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\title{Lecture 4-24: Ext groups, module extensions, and group cohomology}
\date{April 24, 2024}
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We will show that Ext groups are well defined (independent of the choice of projective resolution) and derive some long exact sequences.  We then define group cohomology, using a particular projective resolution of a trivial $G$-module.
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We first prove
\begin{alertblock}{Proposition (DF, p. 782)}
Using the same notation as last time, given a homomorphism $f:A\rightarrow A'$ of $R$-modules, the induced maps $\phi_n:\,$Ext$_R^n(A',D)\rightarrow\,$Ext$_R^n(A,D)$ depend only on $f$, not on the choice of lifts $f_n$ used to define these maps.
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\begin{proof}
For this it is enough to show that if $f$ is the zero map, then the induced maps are all zero as well.  If $f=0$ then by projectivity of the $P_i$ one can inductively define homomorphisms $s_n:P_n\rightarrow P_{n+1}' $ such that $f_n=d_{n+1}'s_n+s_{n-1}d_n$.  The collection of maps $\{s_n\}$ is called a \alert{chain homotopy} between the chain homomorphism given by the $f_n$ and the zero homomorphism.  Taking homomorphisms into $D$ we get the diagram used to determine the $\phi_n$, together with additional diagonal arrows representing a chain homotopy between cochain complex homomorphisms.  Using these diagonal maps, it is easy to see that any element in $\hom_R(P_n',D)$ representing a coset in Ext$_R^n(A',D)$ maps to the zero coset in Ext$_R^n(A,D)$, as claimed.
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Now we can finally prove the independence result promised last time.  
\begin{alertblock}{Theorem (DF, p. 782)}
The groups Ext$_R^n(A,D)$ depend only on $A$ and $D$.
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Indeed, given two projective resolutions $\{P_i\},\{P_i'\}$ of $A$ and the lifts $f_n$ of the identity map $f:A\rightarrow A'=A$, the resulting maps $\phi_n$ from Ext$_R^n(A,D)$ (computed via the $P_i$) and Ext$_R^n(A',D)$ (computed via the $P_i'$) admit maps $\phi_n:\,$Ext$_R^n(A',D)\rightarrow\,$Ext$_R^n(A,D)$ such that the composites $\phi_n'\phi_n$ are induced from the identity map from $A$ to $A$.  These composites are also identity maps, by the previous result, and the assertion follows.
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Given a short exact sequence $0\rightarrow L\rightarrow M\rightarrow N\rightarrow 0$ of $R$-modules and projective resolutions of $L$ and $N$, say by $\{P_i\}$ and $\{P_i'\}$, respectively, it is not difficult to construct a projective resolution of $M$ by $\{Q_i=P_i\oplus P_i'\}$ (DF p. 783) such that the resolutions of $L,M,N$ fit into a short exact sequence of chain complexes.  Taking homomorphisms into another $R$-module $D$ and the corresponding long exact sequence in cohomology, we get a sequence
\[
0\rightarrow\hom_R(D,L)\rightarrow\hom_R(D,M)\rightarrow\hom_R(D,N)\raise .8ex\hbox{$\genfrac{}{}{0pt}{2}{\gamma_0} \longrightarrow$}\,\text{Ext}_R^1(D,L)
\]
\[
\rightarrow\text{Ext}_R^1(D,M)\rightarrow\,\text{Ext}_R^1(D,N)\rightarrow\cdots
\]
\noindent called the \alert{long exact sequence for Ext}.
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In particular, it follows from this sequence and the trivial projective resolution $\cdots\rightarrow0\rightarrow P\rightarrow P\rightarrow 0$ of any projective module $P$ that \alert{an $R$-module $P$ is projective if and only if Ext$_R^1(P,B)=0$ for all $R$-modules $B$, or if and only if Ext$_R^n(P,B)=0$ for all $R$-modules $B$ and all $n\ge1$}.  
