Mixing
Dimension of the space of homogeneous polynomials in a quotient ring
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Consider the ring $k[x_1,x_2,x_3,x_4]/(x_1x_3,x_1x_4,x_2x_3,x_2x_4)$. How do I find the dimension of the vector space of homogeneous representatives of the quotient ring of each degree?
I've only managed to exhibit bases for low degrees:
0: $1$; dimension 1
1: $x_1,x_2,x_3,x_4$; dimension 4
2: $x_1^2,x_2^2,x_3^2,x_4^2,x_1x_2,x_3x_4$; dimension 6
3: $x_1^3,x_2^3,x_3^3,x_4^3, x_1^2x_2,x_1x_2^2, x_3^2x_4,x_3x_4^2$; dimension 8
So my conjecture is that the dimension of polynomials of degree $k>0$ is $2k+2$. Is it true? How to prove it?
linear-algebra abstract-algebra combinatorics polynomials vector-spaces
edited Jan 21 at 20:09
user26857
39.3k124183
asked Jan 21 at 16:39
user437309user437309
722313
$endgroup$
add a comment |
$begingroup$
Consider the ring $k[x_1,x_2,x_3,x_4]/(x_1x_3,x_1x_4,x_2x_3,x_2x_4)$. How do I find the dimension of the vector space of homogeneous representatives of the quotient ring of each degree?
I've only managed to exhibit bases for low degrees:
0: $1$; dimension 1
1: $x_1,x_2,x_3,x_4$; dimension 4
2: $x_1^2,x_2^2,x_3^2,x_4^2,x_1x_2,x_3x_4$; dimension 6
3: $x_1^3,x_2^3,x_3^3,x_4^3, x_1^2x_2,x_1x_2^2, x_3^2x_4,x_3x_4^2$; dimension 8
So my conjecture is that the dimension of polynomials of degree $k>0$ is $2k+2$. Is it true? How to prove it?
linear-algebra abstract-algebra combinatorics polynomials vector-spaces
edited Jan 21 at 20:09
user26857
39.3k124183
asked Jan 21 at 16:39
user437309user437309
722313
$endgroup$
add a comment |
$begingroup$
Consider the ring $k[x_1,x_2,x_3,x_4]/(x_1x_3,x_1x_4,x_2x_3,x_2x_4)$. How do I find the dimension of the vector space of homogeneous representatives of the quotient ring of each degree?
I've only managed to exhibit bases for low degrees:
0: $1$; dimension 1
1: $x_1,x_2,x_3,x_4$; dimension 4
2: $x_1^2,x_2^2,x_3^2,x_4^2,x_1x_2,x_3x_4$; dimension 6
3: $x_1^3,x_2^3,x_3^3,x_4^3, x_1^2x_2,x_1x_2^2, x_3^2x_4,x_3x_4^2$; dimension 8
So my conjecture is that the dimension of polynomials of degree $k>0$ is $2k+2$. Is it true? How to prove it?
linear-algebra abstract-algebra combinatorics polynomials vector-spaces
edited Jan 21 at 20:09
user26857
39.3k124183
asked Jan 21 at 16:39
user437309user437309
722313
$endgroup$
Consider the ring $k[x_1,x_2,x_3,x_4]/(x_1x_3,x_1x_4,x_2x_3,x_2x_4)$. How do I find the dimension of the vector space of homogeneous representatives of the quotient ring of each degree?
I've only managed to exhibit bases for low degrees:
0: $1$; dimension 1
1: $x_1,x_2,x_3,x_4$; dimension 4
2: $x_1^2,x_2^2,x_3^2,x_4^2,x_1x_2,x_3x_4$; dimension 6
3: $x_1^3,x_2^3,x_3^3,x_4^3, x_1^2x_2,x_1x_2^2, x_3^2x_4,x_3x_4^2$; dimension 8
So my conjecture is that the dimension of polynomials of degree $k>0$ is $2k+2$. Is it true? How to prove it?
