Suppose $M$ is a line bundle on a smooth projective curve $C$ and $N$ is a vector bundle on $C$. Suppose the degree of $M$ is sufficiently big so that ${\rm Hom}(M, N)=0$ and ${\rm Ext}^1(N, M)=0$. Suppose $x$ is a point of $C$. Question: what are extensions of the form $0 \to M \oplus N \to X \to {\mathcal O}\_x \to 0$ parametrized by, where $X$ has the form $M\_1 \oplus N$ such that $M\_1$ is a line bundle of degree larger by $1$ than the degree of $M$? Are such extensions parametrized by ${\rm Ext}^1({\mathcal O}\_x, M)$ or by something else? 



Parametrization of extensions of vector bundles
Since in the map $M\oplus N\to M\_1\oplus N$, one has no non-zero maps from $M\to N$, one must have an exact sequence $0\to M\to M\_1\to O\_x\to 0$. The rest should be easy.


This is just an elaboration of the above asked by the OP. Given an exact sequence $0\to M\oplus N\to M\_1\oplus N\to O\_x\to 0$ as in the question, we see that the left map $M\to M\_1\oplus N$ is given by $(i,0)$ and $N\to M\_1\oplus N$ by $(\alpha,\beta)$. Then it is clear that $\beta$ must be an isomorphism. Then, take the isomorphism $M\_1\oplus N\to M\_1\oplus N$, by $(Id\_{M\_1}, \beta^{-1}) $ and then one sees that the original sequence is equivalent to one given by $M\to M\_1\oplus N$ by $(i,0)$ and $N\to M\_1\oplus N$ by $(\alpha, Id\_N)$, (possibly for a different $\alpha)$. Now, one can take the isomorphism $M\_1\oplus N$ to itself by $M\_1\to M\_1\oplus N$, given by $(Id\_{M\_1}, 0)$ and $N\to M\_1\oplus N$ by $(-\alpha, Id\_N)$ and then you get an equivalent extension where $M\to M\_1\oplus N$ is $(i,0)$ and $N\to M\_1\oplus N$ is $(0,Id\_N)$ and the rest should be clear.


