\documentclass[12pt]{amsart}
\title{Lecture 19}
\author{}

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\newtheorem*{rem}{Remark}
\newtheorem*{rems}{Remarks}
\newtheorem*{ex}{Example}

\begin{document}
\maketitle

\PP{19.0}{Definition of $\Ext$ and $\Tor$}
%-----------------------------------------
Let $R$ be a unital commutative ring, and $\Rmod{R}$ be the category
of $R$--modules.
Given $M,N\in\Rmod{R}$, recall that we have the following functors
(see Theorems 10.4 and 12.5).\\

\begin{itemize}
\item $\h^N = \Hom_R(-,N) : \Rmod{R}\to \Rmod{R}$. It is a left exact,
contravariant functor.\\

\item $\h_M = \Hom_R(M,-) : \Rmod{R}\to \Rmod{R}$. It is a left exact,
covariant functor.\\

\item $T_N = -\otimes_R N : \Rmod{R}\to \Rmod{R}$. It is a right exact,
covariant functor. Note that since $M\otimes_R N\cong N\otimes_RM$,
$-\otimes_RN$ is naturally isomorphic to $N\otimes_R-$.\\
\end{itemize}

Recall that given a functor between two abelian categories
$F:\CalA\to\CalB$, its derived functors are constructed as follows.\\

\begin{itemize}
\item {\em Left exact, covariant case.} Given $A\in\CalA$,
choose an injective resolution $I^\bullet_A\in \CoComp{\CalA}$. Then:
\[
R^kF(A) := k^{\sth} \text{ cohomology of the cochain complex } F(I^\bullet_A).
\]

\item {\em Left exact, contravariant case.} Given $A\in\CalA$,
choose a projective resolution $P_\bullet^A\in \Comp{\CalA}$. Then:
\[
R^kF(A) := k^{\sth} \text{ cohomology of the cochain complex } F(P_\bullet^A).
\]

\item {\em Right exact, covariant case.} Given $A\in\CalA$,
choose a projective resolution $P_\bullet^A\in \Comp{\CalA}$. Then:
\[
L_kF(A) := k^{\sth} \text{ homology of the chain complex } F(P_\bullet^A).
\]

(similarly for right exact, contravariant case - but we are not
going to consider it, so it is omitted here)\\
\end{itemize}

\begin{rems}
\begin{enumerate}
\item According to Theorem 17.3 and its corollary, injective and projective
resolutions are unique up to homotopy. This, combined with the
fact that $H^k$ is the same for two homotopic morphisms (Proposition
14.3) implies that derived functors do not depend on the choice
of a resolution.\\

\item Derived functors are additive. To see this, let us assume $F$ is
left exact and covariant (to fix ideas). We can realize $R^kF$
as a composition of additive functors, if we replace $\CoComp{\CalA}$
by its {\em homotopy version}, denoted here by
$\CoComp{\CalA}_h$.\\ 

The objects of $\CoComp{\CalA}_h$ are same
as those of $\CoComp{\CalA}$ (namely cochain complexes), but the
morphisms are changed to:
\[
\Hom_{\CoComp{\CalA}_h}(C^\bullet,D^\bullet) = 
\Hom_{\CoComp{\CalA}}(C^\bullet,D^\bullet)/\Hom^0_{\CoComp{\CalA}}(C^\bullet,D^\bullet),
\]
where $\Hom^0_{\CoComp{\CalA}}(C^\bullet,D^\bullet)$ is the subgroup
of null--homotopic morphisms (see \S 14.3). By Proposition
14.3, we get that $\CoComp{\CalA}_h$ is an additive category,
and by Theorem 17.3, $M\mapsto I_M^\bullet$ (choice of an injective
resolution) is an additive functor $\mathbb{I}:\CalA\to\CoComp{\CalA}_h$.
Thus, $R^kF:\CalA\to\CalB$ is the following composition of additive
functors, hence additive:
\[
\CalA \morph{\mathbb{I}} \CoComp{\CalA}_h \morph{F} \CoComp{\CalB}_h 
\morph{H^k} \CalB.
\]
\end{enumerate}
\end{rems}

