Saturated models and $\kappa=\kappa^{<\kappa}$

Do not assume GCH. Can you characterize the cardinals $\kappa$ such that every theory $T$ with an infinite model has a saturated model of cardinality $\kappa$?

I guess these are the cardinals such that $\kappa=\kappa^{<\kappa}$. (Correct?) Do these cardinals have a name (are they googlable cardinals)? Do they exist in all models of ZFC? If not, how strong is their existence?

• The last question, I think, should be phrased in the peculiar and often more intriguing way. How strong is their nonexistence. – Asaf Karagila May 24 '15 at 7:37
• The problem of describing when a complete theory $T$ has a saturated model of cardinality $\kappa$ has been solved completely. This happens in two cases: If $\kappa = \kappa^{<\kappa} \geq |D(T)|$, where $D(T)$ is the set of all complete types in finitely many variables over the empty set, or if $T$ is stable in $\kappa$. I believe the section 4 of the eight chapter of the revised edition of Shelah's book has these results. – tci May 24 '15 at 22:37
• @tci Your comment should really be an answer! – Alex Kruckman May 25 '15 at 15:05

Unfortunately, it is consistent that if $\kappa>\aleph_0$, then $\kappa<\kappa^{<\kappa}$.
It is consistent from large cardinals (one and two) that for every infinite cardinal, $2^\kappa=\kappa^{++}$. And then we have the following:
1. If $\kappa=\lambda^+$, then $\kappa^{<\kappa}=\kappa^\lambda=\lambda^{++}=\kappa^+>\kappa$.
2. If $\kappa$ is singular, then $\lambda=\operatorname{cf}(\kappa)<\kappa$ and $\kappa^\lambda>\kappa$.
Note that the large cardinals enter the picture when blowing up the continuum of a singular cardinal, if we try to resort to $\sf ZFC$ assumptions only (read: not assuming large cardinals at all), then $\beth_{\omega+1}=(\beth_\omega)^+=\kappa$ will satisfy $\kappa^{<\kappa}=\kappa$.