Let $ I \neq {\{0}\}$ be a monomial ideal of $\mathbb{Q}[X_1, ..., X_n]$. Show that $I$ contains a monomial $m$ such that $m$ is divisible by exactly $2015$ other monomials contained in $I$.
Now, at first glance, surely this has something to do with Dickson's lemma; where there exists a finite subset $T$ of $S$, which is a subset of all monomials in $X_1, ..., X_n$, such that every monomial in $S$ is divisible by at least one of the monomials in $T$.
I can't seem to apply it here, though.
Any suggestions would be appreciated!
It will be easier if we look at the Newton polyhedron of $I$. Just recall that the Newton polyhedron of $I$ is the convex hull of all exponents of monomials in $I$. It is well-known that Newton polyhedron of $I$ can be written as the Minkowski sum of the Newton polytope of $I$ (convex hull of exponents of minimal generators) and the (recession) cone $\mathbb{Q}^n$.
Now, take any integral vertex $v$ of the Newton polytope and consider some extreme ray (from the recession cone) passes through it. Then any point of the Newton polyhedron in this extreme ray can only be written as positive multiple of the vertex. Therefore, on this ray, pick the $2005$-th integral point. This will give the exponent to the monomial that we want to find.