Check: an additive functor $F:\CalA\to\CalB$ applied term-by-term
to cochain complexes $F:\CoComp{\CalA}\to\CoComp{\CalB}$ sends null--homotopic
morphisms to null--homotopic morphisms.\\

Thus we obtain sequences of additive functors $\{R^k\h^N\}, \{R^k\h_M\}$ and $\{L_kT_N\}$,
from $\Rmod{R}$ to itself (here $k\in\Z_{\geq 0}$), 
whose definitions and properties are the
main theme of this lecture.\\

\PP{19.1}{$\Ext$ functors}
%------------------------
The abstract construction of derived functors, for the case of
$\h^N$ and $\h_M$ can be rewritten as follows.

\begin{defn}
Given $M,N\in\Rmod{R}$, $R^k\h^N(M)$ is defined as follows.
\begin{itemize}
\item Choose a projective resolution of $M$:
\[
\cdots \morph{d_2} P_2\morph{d_1} P_1\morph{d_0} P_0\to 0.
\]
\item Apply $\Hom_R(-,N)$ to get a complex:
\[
0\to \Hom_R(P_0,N)\morph{d^0} \Hom_R(P_1,N)\morph{d^1} \Hom_R(P_2,N)
\morph{d^2} \cdots
\]
where $d^j = -\circ d_j$.
\item Take cohomology: $\ds\boxed{R^k\h^N(M) = \Ker(d^k)/\Im(d^{k-1})}$
\end{itemize}

Similarly, $R^k\h_M(N)$ is defined as follows.
\begin{itemize}
\item Choose an injective resolution of $N$:
\[
0\to I^0 \morph{f^0} I^1\morph{f^1} I^2\morph{f^2} \cdots
\]
\item Apply $\Hom_R(M,-)$ to get a complex:
\[
0\to \Hom_R(M,I^0)\morph{d^0} \Hom_R(M,I^1)\morph{d^1} \Hom_R(M,I^2)
\morph{d^2} \cdots
\]
where $d^j = f^j\circ -$.
\item Take cohomology: $\ds\boxed{R^k\h_M(N) = \Ker(d^k)/\Im(d^{k-1})}$
\end{itemize}
\end{defn}

It will be shown next week that $\ds\boxed{R^k\h^N(M)\cong R^k\h_M(N)}$.
For now, we are going to assume this, and define:
\[
\boxed{
\Ext^k_R(M,N) := R^k\h^N(M) = R^k\h_M(N)
}
\]

\begin{thm}
Let $N\in\Rmod{R}$. Then we have additive, contravariant functors:
\[
\Ext^k_R(-,N) : \Rmod{R} \to \Rmod{R},\qquad k\in\Z_{\geq 0}.
\]
This sequence of functors has the following properties:
\begin{enumerate}
\item $\Ext_R^0(-,N) = \Hom_R(-,N)$.\\

