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

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\begin{document}
\maketitle

\PP{12.0}{Tensor products}
%------------------------
Recall that last time we defined and constructed the 
tensor product of two modules $M$ and $N$ over
a commutative ring $A$, denoted by $M\otimes_AN$. Abstractly,
it is the unique $A$--module which comes together
with an $A$--bilinear map $i:M\times N\to M\otimes_AN$, satisfying
a universal property:\\

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:
\[
\xy
(0,20)*{M\times N}="a";
(30,20)*{M\otimes_AN}="b";
(30,0)*{P}="c";
{\ar^i "a"; "b"};
{\ar_f "a"; "c"};
{\ar^{\wt{f}} "b"; "c"};
\endxy
\]

In less words, $M\otimes_AN$ is defined by:
\[
\Hom_A(M\otimes_AN, P) \isom \Bil_A(M,N;P).
\]

Concretely, $M\otimes_A N$ 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}

Now we will see some examples, and functorial properties of tensor
products.\\

\PP{12.1}{Elementary properties of tensor product}
%------------------------------------------------
We will always use the universal property of tensor product
to prove its properties.\\

\begin{prop}\hfill
\begin{enumerate}
\item For every $M,N\in\Rmod{A}$, $M\otimes_AN \cong N\otimes_AM$.
\item $A\otimes_AN \cong N$ for every $N\in\Rmod{A}$.
\item If $\ga\subset A$ is an ideal of $A$, then 
\[
A/\ga \otimes_A N \cong N/\ga N.
\]
\end{enumerate}
\end{prop}

\begin{pf}
(1). Consider the flip map $\sigma: M\times N \to N\times M$. Let
$i:M\times N\to M\otimes_AN$ and $j:N\times M\to N\otimes_A M$ be
the canonical $A$--bilinear maps. It is clear that $j\circ \sigma:
M\times N \to N\otimes_AM$ is $A$--bilinear, hence gives rise
to an $A$--linear map $f:M\otimes_AN \to N\otimes_AM$ given by:
$f(m\otimes n) = n\otimes m$. Similarly, using $i\circ \sigma^{-1}:
N\times M \to M\otimes_AN$, we conclude that there is a unique
$A$--linear map $g:N\otimes_AM \to M\otimes_AN$, given by
$g(n\otimes m) = m\otimes n$. Clearly $f$ and $g$ are inverse
to each other, hence $M\otimes_AN \cong N\otimes_AM$.\\

(2). Note that if $f:A\times N \to P$ is any $A$--bilinear map,
then we get an $A$--linear map $\ol{f}:N\to P$ as
$\ds \ol{f}(n) = f(1,n)$. The bilinear map $f$ is uniquely
determined by $\ol{f}$, since $f(a,n) = af(1,n) = a\ol{f}(n)
= \ol{f}(an)$. Hence,
\[
\Bil_A(A,N;P) \isom \Hom_A(N,P), \qquad f\mapsto \ol{f}.
\]
So, $N$ satisfies the universal property of $A\otimes_AN$.\\

(3). The argument is same as the one for (2). Given a bilinear
map $f:A/\ga \times N \to P$, we use it to define $\ol{f}:N\to P$
as $\ol{f}(n) = f(1+\ga,n)$. It remains to be shown that
$\ol{f}$ factors through the submodule 
\[
\ga N = \{an : a\in \ga,\ n\in N\} \subset N.
\]
That is, $\ol{f}(an) = 0$ for every $a\in \ga$ and $n\in N$.
This is true because (we are using $\ol{x} = x+\ga \in A/\ga$):
\[
\ol{f}(an) = f(\ol{1},an) = af(\ol{1},n) = f(\ol{a},n) = 
f(0,n) = 0.
\]
Denoting by the same symbol, the resulting $A$--linear map
$\ol{f} : N/\ga N \to P$, we obtain the isomorphism (same argument
as before, hence omitted):
\[
\Bil_A(A/\ga,N;P) \isom \Hom_A(N/\ga N,P), \qquad f\mapsto \ol{f},
\]
which proves (3).
\end{pf}

\PP{12.2}{Examples}
%-----------------
The following examples of tensor product follow immediately from
Proposition 12.1.\\

\noindent {\bf I.} Let $m,n\in\Z_{\geq 2}$. Then
$\ds (\Z/m\Z)\otimes_\Z (\Z/n\Z) \cong \Z/\gcd(m,n)\Z$. This
is because we use (3) of the proposition above, and: 
\[
(\Z/n\Z)/m(\Z/n\Z) \cong \Z/(m,n) = \Z/\gcd(m,n)\Z.
\]
Here $(m,n)$ is the ideal of $\Z$ generated by $m$ and $n$, which
is same as $\gcd(m,n)\Z$ by Euclidean algorithm.\\

\noindent {\bf II.} Let $K$ be a field and consider the commutative
ring $K[x,y]$. Let
$K[x]$ be viewed as an $A$--module via $K[x] \cong K[x,y]/(y)$
(similarly $K[y]$). Then:
\[
K[x] \otimes_{K[x,y]} K[y] \cong K = K[x,y]/(x,y).
\]

