We recall two definitions. Let $A$ be a linearly topologized ring which is complete and Hausdorff.



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> We say that $A$ is *pseudo-compact* if, for every open ideal $I\subset A$, the ring $A/I$ is artinian. We say that $A$ is *admissible* if $A$ has an ideal of definition (an open ideal $I\subset A$ such that every neighbourhood of $0$ contains $I^n$ for some $n$).
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Both seem to be used in the definition of formal schemes in algebraic geometry. **I wonder if one of them implies the other.**



What's the relation between pseudo-compact and admissible rings?
Neither of these properties implies the other:


**There is an admissible algebra that is not pseudo-compact**


If $A$ is a ring with the discrete topology that is not artinian then it is admissible but not pseudocompact.


Less degenerately if $A$ is a noetherian ring and $I$ is an ideal such that $A/I$ is not artinian then the $I$-adic completion of $A$ is admissible but not pseudocompact --- the first example is simply the case $I=0$.


**There is a pseudo-compact ring that is not admissible**


If $G$ is any profinite group then the complete group algebra $K[[G]]$ over an algebraically closed field $K$ of characteristic zero, in the sense of <https://www.ams.org/journals/bull/1966-72-02/S0002-9904-1966-11513-6/S0002-9904-1966-11513-6.pdf>, is pseudocompact.


I guess you are only interested in commutative rings but if we take $G$ to be the profinite completion of the integers then $K[[G]]$ is commutative.


Now for every prime $p$, there is a quotient $C\_p$ of $G$ that is cyclic of order $p$. Moreover there is non-trivial homomorphism $\theta\_p\colon K[[G]]\to K$ that factors through $C\_p$. For each $p$, $\ker \theta\_p$ is an open (maximal) ideal. Moreover if $p\neq q$ then $\ker \theta\_p\neq \ker \theta\_q$.


If an open ideal $I$ has the property that for every $p$ there is some $n$ such that $I^n\subset \ker \theta\_p$, then since each $\ker \theta\_p$ is maximal, $I$ is contained in the intersection of all the $\ker \theta\_p$. It follows that $K[[G]]/I$ is not artinian since it has infinitely many maximal ideals.


Thus $K[[G]]$ is pseudo-compact but not admissible.