\item For a projective $P\in\Rmod{R}$, $\Ext_R^k(P,N)=0$, for every $k\geq 1$.\\

\item Given a short exact sequence $0\to M_1\to M_2\to M_3\to 0$, there
exist connecting morphisms $\delta^k : \Ext_R^k(M_1,N)\to \Ext_R^{k+1}(M_3,N)$,
such that the following sequence is exact:
\[
\xy
(0,40)*{0}="i";
(20,40)*{\Hom(M_3,N)}="fx"; {\ar "i"; "fx"};
(60,40)*{\Hom(M_2,N)}="fy"; {\ar "fx"; "fy"};
(100,40)*{\Hom(M_1,N)}="fz"; {\ar "fy"; "fz"};
{\ar (112,40); (120,40)};
(120,40); (120,30) **\crv{(130,40)&(130,30)};
{\ar@{-} (120,30); (0,30)}; (55,32)*{\scriptstyle{\delta^0}};
(0,30); (0,20) **\crv{(-10,30)&(-10,20)};
(20,20)*{\Ext^1(M_3,N)}="rfx"; {\ar (0,20); "rfx"};
(60,20)*{\Ext^1(M_2,N)}="rfy"; {\ar "rfx"; "rfy"};
(100,20)*{\Ext^1(M_1,N)}="rfz"; {\ar "rfy"; "rfz"};
{\ar (112,20); (120,20)};
(120,20); (120,10) **\crv{(130,20)&(130,10)};
{\ar@{-} (120,10); (0,10)}; (55,12)*{\scriptstyle{\delta^1}};
(0,10); (0,0) **\crv{(-10,10)&(-10,0)};
(20,0)*{\Ext^2(M_3,N)}="rrfx"; {\ar (0,0); "rrfx"};
(60,0)*{\Ext^2(M_2,N)}="rrfy"; {\ar "rrfx"; "rrfy"};
(100,0)*{\Ext^2(M_1,N)}="rrfz"; {\ar "rrfy"; "rrfz"};
(120,0)*{\cdots}
\endxy
\]

\item Given two short exact sequences and morphisms making each
square in the following diagram commute:
\[
\xy
(0,20)*{0}="i1";
(20,20)*{M_1}="x1"; {\ar "i1"; "x1"};
(40,20)*{M_2}="y1"; {\ar "x1"; "y1"};
(60,20)*{M_3}="z1"; {\ar "y1"; "z1"};
(80,20)*{0}="t1"; {\ar "z1"; "t1"};
(0,0)*{0}="i2";
(20,0)*{M'_1}="x2"; {\ar "i2"; "x2"};
(40,0)*{M'_2}="y2"; {\ar "x2"; "y2"};
(60,0)*{M'_3}="z2"; {\ar "y2"; "z2"};
(80,0)*{0}="t2"; {\ar "z2"; "t2"};
{\ar "x1"; "x2"};
{\ar "y1"; "y2"};
{\ar "z1"; "z2"};
\endxy
\]

we get the following commutative diagram, for each $k\geq 0$:

\[
\xy
(0,20)*{\Ext^k(M_1,N)}="a";
(50,20)*{\Ext^{k+1}(M_3,N)}="b";
(0,0)*{\Ext^k(M'_1,N)}="c";
(50,0)*{\Ext^{k+1}(M'_3,N)}="d";
{\ar "c"; "a"}; {\ar "d"; "b"}; {\ar "a"; "b"};
{\ar "c"; "d"};
\endxy
\]

\end{enumerate}
\end{thm}

\noindent {\bf Exercise.} Write the statement of the analogous theorem for
$\Ext^k(M,-)$ functors.\\

\PP{19.2}{$\Tor$ functors}
%-------------------------
Let us again write in detail the construction of left derived functors of
a right exact, covariant functor, in the case of $T_N = -\otimes_R N$.\\

\begin{defn}
Given $M,N\in\Rmod{R}$, the $R$--module $\Tor_k^R(M,N)$ is defined as
follows.
\begin{itemize}
\item Choose a projective resolution of $M$:
\[
\cdots \morph{f_2} P_2\morph{f_1} P_1\morph{f_0} P_0\to 0.
\]
\item Tensor with $N$ to get a (chain) complex:
\[
\cdots \morph{d_2} P_2\otimes N \morph{d_1} P_1\otimes N\morph{d_0}
P_0\otimes N \to 0,
\]
where $d_j = f_j\otimes\Id_N$.
\item Take its $k^{\sth}$ homology:
\[
\boxed{
\Tor_k^R(M,N) := \Ker(d_{k-1})/\Im(d_k)
}
\]
\end{itemize}
\end{defn}

