A duplicate of this question in MO: I'm reading this paper: Brochard, Iyengar and Khare: Wiles defect for modules and criteria for freeness. In lemma 4.5, there is an isomorphism $\operatorname{Ext}_A^1(k,A) \cong \frac{I_A}{\varpi I_A} $, but I cannot understand why it is true, any hints?
Edit: $A$ is a Gorenstein local ring with maximal ideal $\mathfrak{m}_A$, together with a surjection $\lambda$ to a DVR $\mathcal{O}$ with kernel $\mathfrak{p}_A$ and $I_A = Ann[\mathfrak{p}_A]$, also $\varpi$ is a uniformizing parameter for $\mathcal{O}$ . Then $R = \frac{A}{(x)}$, where $x$ is a nonzerodivisor of $A$ such that $\lambda (x)$ is a uniformizing parameter for $\mathcal{O}$. Clearly $\mathfrak{m}_A R$ is the maximal ideal of $R$ and $k$ its residue field.
Since I found an answer for this, I might as well share it: Applying the contravariant left-exact functor $\operatorname{Hom}_A(-,A)$ to the exact sequence $0 \rightarrow \mathcal{O} \rightarrow \mathcal{O} \rightarrow \mathcal{O}/(\varpi) = k \rightarrow 0$, where the first nontrivial map is multiplication by $\varpi$, gives rise to the long exact sequence
$ 0 \rightarrow \operatorname{Hom}_A(k,A) \rightarrow \operatorname{Hom}_A(\mathcal{O},A) \xrightarrow{\cdot \varpi} \operatorname{Hom}_A(\mathcal{O},A) \rightarrow \operatorname{Ext}_A^1(k,A) \rightarrow \operatorname{Ext}_A^1(\mathcal{O},A) \rightarrow ...$
Noticing that $\operatorname{Hom}_A(\mathcal{O},A) = I_A$ and that $\operatorname{Ext}_A^1(\mathcal{O},A)$ vanishes, since $A$ is Gorenstein, we get the desired result.