\documentclass[12pt]{amsart}
\title{Lecture 11}
\author{}

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\begin{document}
\maketitle

\PP{11.0}{Left and right exact functors}
%---------------------------------------
Recall that last time we introducted the notion of an exact sequence,
and defined left/right exact functors $F:\CalA\to\CalB$ between two
abelian categories $\CalA$ and $\CalB$. Thus, for every short
exact sequence $0\to X\morph{f} Y\morph{g} Z\to 0$ in $\CalA$:

\begin{itemize}
\item {\em Covariant case.}
\[
\ \ \text{\bf Left exact:}\qquad 0\to F(X)\morph{F(f)}F(Y)\morph{F(g)} F(Z)
\text{ is exact}.
\]
\[
\text{\bf Right exact:}\qquad F(X)\morph{F(f)}F(Y)\morph{F(g)} F(Z) \to 0
\text{ is exact}.
\]

\item {\em Contravariant case.}
\[
\ \ \text{\bf Left exact:}\qquad 0\to G(Z)\morph{G(g)}G(Y)\morph{G(f)} G(X)
\text{ is exact}.
\]
\[
\text{\bf Right exact:}\qquad G(Z)\morph{G(g)}G(Y)\morph{G(f)} G(X) \to 0
\text{ is exact}.
\]
\end{itemize}

We prove that for any $X\in \CalA$ (abelian category), $\h^X$ and $\h_X$
are both left exact functors $\CalA\to\Ab$. We saw some examples to 
demonstrate that they are not necessarily (right) exact.\\

\PP{11.1}{Tensor product I}
%-------------------------
The most typical example of a functor which is right exact, but
not necessarily left exact comes from {\em tensor products}.
In this course, we will mostly be concerned with tensor product
of modules over a commutative ring. \\

Let $A$ be a unital, commutative ring. Let $\Rmod{A}$ be the
category of $A$--modules. It is an abelian category (see \S 9.4). 
Recall that for a commutative ring,
we do not distinguish between left/right modules. Note
that the $\Hom$--functors for this category take values in
$\Rmod{A}$ again. That is, given $M,N\in\Rmod{A}$, $\Hom_A(M,N)$
is naturally an $A$--module, with scalar multiplication given by:
\[
f\in\Hom_A(M,N), a\in A \qquad \rightsquigarrow \qquad
(a\cdot f)(m) := f(am) = af(m),\ \forall\ m\in M.
\]

\begin{defn}
Let $M,N,P$ be three $A$--modules. An {\em $A$--bilinear map} $f:M\times N\to P$
is a set map such that (i) for every $m\in M$, $f(m,-):N\to P$ is
an $A$--module homomorphism, and (ii) for every $n\in N$, $f(-,n):M\to P$
is an $A$--module homomorphism. More explicitly, the following
equations must hold for every $a_1,a_2\in A$, $m,m_1,m_2\in M$
and $n,n_1,n_2\in N$:

\[
f(a_1m_1+a_2m_2,n) = a_1f(m_1,n)+a_2f(m_2,n), \qquad
f(m,a_1n_1+a_2n_2) = a_1f(m,n_1)+a_2f(m,n_2).
\]

Let us denote the set of all $A$--bilinear maps $M\times N\to P$ by
$\Bil_A(M,N;P)$. This set also has a structure of an $A$--module
(see Remark below).\\
\end{defn}

\noindent {\bf Warning.} $\Bil_A(M,N;P) \neq \Hom_A(M\times N,P)$.
Think of the dot product as a typical example of an $\R$--bilinear map
$\R^n\times \R^n \to \R$.\\

\begin{rem}
We have the following equality:
\[
\Bil_A(M,N;P) = \Hom_A(M,\Hom_A(N,P)).
\]
This makes it clear that the set $\Bil_A(M,N;P)$ has an $A$--module
structure. Also, note that for a bilinear $f:M\times N\to P$
and an $A$--linear $g:P\to Q$, the composition $g\circ f : M\times N
\to Q$ is again $A$--bilinear.
\[
\Bil_A(M,N;P)\times \Hom_A(P,Q) \stackrel{\circ}{\longrightarrow}
\Bil_A(M,N;Q).
\]
\end{rem}
\ \\

\PP{11.2}{Tensor product II}
%--------------------------------------------
Let
$M,N\in\Rmod{A}$, and consider the following (covariant) functor:
\[
T_{M,N} : \Rmod{A} \to \Rmod{A}, \qquad P\mapsto T_{M,N}(P) :=
\Bil_A(M,N;P).
\]
On morphisms: $g:P\to Q$ is mapped to $T_{M,N}(g) = g\circ -$.\\

\begin{defn}
If $T_{M,N}$ is representable, the object representing it
is denoted by $M\otimes_A N$ called {\em the tensor product
of} $M$ and $N$ (over $A$). Often, we omit the subscript $A$,
and just write $M\otimes N$, if the ring in question is clear
from the context.\\
\end{defn}

