If we consider the ideal $I=(x,y)$ as a module over the ring $\mathbb{Q}[x,y]$, how can we find a free resolution of $I$?
That is, a sequence $$...P_2 \to P_1 \to P_0 \to I \to 0$$ with $P_i$ free and the sequence being exact. I dont know how one in general constructs free resolution so I have just been brute-forcing. Any hints?
$(x,y)$ is a regular sequence in $R=\mathbf Q[x,y]$ , hence the Koszul complex: \begin{alignat*}{3}0\longrightarrow R&\xrightarrow{\begin{bmatrix}x\!&\!\!y\end{bmatrix}} R^2&\xrightarrow{\smash[t]{\begin{bmatrix}-y\\x\end{bmatrix}}}&R\longrightarrow R/(x,y)\longrightarrow 0 \\ t&\longmapsto \rlap{(xt, yt)}\\ &(u,v)&\longmapsto & -uy+vx \end{alignat*} is a free resolution of $R/(x,y)$, hence you have a free resolution of the kernel of the last non-zero map: $$ 0\longrightarrow R\xrightarrow{\begin{bmatrix}x\! &\!\!y\end{bmatrix}} R^2\xrightarrow {\smash[t]{\begin{bmatrix}-y\\x\end{bmatrix}}}(x,y)\longrightarrow 0 $$