TITLE: Elementary $\mathrm{Ext}^1$ intuition
QUESTION [17 upvotes]: $\DeclareMathOperator{\Hom}{\operatorname{Hom}}\DeclareMathOperator{\Ext}{\operatorname{Ext}}$I am wondering what sort of basic basic intuitive meaning $\Ext^1(M,N)$ has.
As a base case: if $M$ and $N$ are say, (finite-dimensional) vector spaces (with a compatible group/algebra action), and $M$ and $N$ are indecomposable inequivalent (so $\Hom(M,N)={0}$), can I somehow conclude that $\Ext^1(M,N)$ is zero?
All I can get from Weibel/Wikipedia is that $\Ext^1(M,N)$ is a group under the Baer sum operation, and is in bijection with the set of solutions $X$ to the short exact sequence $0\to N\to X\to M\to 0$.  I don't know how to use this second meaning, but it seems the most hands-on.
Full disclosure: If this sounds like a homework exercise, it (almost) was -- past tense.  Although, the problem/text/instructor had no desire for use of $\Ext$ or $\Hom$, I just want to know how to use these functors (better).  I could give character references (even from some past Berkeley grads) to allay fears....
I would be happy to have a good reference to look this up myself.  I hear Rotman's first book was good, but I've only read negative responses to the new edition (and the old one isn't for sale anywhere I've seen), and Weibel is apparently too abstract for me, in some way.  I'll post a separate question for that, in fact.

REPLY [5 votes]: It is fruitful to think of Ext^1 as a generalization of Hom.  If A and B are objects of an abelian category, the set Hom(A,B) has the structure of an abelian group.  Here is one way to define the group law: any two maps A --> B induce a map A x A --> B x B.  This induces a sequence
A --> A x A --> B x B --> B

where the first map is the diagonal and the last is the codiagonal (the addition map).  The composition is the sum.  (What we are using here is that finite products and finite direct sums coincide in an abelian category; in fact, this can be taken as the definition of an abelian category.)
We have used three properties of Hom to construct the addition map: Hom is covariant in the second variable, contravariant in the first, and comes with a canonical map 
Hom(A,B) x Hom(A',B') --> Hom(A x A', B x B')

(plus some compatibilities which I have suppressed).  If we define Ext^1(A,B) to be the category of extensions 
0 --> B --> E --> A --> 0

we see that it also has these properties (using base change for the variable A and pushout for the variable B).  Thus we get a group law on the category Ext^1(A,B).  Passing to isomorphism classes gives the usual group law on the set Ext^1(A,B).
The zero object of Ext^1(A,B) also corresponds nicely to the zero object of Hom(A,B).  In the latter case it is constructed by pushing forward the unique map A --> 0 via the unique map 0 --> B.  For Ext^1(A,B), we push forward the unique extension of A by 0 
0 --> 0 --> A --> A --> 0

by the unique map 0 --> B and get
0 --> B --> B x A --> A --> 0.

So an extension of A by B is zero if and only if it admits a splitting.  Thus in the category of vector spaces over a field, Ext^1(A,B) is always zero because can always be split.