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\begin{document}

\noindent
{\large \bf A Sample HW Solution.}

\bigskip

\noindent
{\bf Additional Problem 1 of HW IV.} 
Let $\m r$ be a nontrivial ring.
Show every $\m r$-module is free iff $\m r$ is a division ring.
(Hint: Part of the argument could include showing 
that if a simple $\m r$-module is free, then $\m r$ is
a division ring.)

\medskip

{\it Solution.} 
For the forward direction, assume that every $\m r$-module is free.
Let $\m m$ be a simple, free, $\m r$-module. $\m m$ cannot
be free over the emptyset, since simple modules have 
exactly {\em two} submodules, and $\m f_{\m r}(\emptyset) = \{0\}$
has only one. $\m m$ cannot be free over a set 
$X = \{x_1, x_2, \ldots\}$ of more than one element,
since this would allow us to define a homomorphism
$\varphi: \m m\rightarrow \m m: x_1\mapsto 0, x_2\mapsto x_2, \ldots$
that is neither injective nor constant. Such a $\varphi$
would contradict the simplicity of $\m m$.
We conclude that $\m m$ is free over a $1$-element set.
But ${}_{\m r}\m r$ is the $1$-generated free $\m r$-module.
Hence ${}_{\m r}\m r\cong \m m$ is a simple module.
Since the submodules of ${}_{\m r}\m r$ are the left ideals
of $\m r$, this
shows that the only left ideals of $\m r$ are $\{0\}$
and $R$.

The left ideal 
generated by an element $r\in R$ is $Rr$.
If $r\in R$ is nonzero, then $Rr$
is nonzero, so $R = Rr$.
Since $1\in R = Rr$ we get that 
$1 = sr$ for some $s\in R$, thereby establishing
that every nonzero element of $R$ has a left inverse.
Now we argue that if $1 = sr$, then $1 = rs$
(so any left inverse is a 2-sided inverse).
Fix $r, s\in R$ with $1 = sr$ and let $e = rs$.
Then $e^2 = (rs)(rs) = r(sr)s = rs = e$.
If $f$ is the left inverse of $e$, then multiplying
$e = e^2$ on the left by $f$ yields $1 = e = rs$. 
This proves that every nonzero element of $\m r$
has a 2-sided inverse, so $\m r$ is a division ring.
\bigskip\bigskip

The other direction of the proof requires us to show
that if $\m d$ is a division ring, then
every vector space over $\m d$ has a basis.
This is usually proved by Zorn's Lemma, as follows.
Let $\m v$ be a $\m d$-space, and let 
$P$ be the poset whose elements are the independent
subsets of $\m v$ ordered by inclusion.

\medskip

\noindent
{\it Claim 1.} $P$ is inductively ordered.

If $I_1 \subseteq I_2 \subseteq \cdots$ is an increasing
chain in $P$, then we will argue that $I = \cup I_j$
is an upper bound of this chain in $P$. For this we must
show that $I$ is independent. Suppose instead that 
$s(x_1,\ldots,x_n) = t(x_1,\ldots,x_n)$ is a 
nontrivial dependence
relation that is satisfied by $a_1,\ldots a_n\in I$.
Then, since all $a_i$ belong to some $I_j\in P$,
we get that $I_j$ is not independent, a contradiction.

\medskip

Zorn's Lemma applies to show that $P$ has a maximal element,
so $\m v$ has a maximal independent set.
(The same argument shows that any algebra has
a maximal independent set.) Denote by 
$\mathcal B$ some maximal independent set of $\m v$.
Our goal is to show that $\mathcal B$ is a 
generating set, hence a basis, for $\m v$.
(Note: this part of the argument must make 
nontrivial use of the fact that $\m v$ is a vector space,
since maximal indendent sets are not usually generating sets.)

To show that $\mathcal B$ is a generating set, 
choose any $a\in V$. If $a\in {\mathcal B}$,
then surely $a$ is in the subalgebra/subspace 
generated by $\mathcal B$.
If $a\not \in {\mathcal B}$, then 
${\mathcal B}\cup \{a\}$ is not independent, so
there is a nontrivial relation $s(\bar{x}) = t(\bar{x})$
that holds among elements of ${\mathcal B}\cup \{a\}$.
It is not hard to see that the derived operations of
any $\m r$-module have the form $s(x_1,\ldots,x_n)=
r_1x_1 + \cdots + r_nx_n$, $r_i\in R$, and also that any
relation $s(\bar{x}) = t(\bar{x})$ is equivalent
to one of the form $w(\bar{x}) = 0$ where 
$w = s-t$. Thus, the only relations for modules 
that one must consider are those of the form 
$$
r_1x_1 + \cdots + r_nx_n = 0.
$$
Now, if ${\mathcal B}\cup \{a\}$ is not independent,
then there exist 
elements $b_i\in {\mathcal B}$ and a nonzero derived operation
$w(\bar{x}) = \sum r_ix_i$ such that 
$$
w(a,b_1,\ldots,b_n) = r_0a + r_1b_1 + \cdots + r_nb_n = 0.
$$
Moreover, since $\mathcal B$ is independent, we cannot have
$r_0= 0$. Therefore we can solve for $a$:
$$
a = (r_0)^{-1}(-r_1b_1-\cdots -r_nb_n) = w'(b_1,\ldots,b_n).
$$
This proves that $a$ is in $\lb{\mathcal B}\rb$.
$\Box$


\vfill

\end{document}

