# Bounds for Betti numbers

Why the graded Betti numbers of ideal $I \subset k[x_1 , \cdots , x_n]$, are bounded by the graded Betti numbers of $\mathrm{gin}(I)$?

(Where $\mathrm{gin}(I)$, is the generic initial ideal of $I$ whit respect to a monomial order)

• Betti numbers always go up under flat degeneration, and stay the same under a linear change of coordinates. $gin(I)$ is the result of applying a generic linear change of coordinates to $I$, followed by a Groebner degeneration. Which parts of this should be explained more? Commented Sep 1, 2014 at 18:04
• @David, I know that there is an open Zariski set $U$, such that for $g \in U$, the ideal $gin(I)=in(g(I))$, is constant, and $\beta _{ij}(S/I) \leq \beta _{ij}(S/in(I))$. can you explain more about your comment.
– A.B.
Commented Sep 2, 2014 at 13:15
• For $g$ in the set $U$ you describe, we have $\beta_{ij}(S/gin(I)) = \beta_{ij}(S/in(g(I))) \geq \beta_{ij}(S/g(I)) = \beta_{ij}(S/I)$. The last equality is because linear change of variables are automorphisms; the other steps are things you already know. Commented Sep 2, 2014 at 13:20
• Do we have the equality $\beta_{ij}(in(I))= \beta_{ij}(in(g(I)))$?
– A.B.
Commented Sep 2, 2014 at 13:29
• There are a few good references for learning about gins, betti numbers, and in general the resolution of in I and how it relates to the resolution of I. Peeva has a book "Graded Syzygies" (available online through Springerlink if your library has a subscription) and Chapter 15 of Eisenbud's Book Commutative Algebra has precise statements about how initial ideal are flat degenerations. Commented Oct 22, 2014 at 13:57

## 1 Answer

This is about the question whether $\beta_{ij}(\mathrm{in}(I)) = \beta_{ij}(\mathrm{in}(g(I)))$. No, they need not be equal. E.g., take $I = (x^2, y^2)$ in $k[x,y]$. Then for a general $g$, $\mathrm{in}(g(I)) = (x^2,xy, y^3)$. Note that $\mathrm{in}(I) = I$ and that $\beta_{03}(I) = 0 \neq 1 = \beta_{03}(\mathrm{in}(g(I)))$.