Code:
\documentclass{article}
\usepackage{amsmath,amssymb,eucal,yfonts,setspace,sectsty,enumitem,dsfont,amscd}
\allsectionsfont{\sffamily\raggedright\underline}
\begin{document}
\vspace*{\fill}
\begin{center}
\section*{Algebra Homework Eleven}
\quad\quad\quad\quad\quad\quad\quad\quad
\begin{minipage}{0.35\textwidth}
\begin{flushleft}
\emph{Author:}\\
Alex \textsc{Youcis}
\end{flushleft}
\end{minipage}
\begin{minipage}{.35\textwidth}
\begin{flushleft}
\emph{Date:} \\
11/14/2011
\end{flushleft}
\end{minipage}
\end{center}
\vspace*{\fill}
\newpage
\noindent\underline{\textbf{D\&F pg. 404, \#8:}} Let $Q$ be a nonzero divisible $\mathbb{Z}$-module. Prove that $Q$ is not a projective $\mathbb{Z}$ module. Deduce that $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module.
\vspace*{5 pt}
\noindent\underline{\textbf{Solution:}} We first claim that if $F$ is a free $\mathbb{Z}$-module then $\displaystyle \bigcap_{n\in\mathbb{N}}nF$ is trivial. Indeed, suppose that $x$ is in the intersection and write it as $\displaystyle \sum_{\alpha\in\mathcal{A}}r_\alpha x_\alpha$ where $\{x_\alpha\}_{\alpha\in\mathcal{A}}$ is a basis for the space. By assumption that $x\in nF$ we know that there exists $\displaystyle s_\alpha\in \mathbb{Z}$ for every $\alpha\in\mathcal{A}$ such that $\displaystyle x=\sum_{\alpha}ns_\alpha x_\alpha$. Comparing this with the orginal representation of $x$ in terms of basis vectors tells us that $r_\alpha=ns_\alpha$. Thus, $r_\alpha$ is divisible by every natural number and thus trivially zero. Thus, $x=0$.
\vspace*{10 pt}
\noindent Now, suppose that $Q$ is projective. We know then that there exists a free module $F$ and another module $M$ such that $F=Q\oplus M$. Clearly then we have that
\vspace*{10 pt}
\begin{center}
$\displaystyle \{0\}=\bigcap_{n\in\mathbb{N}}nF=\bigcap_{n\in\mathbb{N}}\left(n(Q\oplus M\right)\supseteq \bigcap_{n\in\mathbb{N}}nQ=\bigcap_{n\in\mathbb{N}}Q=Q$
\end{center}
\vspace*{10 pt}
so that $Q$ is trivial, contradictory to assumption. Thus, $Q$ is not projective.
\vspace*{10 pt}
Since $\mathbb{Q}$ is a divisible abelian group clearly it isn't projective.
\newpage
\noindent\underline{\textbf{D\&F pg. 404, \#11:}} Let $R$ and $S$ be rings with $1$ and let $M$ and $N$ be left $R$-modules. Assume that $N$ is also a $(R,S)$-bimodule.
\begin{enumerate}[label=\textbf{\roman{*})}, ref=(\roman{*})]
\item For $s\in S$ and for $\varphi\in\text{Hom}_R(M,N)$ define $(\varphi s):M\to N$ by $(\varphi s)(m)=\varphi(m)s$. Prove that $\varphi s$ is a homomorphism of left $R$-modules, and that this action of $S$ on $\text{Hom}_R(M,N)$ makes it into a right $S$-module. Deduce that $\text{Hom}_R(M,R)$ is a right $R$-module or any $R$-module $M$--called the dual module.
\item Let $N=R$ be considered as an $(R,R)$-bimodule as usual. Under the action defined in part \textbf{i)} show that the map $r\mapsto\varphi_r$ is an isomorphism of right $R$ modules, so that $\text{Hom}_R(R,R)\cong R$, where $\varphi_r$ is the homomorphism that takes $1$ to $r$. Deduce that if $M$ is a finitely generated free left $R$-module, then $\text{Hom}_R(M,R)$ is a finitely generated free right $R$-module of the same rank.
