Example of calculating $\text{Tor}_1(M,N)$ for modules $M$ and $N$ I am having some difficulty with calculating $\text{Tor}_1(M,N)$ and suspect that my approach is going wrong somewhere. As an example, I have tried to calculate $\text{Tor}_1^A(K/A,K/A)$, where $K$ is the field of Hahn series in an indeterminate $t$ with exponents in $\mathbb{R}$ and coefficients in $\mathbb{F}_2$, and $A$ is the subring of Hahn series with no negative exponents.
Let $\{e_i\}_{i\geq 0}$ be the standard basis for $\bigoplus_{i=0}^\infty A$. $K/A$ has projective resolution $$0\to\bigoplus_{i=0}^\infty A\to\bigoplus_{i=0}^\infty A\to K/A\to 0.$$ The morphisms in the projective resolution are $d:\bigoplus_{i=0}^\infty A\to\bigoplus_{i=0}^\infty A$ given by $d(e_0):=e_0$ and $d(e_i):=e_{i-1}+te_i$, $i\geq 1$, and $\epsilon:\bigoplus_{i=0}^\infty A\to M$ is given by $\epsilon(e_i):=t^{-i}+A$. In order to calculate $\text{Tor}_1$, we form the chain complex $$0\to\left(\bigoplus_{i=0}^\infty A\right)\otimes (K/A)\to\left(\bigoplus_{i=0}^\infty A\right)\otimes(K/A)\to 0,$$ where $d\otimes\text{id}_{K/A}$ maps $\sum_i e_i\otimes x_i$ to $\sum_i d(e_i)\otimes x_i$. This is equivalent to the chain complex $$0\to\bigoplus_{i=0}^\infty K/A\to\bigoplus_{i=0}^\infty K/A\to 0,$$ with the morphism $d_{K/A}$ taking $\sum_i x_i$ to $\sum_i x_i'$, where $\sum_i d(e_i)\otimes x_i=\sum_i e_i\otimes x_i'$. Since these two chain complexes are equivalent, they have the same homology groups, so $\text{Tor}_1(K/A,K/A)=\text{ker}(d_{K/A})$.
Now if $\sum_i x_i\in\text{ker}(d_{K/A})$, then $x_i'=0$ for all $i$. Each $x_i'$ is a function of the $x_j$'s, so this gives a system of equations in the $x_i$'s and the kernel is the set of $\sum_i x_i$ solving this system. However this approach keeps giving me $0$ when it should not be, so I am confused. Have I approached this wrong?
 A: In this specific case you can consider the short exact sequence $$0\to A\to K\to K/A\to 0.$$ Applying the functor $\def\Tor{\operatorname{Tor}}\Tor_*^A(-,A/K)$ we obtain a long exact sequence.
As $A$ is free and $K$ is flat, we have that $\Tor_n^A(A,X)=\Tor_n^A(K,X)=0$ for all modules $X$ and all $n\geq1$. The long exact sequence has segments that read $$\Tor_{n+1}^A(K,K/A)\to\Tor_{n+1}^A(K/A,K/A)\to\Tor_n^A(A,K/A)$$ and we can therefore conclude that $\Tor_n^1(K/A,K/A)=0$ for all $n\geq2$. The rest of the long exact sequence looks like $$0\to\Tor_1(K/A,K/A)\to\Tor_0(A,K/A)\to\Tor_0(K,K/A)\to \Tor_0(K/A,K/A)\to 0$$ As $\Tor_0$ is just the tensor product, this is
$$0\to\Tor_1(K/A,K/A)\to A\otimes_A K/A\to K\otimes_A K/A\to K/A\otimes_A K/A\to 0$$
The module $K\otimes_AK/A$ is zero: if $a/b$ and $c/d$ are two fractions in $K$, then $$(a/b)\otimes(c/d+A)=(ad/bd)\otimes(c/d+A)=(a/bd)\otimes(dc/d+A)=0.$$ It follows from this that the map $\Tor_1(K/A,K/A)\to A\otimes_AK/A=K/A$ is an isomorphism.