Note that we may as well take a projective resolution of $N$ and tensor
(on the left) with $M$. The answer will be the same since $\otimes$ is commutative.\\

\begin{thm}
Let $N\in\Rmod{R}$. Then we have a sequence of additive, covariant functors:
\[
\Tor_k^R(-,N) : \Rmod{R}\to \Rmod{R},\qquad
k\in\Z_{\geq 0}.
\]

\begin{enumerate}
\item $\Tor_0^R(-,N) = -\otimes_R N$.\\

\item If $P$ is projective, then $\Tor_k^R(P,N)=0$, for every $k\geq 1$.\\

\item For every short exact sequence $0\to M_1\to M_2\to M_3\to 0$,
we have connecting morphisms $\delta_k : \Tor_{k+1}(M_3,N)
\to \Tor_k(M_1,N)$, for every $k\geq 0$, such that the following
sequence is exact:

\[
\xy
(-10,40)*{\cdots}="i";
(20,40)*{\Tor_2(M_1,N)}="fx"; {\ar "i"; "fx"};
(60,40)*{\Tor_2(M_2,N)}="fy"; {\ar "fx"; "fy"};
(100,40)*{\Tor_2(M_3,N)}="fz"; {\ar "fy"; "fz"};
{\ar (112,40); (120,40)};
(120,40); (120,30) **\crv{(130,40)&(130,30)};
{\ar@{-} (120,30); (0,30)}; (55,32)*{\scriptstyle{\delta_1}};
(0,30); (0,20) **\crv{(-10,30)&(-10,20)};
(20,20)*{\Tor_1(M_1,N)}="rfx"; {\ar (0,20); "rfx"};
(60,20)*{\Tor_1(M_2,N)}="rfy"; {\ar "rfx"; "rfy"};
(100,20)*{\Tor_1(M_3,N)}="rfz"; {\ar "rfy"; "rfz"};
{\ar (112,20); (120,20)};
(120,20); (120,10) **\crv{(130,20)&(130,10)};
{\ar@{-} (120,10); (0,10)}; (55,12)*{\scriptstyle{\delta_0}};
(0,10); (0,0) **\crv{(-10,10)&(-10,0)};
(20,0)*{M_1\otimes N}="rrfx"; {\ar (0,0); "rrfx"};
(60,0)*{M_2\otimes N}="rrfy"; {\ar "rrfx"; "rrfy"};
(100,0)*{M_3\otimes N}="rrfz"; {\ar "rrfy"; "rrfz"};
{\ar "rrfz"; (130,0)*{0}};
\endxy
\]

\item Given two short exact sequences and morphisms making each
square in the following diagram commute:
\[
\xy
(0,20)*{0}="i1";
(20,20)*{M_1}="x1"; {\ar "i1"; "x1"};
(40,20)*{M_2}="y1"; {\ar "x1"; "y1"};
(60,20)*{M_3}="z1"; {\ar "y1"; "z1"};
(80,20)*{0}="t1"; {\ar "z1"; "t1"};
(0,0)*{0}="i2";
(20,0)*{M'_1}="x2"; {\ar "i2"; "x2"};
(40,0)*{M'_2}="y2"; {\ar "x2"; "y2"};
(60,0)*{M'_3}="z2"; {\ar "y2"; "z2"};
(80,0)*{0}="t2"; {\ar "z2"; "t2"};
{\ar "x1"; "x2"};
{\ar "y1"; "y2"};
{\ar "z1"; "z2"};
\endxy
\]

we get the following commutative diagram, for each $k\geq 0$:

\[
\xy
(0,20)*{\Tor_{k+1}(M_3,N)}="a";
(50,20)*{\Tor_k(M_1,N)}="b";
(0,0)*{\Tor_{k+1}(M'_3,N)}="c";
(50,0)*{\Tor_k(M'_1,N)}="d";
{\ar "a"; "c"}; {\ar "b"; "d"}; {\ar "a"; "b"};
{\ar "c"; "d"};
\endxy
\]