\item Show that if $M$ is a finitely gnereated projective $R$ module, then $\text{Hom}_R(M,R)$ is projective
\end{enumerate}
\vspace*{5 pt}
\noindent\underline{\textbf{Solution:}} For the sake of notational convenience allow me to denote the dual module of $M$ by $M^\vee$.
\begin{enumerate}[label=\textbf{\roman{*})}, ref=(\roman{*})]
\item Evidently $\varphi s$ is a homomorphism since $(\varphi s)(rm+n)=\varphi(rm+n)s=(r\varphi(m)+n)s=r(\varphi(m)s)+ns=r(\varphi s)(m)+(\varphi s)(n)$. The fact that this notion of right $S$-multiplication makes $\text{Hom}_R(M,N)$ into a right $S$-module is obvious since, by definining all the operations pointwise, all the axioms hold true because $N$ is a right $S$-module. Since $R$ is always a $(R,R)$-bimodule the fact that the dual module can be made into a right $R$-module is clear.
\item Clearly the map $f:r\mapsto \varphi_r$ is an $R$-map since $\varphi_{rr'+s}(x)=x(rr'+s)=xrr'+sx=\varphi_r(x)r'+\varphi_s(x)$ for all $x\in R$ and so $\varphi_{rr'+s}=\varphi_r r'+\varphi_s$. Clearly $f$ is an injection because if $\varphi_r=\varphi_s$ then $r=\varphi_r(1)=\varphi_s(1)=s$. To see that $f$ is surjective one must merely note that if $\varphi\in\text{Hom}_R(R,R)$ then $\varphi(r)=r\varphi(1)=\varphi_{\varphi(1)}(r)$ and so $\varphi=\varphi_{\varphi(1)}$ from where $f$'s surjectivity follows--thus $f$ is an isomorphism. To deduce the result for general finintely generated free modules we merely note that the previous sentence concluded that $R^\vee\cong R$. Now, since any finitely generated $R$-module is of the form $R^n$ for some $n\in\mathbb{N}$ and Hom distributes nicely over finite products we may conclude that $(R^n)^\vee\cong (R^\vee)^n\cong R^n$ and so the conclusion follows.
\item Since $M$ is finitely generated an projective we can find a finitely generated free $R$-module $F$ and some module $N$ for which $M\oplus N=F$. We see then that $M^\vee\oplus N^\vee\cong(M\oplus N)^\vee\cong F^\vee\cong F$ (by the previous result) and so $M^\vee$ is naturally the direct summand of a free $R$-module. Thus, $M^\vee$ is projective.
\end{enumerate}
\newpage
\noindent\underline{\textbf{D\&F pg. 404, \#14:}} Let $C:0\to L\xrightarrow{\psi}M\xrightarrow{\varphi}N\to0$ be a sequence of $R$-modules.
\begin{enumerate}[label=\textbf{\roman{*})}, ref=(\roman{*})]
\item Prove that the associated sequence
\vspace*{10 pt}
\begin{center}
$C^\ast:0\to\text{Hom}_R(D,L)\xrightarrow{\psi^\ast}\text{Hom}_R(D,M)\xrightarrow{\varphi^\ast}\text{Hom}_R(D,N)\to0$
\end{center}
\vspace*{10 pt}
is a short exact sequence of abelian groups for all $R$-modules $D$ if and only if the original sequence is a split short exact sequence.
\item Prove that the associated sequence
\vspace*{10 pt}
\begin{center}
$\displaystyle C^{\ast\ast}:0\to\text{Hom}_R(N,D)\xrightarrow{\varphi^\ast}\text{Hom}_R(M,D)\xrightarrow{\psi^\ast}\text{Hom}_R(L,D)\to0$
\end{center}
\vspace*{10 pt}
is a short exact sequence of abelian groups or all $R$-modules $D$ if and only if the original short exact sequence splits.
