# if $\lambda (I)= \dim R$, can one claim that $I$ is an $m$-primary ideal?

definition from Bruns-Herzog:

It is easy to see that if $I$ is a $m$-primary ideal of $R$ then $\lambda (I)= \dim R$. I wonder if the converse is true:

if $\lambda (I)= \dim R$, can one claim that $I$ is an $m$-primary ideal? if not;
1- is there an explicit counterexample? (not using algebraic geometry)
2- is there a known fact that suggests ideals, $I$, such that $\lambda (I)= \dim R$?

thanks.

• @user26857 Do you remember what you said? – Todd Trimble Dec 28 '14 at 11:01
• @ToddTrimble My comment said the following: "look for the analytic spread of determinantal ideals". (Google shows right a way that the analytic spread of determinantal ideals $I_t(X)$ generated by all $t$-minors of a generic $m\times n$ matrix is $mn$ for $t<\min(m,n)$.) – user26857 Jan 11 '15 at 10:50

Let $A$ be a polynomial ring in $n$ variables over a field, and let $I$ be a monomial ideal. One may represent $I$ as the subset of the nonnegative orthant of $\mathbb R^n$ given by the exponents of monomials in $I$. Let $c(I)$ be the maximum among the dimensions of the compact faces of the convex hull of this subset of $\mathbb R^n$. Bivia-Ausina showed in a Communications in Algebra article from 2003 ("The analytic spread of monomial ideals") that in this context (and somewhat more generally for so-called Newton non-degenerate ideals), $\lambda(I) = c(I)+1$.
In particular, $I$ has maximal analytic spread if and only if some compact face of its Newton polyhedron has dimension $n-1$. This is a very easy criterion to check in low dimension, given a minimal generating set of monomials.
The smallest example I know of is $A=k[x,y]$ and $I=(xy^2, x^2y)$. Then the boundary of the Newton polyhedron consists of the vertical upward-pointing ray starting at $(1,2)$, the horizontal rightward-pointing ray starting at $(2,1)$, and the line segment between $(1,2)$ and $(2,1)$. The last of these is the lone compact face, and it has dimension $1$, whence the analytic spread is $2$, which is the dimension of $A$.