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The notation Ext here stands for \lq\lq extension".  An \alert{extension} of an $R$-module $N$ by another one $L$ is a short exact sequence $0\rightarrow L\rightarrow M\rightarrow N\rightarrow0$ of $R$-modules.  Given two such extensions, say
\[
0\rightarrow L\rightarrow M\rightarrow N\rightarrow0
\]
\[ 
0\rightarrow L\rightarrow M'\rightarrow N\rightarrow0
\]
\noindent we say they are \alert{equivalent} if there is an isomorphism $f:M\rightarrow M'$ making the above diagram commute when supplemented by the identity maps from $L$ and $N$ to themselves (DF p. 381).
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For example, taking $R=\mathbb Z$ and letting $p$ be a prime, there is just one equivalence class of extensions of $L=\mathbb Z/(p)$ by itself with middle term isomorphic to $E=\mathbb Z/(p)\oplus\mathbb Z/(p)$, since the automorphism group of $E$, consisting of all $2\times 2$ invertible matrices over $\mathbb Z/(p)$, is large enough to identify any two such extensions.  On the other hand, there are $p-1$ inequivalent extensions of $L$ with middle term $E'=\mathbb Z/(p^2)$, for if we fix the embedding of $L$ into $E'$ sending the coset of $i$ mod~$p$ to that of $pi$ mod~$p^2$, then there are $p-1$ inequivalent ways to map $E'$ onto $L$ so as to complete the short exact sequence.  These are indexed by the nonzero elements $i$ of $L$, the $i$th one sending the coset of $x$ to that of $ix$ mod~$p$.  
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Thus there are in all $p$ inequivalent extensions of $L$ by itself.  It is no coincidence that Ext$_{\mathbb Z}^1(L,L)$ is cyclic of order $p$ (by a previous calculation), as it is known in general that \alert{there is a bijection between Ext$_R^1(N,L)$ and equivalence classes of extensions of $N$ by $L$} (Theorem 12, DF, p. 787).  In this bijection the zero element of the Ext group corresponds to the \alert{split extension} $L\oplus N$.  More generally, for any $n\ge1$ there is a notion of equivalence class of $n$-extensions of $N$ by $L$, that is, equivalence classes of exact sequences $0\rightarrow L\rightarrow M_n\cdots\rightarrow M_1\rightarrow N\rightarrow0$ and a bijection due to Yoneda between the $n$th Ext group Ext$_R^n(N,L)$ and such extensions.  There is also the structure of an additive group on equivalence classes of extensions corresponding to the additive group structure on Ext$^1$.
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We conclude with a quick definition of group cohomology, so that you can get started on one of the problems in this week's homework.  Given a finite group $G$, let $F=\mathbb ZG$ be its \alert{integral group ring}, defined in the same way as the group algebra $\mathbb CG$, except that integers rather than complex numbers are used as coefficients.  Then an $F$-module is just an abelian group $A$ (or $\mathbb Z$-module) on which $G$ acts by automorphisms.  There is a standard resolution of $\mathbb Z$, the trivial $G$-module, with $n$th term $F_n = \otimes^{n+1}F$, the $(n+1)$st tensor power of $F$, which is a free $F$-module with basis $1\otimes g_1\otimes\cdots\otimes g_n$ as the $g_i$ range over $G$.   Here $G$ acts by multiplication on the leftmost copy of $F$.
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Denoting this basis element by $(g_1,\ldots,g_n)$ we have the following recipe for the $n$th coboundary operator $d_n$ on $F_n$:
\begin{align*}
d_n(g_1,\ldots,g_n) =\,&g_1\cdot (g_2,\ldots,g_n)\\
                                              &+\sum\limits_{i=1}^{n-1} (-1)^i(g_1,\ldots,g_{i-1},g_ig_{i+1},\ldots,g_n)\\
                                              &+(-1)^n(g_1,\ldots,g_{n-1})
 \end{align*}
In Exercise 1 on p. 809 of this week's HW you will use a {\sl contracting homomorphism}  and the {\sl augmentation map} defined there to show that the sequence $\cdots\rightarrow F_n\rightarrow F_{n-1}\rightarrow\cdots\rightarrow F_0\rightarrow \mathbb Z\rightarrow 0$ is exact, so that this is indeed a projective resolution of $\mathbb Z$, which can be used to compute Ext$_F^n(\mathbb Z,A)$.  This last group is by definition the $n$th cohomology group $H^n(G,A)$.  More to come!
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