = 0 = \varphi_2\psi_1$ and
    $\psi_1\varphi_1 + \psi_2\varphi_2 = 1$.  Use this to establish
    the result.

  \item Show that $R \cong R^2$, and deduce $R^m \cong R^n$ for any
    $m,n \ge 1$.
  \end{enumerate}

\item Let $R$ be a ring with identity, and $\{M_i\}_{i\in I}$ a
  collection of left $R$-modules.  For each direction of the
  statements below, provide a proof or a counterexample.
  \begin{enumerate}
  \item $\oplus_{i\in I}M_i$ is free if and only if $M_i$ is free for
    each $i \in I$.
  \item $\oplus_{i\in I}M_i$ is projective if and only if $M_i$ is
    projective for each $i \in I$.
  \end{enumerate}

\item Let $R$ be a commutative ring with identity, and $S \subset R$ a
  multiplicatively closed set containing 1, $0 \notin S$, and let
  $i_R: R \to S^{-1}R$ be the ring homomorphism taking $r
  \mapsto r/1$. For an ideal $I \subseteq R$ define $S^{-1}I = \{i/s
  \mid i\in I, s\in S\}$.
  \begin{enumerate}
  \item Show that $S^{-1}I$ is an ideal in $S^{-1}R$ which is proper
    iff $I \cap S = \emptyset$.  By example show that it is not the
    case that if $a/s \in S^{-1}I$ we must have $a \in I$.
  \item Show that every ideal $J \subset S^{-1}R$ is of the form
    $S^{-1}I$ for an ideal $I \subseteq R$.
  \item Show that there is a one-to-one correspondence between the
    prime ideals of $S^{-1}R$ and the prime ideals of $R$ which are
    disjoint from $S$.
  \end{enumerate}

\item Let $R$ be a commutative ring with identity, and $S \subset R$ a
  multiplicatively closed set containing 1, $0 \notin S$.
  \begin{enumerate}
  \item Show that $S^{-1}$ is an exact functor, that is, given an
    exact sequence of left $R$-modules
    \[\xymatrix{%
      L \ar@{>}[r]^\varphi &M \ar@{>}[r]^\psi&N}
    \]
    show that
    \[\xymatrix{%
      S^{-1}L \ar@{>}[r]^{S^{-1}\varphi} &S^{-1}M
      \ar@{>}[r]^{S^{-1}\psi}&S^{-1}N}
    \]
    is an exact sequence of $S^{-1}R$-modules.

  \item Show that for $R$-modules $A, B$, $S^{-1}(A\oplus B) \cong
    S^{-1}A \oplus S^{-1}B$ (as $S^{-1}R$-modules).
  \end{enumerate}
   
  \newpage
\item Let $R$ be a commutative ring with identity, and $\fp$ a prime
  ideal in $R$ (a proper ideal with the property that $ab \in \fp$
  implies $a\in \fp$ or $b \in \fp$). With $S = R\setminus \fp$, we
  call $S^{-1}R$ the localization of $R$ at the prime ideal $\fp$,
  usually denoted $R_\fp$.
  \begin{enumerate}
  \item Show that $R_\fp$ is a local ring, that is a ring containing a
    unique maximal ideal.
  \item Also in this context, when $M$ is an $R$-module we usually
    denote $S^{-1}M$ by $M_\fp$. Show the following are equivalent:
    \begin{enumerate}
    \item $M = 0$.
    \item $M_\fp = 0$ for all prime ideals $\fp$.
    \item $M_\fM = 0$ for all maximal ideals $\fM$.  \textit{Hint}: To
      show (iii) implies (i), suppose that $M \ne 0$ and choose $m \in
      M$, $m \ne 0$.  Let $I = \{ r \in R\mid rm = 0$, the annihilator
      of $m$ in $R$. It is a proper ideal (why?)  hence contained in a
      maximal ideal, $\fM$.  Now consider $m/1 \in M_\fM$.
    \end{enumerate}
  \end{enumerate}

\item Let $R$ be a commutative ring with identity, and let $\varphi:
  M\to N$ be an $R$-linear map.
  \begin{enumerate}
  \item Show that the following are equivalent:
    \begin{enumerate}
    \item $\varphi$ is injective.
    \item $\varphi_\fp\ (= S^{-1}\varphi): M_\fp \to N_\fp$ is
      injective for all prime ideals $\fp \subset R$.
    \item $\varphi_\fM\ (= S^{-1} \varphi): M_\fM \to N_\fM$ is
      injective for all maximal ideals $\fM \subset R$.
    \end{enumerate}
    Remark: The analogous result is true if we replace injective by
    surjective.
  \item Show that if $M$ is a projective module, then so is
    $M_\fp$ for all prime ideals $\fp$.  The converse is also true,
    but we need just a bit more to prove it.
  \end{enumerate}
  
  
\end{enumerate}


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