Conditions for null-homotopic chain map in a specific example I am trying to calculate the ring $Ext^{\bullet}_R(k,k)$ where $R=k[x,y]/(xy)$ and $k$ is regarded as an $R$-module via $x$ and $y$ acting as zero. I thought I was done but then to my demise I found a hole in my argument. The way I'm going about it is by identifying $Ext^n_R(k,k)$ with chain maps $P_\bullet\longrightarrow P_\bullet[-n]$ modulo null-homotpy, where $P_\bullet$ is the following projective resolution of $k$.
$$\dots\overset{\phi}\rightarrow R\oplus R\overset{\theta}\rightarrow R\oplus R\overset{\phi}\rightarrow R\oplus R\overset{\psi}\rightarrow R\rightarrow k\rightarrow 0$$ where $\phi$ and $\theta$ alternate, $\phi:(1,0)\mapsto (y,0),\ (0,1)\mapsto (0,x)$, $\theta:(1,0)\mapsto (x,0),\ (0,1)\mapsto (0,y)$, $\psi:(1,0)\mapsto x,\ (0,1)\mapsto y$
I am going to describe my question just for the case where $n$ is odd. Let $f_\bullet:P_\bullet\longrightarrow P_\bullet [-n]$ be a chain map. Then have $$f_n(1,0)y=f_{n+1}^{(1)}(1,0)x+f_{n+1}^{(2)}(1,0)y\ \ \ \ f_n(0,1)x=f_{n+1}^{(1)}(0,1)x+f_{n+1}^{(2)}(0,1)y$$ $$(x,x)f_{n+i}(1,0)=(y,x)f_{n+i+1}(1,0)\ \ \ \ (y,y)f_{n+i}(0,1)=(y,x)f_{n+i+1}(0,1)\ \ \ \text{for $i\geq 1$ odd}$$ $$(y,y)f_{n+i}(1,0)=(x,y)f_{n+i+1}(1,0)\ \ \ \ (x,x)f_{n+i}(0,1)=(x,y)f_{n+i+1}(0,1)\ \ \ \text{for $i\geq 1$ even} $$ Now $h_\bullet$ is a null-homotpy of $f_\bullet$ iff $$f_n(1,0)=h_{n-1}(1,0)x+h_n^{(1)}(1,0)x+h_n^{(2)}(1,0)y\ \ \ f_n(0,1)=h_{n-1}(0,1)y+h_n^{(1)}(0,1)x+h_n^{(2)}(0,1)y$$ $$f_{n+i}(1,0)=(y,y)h_{n+i-1}(1,0)+(y,x)h_{n+i}(1,0)\ \ \ f_{n+i}(0,1)=(x,x)h_{n+i-1}(0,1)+(y,x)h_{n+i}(0,1)\ \text{for $i\geq 1$ odd}$$ $$f_{n+i}(1,0)=(x,x)h_{n+i-1}(1,0)+(x,y)h_{n+i}(1,0)\ \ \ f_{n+i}(0,1)=(y,y)f_{n+i-1}(0,1)+(x,y)h_{n+i}(0,1)\ \text{for $i\geq 1$ even}$$ We see that if $f_\bullet$ is null-homotopic then we have the following conditions (C):$$f_n(1,0),\ f_n(0,1)\in <x,y>\vartriangleleft R$$ $$f_{n+i}(1,0)\in <(y,y),(y,x)>\vartriangleleft R\oplus R\ \ \ f_{n+i}(0,1)\in <(x,x),(y,x)>\vartriangleleft R\oplus R\ \text{for $i\geq 1$ odd}$$ $$f_{n+i}(1,0)\in <(x,x),(x.y)>\vartriangleleft R\oplus R\ \ \ f_{n+i}(0,1)\in <(y,y),(x,y)>\vartriangleleft R\oplus R\ \text{for $i\geq 1$ even}$$ At first I thought I showed that conditions (C) are also sufficient for $f_\bullet$ to be null-homotopic but I found an error that I cannot fix and I'm sensing that these conditions might actually not be sufficient. How can I come up with sufficient conditions for $f_\bullet$ to be null-homotopic?
Also if these conditions are in fact sufficient (and the corresponding ones for the case $n$ is even) I get that the ring in question is $R$
 A: Approach 1: minimal resolution: consider a building a resolution of $\mathbb k$ by hand, which you can choose to be a periodic Koszul resolution, as you did:
$$\xrightarrow{\phantom{m}\cdots\phantom{m}} R^2 \xrightarrow{\left[\begin{matrix} 0 & y\\x & 0 \end{matrix}
\right]} R^2 \xrightarrow{\left[\begin{matrix} 0 & y\\x & 0 \end{matrix}
\right]}  R^2 \xrightarrow{\left[\begin{matrix} x \\y \end{matrix}
\right]}  R \longrightarrow 0 $$
This resolution is minimal, in the sense that taking $\mathrm{Hom}$
with $\mathbb k$ yields a zero differential. Since the resulting complex is just
$$\xrightarrow{\phantom{m}0\phantom{m}} \mathbb k^2 \xrightarrow{\phantom{m}0\phantom{m}} \mathbb k^2 \xrightarrow{\phantom{m}0\phantom{m}} \mathbb k^2 \xrightarrow{\phantom{m}0\phantom{m}} \mathbb k\longrightarrow 0 $$
this immediately tells you that $\mathrm{Ext}_R^n(\mathbb k,
\mathbb k)$ is two dimensional for all $n\geqslant 2$. Moreover,
you can show that the augmentation $\varepsilon:P\longrightarrow \mathbb k$ induces a quasi-iso $\varepsilon_*:\mathrm{End}(P) \to \mathrm{Hom}(P,\mathbb k)$, so you don't have to compute the homology of the endomorphism complex by hand.
Approach 2: Koszul duals.
There is a simpler approach to computing $\mathrm{Ext}$ in this case.
Namely, if you consider $A = \mathbb k[x,y]/(xy)$, then this is a monomial associative algebra presented by $x$ and $y$ modulo two relations $xy=0$ and $yx=0$ (note that $xy-yx=0$ can be replaced by the second once since $xy=0$). That is, $A= \mathbb k\langle x,y\rangle / (xy,yx)$.
In this case, the normal forms (things not divisible by $xy$ or $yx$) are $1$ and $x^n,y^n$ for $n\geqslant 1$, so the kernel of multiplication by $x$ is the ideal generated by $y$, and the kernel of multiplication by $y$ is the ideal generated by $x$.
Having said this, quadratic monomial algebras are Koszul, the Koszul dual in this case is $\mathbb k\langle x,y\rangle/(x^2,y^2)$, since all binomials are $xy,yx,y^2,x^2$, and we merely need to pick those not appearing in the presentation of $A$. Hence, $\mathrm{Ext}_A(\mathbb k,\mathbb k)$ as an algebra is isomorphic to this algebra.
Approach 3: Anick chains: For monomial algebras, there is a completely combinatorial way to compute the Yoneda (co)algebra.
In this case, the monomial relations are $xy$ and $yx$, and the $n$-chains are equal to the two element set
$$C_n = \{ X_n = x(yx)^{n-1}, Y_n = y(xy)^{n-1}\}$$
for all $n\geqslant 1$. The Anick resolution then is of the form
$(R\otimes C_\bullet ,d)$ so that $R\otimes C_n = R^2$ for all $n\geqslant 1$ and the differential is $dX_n = x\otimes Y_{n-1}$ and $dY_n = y\otimes X_{n-1}$, i.e. this is isomorphic to the resolution you found, and the resolution given by Koszul duality (because one is dealing with quadratic monomial algebras here).
