\documentclass{amsart}
\title{Algebra 2. Problem Set 6}
\author{}

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\begin{document}
\maketitle
\thispagestyle{empty}

\begin{center}
{\em In all problems below, $R$ is a commutative ring (with $1\neq 0$).\\
$\Rmod$ is the category of $R$--modules.}
\end{center}

\vskip 0.2in

\hrule\hrule

\vskip 0.2in

\problem{1} Let $P\in \Rmod$. Prove that $P$ is projective if, and only
if, there exists $P'\in \Rmod$ such that $P\oplus P'$ is a free $R$--module.\\

\problem{2} Let $p\in R$ be such that $p^2=p$. Prove that the ideal $Rp$
generated by $p$ is a projective $R$--module.\\

\problem{3} For a $P\in \Rmod$, prove that the following conditions are
equivalent.

\begin{enumerate}
\item $P$ is projective.
\item For every $M\in \Rmod$ and $k\geq 1$, $\Ext^k_R(P,M)=0$.
\item For every $M\in \Rmod$, $\Ext^1(P,M)=0$.
\end{enumerate}

\problem{4} Consider the following two short exact sequences of $R$--modules,
where $P_1$ and $P_2$ are projective.\\

\[
\xy
(0,10)*{0}="l1"; (15,10)*{N_1}="n1"; (30,10)*{P_1}="p1"; (45,10)*{M}="m1"; (60,10)*{0}="r1";
{\ar "l1"; "n1"};
{\ar "n1"; "p1"};
{\ar "p1"; "m1"};
{\ar "m1"; "r1"};
(0,0)*{0}="l2"; (15,0)*{N_2}="n2"; (30,0)*{P_2}="p2"; (45,0)*{M}="m2"; (60,0)*{0}="r2";
{\ar "l2"; "n2"};
{\ar "n2"; "p2"};
{\ar "p2"; "m2"};
{\ar "m2"; "r2"};
\endxy
\]

Prove that $P_1\oplus N_2$ is isomorphic to $P_2\oplus N_1$.\\

\problem{5} Let $R = K[x,y,z]$ be the polynomial ring in three variables over
a field $K$, and let $M = K$ with the natural $R$--action (see Lecture 18, pages
5--6). Prove that the sequence of morphisms written at the end of Lecture 18
page 6, is a free resolution of the $R$--module $M$.\\

\problem{6} Prove that an arbitrary direct sum of projective modules
is projective. Prove that an arbitrary direct product of injective
modules is injective.\\

\problem{7} Let $N\in \Z$, $N\geq 2$. Prove that $\Z/N\Z$ is injective 
as an $\Z/N\Z$ module. ({\em Warning: the ring under consideration is not
a domain, so Corollary 17.2 page 4 does not apply.})\\

\problem{8} Give an example of a domain $R$ and an $R$--module $M$, such
that $M$ is divisible but not injective.\\

\problem{9} Assume that $R$ is injective as an $R$--module, and a domain.
Prove that $R$ is a field.\\


\problem{10} For $M,N\in \Rmod$ and $Q\in \AB$, prove that we have an isomorphism
of $R$--modules:
\[
\Hom_R\lp M, \Hom_{\Z}(N,Q)\rp \cong
\Hom_\Z\lp M\otimes_R N, Q\rp
\]
where, recall that, for an $R$--module $X$, and an abelian group $Y$, we defined
an $R$--module structure on $\Hom_{\Z}(X,Y)$ by:
\[
(r\cdot \xi)(x) = \xi(rx) \text{ for every } r\in R,\; x\in X,\; \xi\in\Hom_{\Z}(X,Y).
\]
\end{document}
