Prove: A basis $G$ is a Grobner basis of an ideal $\iff$ for every element $S$ in a homogeneous basis for the syzygies $S(G)$ we have $S.G \to_{G}0. $

algebraic-geometrycommutative-algebragroebner-basisidealsmultivariate-polynomial

A basis $G=(g_1,…,g_t)$ for an ideal I is a Grobner basis $\iff$ for every element $S=(h_1,…h_t)$ in a homogeneous basis for the syzygies $S(G)$ we have $S.G = \Sigma_{i=1}^{t} h_ig_i \to_{G}0. $

This is theorem 9 in chapter 2.9 of Ideals, Varieties and Algorithms by Cox, Little and O'Shea. The proof given there is not clear to me.

Definitions in the text:

Let $F=(f_1,…,f_s)$. A syzygy on the on the leading terms $LT(f_1),…,LT(f_s)$ of $F$ is an s-tuple of polynomials $S=(h_1,…,h_s)\in (k[x_1,…,k_n])^s$ such that $\Sigma_{i-1}^{s} h_iLT(f_i)=0$. $S(F)$ is a collection of all syzygies on $F$.

An element $S \in S(F)$ is homogeneous of multidegree $\alpha$ where $\alpha \in Z^{n}_{\geq0}$ provided that $S=(c_1x^{\alpha_1},…,c_sx^{\alpha_s})$ where $c_i\in k$ and $\alpha_i + \ multideg(f_i) = \alpha$ whenever $c_i \neq 0$

We say that $f \in [x_1,..,x_n]$ reduces to zero modulo $G = \{g_1,…,g_t\} \subset k[x_1,…,x_n]$ i.e. $f \to_{G} 0$ if $f$ can be written in the form $f = a_1g_1+…+a_tg_t$ such that whenever $a_ig_i \neq 0$, we have multideg(f) $\geq$ multideg($a_ig_i$).

My Attempt:

($\implies$) Given $G$ to be a Grobner basis and $S=(h_1,…,h_t)$ a syzygy of $G$ that is homogeneous we know that $\Sigma_{i=1}^t h_iLT(g_i) = 0$ and that $LT(h_i).LT(g_i)$ have the same multidegree for all $i$.

To prove RHS we need to show that multideg($\Sigma h_ig_i$) $\geq$ multideg($h_ig_i$) for all $i$ which is equivalent to showing $\Sigma_{i=1}^t LT(h_i)LT(g_i) \neq 0 $. But the above would imply that $\Sigma_{i=1}^t LT(h_i)LT(g_i)=0$.

What is wrong here ?

Help also in proving the other direction.

Best Answer

$ S.G \rightarrow _G 0$ means that $S.G$ can be written as $\Sigma a_ig_i$ where multideg($S.G$) > multideg($a_ig_i$) whenever $a_ig_i \neq 0$. Here $a_i$ need not equal $h_i$.

This is all cleared up in the fourth version of the book.

Related Solutions

Coefficients such that linear combination lies in an ideal

Suppose $I = \langle f_1, \ldots, f_r\rangle$ and consider the map $R^r \oplus R^m \twoheadrightarrow I + J$ that sends $(b,c) \mapsto \sum_j b_j f_j + \sum_i c_i g_i$.

Its kernel $K$ is the first module of syzygies for $I+J$. The projection of $K$ to $R^m$ is exactly the $R$-module $T$ that you want.

Proof. The projection of $K$ to $R^m$ consists of those $c \in R^m$ for which there exists $b$ such that $\sum c_i g_i = -\sum b_j f_j \in I$, and for every $c \in R^m$ with $\sum c_i g_i \in I$ there exists such a $b$.

So the algorithm is: calculate the first module of syzygies for $I+J$ and project to $R^m$.

Example: $I = \langle f \rangle = \langle x^2 + y^2 - 1\rangle$ and $J = \langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\rangle = \langle 2x, 2y \rangle$. In SageMath:

sage: R.<x,y> = PolynomialRing(QQ,2)
sage: I = R.ideal(x^2+y^2-1)
sage: J = R.ideal(2*x, 2*y)
sage: T = (I+J).syzygy_module()[:,len(I.gens()):]; T
[           -y             x]
[          x*y       y^2 - 1]
[x^2 + y^2 - 1             0]

Here the projection to $R^m$ is achieved by slicing the matrix of generators of $K$.

The rows of the resulting matrix are generators of $T$.

In the example we see that $x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}$ is a polynomial vector field on the circle and furthermore $IR^m \subset T$.

One might be more interested in a (smaller) set of generators for the quotient $T/IR^m$.

To this end we can find the module of syzygies for $T + IR^m$ and project to the space of coefficients for generators of $T$; these are syzygies for $T$ modulo $IR^m$. For every such syzygy modulo $IR^m$ where one of the coefficients is an invertible constant, remove the respective generator (because it can be expressed in terms of the others, modulo $IR^m$).

Continuing the example in Singular:

> ring r=0,(x,y),dp;
> module n = [-y, x], [xy, y2-1], [x2+y2-1,0], [x2+y2-1,0], [0,x2+y2-1]; // T + IR^m
> LIB "matrix.lib";
...
> print(submat(transpose(syz(n)),1..3,1..3));
0,   0, -1,
x,   1, 0, 
y2-1,-x,y

Here we see that columns two and three contain invertible constants, so the second and third generators can be removed.

So in this example $T/IR^m$ is generated by $(-y, x)$.

For completeness, the calculation above can be achieved (in general) through the SageMath interface to Singular:

sage: import itertools
sage: m = len(J.gens())
sage: IRm = matrix([[i if j == b else 0 for j in range(0,m)]
                    for (i,b) in itertools.product(I.gens(),range(0, m))])
sage: r = R._singular_()
sage: n = singular(map(list, list(T) + list(IRm)), type='module') # T + IR^m
sage: singular('transpose(syz(%s))' % n.name(), type='matrix').sage()[:,0:T.nrows()]
[      0       0      -1]
[      x       1       0]
[y^2 - 1      -x       y]

A final useful remark is that the set of $\sum c_i g_i$, where $c$ ranges over the generators of $T$, is a set of generators for $I \cap J$. In SageMath:

sage: R.ideal([sum(map(prod,zip(t,J.gens()))) for t in T]) == I.intersection(J)
True

The proof attempt in the question was an attempt to reverse this process.

Best Answer

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Coefficients such that linear combination lies in an ideal

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