Discussion between A. B. and David Speyer

1:48 PM

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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)

David Speyer

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?

A. B.

@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.

David Speyer

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.

A. B.

Do we have the equality $\beta_{ij}(in(I))= \beta_{ij}(in(g(I)))$?

David Speyer

That shouldn't be true, although I don't have a counterexample ready at the moment.

A. B.

1:48 PM

What about this equality of Hilbert functions,$H(S/I , m)=H(S/gin(I) , m) $?

David Speyer

Yes, of course. Flat degeneration preserves Hilbert function, as does linear change of basis.

Okay, I'm chatting.

I can think of an issue which might be confusing you in your last comment, but I don't want to bring it up if it isn't what is confusing you:

For $I$ any graded ideal, I am taking the Hilbert function $h(S/I,m)$ to be the dimension of the degree $m$ piece of $S/I$. And (I think everyone will agree on this) $g(I)$ means $I$ after the linear change of variables $g$, $in(I)$ means the Groebner degeneration of $I$ and $gin(I)$ means $in(g(I))$ for generic $g$.

In this sense, $h(S/I,m) = h(S/g(I), m) = h(S/in(I), m) = h(S/in(g(I)), m) = h(S/gin(I), m)$.

Now, if $X$ is the projective variety corresponding to $I$, and $in(X)$ the projective variety corresponding to $in(I)$, it isn't necessarily true that $H^0(X, \mathcal{O}(m)) = H^0(in(X), \mathcal{O}(m))$. That's because, even if $I$ is saturated, $in(I)$ might not be, and $H^0(\mathcal{O}(m))$ is only $(S/I)_m$ for $I$ saturated.

But, if you are defining everything in terms of commutative algebra the way I state a few sentences above, then you don't need to think about this.

If that's not what is confusing you, I'm not sure what is.

A. B.

I need some minutes to think more...

But first, what's a Flat degeneration?

David Speyer

I'm not sure exactly how broadly I want to answer that, but:

Here is the case we care about today.

Let $I \subset k[x_1, x_2, \ldots, x_n, t]$ be an ideal which is graded with respect to the $x$-variables and saturated with respect to the $t$ variable.

Saturated meaning that if $f \in I$, then $tf \in I$.

Bleh, not being able to go back and edit earlier parts of a chat.

Let's start with any $I \subset k[x_1, \ldots, x_n, t]$, graded in the $x$ variables, but not yet assumed saturated.

For any $a$, we can talk about the ideal $I_a$.

So we get a hilbert function $h_a(m) = \dim (S/I_a)_m$ at every point $a$ of $k$.

There will be a big Zariski dense open subset of $k$ where $h_a(m)$ is independent of $a$.

Those are the $a$ where the family $I_a$ is flat.

Typo in previous line, should say these are the $a$ where $k[x_1, \ldots, x_n, t]
/I$ is flat.

When we are talking about families parametrized by the affine line, as here, "flat at $a$" is equivalent to "$(t-a)$-saturated at $a$". The general notion of flatness is really needed in when we want to talk other parameter spaces.

I want to now talk about how this fits into the context of Grobner degenerations, but I should probably pause to see if this makes sense.

David Speyer

2:27 PM

Well, I guess I'll keep typing. The awesome thing about families over the affine line is that they can ALWAYS be filled in in a flat way. Suppose we have any ideal $I \subset k[x_1, \ldots, x_n, t]$. We can always saturate it with respect to $t$ and get a new ideal, which will be flat at $0$, and will leave all other $I_a$ unchanged.

You know, I'm going to write a lot here, and I might be coming at this from the wrong angle for you. Let me start over for a bit.

Have you seen flatness at all? Do you think about initial ideals as one parameter limits? Have you taken a Hartshorne based (or equivalent) algebraic geometry course?

A. B.

I studied some chapters of  "Igor R. Shafarevich's book   Basic Algebraic Geometry 1" and first chapter of Hartshorn's book.

David Speyer

Okay, so you haven't seen much flatness. Let me think about where I want to start then.

Let's fill in the basic facts before we talk about any higher context in which to understand them. Let $I$ be a graded ideal in $k[x_1, \ldots, x_n]$. Choose some term order. The monomials which are not in $in(I)$ are a basis for $k[x_1, \ldots, x_n]/I$.

These are called the "standard monomials", and one of the things we do with a Grobner basis is write things in a standard form, using this basis.

Is this something you know?

A. B.

2:54 PM

Yes.

I had a course in Grobner basis.

David Speyer

Okay. Then it is clear that $\dim (S/I)_m = \dim (S/in(I))_m$: They are both the number of standard monomials of degree $m$.

A. B.

Yes.

David Speyer

And $\dim (S/I)_m = \dim (S/g(I))_m$, because $g$ is a (grade preserving) automorphism.

Putting those two together, $\dim (S/I)_m = \dim (S/gin(I))_m$.

A. B.

Yes.

David Speyer

Okay, so that's the question about the Hilbert function. The remaining question is about Betti numbers.

