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\title{Lecture 22}
\author{}

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\begin{document}
\maketitle

\PP{22.0}{Review}
%---------------
Let $R$ be a unital, commutative ring. We defined {\em flat modules}
as: $N\in\Rmod{R}$ is said to be flat if $-\otimes N$ is an exact
functor. Last time we prove the following two results:\\

\begin{itemize}
\item The following three assertions are equivalence.
	\begin{itemize}
	\item $N$ is flat.
	\item $N_{\gp}$ is flat for every prime ideal $\gp\subsetneq R$.
	\item $N_{\gm}$ is flat for every maximal ideal $\gm\subsetneq R$.
	\end{itemize}

\item If $(A,\gm)$ is a local ring, and $N$ is a finitely presented
flat $A$--module, then $N$ is free.\\
\end{itemize}

\begin{rem}
Geometrically, flat morphisms correspond to {\em locally trivial fibrations}
where dimensions (or number of elements) of fibers does not change. For instance,
if $X$ and $Y$ are connected, complex manifolds, and $f:X\to Y$ is an analytic
map, then $f$ being flat implies $f(X)\subset Y$ is open, and for
every $y\in f(X)$, $X_y = f^{-1}(y)$ has the same dimension 
(independent of $y$).\\ 
\end{rem}

\PP{22.1}{$\Ext$, $\Tor$ and loose ends}
%--------------------------------------
Recall that we started (since Lecture 13) with an abstract notion
of derived functors. Let $F:\CalA \to \CalB$ is an additive functor
between two abelian categories.

\begin{itemize}
\item If $F$ is left exact (covariant or contravariant), we get
a sequence of functors $\{R^kF : \CalA\to \CalB\}_{k\geq 0}$ (same variance
as $F$) called {\em right derived functors of $F$}.\\

\item If $F$ is right exact, we get $\{L_kF:\CalA\to\CalB\}_{k\geq 0}$
called {\em left derived functors of $F$}.\\
\end{itemize}


When $\CalA=\CalB =\Rmod{R}$, and $M,N\in\Rmod{R}$, we get the
following derived functors.\\

$\Ext^k(M,N)$ can be viewed as $R^k\h_M(N)$ or $R^k\h^N(M)$. Here,
$\h_M=\Hom(M,-)$ and $\h^N=\Hom(-,N)$ are covariant and contravariant
$\Hom$ functors. They are both left exact.\\

$\Tor_k(M,N)$ can be viewed as $L_kT_M(N)$ or $L_kT^N(M)$. Here,
$T_M=M\otimes -$ and $T^N = -\otimes N$ are both covariant, right
exact functors (since $\otimes$ is commutative, $T^N = T_N$).\\

We still have to prove that the two ways of defining $\Ext$
and $\Tor$ give the same result. This is the main theorem of
these notes.\\

\PP{22.2}{The case of $\Ext$}
%---------------------------
For definiteness, let us start with $R^k\h_M(N)$ and show that
it is the same as $R^k\h^N(M)$. In more detail, we will take the following
definition as the starting point, and call the resulting
functor $E^k(M,N)$.\\

Given $N\in\Rmod{R}$, choose an injective resolution of $N$:
\[
0\to I^0 \morph{i^0} I^1 \morph{i^1} I^2\morph{i^2} \cdots \qquad \text{exact at } I^k,\ 
\forall k\in\Z_{\geq 1},\ \text{and}\ \Ker(i^0)=N.
\]
Define $E^k(M,N)$ as $k^{\sth}$ cohomology of the complex
$\Hom(M,I^\bullet)$. It is a covariant functor in the variable $N$.

\[
\boxed{
E^k(M,N) = H^k(\Hom(M,I^\bullet))
}
\]

Recall the following properties of $E^k(M,N)$, which will be needed for
our proof.\\

\begin{enumerate}
\item\label{P:1} $E^0(M,N) = \Hom(M,N)$. This follows from the left exactness
of $\Hom(M,-)$.\\