linear-algebra abstract-algebra combinatorics polynomials vector-spaces
linear-algebra abstract-algebra combinatorics polynomials vector-spaces
edited Jan 21 at 20:09
user26857
39.3k124183
asked Jan 21 at 16:39
user437309user437309
722313
edited Jan 21 at 20:09
user26857
39.3k124183
asked Jan 21 at 16:39
user437309user437309
722313
edited Jan 21 at 20:09
user26857
39.3k124183
edited Jan 21 at 20:09
user26857
39.3k124183
edited Jan 21 at 20:09
user26857
39.3k124183
39.3k124183
asked Jan 21 at 16:39
user437309user437309
722313
asked Jan 21 at 16:39
user437309user437309
722313
asked Jan 21 at 16:39
user437309user437309
722313
722313
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The $k$-vector subspace of $k[x_1,x_2,x_3,x_4]$ of homogeneous polynomials of degree $n$ is spanned by the monomials of degree $n$, i.e. terms of the form $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ with $d_1+d_2+d_3+d_4=n$. The kernel of the quotient map is the $k$-vector space spanned by the monomials that satisfy either
$$d_1d_3>0,qquad d_1d_4>0,qquad d_2d_3>0,qquadtext{ or }qquad d_2d_4>0.$$
Hence the $k$-vector subspace of $k[x_1,x_2,x_3,x_4]/(x_1x_2,x_1x_3,x_2x_3,x_2x_4)$ of homogeneous polynomials of degree $n$ is spanned by the monomials $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ that satisfy
$$d_1+d_2+d_3+d_4=k,qquad d_1d_3=d_1d_4=d_2d_3=d_2d_4=0.tag{1}$$
For $k=0$ clearly $d_1=d_2=d_3=d_4=0$ is the unique solution. For $kneq0$ we must have $d_ineq0$ for some $i$. If $d_1neq0$ or $d_2neq0$ then $d_3=d_4=0$, and if $d_3neq0$ or $d_4neq0$ then $d_1=d_2=0$. So the number of solutions to $(1)$ is the same as the number of solutions to
$$d_1+d_2=k,quad d_3=d_4=0qquadtext{ or }qquad d_3+d_4=k,quad d_1=d_2=0,$$
which is precisely twice the number of solutions to $a+b=k$ with $a,binBbb{N}$. This is of course precisely twice $k+1$, as you conjecture.
Earlier version of this answer:
In the quotient, the product of any pair of variables vanishes except the products $x_1x_2$ and $x_3x_4$. This means a basis is given by the monomials of the form
$$x_1^n,qquad x_2^n,qquad x_3^n,qquad x_4^n,qquad x_1^mx_2^n,qquad x_3^mx_4^n.$$
Of course the first four are also of the form $x_1^mx_2^n$ or $x_3^mx_4^n$, simply with $m=0$ or $n=0$.
So for a given degree $k$, you want to count the number of monomials $x_1^mx_2^n$ and $x_3^mx_4^n$ with $m+n=k$. There are $k+1$ choices for $m$, and then $n=k-m$. So there are $k+1$ of monomials of each type, so a total of $2k+2$ monomials.
answered Jan 21 at 17:16
ServaesServaes
27.5k34098
$endgroup$
$begingroup$
Why have you updated your answer? I thought the previous version was fine too.
$endgroup$
– user437309
Jan 21 at 23:21
$begingroup$
I prefer the new version, but I have included the original now.
$endgroup$
– Servaes
Jan 21 at 23:38
add a comment |
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Here's a silly overkill way to find the Hilbert function. We compute a graded resolution for the ring $M := k[x_0,x_1,x_2,x_3]/(x_0x_2, x_0x_3, x_1x_2, x_1x_3)$ and then use additivity.
First let's find a resolution for the ideal $I := (x_0 x_2, x_0 x_3, x_1 x_2, x_1 x_3)$. Let $S = k[x_0, x_1, x_2, x_3]$. $I$ has four generators $g_1 := x_0 x_2, g_2 := x_0 x_3, g_3 := x_1 x_2, g_4 := x_1 x_3$, all of degree $2$, so our resolution begins $S(-2)^{oplus 4} overset{varphi_0}{to} I to 0$ defined by $e_i mapsto g_i$, where the $e_i$ are the standard basis vectors. We see that $varphi_0$ has kernel generated by
$$
x_1 e_1 - x_0 e_3, x_3 e_1 - x_2 e_2, x_1 e_2 - x_0 e_4, x_3 e_3 - x_2 e_4 , .
$$
We continue our resolution by defining a map $varphi_1: S(-3)^{oplus 4} to S(-2)^{oplus 4}$ onto these generators for $ker(varphi_0)$. (The twist by $-3$ is necessary to make the map have degree $0$.) This can be written as left multiplication by the matrix
$$
begin{pmatrix}
x_1 & x_3 & 0 & 0\
0 & -x_2 & x_1 & 0\
-x_0 & 0 & 0 & x_3\
0 & 0 & -x_0 & -x_2
end{pmatrix} , .
$$
Denoting the standard basis vectors of $S(-3)^{oplus 4}$ by $f_1, f_2, f_3, f_4$, then $varphi_1$ has kernel generated by $x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$. So the final map in our sequence is $varphi_2: S(-4) to S(-3)^{oplus 4}$, $g_1 mapsto x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$. Thus we have constructed the free resolution
$$
0 to S(-4) underset{varphi_2}{overset{begin{pmatrix} x_3\ -x_1\ -x_2\ x_0end{pmatrix}}{longrightarrow}} S(-3)^{oplus 4} xrightarrow[varphi_1]{begin{pmatrix}
x_1 & x_3 & 0 & 0\
0 & -x_2 & x_1 & 0\
-x_0 & 0 & 0 & x_3\
0 & 0 & -x_0 & -x_2
end{pmatrix}} S(-2)^{oplus 4} underset{varphi_0}{to} I to 0
$$
We can tack on $M$ to the righthand side in order to obtain a resolution for $M$:
$$
0 to S(-4) to S(-3)^{oplus 4} to S(-2)^{oplus 4} to S overset{pi}{to} M to 0
$$
where $pi: S to S/I = M$ is the quotient map.
