It is well-known that an uncountable regular cardinal $\kappa$ is strongly compact if and only if every $\kappa$-complete filter on any set extends to a $\kappa$-complete ultrafilter on that set. The usual proof of this, in the one direction, uses $\theta$-strong compactness to handle filters of size $\theta$, even when those filters concentrate on base sets of size less than $\theta$. My question concerns the nature of the limitation of the size of the base set in this equivalence, and in particular, what is the strength of the assumption in the case of filters on $\kappa$ itself.

**Question.** What is the large cardinal strength of the assumption that $\kappa$ is an uncountable regular cardinal for which every $\kappa$-complete filter on $\kappa$ extends to a $\kappa$-complete ultrafilter on $\kappa$?

For a lower bound, every such $\kappa$ is easily seen to be a measurable cardinal, since the club filter on $\kappa$ is $\kappa$-complete and so the property gives a measure on $\kappa$.

For an upper bound, if $\kappa$ is $2^\kappa$-strongly compact, then the property holds by the usual characterization of strong compactness I alluded to above. Namely, if $F$ is a $\kappa$-complete filter on $\kappa$, then let $j:V\to M$ be a $2^\kappa$-strong compactness embedding. Since $F$ has size at most $2^\kappa$, the strong compactness cover property ensures that there is some $s\in M$ with $j^{\prime\prime}F\subset s$ and $|s|^M<j(\kappa)$. We may assume $s\subset j(F)$, and so $\bigcap s\in j(F)$ by $j(\kappa)$-completeness in $M$. Pick any $\alpha\in \bigcap s$, and it follows that $F\subset\mu$ where $X\in\mu\leftrightarrow\alpha\in j(X)$, which the standard arguments show is a $\kappa$-complete ultrafilter on $\kappa$.

So the property is trapped between $\kappa$ being measurable and $\kappa$ being $2^\kappa$-strongly compact.

Further refined questions would be:

If there is a cardinal $\kappa$ with the property, then can one undertake the construction of inner models with stronger than measurable cardinals? For example, can one construct an inner model with a Woodin cardinal?

Does the measurable cardinal in the canonical inner model $L[\mu]$ fail to have the property?

Can one force an instance of a measurable cardinal without the property?

Can one force a cardinal $\kappa$ to have the property, but not be $2^\kappa$-strongly compact?

I suspect that the answers to all these questions is affirmative.

More generally,

**Question.** What is the strength of the assumption that every $\kappa$-complete filter on a set of size $\theta$ extends to a $\kappa$-complete ultrafilter on that set?

As above, any $2^\theta$-strongly compact cardinal has this property, and this property implies that there are uniform $\kappa$-complete ultrafilters on every regular cardinal up to and including $\theta$, which by a result of Ketonen implies that $\kappa$ is strongly compact up to that degree. So when $\theta$ is regular, this property is trapped between $\theta$-strong compactness and $2^\theta$-strong compactness.

This question grew out of an issue arising in an answer by Noah S to a previous question on MO.