Maximal ideal in local field and density of absolute value Let $\mathbb{K}$ a non archimedean local field, and $\mathfrak{M}$ the maximal ideal of its integers ring. I have to show that $\mathfrak{M}^2\subset\mathfrak{M}$ implies that the absolute value $|\cdot|_\mathbb{K}$ of the field is discrete.
I just started working with non-archimedean absolute values and valutations so the answer is probably easy but I can't really see it. Any hint?

As said in the comment, assume $|\cdot|$ is not discrete. Pick $\pi\in\mathfrak{M}-\mathfrak{M}^2$ and for density there is a $\rho\in\mathfrak{M}$ such that $|\pi|<|\rho|<1$. So both $\rho$ and $\pi/\rho$ are in $\mathfrak{M}$ and $\pi=\rho\cdot\pi/\rho\in\mathfrak{M}\cdot\mathfrak{M}\subseteq \mathfrak{M}^2$, a contradiction.
