The following question was posted to MathStackExchange (original here). As there were no comments/answers on the original, I have ported it unedited.

I am interested in determining the cardinality of $j(\lambda)$ when $j\colon V\to M$ is an elementary embedding arising from an ultrapower embedding obtained through $\lambda$, with $\lambda$ an $\aleph_1$-strongly compact cardinal.

We say that $\lambda$ is $\aleph_1$-strongly compact if for all $\kappa\geq\lambda$ there is a fine $\sigma$-complete ultrafilter $\mathcal{U}$ on $\mathscr{P}_\lambda(\kappa)=\{x\subseteq\kappa\mid|x|<\lambda\}$. Here, $\mathcal{U}$ is fine if for all $\alpha<\kappa$, $\{x\in\mathscr{P}_\lambda(\kappa)\mid\alpha\in x\}\in\mathcal{U}$. Here I am taking $\kappa$ to have cofinality at least $\lambda$, so $|\mathscr{P}_\lambda(\kappa)|=\kappa$.

When taking the ultrapower embedding, so $j\colon V\to M$ obtained from $\mathcal{U}$, we have that $j``\kappa\subseteq[\operatorname{id}]\in M$, and $M\vDash|[\operatorname{id}]|<j(\lambda)$, so certainly $\kappa\leq j(\lambda)$. