Preliminaries:
Given two modules $B,C$ we seek a module $A$ such that $B$ contains an isomorphic copy of $A$ such that resulting quotient module $B/A$ is isomorphic to $C$.
Clearly, $B$ contains an isomorphic copy of $A$ is same as saying that there is an injective homomorphism $\psi:A \rightarrow B$. This can be expressed as
\begin{equation}
A \equiv \psi(A) \subset B
\end{equation}
To say $C$ is isomorphic to quotient means that there is a surjective homomorphism $\phi: B\rightarrow C$ with $ker\;\phi=\psi(A)$.
This gives us a pair of homomorphisms
\begin{equation}
\label{eq:hom}
A \xrightarrow{\psi} B \xrightarrow{\phi} C
\end{equation}
such that $im\;\psi=ker\;\phi$.
These homorphisms such that above holds are known as "exact".
Examples:
Using direct sum of modules $A,C$ with $B=A \oplus C$, the following exact sequence can be constructed.
\begin{equation}
0 \rightarrow A \xrightarrow{i} A \oplus C \xrightarrow{\pi} C \rightarrow 0
\end{equation}
where $i(a)=(a,0)$ and $\pi(a,c) = c$. Notice that $pi \circ i=\pi(a,c)\circ(a,0)=\pi(a,0)=0$. Thus $\partial^2=0$ map is satisfied.
When $A=Z$ a $Z$ module with $C=Z/nZ$, above sequence becomes
\begin{eqnarray}
0 \rightarrow Z \xrightarrow{i} Z \oplus Z/nZ \xrightarrow{\pi} Z/nZ \rightarrow 0 \\
0 \rightarrow Z \xrightarrow{n} Z \xrightarrow{\pi} Z/nZ \rightarrow 0
\end{eqnarray}
If we consider $A=Z$ and $C=Z/nZ$, we can consider these as extension of $C$ by $A$.
For homomorphism $\phi$, we may form the following.
\begin{equation}
0 \rightarrow K \xrightarrow{i} F(S) \xrightarrow{\phi} M \rightarrow 0
\end{equation}
Here $\phi$ is a unique $R$-module homomorphism which is identity in $S$ - set of generators for $M$ an $R$-module.k
Saturday, August 8, 2020
Cohomolgy-Some Module theory
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