Recall that $S(a)$ has Hilbert function $H_{S(a)}(d) = binom{3 + a + d}{3}$. (In general for $S = k[x_0, ldots, x_r]$ we have $H_{S(a)}(d) = binom{r + a + d}{r}$ by "stars and bars".) By additivity, then
begin{align*}
H_M(d) &= binom{d+3}{3} - 4 binom{3 - 2 + d}{3} + 4 binom{3 - 3 + d}{3} - binom{3 - 4 + d}{3}\
&= binom{d+3}{3} - 4 binom{d+1}{3} + 4 binom{d}{3} - binom{d-1}{3} , .
end{align*}
This gives $H_M(0) = 1, H_M(1) = 4, H_M(2) = 6$ and $H_M(3) = 8$, and reduces to $H_M(d) = 2d+2$ for $d geq 4$.
answered Jan 23 at 4:22
André 3000André 3000
12.7k22243
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2 Answers
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2 Answers
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$begingroup$
The $k$-vector subspace of $k[x_1,x_2,x_3,x_4]$ of homogeneous polynomials of degree $n$ is spanned by the monomials of degree $n$, i.e. terms of the form $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ with $d_1+d_2+d_3+d_4=n$. The kernel of the quotient map is the $k$-vector space spanned by the monomials that satisfy either
$$d_1d_3>0,qquad d_1d_4>0,qquad d_2d_3>0,qquadtext{ or }qquad d_2d_4>0.$$
Hence the $k$-vector subspace of $k[x_1,x_2,x_3,x_4]/(x_1x_2,x_1x_3,x_2x_3,x_2x_4)$ of homogeneous polynomials of degree $n$ is spanned by the monomials $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ that satisfy
$$d_1+d_2+d_3+d_4=k,qquad d_1d_3=d_1d_4=d_2d_3=d_2d_4=0.tag{1}$$
For $k=0$ clearly $d_1=d_2=d_3=d_4=0$ is the unique solution. For $kneq0$ we must have $d_ineq0$ for some $i$. If $d_1neq0$ or $d_2neq0$ then $d_3=d_4=0$, and if $d_3neq0$ or $d_4neq0$ then $d_1=d_2=0$. So the number of solutions to $(1)$ is the same as the number of solutions to
$$d_1+d_2=k,quad d_3=d_4=0qquadtext{ or }qquad d_3+d_4=k,quad d_1=d_2=0,$$
which is precisely twice the number of solutions to $a+b=k$ with $a,binBbb{N}$. This is of course precisely twice $k+1$, as you conjecture.
Earlier version of this answer:
In the quotient, the product of any pair of variables vanishes except the products $x_1x_2$ and $x_3x_4$. This means a basis is given by the monomials of the form
$$x_1^n,qquad x_2^n,qquad x_3^n,qquad x_4^n,qquad x_1^mx_2^n,qquad x_3^mx_4^n.$$
Of course the first four are also of the form $x_1^mx_2^n$ or $x_3^mx_4^n$, simply with $m=0$ or $n=0$.
So for a given degree $k$, you want to count the number of monomials $x_1^mx_2^n$ and $x_3^mx_4^n$ with $m+n=k$. There are $k+1$ choices for $m$, and then $n=k-m$. So there are $k+1$ of monomials of each type, so a total of $2k+2$ monomials.
answered Jan 21 at 17:16
ServaesServaes
27.5k34098
$endgroup$
$begingroup$
Why have you updated your answer? I thought the previous version was fine too.
$endgroup$
– user437309
Jan 21 at 23:21
$begingroup$
I prefer the new version, but I have included the original now.
