Explicit computation $\operatorname{Tor}(M,N)$ Let $R=\mathbb{C}[t]/t^2$ the ring of dual numbers. Using the homomorphism $\phi:R \to \mathbb{C}=R/(t)$ we have that $\mathbb{C}$ is a $R$-module, infact we have
$$\psi: \mathbb{C} \times \mathbb{C}[t]/t^2 \to \mathbb{C} $$
taking $\psi(a,b)=a\phi(b)$. So we have that $\mathbb{C}$ is a $R$-module and it is obvious that $R$ is a $R$-module. I'd like to compute $\operatorname{Tor}_i(R,\mathbb{C})$ for all $i \ge 1$. In order to do this I have to find a free resolution of $R$ ($\cdots \to P_1 \to P_0 \to R \to 0$). I think that we can take $P_i=R$:
$$ \cdots \to R \to R\to R \to 0 .$$
If I had taken the correct resolution we have to consider the tensor product $- \otimes_{\mathbb{C}[t]/(t^2)} \mathbb{C}$ and have
$$ R \otimes_{R} \mathbb{C} .$$ 
So I have to calculate the homology of the complex having $\operatorname{Tor}_i(R,\mathbb{C})=\mathbb{C}$ for all $i \ge 0$. Is it correct? Thanks!  
 A: For modules over any ring $R$ we have
$$
\operatorname{Tor}_i^R(M,N)=0 \quad(i>0)
$$
whenever either of $M$ and $N$ is flat, in particular projective (or free).
If $M$ is projective, it's quite easy to show: a projective resolution of $M$ is $$\dots\to0\to0\to M\to M\to 0,$$ so removing the last $M$ and tensoring with $N$ gives the complex
$$
\dots\to 0\to 0 \to M\otimes_R N\to 0.
$$
Therefore the homology is zero on all degrees $>0$.
Thus you need nothing else to show that $\operatorname{Tor}_i^R(R,\mathbb{C})=0$ for $i>0$.
A: Let's denote $\mathbb{C}[t]/t^2$ with $\mathbb{C}[\epsilon]$ and calculate $Tor^\mathbb{C}_i(R,\mathbb{C})$. We have the resolution
$$ \to R \stackrel{\epsilon}{\to} R \stackrel{\epsilon}{\to} R \to \mathbb{C} \to 0 .$$
After tensoring we have 
$$ \to \mathbb{C} \otimes_R R \to \mathbb{C} \otimes_R R \to \cdots $$
we can easily observe that $R \otimes_R \mathbb{C} \simeq \mathbb{C}$, so we have the complex
$$ \cdots \to \mathbb{C} \to \mathbb{C} \to \mathbb{C} \to 0 ,$$ where every map is the null homomorphsm. So the homology of this complex is $\mathbb{C}$ in every degree and we can conclude that $Tor^\mathbb{C}_i(R,\mathbb{C}) \simeq \mathbb{C}$ for all $i \ge 1$.
