Hi everyone,

I've got a question about explicitly lifting regular sequences. Let $I$ be an ideal in a polynomial ring $S$ with some term order. We'll denote the initial ideal by $in(I)$. It is false in general that a regular sequence on $S/in(I)$ is regular on $S/I$. For example consider $I=(x+y)$, with $x>y$ Then $x+y$ is a regular element on $S/in(I)$ but is not regular on $S/I$. However, $3x-y$ IS a regular element mod $I$.

My question is: Can we can do this in general? i.e. Given a regular sequence on $S/in(I)$, can we obtain a regular sequence on $S/I$ by just replacing all the coefficients in all the elements with generic coefficients?

We know that the depth of $S/in(I)$ is at most the depth of $S/I$, but I haven't actually seen too many proofs of this written down. The ones I've seen first show a bound on Betti numbers and then use the Auslander-Buchsbaum formula. I was wondering if one could prove this fact by answering the question above, and if anyone has a reference. I think one might be able to use a flat family argument. In general it would be nice to have an explicit way of going back and form between regular sequences on $I$ and $in(I)$. Any reference or suggestions would be greatly appreciated.

Thanks so much for your help!


  • $\begingroup$ What does in(I) mean? $\endgroup$
    – Olivier
    Commented Apr 3, 2011 at 12:06
  • $\begingroup$ in(I) denotes the initial ideal with respect to the term order. I'll update my post to indicate. Thanks! $\endgroup$ Commented Apr 3, 2011 at 15:44

1 Answer 1


Hi Adam,

What you are asking is true. In fact it is actually possible to "lift" not only regular sequences but also filter regular sequences. The proof is a simple modification of the usual argument that shows the inequality between the graded Betti numbers of $I$ and the ones of $in(I).$ Say that $in(I)=in_w(I)$ for some weight $w.$ Assume that the field $K$ in infinite. We know that there is a flat family, parametrized say by $t$, induced by homogenizing using $t$ w.r.t. the grading induced by $w$. At the special point $t=0$ we get $in(I)$ and for $t=1$ we get $I$. In general for $t=\alpha \not =0$ we would get $D_{\alpha}I$ where $D_{\alpha}$ is a diagonal change of coordinate. Fix a degree $d,$ and let $J$ be a homogeneous ideal of $S$ (for instance generated by the regular sequence of $r$ linear forms you want to lift) then $dim_K (S/(in I+J))_d \geq dim_K (S/(D_{\alpha}I+J))_d$ for $\alpha$ in a neighborhood of $0.$ Now the right hand side is isomorphic to $dim_K (S/(I+D_{\alpha}^{-1}J))_d.$ There is actually a whole non-empty Zariski open set of such $\alpha$'s, call it $U_d.$
In your case $J$ is generated by a regular sequence of $r$ linear forms and $dim(S/(in I+J))_d$ is the "smallest possible" given the Hilbert function of $S/I$ and the fact that $J$ is generated by $r$- linear forms. The inequality above is therefore forced to be an equality. By intersecting all the the $U_i's$ for $1\leq i\leq d$ , $d$ sufficiently large, you find an a $D_{\alpha}$ Such that $D_{\alpha}^{-1}(J)$ is the lift. The proof of the fact that one needs only to intersect finitely many $U_i$ depends on the ascending chain condition and is quite similar to the analogous step needed to show that the generic initial ideal is well defined. Regarding the above inequality, with the proper modifications one can show, more generally, that $dim_K Tor_i(S/inI,S/J)_d \geq dim_K Tor_i(S/I,S/DJ)_d$ for some diagonal change of coordinates $D.$


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