$endgroup$
– Servaes
Jan 21 at 23:38
add a comment |
$begingroup$
The $k$-vector subspace of $k[x_1,x_2,x_3,x_4]$ of homogeneous polynomials of degree $n$ is spanned by the monomials of degree $n$, i.e. terms of the form $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ with $d_1+d_2+d_3+d_4=n$. The kernel of the quotient map is the $k$-vector space spanned by the monomials that satisfy either
$$d_1d_3>0,qquad d_1d_4>0,qquad d_2d_3>0,qquadtext{ or }qquad d_2d_4>0.$$
Hence the $k$-vector subspace of $k[x_1,x_2,x_3,x_4]/(x_1x_2,x_1x_3,x_2x_3,x_2x_4)$ of homogeneous polynomials of degree $n$ is spanned by the monomials $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ that satisfy
$$d_1+d_2+d_3+d_4=k,qquad d_1d_3=d_1d_4=d_2d_3=d_2d_4=0.tag{1}$$
For $k=0$ clearly $d_1=d_2=d_3=d_4=0$ is the unique solution. For $kneq0$ we must have $d_ineq0$ for some $i$. If $d_1neq0$ or $d_2neq0$ then $d_3=d_4=0$, and if $d_3neq0$ or $d_4neq0$ then $d_1=d_2=0$. So the number of solutions to $(1)$ is the same as the number of solutions to
$$d_1+d_2=k,quad d_3=d_4=0qquadtext{ or }qquad d_3+d_4=k,quad d_1=d_2=0,$$
which is precisely twice the number of solutions to $a+b=k$ with $a,binBbb{N}$. This is of course precisely twice $k+1$, as you conjecture.
Earlier version of this answer:
In the quotient, the product of any pair of variables vanishes except the products $x_1x_2$ and $x_3x_4$. This means a basis is given by the monomials of the form
$$x_1^n,qquad x_2^n,qquad x_3^n,qquad x_4^n,qquad x_1^mx_2^n,qquad x_3^mx_4^n.$$
Of course the first four are also of the form $x_1^mx_2^n$ or $x_3^mx_4^n$, simply with $m=0$ or $n=0$.
So for a given degree $k$, you want to count the number of monomials $x_1^mx_2^n$ and $x_3^mx_4^n$ with $m+n=k$. There are $k+1$ choices for $m$, and then $n=k-m$. So there are $k+1$ of monomials of each type, so a total of $2k+2$ monomials.
answered Jan 21 at 17:16
ServaesServaes
27.5k34098
$endgroup$
$begingroup$
Why have you updated your answer? I thought the previous version was fine too.
$endgroup$
– user437309
Jan 21 at 23:21
$begingroup$
I prefer the new version, but I have included the original now.
$endgroup$
– Servaes
Jan 21 at 23:38
add a comment |
$begingroup$
The $k$-vector subspace of $k[x_1,x_2,x_3,x_4]$ of homogeneous polynomials of degree $n$ is spanned by the monomials of degree $n$, i.e. terms of the form $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ with $d_1+d_2+d_3+d_4=n$. The kernel of the quotient map is the $k$-vector space spanned by the monomials that satisfy either
$$d_1d_3>0,qquad d_1d_4>0,qquad d_2d_3>0,qquadtext{ or }qquad d_2d_4>0.$$
Hence the $k$-vector subspace of $k[x_1,x_2,x_3,x_4]/(x_1x_2,x_1x_3,x_2x_3,x_2x_4)$ of homogeneous polynomials of degree $n$ is spanned by the monomials $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ that satisfy
$$d_1+d_2+d_3+d_4=k,qquad d_1d_3=d_1d_4=d_2d_3=d_2d_4=0.tag{1}$$
For $k=0$ clearly $d_1=d_2=d_3=d_4=0$ is the unique solution. For $kneq0$ we must have $d_ineq0$ for some $i$. If $d_1neq0$ or $d_2neq0$ then $d_3=d_4=0$, and if $d_3neq0$ or $d_4neq0$ then $d_1=d_2=0$. So the number of solutions to $(1)$ is the same as the number of solutions to
$$d_1+d_2=k,quad d_3=d_4=0qquadtext{ or }qquad d_3+d_4=k,quad d_1=d_2=0,$$
which is precisely twice the number of solutions to $a+b=k$ with $a,binBbb{N}$. This is of course precisely twice $k+1$, as you conjecture.
Earlier version of this answer:
In the quotient, the product of any pair of variables vanishes except the products $x_1x_2$ and $x_3x_4$. This means a basis is given by the monomials of the form
$$x_1^n,qquad x_2^n,qquad x_3^n,qquad x_4^n,qquad x_1^mx_2^n,qquad x_3^mx_4^n.$$
Of course the first four are also of the form $x_1^mx_2^n$ or $x_3^mx_4^n$, simply with $m=0$ or $n=0$.
