# 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

-
This means that I and J are in general position. This is a quite strong restriction which implies that I and J are both complete intersections. You could think about linearly independent linear forms. Basically, what happens algebraically is that Serre´s tor formula for the intersection of two subvarieties simplifies because the higher Tor´s vanish (as you mentioned). (Note that I and J intersect properly if they are in general position. Krull's hauptidealsatz or something..) –  Ariyan Javanpeykar Sep 30 '10 at 8:58
One should point out that in the geometric situation, if one additionally assumes that $x_1, \dots, x_n$ form a set of generators for the maximal ideal, then this means that $\Spec R/I$ and $\Spec R/J$ intersect (only at the origin) as transversally as possible. –  Karl Schwede Sep 30 '10 at 16:19

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.