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\title{\vspace{-50pt}Math 210A: Modern Algebra\\
\large Thomas Church ({\tt tfchurch@stanford.edu})\\
\href{http://math.stanford.edu/~church/teaching/210A-F17/}{\nolinkurl{http://math.stanford.edu/~church/teaching/210A-F17}}}
\author{}
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{\huge \bf Homework 1}\\\vspace{8pt}
{\Large Due Saturday, September 30 by 5:00pm}\\


\end{center}
\noindent Do all the 
\textbf{unstarred questions}~below. (Starred questions are important and you really should do them, but they will not be collected~or~graded.)

{\small Reminder: when we just say ``$R$-module'' we mean ``left $R$-module'', and we allow $R$ to not be commutative. But in these questions, it doesn't really matter whether $R$ is commutative or not.}

The material needed for this assignment ($R$-modules, homomorphisms, submodules) is covered~in: Lang p117; or Atiyah--Macdonald pp17--18; or Dummit--Foote p337.

\begin{question}
Given an $R$-module $M$ and a subset $S\subset M$, prove that the following are equivalent.
\begin{enumerate}[label=(\Alph*)]
\item ``\emph{Every element of $M$ is an $R$-linear combination of elements of $S$}'':\newline For all $m\in M$ there exist $r_1,\ldots,r_k\in R$ and $s_1,\ldots,s_k\in S$ such that $m=\sum_{i=1}^k r_is_i$.
\item ``\emph{Homomorphisms are determined by their value on $S$}'':\newline For any $R$-module $N$ and any homomorphisms $f\colon M\to N$ and $g\colon M\to N$,
\[f|_S = g|_S \quad\implies\quad f = g.\]
\item ``\emph{Any homomorphism whose image contains $S$ is a surjection}'':\newline For any $R$-module $L$ and any homomorphism $h\colon L\to M$,
\[h(L)\supseteq S\quad\implies\quad h(L)=M.\]
\end{enumerate}
When these equivalent conditions hold, we say that \emph{$S$ generates $M$} ({\small or \emph{$S$ spans $M$}, or \emph{$M$ is generated by $S$}, or \emph{$M$ is spanned by $S$}}).
\end{question}

\begin{question}
We say an $R$-module $M$ is \emph{finitely generated} if there exists a finite set $S\subset M$ that generates $M$. Prove that if $M$ is finitely generated, then it has the following property:

For any infinite chain $N_1\subseteq N_2\subseteq \cdots \subseteq M$ of submodules of $M$ whose union $\bigcup_{i\in \N} N_i=M$ is equal to $M$,  there exists $k\in \N$ such that $N_k=M$.

\noindent {\small (You can think about whether this condition is \emph{equivalent} to finite generation, but you do not have to~prove~it.)}
\end{question}

\begin{question}
Prove that $\Z$-modules and abelian groups are the same thing. Specifically, prove that
\begin{enumerate}
\item[(a*)] Every abelian group $A$ admits one and only one structure of a $\Z$-module (i.e. there is a unique ``multiplication map''\quad $\cdot\colon \Z\times A\to A$ making $A$ into a $\Z$-module).
\item[(b*)] For any abelian groups $A$ and $B$, the set of $\Z$-module homomorphisms $f\colon A\to B$ is exactly the set of abelian-group homomorphisms $f\colon A\to B$.
\item[(c)] Please write one sentence summarizing what makes (a) and (b) happen (i.e.\ what's the key idea of your proof?). \end{enumerate}
{\footnotesize [Note (c) is the only part that you need to submit! You should work out (a)+(b) but  don't need to write them~up.]}
\end{question}\pagebreak

\begin{question}
Let $R$ be a ring, and let $\{M_i\}_{i\in I}$ be a family of $R$-modules indexed by some set $I$.

Define the \emph{direct product} $\prod_{i\in I} M_i$ to be the Cartesian product, i.e.\ the set of families $(m_i)_{i\in I}$ with $m_i\in M_i$. This becomes an $R$-module with the component-wise addition and multiplication:
\[(m_i)_i + (n_i)_i = (m_i+n_i)_i\qquad\qquad r\cdot (m_i)_i = (r\cdot m_i)_i\]

Define the \emph{direct sum} $\bigoplus_{i\in I} M_i$ to be the submodule consisting of elements where $m_i=0$ for all but finitely many $i$. {\small (You do not have to prove this is a submodule, but you should understand why it is.)} Note that when $I$ is finite $\bigoplus_{i\in I} M_i$ is the same as $\prod_{i\in I} M_i$, but in general it is a proper submodule.\\[15pt]

For readability, let $P=\prod_{i\in I} M_i$ and $S=\bigoplus_{i\in I} M_i$.
\begin{enumerate}[label=(\alph*),wide]
\item Show that \[\text{``a map to $P=\prod_{i\in I} M_i$ is the same as a family of maps to $M_i$'',}\] by proving the following.
Let $\pi_i\colon P\to M_i$ be the projection taking $(m_i)_{i\in I}\mapsto m_i$. Prove that for any $R$-module $L$, given homomorphisms $f_i\colon L\to M_i$ there exists a unique homomorphism $f\colon L\to P$ such that $f_i=\pi_i\circ f$ for all $i$.
\item Show that  \[\text{``a map \emph{from} $S=\bigoplus_{i\in I} M_i$ is the same as a family of maps \emph{from} $M_i$'',}\] by \textbf{formulating} and \textbf{proving} a separate universal property along similar lines to (a).\newline {\small (Hint: consider the inclusions $j_i\colon M_i\to S$ given by $m_i\mapsto (0,\ldots,0,m_i,0,\ldots,0)$ [0 except in the $M_i$ factor].)}
\end{enumerate}
\end{question}

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