So for a given degree $k$, you want to count the number of monomials $x_1^mx_2^n$ and $x_3^mx_4^n$ with $m+n=k$. There are $k+1$ choices for $m$, and then $n=k-m$. So there are $k+1$ of monomials of each type, so a total of $2k+2$ monomials.
answered Jan 21 at 17:16
ServaesServaes
27.5k34098
$endgroup$
The $k$-vector subspace of $k[x_1,x_2,x_3,x_4]$ of homogeneous polynomials of degree $n$ is spanned by the monomials of degree $n$, i.e. terms of the form $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ with $d_1+d_2+d_3+d_4=n$. The kernel of the quotient map is the $k$-vector space spanned by the monomials that satisfy either
$$d_1d_3>0,qquad d_1d_4>0,qquad d_2d_3>0,qquadtext{ or }qquad d_2d_4>0.$$
Hence the $k$-vector subspace of $k[x_1,x_2,x_3,x_4]/(x_1x_2,x_1x_3,x_2x_3,x_2x_4)$ of homogeneous polynomials of degree $n$ is spanned by the monomials $x_1^{d_1}x_2^{d_2}x_3^{d_3}x_4^{d_4}$ that satisfy
$$d_1+d_2+d_3+d_4=k,qquad d_1d_3=d_1d_4=d_2d_3=d_2d_4=0.tag{1}$$
For $k=0$ clearly $d_1=d_2=d_3=d_4=0$ is the unique solution. For $kneq0$ we must have $d_ineq0$ for some $i$. If $d_1neq0$ or $d_2neq0$ then $d_3=d_4=0$, and if $d_3neq0$ or $d_4neq0$ then $d_1=d_2=0$. So the number of solutions to $(1)$ is the same as the number of solutions to
$$d_1+d_2=k,quad d_3=d_4=0qquadtext{ or }qquad d_3+d_4=k,quad d_1=d_2=0,$$
which is precisely twice the number of solutions to $a+b=k$ with $a,binBbb{N}$. This is of course precisely twice $k+1$, as you conjecture.
Earlier version of this answer:
In the quotient, the product of any pair of variables vanishes except the products $x_1x_2$ and $x_3x_4$. This means a basis is given by the monomials of the form
$$x_1^n,qquad x_2^n,qquad x_3^n,qquad x_4^n,qquad x_1^mx_2^n,qquad x_3^mx_4^n.$$
Of course the first four are also of the form $x_1^mx_2^n$ or $x_3^mx_4^n$, simply with $m=0$ or $n=0$.
So for a given degree $k$, you want to count the number of monomials $x_1^mx_2^n$ and $x_3^mx_4^n$ with $m+n=k$. There are $k+1$ choices for $m$, and then $n=k-m$. So there are $k+1$ of monomials of each type, so a total of $2k+2$ monomials.
answered Jan 21 at 17:16
ServaesServaes
27.5k34098
edited Jan 21 at 23:37
answered Jan 21 at 17:16
ServaesServaes
27.5k34098
answered Jan 21 at 17:16
ServaesServaes
27.5k34098
answered Jan 21 at 17:16
ServaesServaes
27.5k34098
27.5k34098
$begingroup$
Why have you updated your answer? I thought the previous version was fine too.
$endgroup$
– user437309
Jan 21 at 23:21
$begingroup$
I prefer the new version, but I have included the original now.
$endgroup$
– Servaes
Jan 21 at 23:38
add a comment |
$begingroup$
Why have you updated your answer? I thought the previous version was fine too.
$endgroup$
– user437309
Jan 21 at 23:21
$begingroup$
I prefer the new version, but I have included the original now.
$endgroup$
– Servaes
Jan 21 at 23:38
$begingroup$
Why have you updated your answer? I thought the previous version was fine too.
$endgroup$
– user437309
Jan 21 at 23:21
$begingroup$
Why have you updated your answer? I thought the previous version was fine too.
$endgroup$
– user437309
Jan 21 at 23:21
$begingroup$
I prefer the new version, but I have included the original now.
$endgroup$
– Servaes
Jan 21 at 23:38
$begingroup$
I prefer the new version, but I have included the original now.
$endgroup$
– Servaes
Jan 21 at 23:38
add a comment |
$begingroup$
Here's a silly overkill way to find the Hilbert function. We compute a graded resolution for the ring $M := k[x_0,x_1,x_2,x_3]/(x_0x_2, x_0x_3, x_1x_2, x_1x_3)$ and then use additivity.
