Tor one of R modulo I and M is not always the I-torsion submodule of M

Every element of $R=k[x][y]$ is uniquely a finite sum ${\sum }_{a,b\ge 0}{c}_{ab}{x}^a{y}^b$. Constant-term evaluation $ϵ:R\to k$ is onto and has kernel $I$: every nonconstant monomial is divisible by $x$ or by $y$. Multiplication by $x$ in $R$, and by $y$ in $k[y]$, is injective because it shifts the corresponding coefficient indices.

Consider $0\to R\stackrel{d_2}{\to }{R}^2\stackrel{d_1}{\to }R\stackrel{ϵ}{\to }k\to 0$, where $d_2(c)=(-yc,xc)$ and $d_1(a,b)=xa+yb$. Its composites are zero; $d_2$ is injective by step 1.1, and $\mathrm{im}{d}_1=I=\ker ϵ$. If $xa+yb=0$, setting $x=0$ gives $yb(0,y)=0$, hence $b(0,y)=0$. Coefficient uniqueness gives $b=xc$ for some $c$, and then $x(a+yc)=0$ implies $a=-yc$. Thus $\ker d_1=\mathrm{im}{d}_2$ and the displayed augmented complex is exact.

The finite free modules in step 2.1 are projective by the lifting definition Projective modules and the lifting property: lift the images of their one or two specified basis vectors through a surjection, and extend linearly. Only finitely many choices are made; no choice axiom is needed. Commutativity gives the same lifting property on the right, so step 2.1 is a supplied right projective resolution of $R\mathrm{/}I=k$.

For $r=1,2$, the mutually inverse maps $R^r{\otimes }_{R}k\to k^r$, $(a_j)\otimes t\mapsto (ϵ(a_j)t)$, and $(t_j)\mapsto {\sum }_j{e}_j\otimes t_j$ are well-defined by Universal property of the tensor product for balanced maps into abelian groups. Tensoring the resolution with $k$ therefore gives $0\to k\stackrel{0}{\to }{k}^2\stackrel{0}{\to }k\to 0$, since $x,y$ act by zero. Its degree-one homology is $k^2$, which is the right-resolution Tor group by Tor from a projective resolution of the right module and represents balanced Tor under DC.

The cotangent calculation also holds explicitly: $I^2$ consists exactly of polynomials all of whose monomials have total degree at least two. Products of two elements of $I$ have this property, and every such monomial is divisible by one of $x^2,xy,y^2$. Consequently every class in $I\mathrm{/}{I}^2$ is uniquely $ax+by$ modulo $I^2$, for $a,b\in k$, and the classes of $x,y$ are a basis. The isomorphism $I{\otimes }_{R}k\to I\mathrm{/}{I}^2$ sends $a\otimes t$ to $ta+I^2$ (view $t$ as a constant); its inverse sends $a+I^2$ to $a\otimes 1$. For $u,v\in I$, $(uv)\otimes 1=u\otimes ϵ(v)=0$, so the inverse is well-defined. Balancing and the inverse composites follow on the generating tensors.

Because $I$ annihilates $k$, its annihilator submodule is all of $k$. There is no $k$-linear isomorphism $k^2\to k$: if the images of its standard vectors are $a,b$, then for $(a,b)\mathrm{\ne }(0,0)$ the nonzero vector $(-b,a)$ lies in the kernel, and for $(a,b)=(0,0)$ the entire map is zero. Every $R$-linear map between these modules is $k$-linear by the constant scalars. Thus the $k^2$ in step 4.1 cannot be the asserted annihilator module.