\end{enumerate}
\end{thm}

\PP{19.3}{Examples of $\Ext$}
%---------------------------

\noindent {\bf I.} Let us take $R=\Z$, $M=\Z/m\Z$ and $N=\Z$. We compute
$\Ext^k_{\Z}(M,N)$ using the two methods:
\begin{enumerate}
\item Projective resolution of $M$: $\ds 0\to \Z\morph{\mu_m}\Z\to 0$,
where $\mu_m$ is multiplication by $m$. Applying $\Hom(-,N)$ turns it
into (using $\Hom(\Z,\Z)=\Z$)
$0\to \Z \morph{\mu_m} \Z\to 0$, whose cohomology gives us:
\[
\Ext^0(\Z/m\Z,\Z) = \{0\},\qquad
\Ext^1(\Z/m\Z,\Z) = \Z/m\Z,
\]
and $\Ext^k(M,N)=0$ for $k\geq 2$.\\

\item Injective resolution of $N$: $\ds 0\to \Q \morph{\pi} \Q/\Z \to 0$,
where $\pi$ is the natural surjection. Apply $\Hom(M,-)$ and use the
fact that $\Hom(\Z/m\Z,\Q)=\{0\}$ to get (the terms are in degrees
$-1,0,1,2$ respectively):
\[
0 \to 0 \to \Hom(\Z/m\Z,\Q/\Z) \to 0.
\]
Its cohomology gives exactly the same answer as before.
\end{enumerate}

\begin{rem}
As we will see next week, the notation $\Ext$ stands for {\em extensions}
and the relation can be seen in the previous example. Namely,
for every $\alpha\in\Z$, we have a short exact sequence:
\[
0\to \Z \morph{i} M_\alpha \morph{p} \Z/m\Z \to 0,
\]
where $M_\alpha$ is the abelian group generated by two elements,
$e_1,e_2$, subject to a relation: $me_2=\alpha e_1$. The morphisms
$i$ and $p$ are given by: $i(1)=e_1$ and $p(e_1)=0,
p(e_2)=\ol{1}$. It turns out that we can find an isomorphism
$M_\alpha\isom M_\beta$ which commutes with the maps $i$ and $p$,
if and only if $\alpha\equiv\beta$ (mod $m$).\\

In general, a short exact sequence $0\to A\to B\to C\to 0$ is also
called {\em an extension of $C$ by $A$}. Thus, there are $m$ (up to isomorphism)
extensions of $\Z/m\Z$ by $\Z$, which is reflected in the
fact that $\Ext^1(\Z/m\Z,\Z)=\Z/m\Z$.\\
\end{rem}

\noindent {\bf II.} $R=\Z$, $M=\Z/m\Z$, $N=\Z/n\Z$. Taking the projective
resolution of $M$ as above, and applying $\Hom(-,N)$, we get the
following complex:
\[
0\to \Z/n\Z \morph{\mu_m} \Z/n\Z \to 0.
\]
Now we can write down the cohomology of this complex as:
\[
\Ext^0(M,N) = \{x\in\Z/n\Z : mx \text{ is divisible by } n\}
\]
Note that this is same as $\Hom(\Z/m\Z,\Z/n\Z)$ as expected. Check:
there are $\gcd(m,n)$ many elements in this hom set.
\[
\Ext^1(M,N) = \Z/(m,n) = \Z/\gcd(m,n)\Z.
\]

\PP{19.4}{Example of $\Tor$}
%----------------------------
The notation $\Tor$ stands for {\em torsion} and is explained by
the following example.
Let $R$ be an integral domain, and let $a\in R$ be a non--zero
element. Let $M = R/(a)$ and $N\in\Rmod{R}$ arbitrary. We begin by writing
down a projective resolution of $M$: $\ds 0\to R\morph{\mu_a} R\to 0$.
Tensoring with $N$, and remembering $R\otimes_R N \cong N$, we get
\[
0\to N\morph{\mu_a} N \to 0,\qquad \text{in degrees } 2,1,0,-1.
\]
Thus, we get:
\[
\Tor_0(M,N) = N/aN = (R/(a))\otimes_R N, \qquad
\Tor_1(M,N) = \{x\in N : ax=0\}.
\]
Now we can clearly see $\Tor_1(R/(a),N)$ consists of elements which
are $a$--{\em torsion} (meaning $n\in N$ so that $an=0$). This
explains the name {\em torsion modules} for $\Tor$'s.
\end{document}