First let's find a resolution for the ideal $I := (x_0 x_2, x_0 x_3, x_1 x_2, x_1 x_3)$. Let $S = k[x_0, x_1, x_2, x_3]$. $I$ has four generators $g_1 := x_0 x_2, g_2 := x_0 x_3, g_3 := x_1 x_2, g_4 := x_1 x_3$, all of degree $2$, so our resolution begins $S(-2)^{oplus 4} overset{varphi_0}{to} I to 0$ defined by $e_i mapsto g_i$, where the $e_i$ are the standard basis vectors. We see that $varphi_0$ has kernel generated by
$$
x_1 e_1 - x_0 e_3, x_3 e_1 - x_2 e_2, x_1 e_2 - x_0 e_4, x_3 e_3 - x_2 e_4 , .
$$
We continue our resolution by defining a map $varphi_1: S(-3)^{oplus 4} to S(-2)^{oplus 4}$ onto these generators for $ker(varphi_0)$. (The twist by $-3$ is necessary to make the map have degree $0$.) This can be written as left multiplication by the matrix
$$
begin{pmatrix}
x_1 & x_3 & 0 & 0\
0 & -x_2 & x_1 & 0\
-x_0 & 0 & 0 & x_3\
0 & 0 & -x_0 & -x_2
end{pmatrix} , .
$$
Denoting the standard basis vectors of $S(-3)^{oplus 4}$ by $f_1, f_2, f_3, f_4$, then $varphi_1$ has kernel generated by $x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$. So the final map in our sequence is $varphi_2: S(-4) to S(-3)^{oplus 4}$, $g_1 mapsto x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$. Thus we have constructed the free resolution
$$
0 to S(-4) underset{varphi_2}{overset{begin{pmatrix} x_3\ -x_1\ -x_2\ x_0end{pmatrix}}{longrightarrow}} S(-3)^{oplus 4} xrightarrow[varphi_1]{begin{pmatrix}
x_1 & x_3 & 0 & 0\
0 & -x_2 & x_1 & 0\
-x_0 & 0 & 0 & x_3\
0 & 0 & -x_0 & -x_2
end{pmatrix}} S(-2)^{oplus 4} underset{varphi_0}{to} I to 0
$$
We can tack on $M$ to the righthand side in order to obtain a resolution for $M$:
$$
0 to S(-4) to S(-3)^{oplus 4} to S(-2)^{oplus 4} to S overset{pi}{to} M to 0
$$
where $pi: S to S/I = M$ is the quotient map.
Recall that $S(a)$ has Hilbert function $H_{S(a)}(d) = binom{3 + a + d}{3}$. (In general for $S = k[x_0, ldots, x_r]$ we have $H_{S(a)}(d) = binom{r + a + d}{r}$ by "stars and bars".) By additivity, then
begin{align*}
H_M(d) &= binom{d+3}{3} - 4 binom{3 - 2 + d}{3} + 4 binom{3 - 3 + d}{3} - binom{3 - 4 + d}{3}\
&= binom{d+3}{3} - 4 binom{d+1}{3} + 4 binom{d}{3} - binom{d-1}{3} , .
end{align*}
This gives $H_M(0) = 1, H_M(1) = 4, H_M(2) = 6$ and $H_M(3) = 8$, and reduces to $H_M(d) = 2d+2$ for $d geq 4$.
answered Jan 23 at 4:22
André 3000André 3000
12.7k22243
$endgroup$
add a comment |
$begingroup$
Here's a silly overkill way to find the Hilbert function. We compute a graded resolution for the ring $M := k[x_0,x_1,x_2,x_3]/(x_0x_2, x_0x_3, x_1x_2, x_1x_3)$ and then use additivity.
First let's find a resolution for the ideal $I := (x_0 x_2, x_0 x_3, x_1 x_2, x_1 x_3)$. Let $S = k[x_0, x_1, x_2, x_3]$. $I$ has four generators $g_1 := x_0 x_2, g_2 := x_0 x_3, g_3 := x_1 x_2, g_4 := x_1 x_3$, all of degree $2$, so our resolution begins $S(-2)^{oplus 4} overset{varphi_0}{to} I to 0$ defined by $e_i mapsto g_i$, where the $e_i$ are the standard basis vectors. We see that $varphi_0$ has kernel generated by
$$
x_1 e_1 - x_0 e_3, x_3 e_1 - x_2 e_2, x_1 e_2 - x_0 e_4, x_3 e_3 - x_2 e_4 , .
$$
We continue our resolution by defining a map $varphi_1: S(-3)^{oplus 4} to S(-2)^{oplus 4}$ onto these generators for $ker(varphi_0)$. (The twist by $-3$ is necessary to make the map have degree $0$.) This can be written as left multiplication by the matrix
$$
begin{pmatrix}
x_1 & x_3 & 0 & 0\
0 & -x_2 & x_1 & 0\
-x_0 & 0 & 0 & x_3\
0 & 0 & -x_0 & -x_2
end{pmatrix} , .