\noindent {\bf III.} If $A=K$ is a field, $M,N\in\Rmod{A}$ are
$K$--vector spaces. Let $\{m_i\}_{i\in I}$ and $\{n_j\}_{j\in J}$
be bases of $M$ and $N$ resp. Then $M\otimes_K N$ is the vector
space with basis $\{m_i\otimes n_j\}_{(i,j)\in I\times J}$.\\

\PP{12.3}{Functorial properties of tensor product}
%-------------------------------------------------
Again, let $A$ be a unital commutative ring.

\begin{lem}
Given $A$--linear morphisms $\ds M\morph{f} M'$ and
$\ds N\morph{g} N'$, there is a unique $A$--linear map
$f\otimes g : M\otimes N \to M'\otimes N'$ such that
$(f\otimes g)(m\otimes n) = f(m)\otimes g(n)$.\\

Moreover, if $\ds M'\morph{f'}M''$ and $\ds N'\morph{g'} N''$
are two more morphisms, then:
\[
(f'\otimes g') \circ (f\otimes g) = (f'\circ f) \otimes (g'\circ g).
\]
\end{lem}

The proof is left as an exercise. All you have to do is apply the universal
property to the composition:
\[
\xy
(0,0)*{M\times N}="a";
(30,0)*{M'\times N'}="b";
(60,0)*{M'\otimes N'}="c";
{\ar^{(f,g)} "a"; "b"};
{\ar^{i'} "b"; "c"};
\endxy
\]
where $i'$ is the canonical bilinear map $M'\times N' \to M'\otimes N'$.\\

Using this lemma, we are able to view tensoring with a fixed module
as a functor. Let $N\in\Rmod{A}$ be fixed.

\begin{cor}
We have an additive functor $-\otimes N : \Rmod{A} \to \Rmod{A}$.
\end{cor}

\begin{pf}
This functor maps $M\in\Rmod{A}$ to $M\otimes N$, and a morphism
$f:M\to M'$ is mapped to $\ds f\otimes\Id_N$. By the lemma above,
it is clear that $\Id_M\otimes \Id_N = \Id_{M\otimes N}$,
and that $(g\circ f)\otimes \Id_N = (g\otimes\Id_N) \circ
(f\otimes\Id_N)$, proving that $-\otimes N$ is a functor.\\

To prove that it is additive, we need to show that
\[
-\otimes\Id_N: \Hom_A(M_1,M_2) \to \Hom_A(M_1\otimes N, M_2\otimes N)
\]
is a group homomorphism. It is clear by the lemma that $0\otimes \Id_N = 0$.
Now let $f,g:M_1\to M_2$. Then for every $m\in M$ and $n\in N$, we
have:
\[
((f_1+f_2)\otimes\Id_N)\ (m\otimes n) = (f_1(m)+f_2(m))\otimes n,
\]
\[
((f_1\otimes \Id_N) + (f_2\otimes \Id_N))\ (m\otimes n)
= f_1(m)\otimes n + f_2(m)\otimes n.
\]
which are equal. Hence $(f_1+f_2)\otimes \Id_N = (f_1\otimes\Id_N)
+ (f_2\otimes \Id_N)$.
\end{pf}
\ \\

\PP{12.4}{Tensor-Hom adjointness}
%-------------------------------
The way we defined the tensor product, the following isomorphism
is clear:

\[
\Hom_A(M\otimes N, P) \isom \Bil_A(M,N;P) = \Hom_A(M,\Hom_A(N,P)).
\]

Let us keep $N\in\Rmod{A}$ fixed, and denote the inverse of this bijection by
$\beta_{M,P}$ (to be consistent with our conventions). Note
that for any $\xi: M \to \Hom_A(N,P)$, $\beta_{M,P}(\xi)$
is given by:
\[
\beta_{M,P}(\xi) : m\otimes n \mapsto \xi(m)(n).
\]

\begin{prop}
$\beta_{M,P}$ is natural in $M$ and $P$. Hence, we have
a pair of adjoint functors $(-\otimes N, \Hom_A(N,-))$.
\end{prop}

\begin{pf}
We only need to check that given any two morphisms $M\morph{f} M'$
and $P'\morph{g} P$ the following diagram commutes:
\[
\xy
(0,20)*{\Hom_{A}(M,\Hom_A(N,P))}="a";
(60,20)*{\Hom_A(M\otimes N,P)}="b";
(0,0)*{\Hom_A(M',\Hom_A(N,P'))}="c";
(60,0)*{\Hom_A(M'\otimes N,P')}="d";
{\ar^{\beta_{M,P}} "a"; "b"};
{\ar_{\beta_{M',P'}} "c"; "d"};
{\ar^L "c"; "a"};
{\ar_R "d"; "b"};
\endxy
\]
So, let $\xi:M'\to \Hom(N,P')$ be given. Following it through
the composition $\beta_{M,P}\circ L$ and evaluating on $m\otimes n$
gives:
\[
\beta_{M,P}(L(\xi))(m\otimes n) = L(\xi)(m)(n) = g(\xi(f(m))(n)).
\]
Similarly, $R\circ\beta_{M',P'}$ sends $\xi$ to the morphism
which evaluated on $m\otimes n$ yields:
\[
R(\beta_{M',P'}(\xi))(m\otimes n) = 
g(\beta_{M',P'}(\xi)(f(m)\otimes n)) = g(\xi(f(m))(n)).
\]
This finishes the proof of naturality of $\beta$.
\end{pf}
\ \\

