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Let $M$ be an algebraic family of isocrystals over a base scheme $R/\mathbb{Q}_p$ (not a rigid analytic space).

The question is: is the set of weakly admissible points (i.e., the points $r\in R$ over which $M$ is weakly admissible) Zariski closed or open (or neither)?

The question seems to (in my case) come down to determining if the points $r\in R$ where $v_p(\mathrm{det} M_r)$ (where $M_r$ is the fiber above $r$) is bounded above, is Zariski closed/open/neither. Here $v_p$ is the $p$-adic valuation.

The answer might be very simple and/or well-known, but I haven't been able to figure it out.

Thanks for any help!

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# Weak admissibility in algebraic families

Let $M$ be an algebraic family of isocrystals over a base scheme $R/\mathbb{Q}_p$ (not a rigid analytic space).

The question is: is the set of weakly admissible points (i.e., the points $r\in R$ over which $M$ is weakly admissible) Zariski closed or open (or neither)?

The question seems to (in my case) come down to determining if the points $r\in R$ where $v_p(\mathrm{det} M_r)$ (where $M_r$ is the fiber above $r$) is bounded above, is Zariski closed/open/neither. Here $v_p$ is the $p$-adic valuation.

The answer might be very simple and/or well-known, but I haven't been able to figure it out.

Thanks for any help!