Return to Answer

It seems unlikely that the Betti numbers can be determined by the data you list (bigraded Hilbert function and ranks of multiplication maps). Consider the following possibility: $\dim M_{1,0} = \dim M_{0,1} = 1$, $\dim M_{1,1} = 2$, and all other $M_{a,b}$ are zero; and the maps $y : M_{1,0} \to M_{1,1}$, $x : M_{0,1} \to M_{1,1}$ each have rank $1$ (all other multiplication maps are apparently zero).

Given this data and nothing else, it's impossible to tell if the maps from $M_{1,0}$ and $M_{0,1}$ to $M_{1,1}$ have the same image in $M_{1,1}$, or not.

An example where they have different images is $$M = (x)/(x^2,y) \oplus (y)/(x,y^2)$$ (where parentheses denote ideals). An example where they have the same image is $$M' = (x,y)/(x^2,y^2) \oplus (xy)/(x^2y,xy^2).$$

$M$ is generated by $x,y$ in bidegrees $(1,0)$ and $(0,1)$, while $M'$ requires generators $x,y$, and an "extra" copy of $xy$, in bidegrees $(1,0)$, $(0,1)$, and $(1,1)$. So, not even the first Betti degrees are determined by the data that you have.

Sometimes you’ll be able to read off some information. If one of the incoming maps is zero or surjective then you’ll know how many generators there are in that bidegree. At the moment I can’t think of any other cases, or anything for higher Betti numbers.

It seems unlikely that the Betti numbers can be determined by the data you list (bigraded Hilbert function and ranks of multiplication maps). Consider the following possibility: $\dim M_{1,0} = \dim M_{0,1} = 1$, $\dim M_{1,1} = 2$, and all other $M_{a,b}$ are zero; and the maps $y : M_{1,0} \to M_{1,1}$, $x : M_{0,1} \to M_{1,1}$ each have rank $1$ (all other multiplication maps are apparently zero).

Given this data and nothing else, it's impossible to tell if the maps from $M_{1,0}$ and $M_{0,1}$ to $M_{1,1}$ have the same image in $M_{1,1}$, or not.

An example where they have different images is $$M = (x)/(x^2,xy^2) \oplus (y)/(x^2y,y^2)$$ (where parentheses denote ideals). An example where they have the same image is $$M' = (x,y)/(x^2,y^2) \oplus (xy)/(x^2y,xy^2).$$

$M$ is generated by $x,y$ in bidegrees $(1,0)$ and $(0,1)$, while $M'$ requires generators $x,y$, and an "extra" copy of $xy$, in bidegrees $(1,0)$, $(0,1)$, and $(1,1)$. So, not even the first Betti degrees are determined by the data that you have.

Sometimes you’ll be able to read off some information. If one of the incoming maps is zero or surjective then you’ll know how many generators there are in that bidegree. At the moment I can’t think of any other cases, or anything for higher Betti numbers.

It seems unlikely that the Betti numbers can be determined by the data you list (bigraded Hilbert function and ranks of multiplication maps). Consider the following possibility: $\dim M_{1,0} = \dim M_{0,1} = 1$, $\dim M_{1,1} = 2$, and all other $M_{a,b}$ are zero; and the maps $y : M_{1,0} \to M_{1,1}$, $x : M_{0,1} \to M_{1,1}$ each have rank $1$ (all other multiplication maps are apparently zero).

Given this data and nothing else, it's impossible to tell if the maps from $M_{1,0}$ and $M_{0,1}$ to $M_{1,1}$ have the same image in $M_{1,1}$, or not.

An example where they have different images is $$M = (R/(x,y^2))[-1,0] \oplus (R/(x^2,y))[0,-1]$$ (where $R=k[x,y]$). An example where they have the same image is $$M' = (x,y)R/(x^2,y^2) \oplus k[-1,-1].$$ (The bracketed things are meant to be bidegree shifts, so $k[-1,-1]$ is supposed to be supported in bidegree $(1,1)$.)

$M$ is generated in bidegrees $(1,0)$ and $(0,1)$, while $M'$ requires generators in bidegrees $(1,0)$, $(0,1)$, and $(1,1)$. So, not even the first Betti degrees are determined by the data that you have.

Sometimes you’ll be able to read off some information. If one of the incoming maps is zero or surjective then you’ll know how many generators there are in that bidegree. At the moment I can’t think of any other cases, or anything for higher Betti numbers.