\PP{12.5}{Exactness of tensor functor}
%------------------------------------
The following simple example shows that tensor is not left exact.\\

\noindent {\bf Example.} Consider the short exat sequence of abelian
groups ($\Z$--modules):
\[
0 \to \Z \morph{\cdot 2} \Z \morph{\pi} \Z/2\Z \to 0.
\]
Tensoring with $-\otimes \Z/2\Z$ yields (recall $A\otimes_A N\cong N$):
\[
0\to \Z/2\Z \morph{0} \Z/2\Z \morph{\Id} \Z/2\Z \to 0
\]
which is not exact at the first $\Z/2\Z$. Basically, tensoring
could turn an injective morphism to zero morphism, which is the
reason it is not left exact.\\

\begin{thm}
For any $N\in\Rmod{A}$, $-\otimes_A N : \Rmod{A} \to \Rmod{A}$ is right exact.
\end{thm}

\begin{pf}
We need to prove that given an exact sequence $\ds 0\to M_1\morph{f}
M_2\morph{g} M_3\to 0$ of $A$--modules, the following sequence
is also exact:
\[
\xy
(0,0)*{M_1\otimes N}="a";
(30,0)*{M_2\otimes N}="b";
(60,0)*{M_3\otimes N}="c"; (80,0)*{0}="d";
{\ar^{f\otimes\Id} "a"; "b"};
{\ar^{g\otimes\Id} "b"; "c"};
{\ar "c"; "d"};
\endxy
\]

\noindent {\em $g\otimes\Id$ is surjective.} This is clear since
each simple tensor $m_3\otimes n\in M_3\otimes N$ is of the
form $g(m_2)\otimes n = (g\otimes\Id)(m_2\otimes n)$ for some
$m_2\in M_2$, since $g$ is assumed to be surjective. Hence,
$m_3\otimes n\in \Im(g\otimes \Id)$. As $M_3\otimes N$ is generated
as an $A$--module by such element we conclude that $g\otimes\Id$
is surjective.\\

\noindent $\Im(f\otimes\Id)\subset\Ker(g\otimes\Id)$. This
is again clear, since by Lemma 12.3, $(g\otimes\Id)\circ
(f\otimes\Id) = (g\circ f)\otimes\Id = 0$.\\

It remains to show that $\Ker(g\otimes\Id) = \Im(f\otimes\Id)$.
For this, note that $g\otimes\Id$ gives rise to a well--defined $A$--linear map
$M_2\otimes N/\Im(f\otimes\Id) \to M_3\otimes N$,
by the previous part, which we will denote by $\wt{g}$. We are going to 
construct an $A$--linear map $\wt{h}:M_3\otimes N
\to M_2\otimes N/\Im(f\otimes \Id)$ and show that $\wt{g}$
and $h$ are inverse to each other.\\

Consider $h:M_3\times N \to M_2\otimes N/\Im(f\otimes\Id)$ given by
\[
h(m_3,n) = m_2\otimes n \text{ (mod $\Im(f\otimes\Id)$)},
\]
where $m_2\in M_2$ is chosen so that $g(m_2)=m_3$. Note that
a different choice $m_2'$ will differ from $m_2$ by an element
of $\Ker(g) = \Im(f)$, so modulo $\Im(f\otimes\Id)$, $m_2\otimes n
\equiv m_2'\otimes n$. This proves that $h$ is well--defined.\\

It is left to the reader to check that $h$ is $A$--bilinear, and 
hence gives rise to $\wt{h}:M_3\otimes N\to M_2\otimes N/\Im(f\otimes\Id)$.
By their definition, one can easily verify that $\wt{g}$
and $\wt{h}$ are inverse to each other. Hence, $\wt{g}$
sets up an isomorphism
\[
\wt{g}:M_2\otimes N/\Im(f\otimes\Id) \isom M_3\otimes N,
\]
proving that $\Im(f\otimes\Id)$ is indeed the kernel of 
$g\otimes\Id$.
\end{pf}
\ \\

\PP{12.6}{A more hands-off proof}
%-------------------------------
Let $\CalA$ and $\CalB$ be two abelian categories, and assume that
we have a pair of adjoint functors $(F,G)$:
$\ds \xy (0,0)*{\CalA}; (10,0)*{\CalB}; {\ar^F (2,1); (8,1)};
{\ar^G (8,-1); (2,-1)}  \endxy$.\\

\noindent {\bf Exercise.} Prove that $F$ is right exact, and $G$ is
left exact.\\

Thus, Theorem 12.5 can be obtained from ``Tensor-Hom adjointness"
Proposition 12.4.
\end{document}