$$
Denoting the standard basis vectors of $S(-3)^{oplus 4}$ by $f_1, f_2, f_3, f_4$, then $varphi_1$ has kernel generated by $x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$. So the final map in our sequence is $varphi_2: S(-4) to S(-3)^{oplus 4}$, $g_1 mapsto x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$. Thus we have constructed the free resolution
$$
0 to S(-4) underset{varphi_2}{overset{begin{pmatrix} x_3\ -x_1\ -x_2\ x_0end{pmatrix}}{longrightarrow}} S(-3)^{oplus 4} xrightarrow[varphi_1]{begin{pmatrix}
x_1 & x_3 & 0 & 0\
0 & -x_2 & x_1 & 0\
-x_0 & 0 & 0 & x_3\
0 & 0 & -x_0 & -x_2
end{pmatrix}} S(-2)^{oplus 4} underset{varphi_0}{to} I to 0
$$
We can tack on $M$ to the righthand side in order to obtain a resolution for $M$:
$$
0 to S(-4) to S(-3)^{oplus 4} to S(-2)^{oplus 4} to S overset{pi}{to} M to 0
$$
where $pi: S to S/I = M$ is the quotient map.
Recall that $S(a)$ has Hilbert function $H_{S(a)}(d) = binom{3 + a + d}{3}$. (In general for $S = k[x_0, ldots, x_r]$ we have $H_{S(a)}(d) = binom{r + a + d}{r}$ by "stars and bars".) By additivity, then
begin{align*}
H_M(d) &= binom{d+3}{3} - 4 binom{3 - 2 + d}{3} + 4 binom{3 - 3 + d}{3} - binom{3 - 4 + d}{3}\
&= binom{d+3}{3} - 4 binom{d+1}{3} + 4 binom{d}{3} - binom{d-1}{3} , .
end{align*}
This gives $H_M(0) = 1, H_M(1) = 4, H_M(2) = 6$ and $H_M(3) = 8$, and reduces to $H_M(d) = 2d+2$ for $d geq 4$.
answered Jan 23 at 4:22
André 3000André 3000
12.7k22243
$endgroup$
add a comment |
$begingroup$
Here's a silly overkill way to find the Hilbert function. We compute a graded resolution for the ring $M := k[x_0,x_1,x_2,x_3]/(x_0x_2, x_0x_3, x_1x_2, x_1x_3)$ and then use additivity.
First let's find a resolution for the ideal $I := (x_0 x_2, x_0 x_3, x_1 x_2, x_1 x_3)$. Let $S = k[x_0, x_1, x_2, x_3]$. $I$ has four generators $g_1 := x_0 x_2, g_2 := x_0 x_3, g_3 := x_1 x_2, g_4 := x_1 x_3$, all of degree $2$, so our resolution begins $S(-2)^{oplus 4} overset{varphi_0}{to} I to 0$ defined by $e_i mapsto g_i$, where the $e_i$ are the standard basis vectors. We see that $varphi_0$ has kernel generated by
$$
x_1 e_1 - x_0 e_3, x_3 e_1 - x_2 e_2, x_1 e_2 - x_0 e_4, x_3 e_3 - x_2 e_4 , .
$$
We continue our resolution by defining a map $varphi_1: S(-3)^{oplus 4} to S(-2)^{oplus 4}$ onto these generators for $ker(varphi_0)$. (The twist by $-3$ is necessary to make the map have degree $0$.) This can be written as left multiplication by the matrix
$$
begin{pmatrix}
x_1 & x_3 & 0 & 0\
0 & -x_2 & x_1 & 0\
-x_0 & 0 & 0 & x_3\
0 & 0 & -x_0 & -x_2
end{pmatrix} , .
$$
Denoting the standard basis vectors of $S(-3)^{oplus 4}$ by $f_1, f_2, f_3, f_4$, then $varphi_1$ has kernel generated by $x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$. So the final map in our sequence is $varphi_2: S(-4) to S(-3)^{oplus 4}$, $g_1 mapsto x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$. Thus we have constructed the free resolution
$$
0 to S(-4) underset{varphi_2}{overset{begin{pmatrix} x_3\ -x_1\ -x_2\ x_0end{pmatrix}}{longrightarrow}} S(-3)^{oplus 4} xrightarrow[varphi_1]{begin{pmatrix}
x_1 & x_3 & 0 & 0\
0 & -x_2 & x_1 & 0\
-x_0 & 0 & 0 & x_3\
0 & 0 & -x_0 & -x_2
end{pmatrix}} S(-2)^{oplus 4} underset{varphi_0}{to} I to 0
$$
We can tack on $M$ to the righthand side in order to obtain a resolution for $M$:
$$
0 to S(-4) to S(-3)^{oplus 4} to S(-2)^{oplus 4} to S overset{pi}{to} M to 0
$$
where $pi: S to S/I = M$ is the quotient map.