It seems unlikely that the Betti numbers can be determined by the data you list (bigraded Hilbert function and ranks of multiplication maps). Consider the following possibility: $\dim M_{1,0} = \dim M_{0,1} = 1$, $\dim M_{1,1} = 2$, and all other $M_{a,b}$ are zero; and the maps $y : M_{1,0} \to M_{1,1}$, $x : M_{0,1} \to M_{1,1}$ each have rank $1$ (all other multiplication maps are apparently zero).

Given this data and nothing else, it's impossible to tell if the maps from $M_{1,0}$ and $M_{0,1}$ to $M_{1,1}$ have the same image in $M_{1,1}$, or not.

An example where they have different images is $$M = (x)/(x^2,y) \oplus (y)/(x,y^2)$$ (where parentheses denote ideals). An example where they have the same image is $$M' = (x,y)/(x^2,y^2) \oplus (xy)/(x^2y,xy^2).$$

$M$ is generated by $x,y$ in bidegrees $(1,0)$ and $(0,1)$, while $M'$ requires generators $x,y$, and an "extra" copy of $xy$, in bidegrees $(1,0)$, $(0,1)$, and $(1,1)$. So, not even the first Betti degrees are determined by the data that you have.

Sometimes you’ll be able to read off some information. If one of the incoming maps is zero or surjective then you’ll know how many generators there are in that bidegree. At the moment I can’t think of any other cases, or anything for higher Betti numbers.

It seems unlikely that the Betti numbers can be determined by the data you list (bigraded Hilbert function and ranks of multiplication maps). Consider the following possibility: $\dim M_{1,0} = \dim M_{0,1} = 1$, $\dim M_{1,1} = 2$, and all other $M_{a,b}$ are zero; and the maps $y : M_{1,0} \to M_{1,1}$, $x : M_{0,1} \to M_{1,1}$ each have rank $1$ (all other multiplication maps are apparently zero).

Given this data and nothing else, it's impossible to tell if the maps from $M_{1,0}$ and $M_{0,1}$ to $M_{1,1}$ have the same image in $M_{1,1}$, or not.

An example where they have different images is $$M = (R/(x,y^2))[-1,0] \oplus (R/(x^2,y))[0,-1]$$ (where $R=k[x,y]$). An example where they have the same image is $$M' = (x,y)R/(x^2,y^2) \oplus k[-1,-1].$$ (The bracketed things are meant to be bidegree shifts, so $k[-1,-1]$ is supposed to be supported in bidegree $(1,1)$.)

$M$ is generated in bidegrees $(1,0)$ and $(0,1)$, while $M'$ requires generators in bidegrees $(1,0)$, $(0,1)$, and $(1,1)$. So, not even the first Betti degrees are determined by the data that you have.

It seems unlikely that the Betti numbers can be determined by the data you list (bigraded Hilbert function and ranks of multiplication maps). Consider the following possibility: $\dim M_{1,0} = \dim M_{0,1} = 1$, $\dim M_{1,1} = 2$, and all other $M_{a,b}$ are zero; and the maps $y : M_{1,0} \to M_{1,1}$, $x : M_{0,1} \to M_{1,1}$ each have rank $1$ (all other multiplication maps are apparently zero).

Given this data and nothing else, it's impossible to tell if the maps from $M_{1,0}$ and $M_{0,1}$ to $M_{1,1}$ have the same image in $M_{1,1}$, or not.

An example where they have different images is $$M = (R/(x,y^2))[-1,0] \oplus (R/(x^2,y))[0,-1]$$ (where $R=k[x,y]$). An example where they have the same image is $$M' = (x,y)R/(x^2,y^2) \oplus k[-1,-1].$$ (The bracketed things are meant to be bidegree shifts, so $k[-1,-1]$ is supposed to be supported in bidegree $(1,1)$.)

$M$ is generated in bidegrees $(1,0)$ and $(0,1)$, while $M'$ requires generators in bidegrees $(1,0)$, $(0,1)$, and $(1,1)$. So, not even the first Betti degrees are determined by the data that you have.

Sometimes you’ll be able to read off some information. If one of the incoming maps is zero or surjective then you’ll know how many generators there are in that bidegree. At the moment I can’t think of any other cases, or anything for higher Betti numbers.