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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 $n$.

How to compute $\mathrm{Ext}^1_A(A/I,A/J)$ in terms of $\nu$ and $\mu$?

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.

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 $n$.

How to compute $\mathrm{Ext}^1_A(A/I,A/J)$ in terms of $\nu$ and $\mu$?

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 $n$.

How to compute $\mathrm{Ext}^1_A(A/I,A/J)$ in terms of $\nu$ and $\mu$?

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.

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Michael Hardy
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How to compute this $Ext^1$$\mathrm{Ext}^1$?

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...\geq\nu_{\ell}\geq0)$$\nu=(\nu_1\geq\nu_2\geq\cdots\geq\nu_{\ell}\geq0)$, $\mu=(\mu_1\geq\mu_2\geq...\geq\mu_m\geq0)$$\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},...,x^{\ell - 1}y^{\nu_{\ell}},x^{\ell})$$I=(y^{\nu_1},xy^{\nu_2},x^2y^{\nu_3},\ldots,x^{\ell - 1}y^{\nu_\ell},x^\ell)$ and $J=(y^{\mu_1},xy^{\mu_2},x^2y^{\mu_3},...,x^{m-1}y^{\mu_{m}},x^{m})$$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 $n$.

How to compute $Ext^1_A(A/I,A/J)$$\mathrm{Ext}^1_A(A/I,A/J)$ in terms of $\nu$ and $\mu$?

How to compute this $Ext^1$?

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 $\nu=(\nu_1\geq\nu_2\geq...\geq\nu_{\ell}\geq0)$, $\mu=(\mu_1\geq\mu_2\geq...\geq\mu_m\geq0)$ be partitions of the same $n\in\mathbb{N}$. Consider the ideals $I=(y^{\nu_1},xy^{\nu_2},x^2y^{\nu_3},...,x^{\ell - 1}y^{\nu_{\ell}},x^{\ell})$ and $J=(y^{\mu_1},xy^{\mu_2},x^2y^{\mu_3},...,x^{m-1}y^{\mu_{m}},x^{m})$ 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 $n$.

How to compute $Ext^1_A(A/I,A/J)$ in terms of $\nu$ and $\mu$?

How to compute this $\mathrm{Ext}^1$?

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 $n$.

How to compute $\mathrm{Ext}^1_A(A/I,A/J)$ in terms of $\nu$ and $\mu$?

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How to compute this $Ext^1$?

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 $\nu=(\nu_1\geq\nu_2\geq...\geq\nu_{\ell}\geq0)$, $\mu=(\mu_1\geq\mu_2\geq...\geq\mu_m\geq0)$ be partitions of the same $n\in\mathbb{N}$. Consider the ideals $I=(y^{\nu_1},xy^{\nu_2},x^2y^{\nu_3},...,x^{\ell - 1}y^{\nu_{\ell}},x^{\ell})$ and $J=(y^{\mu_1},xy^{\mu_2},x^2y^{\mu_3},...,x^{m-1}y^{\mu_{m}},x^{m})$ 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 $n$.

How to compute $Ext^1_A(A/I,A/J)$ in terms of $\nu$ and $\mu$?