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Elementary $Ext^1$$\mathrm{Ext}^1$ intuition

I$\DeclareMathOperator{\Hom}{\operatorname{Hom}}\DeclareMathOperator{\Ext}{\operatorname{Ext}}$I am wondering what sort of basic basic intuitive meaning $Ext^1(M,N)$$\Ext^1(M,N)$ has.
As a base case: if $M$ and $N$ are say, (finite-dimensional) vector spaces (with a compatible group/algebra action), and $M$ and $N$ are indecomposable inequivalent (so $Hom(M,N)={0}$$\Hom(M,N)={0}$), can I somehow conclude that $Ext^1(M,N)$$\Ext^1(M,N)$ is zero?

All I can get from Weibel/Wikipedia is that $Ext^1(M,N)$$\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$$\Ext$ or $Hom$$\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.

Elementary $Ext^1$ intuition

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.

Elementary $\mathrm{Ext}^1$ intuition

$\DeclareMathOperator{\Hom}{\operatorname{Hom}}\DeclareMathOperator{\Ext}{\operatorname{Ext}}$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.

elementary Ext^1 Elementary $Ext^1$ intuition

I am wondering what sort of basic basic intuitive meaning Ext1(M,N)$Ext^1(M,N)$ has.
As a base case: if M$M$ and N$N$ are say, (finite-dimensional) vector spaces (with a compatible group/algebra action), and M$M$ and N$N$ are indecomposable inequivalent (so Hom(M,N)={0}$Hom(M,N)={0}$), can I somehow conclude that Ext1(M,N)$Ext^1(M,N)$ is zero?

All I can get from Weibel/Wikipedia is that Ext1(M,N)$Ext^1(M,N)$ is a group under the Baer sum operation, and is in bijection with the set of solutions {X}$X$ to the short exact sequence 0 --> N --> X --> M --> 0$0\to N\to X\to M\to 0$. 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$Ext$ or Hom$Hom$, 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.

elementary Ext^1 intuition

I am wondering what sort of basic basic intuitive meaning Ext1(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 Ext1(M,N) is zero?

All I can get from Weibel/Wikipedia is that Ext1(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 --> N --> X --> M --> 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.

Elementary $Ext^1$ intuition

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.

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