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Revision difference : Ext
Version 5 Version 4
For a ring $R$, and $R$-module $A$, we have a covariant functor $\Hom A-R$. The functions $\Ext^n_R(A,-)$ are defined to be the right derived functors of $\Hom A-R: \Ext^n_R(A,-)=R^n\Hom A-R$. For a ring $R$, and $R$-module $A$, we have a covariant functor $\Hom A-R$. The functions $\Ext^n_R(A,-)$ are defined to be the right derived functors of $\Hom A-R: \Ext^n_R(A,-)=R^n\Hom A-R$.
$\Ext$ gets its name from the existence of a natural bijection between elements of $\Ext^1_R(A,B)$ and extensions of $B$ by $A$ up to isomorphism of short exact sequences, where an extension of $B$ by $A$ is an exact sequence $$0\to B\to C\to A\to 0.$$ For example, $$\Ext^1_\Z(\Z/n\Z,\Z)\cong\Z/n\Z,$$ with $0$ corresponding to the trivial extension $0\to\Z\to\Z\oplus\Z/n\Z\to 0$, and $m\neq 0$ corresponding to $\Ext$ gets its name from the existence of a natural bijection between elements of $\Ext^1_R(A,B)$ and extensions of $B$ by $A$ up to isomorphism of short exact sequences, where an extension of $B$ by $A$ is an exact sequence $$0\to B\to C\to A\to 0.$$ For example, $$\Ext^1_\Z(\Z/n\Z,\Z)\cong\Z/n\Z,$$ with $0$ corresponding to the trivial extension $0\to\Z\to\Z\oplus\Z/n\Z\to 0$, and $m\neq 0$ corresponding to
$$\xymatrix{0 \ar [r] & \Z \ar [r]^n & \Z \ar [r]^m & \Z/n\Z \ar [r]&0}.$$ $$\xymatrix{0 \ar [r] & \Z \ar [r]^n & \Z \ar [r]^m & \Z/n\Z \ar [r]&0}.$$
A more modern interpretation of the $\Ext$ functors was given by S. Mac Lane, namely that there is a correspondence between $\Ext^n(A,B)$ with equivalence classes of exact sequences A more modern interpretation of the $\Ext$ functors was given by S. Maclane, namelt that there is a corresopndence between $\Ext^n(A,B)$ with equivalence classes of exact sequences
\begin{align*} \begin{align*}
0\rightarrow B\rightarrow C_n\rightarrow C_{n-1}\rightarrow\cdots\rightarrow C_1\rightarrow A\rightarrow 0 0\rightarrow B\rightarrow C_n\rightarrow C_{n-1}\rightarrow\cdots\rightarrow C_1\rightarrow A\rightarrow 0
\end{align*} \end{align*}