About a month ago I asked this question on math.stackexchange and unfortunately there was no response. Perhaps someone here knows the answer.


Let $A \in \mathbb{Z}^{m \times n}$ be a matrix of full row rank and $m < n$. Let $\ker A = \{ u \in \mathbb{Z}^n \; | \; Au = 0\}$. For any $u \in \ker A$ vectors $u\_+$ and $u\_-$ can be defined as follows:
$$u\_+ = \sum\_{u\_i > 0} u\_i e\_i,$$
$$u\_- = - \sum\_{u\_i < 0} u\_i e\_i.$$
The ideal $I\_A = \langle \{ x^{u\_+} - x^{u\_-} \; | \; u \in \ker A \} \rangle$ is no doubt toric. The well-known way to construct $I\_A$ is the following one. Let $L = \{ l^1,...,l^r \}$ $(r = n - m)$ be a basis for $\ker A$ and
$$I\_L = \langle \{ x^{l^i\_+} - x^{l^i\_-} \; | \; i = \overline{1,r}\; \} \rangle$$
then
$$I\_A = I\_L : (x\_1 \cdots x\_n)^\infty.$$
My questions are the following:


1) Is it right that $I\_L$ is always $I\_A$-primary (i.e. $I\_A = \sqrt{I\_L}$)? If so is there any relatively easy proof of it? If not what is the case when this statement may fail?


2) Are there any reasonable bounds on the size of a reduced Groebner basis of $I\_A$ (for example for grevlex ordering)? Possible bounds in terms of matrix $A$ (or equivalently in terms of its basis $L$) would be much appreciated.


Any hint or reference would help a lot.


Thanks in advance.



Computing toric ideals via saturation and Groebner bases of toric ideals
The answer to your first question is "No". I think it's almost always No. You can already see this in the well-known twisted cubic example. Here 
$A = \begin{pmatrix}1&
 1&
 1&
 1\\
 0&
 1&
 2&
 3\\
 \end{pmatrix}$ and take, for example, the lattice basis {(1,-2,1,0),(0,1,-2,1)}. The corresponding binomial ideal $(-b^{2}+a c,-c^{2}+b d)$ is not primary. It has two components: $I\_L$ and the ideal $(b,c)$


