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\begin{document}

\noindent
{\large \bf Additional notes \#3.}

\bigskip

\noindent
{\bf Some Facts About Free Algebras.}

\bigskip

\begin{thm} {\rm (Equivalence of two definitions.)}
Let $\vr V$ be a variety. An algebra $\m f\in \vr V$
is free over $X$ (that is, 
$\m f$ satisfies the universal
mapping property with respect to $X$) if and only if
$X$ is an independent generating set for $\m f$.
\end{thm}

\smallskip

{\it Sketch of proof.}
Show that the existence part of the definition 
of the universal mapping property proves that 
$X$ is independent, and the uniqueness part of
the statement implies that $X$ generates.
Conversely, assume that $\varphi:X\rightarrow A$
is a function for some $\m a\in {\vr V}$.
If $X$ is a generating set for $\m f$, then there
is at most one way to define an
extension of $\varphi$ to a homomorphism
$\overline{\varphi}:\m f\rightarrow \m a$.
Use the definition of independence to prove
that this definition yields a well-defined
homomorphism.
\bigskip

\begin{thm} {\rm (Existence.)}
Let $\vr V$ be a variety and let $X$
be a set. There is an algebra $\m f\in \vr V$
that is free over $X$. {\rm (Written $\m f_{\vr V}(X)$.)}
\end{thm}

\smallskip

{\it Sketch of proof.} The full proof is not short, but a rough
idea is this: Let $X = \{x_1, x_2, \ldots\}$ be any set.
Thinking of this set as a set of variables, construct the
set $F = \{f(x_{i_1},\ldots,x_{i_n})\}$ of all
derived operations of $\vr V$ that involve
 the variables $X$.
Show that the operations of $\vr V$ operate
naturally on $F$, producing an algebra
that satisfies the equations of $\vr V$. 
Show that the constructed algebra is free over $X$.
(Take the insertion of generators to be the 
function from $X$ to $F$ that sends $x_i$ to
the operation that is projection onto the $i$-th variable.)

\bigskip

\begin{thm}
Let $\vr V$ be a variety. Any $\m a\in {\vr V}$
is a homomorphic image of a free algebra.
\end{thm}

\smallskip

{\it Sketch of proof.} The algebra $\m a\in {\vr V}$
is a homomorphic image of the free algebra
$\m f_{\vr V}(A)$. To see this, use the universal 
mapping property to extend the identity function
${\rm id}_A: A\rightarrow A$ to a homomorphism 
$\overline{{\rm id}_A}:\m f_{\vr V}(A)\rightarrow \m a$.
This homomorphism is onto since ${\rm id}_A$ is onto.

\bigskip

\begin{thm}
If $|X| = |Y|$, then $\m f_{\vr V}(X)\cong \m f_{\vr V}(Y)$.
\end{thm}

\smallskip

{\it Sketch of proof.}
Let $\varphi:X\rightarrow Y$ be a bijection with inverse
$\varphi^{-1}$. Use the universal mapping property to
produce homomorphisms $\overline{\varphi}:\m f_{\vr V}(X)\rightarrow 
\m f_{\vr V}(Y)$ and $\overline{\varphi^{-1}}:\m f_{\vr V}(Y)\rightarrow 
\m f_{\vr V}(X)$. What remains to show is that these
homomorphisms are inverses of each other.
To see that $\overline{\varphi^{-1}}\circ \overline{\varphi}$
equals the identity function ${\rm id}$, observe that 
both $\overline{\varphi^{-1}}\circ \overline{\varphi}$
and ${\rm id}$ extend the function $i:X\rightarrow F_{\vr V}(X)$
(insertion of generators), so by the uniqueness of extensions 
we indeed have 
$\overline{\varphi^{-1}}\circ \overline{\varphi}={\rm id}$.
The same argument proves 
$\overline{\varphi}\circ \overline{\varphi^{-1}}={\rm id}$.
\bigskip

\begin{thm}
$\m f_{\vr V}(X)\oplus\m f_{\vr V}(Y) = \m f_{\vr V}(X\oplus Y)$.
In particular, $\m f_{\vr V}(n)$ is the $n$-th
copower of $\m f_{\vr V}(1)$.
\end{thm}

\smallskip

{\it Sketch of proof.}
Prove that $\m f_{\vr V}(X\oplus Y)$ satisfies the 
defining property for $\m f_{\vr V}(X)\oplus\m f_{\vr V}(Y)$.
Here the canonical injections 
$i_1: \m f_{\vr V}(X)\rightarrow \m f_{\vr V}(X\oplus Y)$ 
and $i_2:\m f_{\vr V}(Y)\rightarrow \m f_{\vr V}(X\oplus Y)$ 
are induced by the inclusions of $X$ and $Y$ into
$X\oplus Y$. To prove that $\m f_{\vr V}(X\oplus Y)$ satisfies the 
defining property for $\m f_{\vr V}(X)\oplus\m f_{\vr V}(Y)$, 
start with two homomorphisms 
$\varphi:\m f_{\vr V}(X)\rightarrow \m a$ and
$\psi:\m f_{\vr V}(Y)\rightarrow \m a$. 
This yields set functions from $X$ and $Y$ into
$A$, and hence a set function from $X\oplus Y$ into $A$.
This yields a unique homomorphism $\varphi\oplus \psi$
from $\m f_{\vr V}(X\oplus Y)$ into $\m a$.
Show that this homomorphism satisfies the properties
necessary to make $\m f_{\vr V}(X\oplus Y)$ the coproduct
of $\m f_{\vr V}(X)$ and $\m f_{\vr V}(Y)$.

\bigskip\bigskip\bigskip

\noindent
Exercise.
\begin{enumerate}
\item[(1)] Verify the details of this weird example.
There is a ring $\m r$ for which the variety
$\vr R$ of all left $\m r$-modules has the following
property: $\m f_{\vr R}(1)\cong \m f_{\vr R}(n)$
for every finite $n$. (Thus the rank of a finitely
generated free module is not uniquely determined.)
To construct $\m r$, begin with the polynomial ring 
${\BZ}\lb a,b,c,d\rb$ in $4$ {\it noncommuting}
variables. Let $I = (ac+bd-1, ca-1, cb, da, db-1)$.
Let $\m r$ be the quotient ${\BZ}\lb a,b,c,d\rb/I$.
Verify that the matrices $[a\;\;b]$ and 
$\left[\begin{array}{c} c\\d\end{array}\right]$
are inverses of one another. Use these matrices to
construct an isomorphism between the modules
$\m f_{\vr R}(1) = \m r$
and $\m f_{\vr R}(2)=\m r\oplus \m r$. Use this isomorphism 
to construct isomorphisms between $\m f_{\vr R}(1)$
and all other $\m f_{\vr R}(n)$.
\end{enumerate}
\end{document}