\end{enumerate}
\vspace*{ 5pt}
\noindent\underline{\textbf{Solution:}}
\begin{enumerate}[label=\textbf{\roman{*})}, ref=(\roman{*})]
\item Assume first that $C$ splits. By the splitting lemma we know then that there exists a backmap $N\xrightarrow{\varphi^{\leftarrow}} M$ (i.e. a an $R$-map $N\to M$ with $\varphi\circ\varphi^{\leftarrow}=\mathds{1}_N$). Define then a map $\text{Hom}_R(D,N)\xrightarrow{f}\text{Hom}_R(D,M)$ by $\text{Hom}_R(D,N)\ni h\mapsto \varphi^{\leftarrow}\circ h$. Clearly $f$ is a well-defined $\mathbb{Z}$-maps and since $\varphi^\ast(f(h))=\varphi\circ\varphi^{\leftarrow}\circ h=h$ for all $h\in\text{Hom}_R(D,N)$ we have that $\varphi^\ast\circ f=\mathds{1}_{\text{Hom}_R(D,N)}$ and so $f$ is backmap for $\varphi^\ast$. In particular, we see that $\varphi^\ast$ has a right inverse. Similarly, since $C$ splits the splitting lemma tells us that there is a backmap $M\xrightarrow{\psi^\leftarrow}L$ such that $\mathds{1}_L=\psi^{\leftarrow}\circ\psi$. It's not hard then to see that if we define $g:\text{Hom}_R(D,M)\to\text{Hom}_R(D,L)$ by $g(h)=\psi^{\leftarrow}\circ h$. Then, $g$ is a left inverse for $\psi^\ast$ and so $\psi^\ast$ is injective. Thus, we know that $C^\ast$ is exact at $\text{Hom}_R(D,L)$ and $\text{Hom}_R(D,N)$, and so it remains to show it's exact at $\text{Hom}_R(D,M)$. To do this we must prove that $\text{im }\psi^\ast=\ker\varphi^\ast$. To see that $\text{im }\psi^\ast\subseteq\ker\varphi^\ast$ we merely note that if $k\in\text{im }\psi^\ast$ then $k=\psi\circ j$ for some $j\in\text{Hom}_R(D,L)$ and so $\varphi^\ast(k)=\varphi\circ\psi\circ j=0$ since $\varphi\circ\psi=0$. Conversely, if $k\in\ker\varphi^\ast$ then we see that $k(x)\in\ker\varphi$ for all $x\in D$ and so, in particular, $\text{im }k\subseteq\ker \varphi=\text{im }\psi$. But, since $\psi$ is a mono we know from the basics of triangle diagram completions that this implies that there exists some $j\in\text{Hom}_R(D,L)$ such that $h=\psi\circ j$, and so $h\in\text{im }\psi^\ast$. Thus, $\ker\varphi^\ast=\text{im }\psi^ast$ and thus all the points of exactness have been verified.
\vspace*{10 pt}
\noindent Conversely, suppose that $C^\ast$ is short exact for all $D$ modules. Letting then $D=N$ we see that
\vspace*{10 pt}
\begin{center}
$\displaystyle 0\to\text{Hom}_R(N,L)\xrightarrow{\psi^\ast}\text{Hom}_R(N,M)\xrightarrow{\varphi^\ast}\text{Hom}_R(N,N)\to0$
\end{center}
\vspace*{10 pt}
In particular, this tells us that $\varphi^\ast$ is surjective, and so there exist some $j\in\text{Hom}_R(N,M)$ with $\varphi\circ j=\varphi^ast(j)=\mathds{1}_N$. Thus, $j:N\to M$ is a backmap for $\varphi$ and so the splitting lemma tells us that $C$ splits.
\item Suppose first that $C$ splits. We then know by the splitting lemma that there exists $\varphi^{\leftarrow}\in\text{Hom}(N,M)$ with $\varphi\circ\varphi^\leftarrow=\mathds{1}_N$. Define $\text{Hom}_R(M,D)\xrightarrow{f}\text{Hom}_R(N,D)$ by $f(h)=h\circ\varphi^\leftarrow$. Clearly $f$ is a well-deined $\mathbb{Z}$-map and $(f\circ\varphi^\ast)(h)=f(h\circ\varphi)h\circ\varphi\circ\varphi^\leftarrow=h$ and so $f\circ\varphi^\ast=\mathds{1}_{\text{Hom}(N,D)}$. Thus, $C^{\ast\ast}$ is exact at $\text{Hom}_R(N,D)$. Similarly, by the splitting lemma we know there exists $\psi^\leftarrow\in\text{Hom}(M,L)$ such that $\psi^\leftarrow\circ\psi=\mathds{1}_L$. Define $\text{Hom}_R(L,D)\xrightarrow{g}\text{Hom}_R(M,D)$ by $g(h)=h\circ\psi^\leftarrow$. This is clearly well-defined $\mathbb{Z}$-map and $(\psi^\ast\circ g)(h)=\psi^\ast(h\circ\psi^\leftarrow_=h\circ\psi^\leftarrow\circ\psi=h$ for all $h$ so that $\psi^\ast\circ g=\mathds{1}_{\text{Hom}(L,D)}$. Thus, $C^{\ast\ast}$ is exact at $\text{Hom}(L,D)$. Thus, it remains to show that $C^{\ast\ast}$ is exact at $\text{Hom}_R(M,D)$. To do this we must show that $\ker\psi^\ast=\text{im }\varphi^\ast$. To prove that $\text{im }\varphi^\ast\subseteq\ker\psi^\ast$ we merely note that for each $h\in\text{Hom}_R(N,D)$ we have that $(\psi^\ast\circ\varphi^\ast)(h)=h\circ\varphi\circ\psi=h\circ0=0$ and so evidently $\text{im }\varphi^\ast\subseteq\ker\psi^\ast$. Conversely, if $h\in\ker\psi^\ast$ then $h(\psi(x))=0$ for all $x\in L$ it thus follows that $\ker\varphi=\text{im }\psi\subset\ker h$. Since $\varphi$ is an epi we may conclude from the basics of triangle diagram completion that there exists some $j:N\to D$ such that $h=j\circ\varphi=\varphi^\ast(j)$ and so $h\in\text{im }\varphi^\ast$. Thus, the equality $\ker\psi^\ast=\text{im }\varphi^\ast$ and thus exactness at $\text{Hom}_R(M,D)$ and so overall exactness of $C^{\ast\ast}$ follows.
