Tor for graded modules over a graded ring I am confused about how this Tor is defined. 

Suppose $R$ is a graded ring, $M,N$ graded modules over $R$. What is $\operatorname{Tor}_{st}^R(M,N)$?

I am confused about the subscripts. I realize there is a lot of grading, so a definition that clarifies the grading and the subscripts would be helpful.
 A: My favorite explanation of the grading on this object is in the preliminaries to (the published version of) Paul Baum's dissertation, linked here.
One first forms a (graded) projective resolution $P_\bullet = (P_s)$ of $M$, with degree-0 maps of graded modules $f_s\colon P_{s} \to P_{s-1}$, then removes the segment $P_0 \overset{f_0}\to M$, replacing it with $P_0 \to 0$. 
Then one tensors the whole thing with $N$ to get a complex $P_\bullet \otimes_R N = (P_s \otimes_R N)$, with maps $f_s \otimes \mathrm{id}_N$. One grades the tensor product by total degree $t$, so 
$$(P_s \otimes_R N)^{(t)} = 
\bigoplus_{a+b = t} \big(P_s^{(a)} \otimes_R N^{(b)}\big).$$ 
The maps $f_s \otimes \mathrm{id}_N$ in this complex take bidegree $(s,t)$ to bidegree $(s-1,t)$. For fixed $t$, the Tor with subscript $(s,t)$ is the $s$th homology of the complex $(P_\bullet \otimes_R N)^{(t)}$; that is,
$$\mathrm{Tor}_{s,t}^R(M,N) = 
\frac{\ker\!\big((f_s \otimes \mathrm{id}_N)^{(t)}\colon \ 
(P_s \otimes_R N)^{(t)} \to (P_{s-1} \otimes_R N)^{(t)}
\big)}
{\mathrm{im}\big((f_{s+1} \otimes \mathrm{id}_N)^{(t)}\colon \ 
(P_{s+1} \otimes_R N)^{(t)} \to (P_s \otimes_R N)^{(t)}\big)}.
$$
If you want to forget the grading on $R$, $M$, and $N$, then, summing along $t$, you can, and you recover
$$\mathrm{Tor}^R_s(M,N) = \mathrm{Tor}^R_{s,\bullet}(M,N) = \bigoplus_{t\geq 0} \mathrm{Tor}^R_{s,t}(M,N),$$
the regular, ungraded $s$th Tor functor. As in the ungraded case, the results turn out to be independent of the resolution chosen.
It's also common to cohomologically grade by swapping $-s$ for $s$ everywhere. One then writes $$\mathrm{Tor}^{-s,t}_R(M,N) = \mathrm{Tor}_{s,t}^R(M,N).$$