In more detail, $M\otimes_AN$ is an $A$--module together with
an $A$--bilinear map $i:M\times N\to M\otimes_AN$, such that
the following universal property holds.\\

\noindent {\em Universal property of tensor product.} For every
$P\in\Rmod{A}$, and an $A$--bilinear map $f:M\times N\to P$,
there exists a unique $A$--linear map $\wt{f}:M\otimes_AN
\to P$ making the following diagram commute:
\[
\xy
(0,20)*{M\times N}="a";
(30,20)*{M\otimes_AN}="b";
(30,0)*{P}="c";
{\ar^i "a"; "b"};
{\ar_f "a"; "c"};
{\ar^{\wt{f}} "b"; "c"};
\endxy
\]

\PP{11.3}{Construction of tensor product}
%---------------------------------------
Now we give a construction of tensor product, to make sure that
it exists. Let $M,N\in\Rmod{A}$. Let $\wt{T}$ be the free $A$--module
generated over the set of symbols $\{t(m,n): m\in M, n\in N\}$.
That is, a typical element of $\wt{T}$ is a finite linear
expression:

\[
\wt{T} = \left\{
\sum_{(m,n)\in M\times N} a_{m,n}t(m,n) : a_{m,n}\in A \text{ are
all zero, except for finitely many}
\right\}.
\]

Thus, $\wt{T}$ is naturally an $A$--module. It is the direct
sum of $M\times N$ many copies of $A$.\\

Let $K\subset\wt{T}$ be the $A$--submodule generated by
following elements:\\

\begin{itemize}
\item $t(a_1m_1+a_2m_2,n)-a_1t(m_1,n)-a_2t(m_2,n)$, for every
$a_1,a_2\in A$, $m_1,m_2\in M$ and $n\in N$.\\

\item $t(m,a_1n_1+a_2n_2)-a_1t(m,n_1)-a_2t(m,n_2)$, for every
$a_1,a_2\in A$, $m\in M$ and $n_1,n_2\in N$.\\
\end{itemize}

Define $T := \wt{T}/K$. We have an $A$--bilinear map $i:M\times N\to T$
sending $(m,n)\mapsto t(m,n) + K$.\\

\begin{prop}
$T$ together with the $A$--bilinear map $i:M\times N\to T$ satisfies
the universal property of the tensor product of $M$ and $N$
over $A$. It will be denoted by $M\otimes_AN$, after the proof
is complete.
\end{prop}

\begin{pf}
Let $P\in\Rmod{A}$ and let $f:M\times N\to P$ be an $A$--bilinear
map. Viewing it as a set map, we get a unique $A$--linear map 
$F : \wt{T} \to P$
such that $F\circ i = f$. Here, we are using the same letter
to denote the (set) map $M\times N\to \wt{T}$, defined as:
$i(m,n) = t(m,n)$. The $A$--linear map $F$ is thus given by:
$F(t(m,n)) = f(m,n)$.\\

It remains to show that $F:\wt{T}\to P$ factors through $K$.
That is $F(k) = 0$ for every $k\in K$. Note that since $F$ is,
by definition, $A$--linear, it suffices to check $F(k)=0$,
when $k$ is one of the generators listed above.\\

When $k=t(a_1m_1+a_2m_2,n) - a_1t(m_1,n)-a_2t(m_2,n)$, we get

\begin{align*}
F(k) &= F(t(a_1m_1+a_2m_2,n)) - a_1 F(t(m_1,n)) - a_2 F(t(m_2,n))\\
&= f(a_1m_1+a_2m_2,n) - a_1 f(m_1,n) - a_2 f(m_2,n) = 0,
\end{align*}

by bilinearity of $f$. Similarly for the other type of generator
of $k$, we can easily see that $F(k)=0$. Thus, $F$ factors through
a unique morphism $\wt{f}:T\to P$, sending the coset
$t(m,n)+K$ to $f(m,n)$. This proves that $T$ satisfies the
universal property written in \S 11.2 above.
\end{pf}
\ \\

\noindent {\bf Remark.}
The module constructed in the previous section will be denoted by
$M\otimes_AN$. We use the notation $m\otimes n$ to mean $i(m,n)$,
where $i:M\times N \to M\otimes_A N$ is the canonical bilinear
map. Informally speaking, $M\otimes_A N$ consists of finite
linear expressions of the form $\ds \sum_{i=1}^p m_i\otimes n_i$,
which are manipulated according to the following rules:
\begin{itemize}
\item $(m_1+m_2)\otimes n = m_1\otimes n + m_2\otimes n$,
for every $m_1,m_2\in M$ and $n\in N$.
\item $m\otimes (n_1+n_2) = m\otimes n_1 + m\otimes n_2$,
for every $m\in M$ and $n_1,n_2\in N$.
\item $(am)\otimes n = m\otimes (an)$, for every $m\in M$,
$n\in N$ and $a\in A$.
\end{itemize}

\end{document}