\vspace*{10 pt}
\noindent Conversely, suppose that $C^{\ast\ast}$ is short exact for all $D$. Taking $D=L$ we see we have the exact sequence
\vspace*{10 pt}
\begin{center}
$\displaystyle 0\to\text{Hom}_R(N,L)\xrightarrow{\varphi^\ast}\text{Hom}_R(M,L)\xrightarrow{\psi^\ast}\text{Hom}_R(L,L)$
\end{center}
In particular, since $\psi^\ast$ is surjective we know there exists some $j\in\text{Hom}_R(L,L)$ with $j\circ\psi=\psi^\ast(j)=\mathds{1}_L$ and so, $j:M\to L$ is a backmap for $\psi$. Thus, $C$ splits by the splitting lemma.
\end{enumerate}
\newpage
\noindent\underline{\textbf{D\&F pg. 405, \#21.:}} Let $R$ and $S$ be rings with $1$ and suppose that $M$ is a right $R$-module, and $N$ is an $(R,S)$-bimodule. If $M$ is flat over $R$ and $N$ is flat over $S$ prove that $M\otimes_R N$ is flat over $S$.
\vspace*{5 pt}
\noindent\underline{\textbf{Solution:}} Let $0\to A\xrightarrow{f} B$ be any short exact sequence of let $S$-modules. By first applying the fact that $N$ is a flat over $S$ $(R,S)$-bimodule and then applying the fact that $M$ is a flat over $R$ right $R$-module we see that we get an exact sequence
\vspace*{10 pt}
\begin{center}
$\displaystyle 0\to M\otimes_R(N\otimes_S A)\xrightarrow{\mathds{1}_M\otimes(\mathds{1}_N\otimes f)}M\otimes_R(N\otimes_S B)$
\end{center}
\vspace*{10 pt}
Letting $T$ and $S$ be the obvious 'associative' isomorphisms we clearly have the following commutative diagram
\vspace*{10 pt}
\begin{center}
$
\begin{CD}
(M\otimes_R N)\otimes_S A @>\mathds{1}_{M\otimes N}\otimes f>> (M\otimes_R N)\otimes_S B\\
@VTVV @AA S A\\
M\otimes_R(N\otimes_S A) @>\mathds{1}_M\otimes(\mathds{1}_N\otimes f)>> M\otimes_R(N\otimes_S B)
\end{CD}$
\end{center}
\vspace*{10 pt}
Since the two vertical maps and the bottom horizontal maps are injections we must clearly have that the top horizontal map is an injection. But, since our initial sequence was arbitrary it follows that $M\otimes_R N$ is flat over $S$ as desired.
\newpage
\noindent\underline{\textbf{D\&F pg. 406, \#24:}} Prove that $A$ is a flat $R$-module if and only if or any left $R$-module $L$ and $M$ where $L$ is finitely generated, then $\psi:L\to M$ injective implies that $\mathds{1}\otimes\psi:A\otimes_R L\to A\otimes_R M$ is injective.
\vspace*{5 pt}
\noindent{\underline{\textbf{Solution:}} Suppose irst that $A$ is flat 'over finitely generated modules' and that $U,V$ are any two modules and $0\to U\xrightarrow{f} V$ is exact. Suppose that $\displaystyle \sum_i a_i\otimes u_i$ is in $\ker\mathds{1}\otimes f$. Taking the space $K$ spanned by the vectors $\{a_i\otimes u_i\}$ we see that $K$ is finitely generated and so the induced chain $0\to A\otimes_R K\xrightarrow{\mathds{1}\otimes f}V$ is exact. Thus, since $\displaystyle \sum_i a_i\otimes u_i$ still goes to zero we may conclude that $\displaystyle \sum_i a_i\otimes u_i$ is zero. The injectivness of $\mathds{1}\otimes \psi$ on $U$ follows.
