Let $L$ be a modular lattice. We say that $x$ is essential in $L$, if $x\wedge y \neq 0$ for all $y\in L$ with $y\neq 0$. Also, we denote the fact $x\unlhd y $, if $x\leq y$ and $x$ is essential in $[0,y]$.
My question is. If $x\unlhd y_1$ and $x\unlhd y_2$ then $x\unlhd y_1\vee y_2$?
We know that the dual statement is true, If $x_1\unlhd y$ and $x_2\unlhd y$ then $x_1\wedge x_2\unlhd y$.
Consider the lattice in the following picture:
It's easy to see that it's modular: to obtain a pentagon, one needs a four-element chain $u < b < c < v$ and an element $w$ with $b \vee w = v$ and $c \wedge w = u$, and that pattern doesn't show in the lattice.
Now, $x\unlhd y_1$ since for $0 < t \leq y_1$, we don't have $x \wedge t = 0$. Likewise $x\unlhd y_2$.
But $x \wedge z = 0$, and $0 \neq z < y_1 \vee y_2$. Hence $x\not\unlhd y_1 \vee y_2$.