Let $A$ be a regular local $\mathbb{C}$-algebra of dimension $2$, such as the localization of $\mathbb{C}[x,y]$ at $(x,y)$, and let $\nu=(\nu\_1\geq\nu\_2\geq\cdots\geq\nu\_{\ell}\geq0)$, $\mu=(\mu\_1\geq\mu\_2\geq\cdots\geq\mu\_m\geq0)$ be partitions of the same $n\in\mathbb{N}$. Consider the ideals $I=(y^{\nu\_1},xy^{\nu\_2},x^2y^{\nu\_3},\ldots,x^{\ell - 1}y^{\nu\_\ell},x^\ell)$ and $J=(y^{\mu\_1},xy^{\mu\_2},x^2y^{\mu\_3},\ldots,x^{m-1}y^{\mu\_m},x^m)$ generated by elements in the maximal ideal (here $x,y$ are the uniformizers). So, $A/I$ and $A/J$ have the same length equal to $n$.



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> How to compute $\mathrm{Ext}^1\_A(A/I,A/J)$ in terms of $\nu$ and $\mu$?
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Edit: I would add that it's probably possible to make the computation using the theory of ADHM data developped e.g. in Nakajima's *Lectures on Hilbert schemes of points on surfaces*. But I would like to get a direct answer that does not use ADHM data and such.



How to compute this $\mathrm{Ext}^1$?
First off, Ext and localisation commute in situations like this (Weibel *An introduction to homological algebra* Proposition 3.3.10 p.76) so we can work with $A=k[x,y]$ if we like. I will drop the assumption that the two partitions have the same size, not least because you get a nice interpretation of $\operatorname{Ext}^1$ when the first partition is (1): it has a basis indexed by addable nodes of the second partition.


Let 
$\lambda=\lambda\_1^{a\_1} \lambda\_2^{a\_2}\cdots$
be a partition of $n$, where
$\lambda\_1>\lambda\_2>\cdots> \lambda\_l >0$.
A minimal $A$-projective resolution of the module you associate to $\lambda$ is
$$0 \to \bigoplus\_{i=1}^l A \otimes (x^{a\_i}\otimes x^{\sum\_{j=0}^{i-1}a\_j } y^{\lambda\_i}-y^{\lambda\_i-\lambda\_{i+1}} \otimes x^{\sum\_{j=0}^i a\_j}y^{\lambda\_{i+1}})
\to \bigoplus \_{i=0}^l A\otimes x^{\sum\_{j=0}^i a\_j}y^{\lambda\_{i+1}}
\to A \to A/I\_\lambda \to 0$$
in which you should interpret $\lambda\_{l+1}$ as zero.


This leads to a direct but awkward description of $\operatorname{Ext}^1\_A(A/I\_\lambda, A/I\_\mu)$ as follows. Consider the Ferrers diagram of $\mu$ drawn in the French style: by this I mean the subset of $\mathbb{Z}\times \mathbb{Z}$ consisting of $(0,0), (0,1), \ldots, (0,m\_1)$, $(1,0), (1,1), \ldots, (1,m\_2)$ and so on, where $\mu=(m\_1,m\_2,\ldots)$.
Pick $0 \leq r \leq s \leq l$ and choose elements $\rho\_s,\ldots,\rho\_s$ of the Ferrers diagram of $\mu$ with the property that 


$\rho\_r + (0,\lambda\_r-\lambda\_{r-1}) \notin \mu$, 


$\rho\_j + (a\_{j+1},0)=\rho\_{j+1} + (0,\lambda\_{j+1}-\lambda\_j) \in \mu$ for each $r<j<s$, and


$\rho\_s + (a\_{s+1},0) \notin \mu$.


The first condition should be dropped if $r=0$ and the last if $s=l$.


Thus these elements form the concave corners in a fragment of the outer rim of the Ferrers diagram for $\lambda$. The Ext-group is the free vector space on these "paths," modulo linear combintations of paths that arise from the entire outer rim of the Ferrers diagram for $\lambda$.


Things look a little nicer in the case $\lambda = (1)$. Then a path with $r=s=0$ is an element of the Ferrers diagram of $\mu$ with nothing directly to its right, a path wtih $r=s=1$ is an element with nothing directly above it, and every path with $r=0,s=1$ is a coboundary. Furthermore an $r=s=1$ path is a boundary if there's an element of the diagram to its left and a $r=s=0$ path is a boundary if there's an element of the diagram above it. It follows the non-bounding paths sit next to the addable nodes of $\mu$.


