Let $R=k[x,y,z]$ be a polynomial ring over a field $k$. Let $m,v$ be monomials in $R$ of degree $d$. Let $I$ be a monomial ideal (i.e. ideal generated by monomials) in $R$ and $m\in I+(v)$ but $m\notin I$. Let $\mathrm{supp}(m)=\{i : x_i\text{ divides } m\}$ and $\mathrm{supp}(v)=\{i : x_i\text{ divides } v\}$.
Q) Is it true that $\mathrm{supp}(m)=\mathrm{supp}(v)$?
I think this is easier than it looks for the following reason: if $m \in I + (v)$, but $m \not \in I$, $m$ must have the form $v^l n$, where $l \in \mathbb N$ and $n \not \in I$. Then $m=v^l n$. But $m$ and $v$ have the same degree $d$, hence $n$ must be a scalar and $l=1$.
So $m=cn$, and so their support is clearly equal.