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 <title>Ext</title>
 <name>Ext</name>
 <created>2003-08-14 20:33:06</created>
 <modified>2007-01-11 21:38:33</modified>
 <type>Definition</type>
 <creator id="2727" name="mathcam"/>
 <author id="2727" name="mathcam"/>
 <author id="988" name="bwebste"/>
 <classification>
	<category scheme="msc" code="18G15"/>
 </classification>
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 <content>
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 \PMlinkname{extensions}{GroupExtension} of $B$ by $A$ up to isomorphism of short exact sequences, where an \PMlinkname{extension}{GroupExtension} 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 \PMlinkname{extension}{GroupExtension} $0\to\Z\to\Z\oplus\Z/n\Z\to 0$, and $m\neq 0$ corresponding to 
$$\xymatrix{0 \ar [r] &amp; \Z \ar [r]^n &amp; \Z \ar [r]^m &amp; \Z/n\Z \ar [r]&amp;0}.$$

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

\begin{align*}
0\rightarrow B\rightarrow C_n\rightarrow C_{n-1}\rightarrow\cdots\rightarrow C_1\rightarrow A\rightarrow 0
\end{align*}</content>
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