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This is an example from Miller, Sturmfels "Combinatorial Commutative Algebra".

Let $\pi:\mathbb{Z}^3 \rightarrow \mathbb{Z}$ be the group homomorphism defined by the matrix $\left(\begin{array}{ccc} 3 & 4 & 5 \end{array} \right)$. Thus

$$ \ker(\pi)=\{(u,v,w) \in \mathbb{Z}^3 \ | \ 3u+4v+5w=0\}. $$

A lattice basis for the kernel is $\mathbb{Z}\{(3,-1,-1),(2,1,-2)\}$. I am trying to understand how one goes from a lattice basis for $\ker(\pi)$ to a generating set for the corresponding lattice ideal

$$ I_L=\langle \mathbf{x}^{\mathbf{u}}-\mathbf{x}^{\mathbf{v}} \ | \ u,v \in \mathbb{N}^3 \ \textrm{with} \ \mathbf{u}-\mathbf{v} \in L \rangle. $$

In this example it is said that $I_L = \langle x^3-yz, \ x^2y-z^2, \ xz-y^2 \rangle$. Why is $I_L \neq \langle x^3-yz, \ x^2y-z^2 \rangle$ i.e. the analogue of the lattice basis above?

I want to be able to go from the lattice basis to a minimal generating set for $I_L$.

RedOrange
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1 Answers1

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One way to go from a lattice basis to a minimal generating set for $I_L$ is described in Lemma 7.6 in Miller/Sturmfels. In particular, a lattice ideal is saturated by the product of all variables.

In your example, the lattice ideal is actually equal to the saturated ideal $$I_L=\langle x^3-yz,x^2y-z^2\rangle : \langle xyz\rangle^\infty$$ A Gröbner basis (with respect to the graded reverse lexicographical ordering) of this ideal is given by the three binomials $x^3-yz,x^2y-z^2,xz-y^2$.

emeu
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  • Thanks, @emeu. How can I compute such a saturation by hand? Or how could I get a Groebner basis for $I_L$? The algorithms that I am familiar with for computing Groebner bases require one to start off with a generating set. – RedOrange Dec 02 '13 at 17:14
  • Saturation can be computed by adding a new variable: if $I$ is an ideal of $k[X]$ and $f\in k[X]$ then by introducing a new variable $T$, there is the relation $I:f^\infty=(I+⟨Tf−1⟩)\cap k[X]$. In your example, you can compute a Gröbner basis G of $⟨x^3−yz,x^2y−z^2,xyzt−1⟩$ for the lexicographical ordering with $x\prec y\prec z\prec t$ (I am not 100% sure, but I believe that the ordering doesn't matter for binomials ideals. In the general case, the monomial ordering has to be an elimination ordering). Then the elements of $G$ which lie in $k[x,y,z]$ will be the Gröbner basis of $I_L$. – emeu Dec 03 '13 at 18:35