Why do we say that any short exact sequence of $R$-modules (for any ring $R$) $0 \to A \to B \to C \to 0$ has an invariant in $\text{Ext}^1(C,A)$?
I don't quite get what this means or how to prove this...
I'm really confused... Can anyone give me a hand?
Thank you so much!
If you're familiar with cohomology or homology at all, the idea of $\operatorname{Ext}$ is very similar (in fact, cohomology and $\operatorname{Ext}$ are two cases of the same derived functor construction). In algebraic topology, one associates to a topological space $X$ a series of abelian groups $\{H^n(X,\Bbb Z)\}_{n\in\Bbb N_0}$, which are related to each other in specific ways, and are functorial (e.g., given a continuous map $X\to Y$, one gets a map $H^n(Y,\Bbb Z)\to H^n(X,\Bbb Z)$). One can think of the elements of the $H^n(X,\Bbb Z)$'s as algebraic invariants of the space $X$; that is, if you have a homeomorphism of spaces $X\to Y$, the induced map on $H^n$ is an isomorphism.
The same thing happens with $\operatorname{Ext}$. It is also functorial, so an isomorphism of $R$-modules will induce an isomorphism of $\operatorname{Ext}$ groups. You might think of elements of $\operatorname{Ext}^1_R(C,A)$ as invariants of $A$ (or $C$). The statement you're asking about is a consequence of the existence of a bijection $$ \operatorname{Ext}^1_R(C,A)\leftrightarrow\{\textrm{Extensions of }C\textrm{ by }A\} / \sim, $$ where an extension of $C$ by $A$ is precisely a short exact sequence $0\to A\to B\to C\to 0$. We want to take these modulo an equivalence relation where two extensions $0\to A\xrightarrow{f} B\xrightarrow{g} C\to 0$ and $0\to A\xrightarrow{f'} B'\xrightarrow{g'} C\to 0$ are equivalent if there exists a map $b : B\to B'$ such that $b\circ f = f'$ and $g'\circ b = g$.
So, given any exact sequence $0\to A\to B\to C\to 0$, one obtains an element $\sigma\in\operatorname{Ext}^1_R(C,A)$ by looking at the element of $\operatorname{Ext}^1_R(C,A)$ corresponding to the equivalence class of the exact sequence $0\to A\to B\to C\to 0$.
To get your element of $\operatorname{Ext}^1_R(C,A)$, take a projective resolution $X^\bullet\to C$ of $C$ (i.e., an exact sequence $\dots\to X^2\to X^1\to X^0\to C\to 0$, where each $X^i$ is projective). Then homological algebra tells you have maps $X^1\to A$ and $X^0\to B$ such that the following diagram commutes: $$\require{AMScd} \begin{CD} X^2 @>>> X^1 @>>> X^0 @>>> C @>>> 0\\ @VVV @VVV @VVV @VVV \\ 0 @>>> A @>>> B @>>> C @>>> 0. \end{CD}$$ You can check that the map $X^1\to A$ is a cocycle, and that its corresponding class in $\operatorname{Ext}^1_R(C,A)$ is independent of any choices made, which gives you your desired invariant. You can find more details here, or in your favorite book on homological algebra.