When are two ideals in a regular local ring generated by a regular sequence? Hello!
Let $R$ be a regular local ring, and let $I,J\subset R$ be ideals. I'd like to understand the "meaning" of the existence of a regular sequence $(x_1,...,x_n)$ in $R$ such that $I$ is generated by $x_1,...,x_k$ and $J$ is generated by $(x_{k+1},...,x_n)$ for some $1\leq k\leq n$.
For example, the existence of such a sequence implies that $\text{Tor}^R_k(R/I,R/J)=0$ for all $k>0$. 
Is it possible to give an equivalent description of the above property in terms of the vanishing of certain Tor and/or Ext terms?
Thank you!
Hanno
 A: This question is a bit vague, but I will try my best. Equivalent conditions involving only vanishing of Tor or Ext over $R$ are unlikely to exist, as they tend to be able to detect only projective dimensions, or depth over $R$. 
Here is an example straight from yours: if $R/(I+J)$ has finite length, then $\text{Tor}_i^R(R/I,R/J)=0$ for all $i>0$ if and only if both $R/I, R/J$ are Cohen-Macaulay. This very neat result can be found in Serre's "Local Algebra".
If you are willing to look at finer data, then it is  not hard to detect complete intersection. For example, $I$ is generated by $R$-sequence iff the conormal module $I/I^2$ is free over $R/I$. This is equivalent to  the vanishing of $\text{Tor}_1^{R/I}
(I/I^2,R/m)$ alone ($m$ is the maximal ideal). 
The last one while great in theory, is not very helpful in practice, since you need to present $I/I^2$ as a $R/I$-module. It is simpler to compute $\text{Tor}_1^{R}(R/I,R/m)$ and compare with the codimension of $I$. If they are equal, $I$ must be complete intersection. 