\item\label{P:2} If $N$ is injective, then
$E^k(M,N)=0$ for every $k\geq 1$, $M\in\Rmod{R}$. This is true because
we can take $0\to N\to 0$ as an injective resolution of $N$.\\

\item\label{P:3} For every short exact sequence
$0\to N_1\to N_2\to N_3\to 0$, we get a connecting homomorphism
$E^k(M,N_3)\to E^{k+1}(M,N_1)$, making the following sequence
exact:
\[
\cdots \to E^k(M,N_1)\to E^k(M,N_2)\to E^k(M,N_3)\to E^{k+1}(M,N_1)
\to E^{k+1}(M,N_2)\to \cdots
\]

%\item\label{P:4} If $M$ is projective, then $E^k(M,N)=0$
%for every $k\in\Z_{\geq 1}$ and $N\in\Rmod{R}$. This is because,
%in this case, $\Hom(M,-)$ is exact.\\
\end{enumerate}

\begin{lem}
Assume that $0\to N\to I\to N'\to 0$ is a short exact sequence
of $R$--modules, where $I$ is injective. Then we have:
\[
E^{k+1}(M,N) = E^k(M,N'),\ \text{ for every } k\geq 1,
\]
and $E^1(M,N) = \Coker(\Hom(M,I)\to \Hom(M,N'))$.
\end{lem}

\begin{pf}
The proof follows easily by the long exact sequence in cohomology.
Namely,
\[
0\to \Hom(M,N)\to \Hom(M,I)\to \Hom(M,N')\to E^1(M,N)\to E^1(M,I)=0.
\]
The last term is zero since $I$ is injective. Thus we get:
\[
E^1(M,N) = \Coker(\Hom(M,I)\to \Hom(M,N')).
\]

Now for $k\geq 1$, the following is a part of the long exact sequence:
\[
0=E^k(M,I)\to E^k(M,N') \to E^{k+1}(M,N)\to E^{k+1}(M,I)=0,
\]
which proves that $E^k(M,N') \cong E^{k+1}(M,N)$.
\end{pf}
\ \\

\PP{22.3}{$R^k\h_M(N) = R^k\h^N(M)$}
%-----------------------------------
Now we are ready to state and prove the desired result. Let
$P_\bullet$ be a projective resolution of $M$. That is,
we have a chain complex:

\[
\cdots \morph{p_2} P_2 \morph{p_1} P_1 \morph{p_0} P_0 \to 0,
\]
which is exact at $P_k$ for every $k\geq 1$, and $\Coker(p_0)=M$.
In addition to the properties of $E^k(M,N)$ listed above, we
are going to use the fact that $\Hom(P,-)$ is exact for
$P$ a projective $R$--module.\\

\begin{thm}
$E^k(M,N) = H^k(\Hom(P_\bullet,N))$.
\end{thm}

\begin{pf}
The proof is an induction argument on $k$. For $k=0$, we have
\[
E^0(M,N) = \Hom(M,N) = H^0(\Hom(P_\bullet,N)), \text{ for every } N\in\Rmod{R},
\]
again by left exactness of $\Hom$ functors.\\

Assume that we know $E^k(M,X) = H^k(\Hom(P_\bullet,X))$ for every
$X\in\Rmod{R}$ and $0\leq k\leq \ell$, where $\ell\in\Z_{\geq 0}$.\\

Consider a short exact sequence of $R$--modules, where $I$ is injective:
\[
0\to N\morph{f} I\morph{g} N'\to 0.
\]

Applying $\Hom(P_\bullet,-)$ to this short exact sequence, and using the
fact that each $P_j$ is projective, so $\Hom(P_j,-)$ is exact, we
get get a short exact sequence of cochain complexes:

\[
0\to \Hom(P_\bullet,N) \to \Hom(P_\bullet, I) \to \Hom(P_\bullet,N') \to 0.
\]

This gives rise to a long exact sequence in cohomology (see Lecture 15,
Theorem 15.1). Now we are ready to carry out the induction step.\\

