Let $k$ be an algebraically closed field of characteristic $0$, and let $I \subset R = k[[x,y]]$ be a non-principal ideal in the power series ring in two variables such that $I \subset (x,y)$. Is it true that $\mathrm{Ext}^1(I,R) \neq 0$ for all $I$?
What I know: From the exact sequence $0 \to I \to R \to R/I \to 0$, it follows that $\mathrm{Ext}^1(I,R) = \mathrm{Ext}^2(R/I,R)$. In the case where $I$ has depth $2$ in $R$ (i.e., the length of a maximal regular sequence of $R$ using elements of $I$ is equal to $2$), it is stated in Matsumura's Commutative Ring Theory that $2$ is the infimum of all $i$ such that $\mathrm{Ext}^i(R/I,R) \neq 0$.
As for the depth-$1$ case, I'm not exactly sure how to handle it, but here is my idea. It seems like the only way that the ideal $I$ has depth $1$ is for every element of $I$ to have some nontrivial common factor, call it $f$. Then the ideal $J = (1/f) \cdot I \subset R$ is isomorphic as an $R$-module to $I$ and has depth $2$ in $R$, so $\mathrm{Ext}^2(R/I,R)= \mathrm{Ext}^2(R/J,R) = 0$ by the argument for the depth 2 case. Is there something amiss with this argument?
Any such ideal $I=fJ$ where $J$ has depth 2. $f$ is the gcd (makes sense in $R$ which is a UFD) of all elements in $I$. Since $I$ is isomorphic to $J$, we have $\mathrm{Ext}^1(I,R)$ is isomorphic to $\mathrm{Ext}^1(J,R)$ and thus you are done.