Recall that $S(a)$ has Hilbert function $H_{S(a)}(d) = binom{3 + a + d}{3}$. (In general for $S = k[x_0, ldots, x_r]$ we have $H_{S(a)}(d) = binom{r + a + d}{r}$ by "stars and bars".) By additivity, then
begin{align*}
H_M(d) &= binom{d+3}{3} - 4 binom{3 - 2 + d}{3} + 4 binom{3 - 3 + d}{3} - binom{3 - 4 + d}{3}\
&= binom{d+3}{3} - 4 binom{d+1}{3} + 4 binom{d}{3} - binom{d-1}{3} , .
end{align*}
This gives $H_M(0) = 1, H_M(1) = 4, H_M(2) = 6$ and $H_M(3) = 8$, and reduces to $H_M(d) = 2d+2$ for $d geq 4$.
answered Jan 23 at 4:22
André 3000André 3000
12.7k22243
$endgroup$
Here's a silly overkill way to find the Hilbert function. We compute a graded resolution for the ring $M := k[x_0,x_1,x_2,x_3]/(x_0x_2, x_0x_3, x_1x_2, x_1x_3)$ and then use additivity.
First let's find a resolution for the ideal $I := (x_0 x_2, x_0 x_3, x_1 x_2, x_1 x_3)$. Let $S = k[x_0, x_1, x_2, x_3]$. $I$ has four generators $g_1 := x_0 x_2, g_2 := x_0 x_3, g_3 := x_1 x_2, g_4 := x_1 x_3$, all of degree $2$, so our resolution begins $S(-2)^{oplus 4} overset{varphi_0}{to} I to 0$ defined by $e_i mapsto g_i$, where the $e_i$ are the standard basis vectors. We see that $varphi_0$ has kernel generated by
$$
x_1 e_1 - x_0 e_3, x_3 e_1 - x_2 e_2, x_1 e_2 - x_0 e_4, x_3 e_3 - x_2 e_4 , .
$$
We continue our resolution by defining a map $varphi_1: S(-3)^{oplus 4} to S(-2)^{oplus 4}$ onto these generators for $ker(varphi_0)$. (The twist by $-3$ is necessary to make the map have degree $0$.) This can be written as left multiplication by the matrix
$$
begin{pmatrix}
x_1 & x_3 & 0 & 0\
0 & -x_2 & x_1 & 0\
-x_0 & 0 & 0 & x_3\
0 & 0 & -x_0 & -x_2
end{pmatrix} , .
$$
Denoting the standard basis vectors of $S(-3)^{oplus 4}$ by $f_1, f_2, f_3, f_4$, then $varphi_1$ has kernel generated by $x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$. So the final map in our sequence is $varphi_2: S(-4) to S(-3)^{oplus 4}$, $g_1 mapsto x_3 f_1 - x_1 f_2 - x_2 f_3 + x_0 f_4$. Thus we have constructed the free resolution
$$
0 to S(-4) underset{varphi_2}{overset{begin{pmatrix} x_3\ -x_1\ -x_2\ x_0end{pmatrix}}{longrightarrow}} S(-3)^{oplus 4} xrightarrow[varphi_1]{begin{pmatrix}
x_1 & x_3 & 0 & 0\
0 & -x_2 & x_1 & 0\
-x_0 & 0 & 0 & x_3\
0 & 0 & -x_0 & -x_2
end{pmatrix}} S(-2)^{oplus 4} underset{varphi_0}{to} I to 0
$$
We can tack on $M$ to the righthand side in order to obtain a resolution for $M$:
$$
0 to S(-4) to S(-3)^{oplus 4} to S(-2)^{oplus 4} to S overset{pi}{to} M to 0
$$
where $pi: S to S/I = M$ is the quotient map.
Recall that $S(a)$ has Hilbert function $H_{S(a)}(d) = binom{3 + a + d}{3}$. (In general for $S = k[x_0, ldots, x_r]$ we have $H_{S(a)}(d) = binom{r + a + d}{r}$ by "stars and bars".) By additivity, then
begin{align*}
H_M(d) &= binom{d+3}{3} - 4 binom{3 - 2 + d}{3} + 4 binom{3 - 3 + d}{3} - binom{3 - 4 + d}{3}\
&= binom{d+3}{3} - 4 binom{d+1}{3} + 4 binom{d}{3} - binom{d-1}{3} , .
end{align*}
This gives $H_M(0) = 1, H_M(1) = 4, H_M(2) = 6$ and $H_M(3) = 8$, and reduces to $H_M(d) = 2d+2$ for $d geq 4$.
answered Jan 23 at 4:22
André 3000André 3000
12.7k22243
edited Jan 23 at 5:18
answered Jan 23 at 4:22
André 3000André 3000
12.7k22243
answered Jan 23 at 4:22
André 3000André 3000
12.7k22243
answered Jan 23 at 4:22
André 3000André 3000
12.7k22243
12.7k22243
add a comment |
add a comment |
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