Are the conditions for completeness over a valued field equivalent?

I am a beginner in learning perfectoid fields, and I have several questions about complete valued fields.

Let $K$ be a nonarchimedian valued field(not necessarily discrete). Let $m_K$ be the maximal ideal of the valuation ring $O_K$, and assume the residue field $k=O_K/m_K$ is of characteristic $p$(which means $p\in m_K$). I want to make sure wether these conditions for completeness are equivalent:

(1)The absolute value $|\cdot|_K$ is complete.

(2)$O_K$ is complete in $m_K$-adic topology.

(3)For some proper ideal $I\subset m_K$, $O_K$ is complete in the $I$-adic topology.(Especially we can take $I=(p)$)

I know when the valuation is discrete these are equivalent. But I don't know the nondiscrete case(that's the case in perfectoid fields.). Could you prove them or give some clearer statements about them? Thanks!

Solution 1:

Assume the norm is nontrivial and fix any $\varpi\in m_K\smallsetminus\{0\}$. If you replace (3) with the following

(3) $O_K$ is $\varpi$-adically complete

then (1) and (3) are equivalent.

You will never be $m_K$-adically complete when non-discretely valued; the issue is separatedness. In fact non-discreteness is equivalent to the assertion $m_K^2=m_K$, so in this case you can never get $\bigcap_n m_K^n=0$ which is what it means to be $m_K$-adically separated.

Edit: I guess it's worth adding the comment that your original statement of (3) is also equivalent to (1), it's just not an ideal formulation. This is because if $I$ is not finitely generated then you can't be $I$-adically separated for the same reasoning as above, so $I$ must be finitely generated but this implies that $I$ is principal (take a finite set of generators and the one with largest absolute value will generate $I$), so you might as well state it with $I$ a principal ideal.