Let \(R = k[x,y]\) where \(k\) is a field, and let \(I=(x,y)R\).
Show that \[\begin{align*} 0 \to R \mapsvia{\phi} R \oplus R \mapsvia{\psi} R \to k \to 0 \end{align*}\] where \(\phi(a) = (-ya,xa)\), \(\psi((a,b)) = xa+yb\) for \(a,b \in R\), is a projective resolution of the \(R\)-module \(k \simeq R/I\).
Show that \(I\) is not a flat \(R\)-module by computing \(\Tor_i^R(I,k)\)
Solution
Write \(\varepsilon:R\to k=R/(x,y)\) for the quotient map.
First consider \[ 0\longrightarrow R\xrightarrow{\phi}R^2\xrightarrow{\psi}R\xrightarrow{\varepsilon}k\longrightarrow0, \] with \[ \phi(a)=(-ya,xa),\qquad \psi(a,b)=xa+yb. \] All three nonzero modules in the resolution are free, hence projective. We verify exactness.
The map \(\phi\) is injective: if \((-ya,xa)=(0,0)\), then \(xa=0\); since \(R=k[x,y]\) is a domain and \(x\neq0\), we get \(a=0\).
Also \(\psi\phi=0\), because \[ \psi(-ya,xa)=-xya+yxa=0. \] Conversely, suppose \((a,b)\in\ker\psi\). Then \[ xa=-yb. \] Since \(x\) is prime in the UFD \(k[x,y]\) and \(x\nmid y\), we have \(x\mid b\); write \(b=xc\). Substituting gives \(xa=-yxc\), hence \(a=-yc\). Therefore \[ (a,b)=(-yc,xc)=\phi(c), \] so \(\ker\psi=\operatorname{im}\phi\).
Finally, \[ \operatorname{im}\psi=(x,y)=I=\ker\varepsilon, \] and \(\varepsilon\) is surjective. Thus the displayed complex is a projective resolution of \(k\).
For the Tor computation of \(I\), truncate the resolution at \(I=\operatorname{im}\psi\): \[ 0\longrightarrow R\xrightarrow{\phi}R^2\xrightarrow{\psi}I\longrightarrow0. \] This is a free resolution of \(I\). Tensoring with \(k=R/I\) gives \[ 0\longrightarrow k\xrightarrow{\phi\otimes 1}k^2\longrightarrow0. \] Because \(x\) and \(y\) act as zero on \(k\), the map \(\phi\otimes1\) is the zero map. Hence \[ \Tor_1^R(I,k)\cong k, \qquad \Tor_0^R(I,k)=I\otimes_Rk\cong k^2, \] and, since the resolution has length one, \[ \Tor_i^R(I,k)=0\qquad(i\ge2). \] In particular \(\Tor_1^R(I,k)\neq0\). A flat module has vanishing \(\Tor_1\) against every module, so \(I\) is not flat.