\vspace*{10 pt}
\noindent Conversely, if $A$ is flat then we know that exact sequences $0\to L\to M$ get carried to exact sequences $0\to A\otimes L\to A\otimes M$ for all $L$, and thus clearly for finitely generated $L$.
\newpage
\noindent\underline{\textbf{D\&F pg. 406, \#26:}} Suppose $R$ is a P.I.D.
\begin{enumerate}[label=\textbf{\roman{*})}, ref=(\roman{*})]
\item Suppose that $A$ is flat over $R$. Prove that the map $\psi_R:A\to A:a\mapsto ra$ is injective, and conclude that $A$ is torsion free.
\item Suppose that $A$ is torsion free. If $I$ is a nonzero ideal of $R$, then $I=rR$ for some nonzero $r\in R$. Show that the map $\psi_r$ in \textbf{i)} induces an isomorphism $R\cong I$ of $R$-modules and that the composite $R\xrightarrow{\psi}I\xrightarrow{\iota}R$ is multiplication by $r$. Prove that the composite $A\otimes_R R\xrightarrow{\mathds{1}\otimes\psi_R}A\otimes_R I\xrightarrow{\mathds{1}\otimes\iota}A\otimes_R R$ corresponds to the map $a\mapsto ra$ under the identification $A\otimes_R R=A$ and taht this composite is injective since $A$ is torsion free. Show taht $\mathds{1}\otimes\psi_r$ is an isomorphism and deduce that $\mathds{1}\otimes\iota$ is injective. Use the previous exercise to conclude that $A$ is flat.
\end{enumerate}
\vspace*{5 pt}
\noindent\underline{\textbf{Solution:}}
\begin{enumerate}[label=\textbf{\roman{*})}, ref=(\roman{*})]
\item Since $R$ is an integral domain we know that the map $R\xrightarrow{\psi_r}R$ (multiplication by $R$) is injective for nonzero $r$. Since $A$ is flat the induced map $A\otimes_R R\xrightarrow{\mathds{1}\otimes\psi_r}A\otimes_R R$ is an injection. To show that the map $\psi_r$ on $A$ (also multiplication) is injective we suppose that $\psi_r(a)=0$. Note then that this says that $ar=0$ and so $ar\otimes 1=0$. Thus, $a\otimes r=0$ and so $(\mathds{1}\otimes\psi_r)(a\otimes 1)=0$. Since $\mathds{1}\otimes\psi_r$ is injective this implies that $a\otimes 1$ is zero. But, by the isomorphism used to prove that $A\otimes_R R\cong A$ we see that this implies that $a=0$. Thus, each map $\psi_r$ is injective on $A$ and so clearly this implies that $A$ is torsion free.
\item The fact that $R\cong I$ is obvious since $\psi_r:R\to I$ is a surjection (since $rR=I$) and it's injective since $R$ is an integral domain--since $\psi_r$ is (as always) an $R$-map the isomorphism follows. Evidently then we also have that $R\xrightarrow{\psi_r}I\hookrightarrow R$ is just $\psi_r$. Tensoring with $A$ gives us the sequence $A\otimes_R R\xrightarrow{\mathds{1}\otimes \psi_R}A\otimes_R I\xrightarrow{\mathds{1}\otimes\iota}A\otimes R$. That said, we note that if $T:A\to A\otimes_R R$ is the usual isomorphism then it's not hard to see the following equality of maps $T\circ (\mathds{1}\otimes\psi_r)\otimes(\mathds{1}\otimes\iota)\otimes T^{-1}=\psi_r$ where the right hand map is left $r$-multiplication on $A$. But, since $A$ is torsion free we know that $\psi_r$ on $A$ is injective and so $T\circ(\mathds{1}\otimes\psi_r)\circ(\mathds{1}\otimes\iota)\otimes T^{-1}$ is injective. But, clearly this implies that $(\mathds{1}\otimes\psi_r)\circ(\mathds{1}\otimes\iota)$ is injective, which by basic set theory implies that $\mathds{1}\otimes\iota$ is injective. By the previous problem (part c) we may conclude that $A$ is flat.
\end{enumerate}
\end{document}