Specifically, why $\beta_{ij}(S/I) \leq \beta_{ij}(S/in(I))$.

I'm trying to think how to present this without mentioning flat families ...

Let's start with an example. We'll look at $\langle x^2, (x+y)^2 \rangle$, with respect to a term order where $x$ is the leading term of $x+y$.

A Grobner basis is $\langle x^2, 2xy+y^2, y^3 \rangle$. (Note that $2xy+y^2 = (x+y)^2-x^2$ and $y^3 = y(2xy+y^2) - 2x y^2$, so these are in the ideal.

The initial ideal is $I_0 = \langle x^2, xy, y^3 \rangle$. A resolution of the initial ideal looks like (give me a moment...)

$S[-3] \oplus S[-4] \to S[-2] \oplus S[-2] \to S[-3] \to S \to S/I_0$.

The map S[-2] \oplus S[-2] \to S[-3] \to S hits the generators x^2, xy and y^3.

The first map is given by the matrix

y -x 0

0 y^2 -x

Corresponding to the syzygies y*x^2-x*xy = 0 and y^2 (xy) - x*y^3 = 0.

Now, we can lift that back to a resolution of the original ideal. (hold on....)

Again, the shape of the resolution is
$S[-3] \oplus S[-4] \to S[-2] \oplus S[-2] \to S[-3] \to S \to S/I$.

and the map S[-2] \oplus S[-2] \to S[-3] \to S hits the Grobner basis x^2, 2xy+y^2, y^3.

The syzygies are (hold on...)

This answer will be clearer if I go back and write the generators of I_0 as x^2, 2xy and y^3.

Then the syzygies for I_0 are

2y -x 0

0 y^2 -2x

And the syzygies when resolving I are

2y -x+y/2 -1/2

Uggh, this working in chat is a pain.

Let me explain how I am thinking first.

Look at the family of ideals <x^2, 2xy+t y^2, y^3 >.

For t=1, this is the original ideal I.

For any t which isn't 0, this is the ideal I after the change of variables (x,y) ---> (x,x+ty). In other words, this is very close to being the ideal < x^2, (x+ty)^2 >.

But it isn't quite the ideal < x^2, (x+ty)^2 >. For example, < x^2, (x+ty)^2 > contains (x+ty)^2 - x^2 = t (2xy+t y^2), but it doesn't contain (2xy+t y^2) itself.

The precise statement is that <x^2, 2xy+t y^2, y^3 > is the saturation of <x^2, (x+ty)^2> with respect to t.

And notice that, when we plug in t=0, we get <x^2, xy, y^3>.

The initial ideal.

This is true of any Grobner degeneration: We can always write it as a family, parametrized by t, where

for t=1, we have the original ideal

for t \neq 0, we have an ideal which is isomorphism to the original ideal under some linear change of variables

and for t=0, we have the initial ideal.

Is this a story you are happy with/wiling to believe in?

A. B.

3:33 PM

This is very good.

David Speyer

Great

Let's look at the syzygies of I_t as t varies.

Once again, for t=0, the generators are x^2, 2xy and y^3, and the syzygies are 2y*x^2-x*(2xy) and y^2 * (2xy) - 2x * y^3

For any t, we can generate the ideal by x^2, 2xy+ty^2 and y^3

and syzygies are (hold on....)

2y*x^2 - (x-t y/2)*(2xy+ty^2)-(t^2/2) y^3 and ...

y^2 * (2xy+t y^2) - (2x+t y) y^3

For t=0, this is a minimal resolution.

But for t \neq 0, this is not minimal, because the coefficient of y^3 in the first syzygy is degree 0 in the (x,y) grading.

So I_0 has minimial resolution of shape $S[-3] \oplus S[-4] \to S[-2] \oplus S[-2] \to S[-3] \to S \to S/I_0$

I has a non-minimal resolution of the same shape

But in the minimal resolution, the S[-3] terms go away and we have

S[-4] \to S[-2] \oplus S[-2] \to S \to S/I

This is an example of why the betti numbers of S/in(I) can be larger than those of S/I.

Does the example work for you?

Okay, I need to get going soon, so I'll write the rest quickly.

For any Grobner degeneration of I in k[x_1, ..., x_n], there is an ideal J in k[x_1, ..., x_n, t] so that

plugging in t=1 gives I

plugging in t \neq 0 gives an ideal isomorphic to I under a change of variables

and plugging in t=0 gives in(I).

If you take a graded free resolution of in(I)

you can extend to a graded free resolution of J (where t is in grade 0).

For t \neq 0, it may no longer be minimal.

So the minimal resolution at t \neq 0 has terms which are a subset of the terms at t=0, and this is why betti numbers can jump up but can't jump down at t=0.

This story works for all t-saturated ideals, not just the ones coming from Grobner degenerations.

If you want to work with situations more general than a single parameter, you need to replace the notion of "t-saturated" with "flat".

A. B.

4:02 PM

In general, how we can obtain the resolution of $I$ form the resolution of $in(I)$?

Thank you David I learned many things.

Transcript for

Discussion between A. B. and David Sp…