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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?

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  • $\begingroup$ Choose a $\pi\in\mathfrak{M}\setminus\mathfrak{M}^2$. If there is a $\rho\in\mathbb K$ such that $1>|\rho|>|\pi|$ then $\rho,\pi/\rho\in\mathfrak M$ and thus $\pi\in\mathfrak M^2$, a contradiction. $\endgroup$ – user8268 Dec 28 '14 at 22:42
  • $\begingroup$ Ok it works. Thanks! $\endgroup$ – Angelo Rendina Dec 28 '14 at 22:51
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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.

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