Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a collection $A:=\{a_1, \ldots ,a_n \}$ of different integer points in $\mathbb{N}^d$, which span an affine hyperplane when viewed in $\mathbb{R}^d$, one can define a toric ideal $I_A$ from a monomial homomophism: \begin{align} \phi_A\colon k[x_1,\ldots, x_n]&\to k[y_1,\ldots, y_d],\\ x_i&\mapsto y^{a_i}:=\prod_{j=1}^{d}y_j^{a_{i,j}} \end{align} according to $I_A:=\ker \phi_A$ (suppose $k$ is algebraically closed of characteristic zero). Let $H$ be a hyperplane in $k^d$ such that $conv(A)$ (when viewed as a polytope in $\mathbb{R}^d$) has integer (or at least rational) vertices, and call $A\cap H$ the resulting point configuration, consisting of the vertices of $A\cap H$. To rule out known situations, suppose $H$ further that intersects the interior of $A$.

Question: Carrying out the construction for the toric ideal with the input point configuration $A\cap H$, is there a relation between $I_{A\cap H}, I_A$ and $H$? Or are there hypothesis that can lead to a relation between the two ideals (or the corresponding rings)?

share|cite|improve this question

1 Answer 1

There may be distinct ways of viewing this topic, but the way I am familiar with it we have that the monomial homomorphism is defined by $$\phi_{A} : k[x_{1},...,x_{n}] \rightarrow k[t_{1},...,t_{d},t_{1}^{-1},...,t_{d}^{-1}] \\ \phi(x_{i}) \mapsto \mathbf{t}^{a_{i}}:=\prod_{j=1}^{d} t_{j}^{a_{j,i}}, \forall 1 \leq i \leq n.$$ My suggestion is to try using the rational polyhedral cone $$\text{pos}_{\mathbb{Q}}((a_{1},...,a_{n})) = \left\{ \sum_{i=1}^{n} \lambda_{i}a_{i} \; | \; \lambda_{i} \in \mathbb{Q}_{\geq 0}\right\}$$ attached to the toric variety $$V(I_{A}) = \{(u_{1},...,u_{n}) \in k^{n} \; | \; F(u_{1},...,u_{n})=0, \forall F \in I_{A}\}$$ where $I_{A}=\ker(\phi_{A})$ is the toric ideal. In particular, my intuitive idea is that you construct $I_{A}$ and then pass to the toric variety $V(I_{A})$ attached to $I_{A}$ (which is in your case an affine monomial curve if you choose $a_{1}<\cdot\cdot\cdot<a_{n}$ as relatively prime positive integers, think of it in this case before you generalize!). Now, constructing the polyhedral cone from the toric variety $V(I_{A})$ will allow you to have some sort of bound to how the points of $A$, and hence of $\text{conv}(A)$ will behave. In particular, I think you will be able to define $A$ and $A \cap H$ as subsets of the polyhedral cone, and that there is an associated height of the cone for which we define the hyperplane $H$ so that $$\text{conv}\left((A \cap H)\cup \bigcup_{i=1}^{n}\chi_{H}(a_{i})\right) \subset \text{pos}_{\mathbb{Q}}((a_{1},...,a_{n}))\big|_{h}$$ where $\chi_{H}: \mathbb{N}^{d} \rightarrow \mathbb{N}^d$ is an indicator function defined in terms of the hyperplane $H$ (and a chosen orientation for a normal) which will equal $a_{i}$ when the point is on the desired side of the hyperplane (bounding the polyhedral cone to a subset with finite metric quantities) and $1$ when the point is on the undesired side of the hyperplane (the unbounded region). I am using the $\big|_{h}$ on the rational polyhedral cone to denote the restriction unto the height $h$ induced by your choice of $H$. Using some of these ideas and intuitions I recommend that you try and construct the toric varieties and rational polyhedral cones attached to $I_{A}$ and $I_{A \cap H}$ in order to understand their relationship as toric ideals.

share|cite|improve this answer
Thanks for your answer. Perhaps I've misread it, but as I defined the integer points $A$ they define an arbitrary affine toric variety, rather than a monomial curve as you say. Anyhow, I have started a bounty since I'm more interested in an answer of the sort "under these hypotheses..., the precise relation is this:..." or "it's a mess and there is no visible relation between both ideals". – Camilo Sarmiento Jan 31 '14 at 18:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.