\noindent {\em Case 1: $\ell=1$}. Using $H^0(\Hom(P_\bullet),X)=\Hom(M,X)$,
the first part of the long exact sequence in cohomology takes the
following form:
\[
0\to \Hom(M,N)\to \Hom(M,I)\to \Hom(M,N')\to H^1(\Hom(P_\bullet,N))
\to H^1(\Hom(P_\bullet,I)).
\]
The last term is zero, since $I$ is injective, so $\Hom(-,I)$ is exact.
This implies that:
\[
H^1(\Hom(P_\bullet,N)) = \Coker(\Hom(M,I)\to \Hom(M,N'))
= E^1(M,N),
\]
by Lemma 22.2 above.\\

\noindent {\em Case 2: $\ell>1$}. Again we focus on a part of
the long exact sequence, and use the fact that $I$ being injective,
we have $H^k(\Hom(P_\bullet,I))=0$ for every $k\geq 1$.\\

\[
0=H^\ell(P_\bullet,I)\to H^\ell(P_\bullet,N') \to H^{\ell+1}(P_\bullet,N)
\to H^{\ell+1}(P_\bullet,I)=0.
\]

Thus, we obtain:
\[
H^{\ell+1}(P_\bullet,N) = H^\ell(P_\bullet,N') = 
E^\ell(M,N') = E^{\ell+1}(M,N),
\]
where in the second equality we have used the induction hypothesis,
and in the third, Lemma 22.2.
\end{pf}
\ \\

\PP{22.4}{The case of $\Tor$}
%---------------------------
Exact same argument as in the proof of the theorem above works 
(details are left as an instructive exercise to the reader) to
prove that:\\

\[
\boxed{
\Tor_k(M,N) = H_k(P^M_\bullet\otimes N) = H_k(M\otimes P^N_\bullet)
}
\]

Here, $P^M_\bullet$ and $P^N_\bullet$ are projective resolutions of
$M$ and $N$ respectively.\\

In fact, since in the proof of the theorem, we only used the fact
that $\Hom(P,-)$ is exact, it allows applies to prove that

\[
\boxed{
\Tor_k(M,N) = H_k(F^M_\bullet\otimes N)
}
\]
where $F^M_\bullet$ is a {\em flat resolution} of $M$.
Compare this argument with Exercise 11 of Homework 7.\\

\begin{ex}
Assume that $R$ is an integral domain, and $K$ is its field
of fractions (that is, $K = (R\setminus\{0\})^{-1}R$). The
following is a flat resolution of $M=K/R$:

\[
0\to R \morph{j} K \to 0,
\]

where $j:R\to K$ is the canonical inclusion. Using this resolution,
we can compute $\Tor_k(K/R,N)$ for $N\in\Rmod{R}$ as follows.
Tensoring the above complex with $N$ (and using $R\otimes_R N = N$) 
gives:\\

\[
0\to N \morph{j_N} K\otimes_R N \to 0,\qquad
\text{degrees } 2, 1, 0, -1 \text{ respectively}.
\]
\ \\

Thus $\Tor_0(K/R,N) = N_K/j_N(N)$ (where $N_K = K\otimes_R N$ is a $K$--vector
space and $j_N(N)$ is its $R$--submodule).\\

\[
\Tor_1(K/R,N) = \Ker(j_N) = \{n\in N : \exists\ 0\neq a\in R \text{ such that }
an=0\} = N_{\tor}.
\]
\ \\

For instance, take $R=\Z$, so that $K=\Q$. A typical finitely--generated
abelian group has the following form:
\[
N = \Z^{\oplus r} \oplus \lp \Z/m_1\Z \oplus\cdots\oplus \Z/m_s\Z\rp.
\]
Then, the computation above gives:
\[
\Tor_0(\Q/\Z,N) = (\Q/\Z)^{\oplus r}, \qquad
\Tor_1(\Q/\Z,N) = \Z/m_1\Z \oplus\cdots\oplus \Z/m_s\Z.
\]
\end{ex}


\end{document}
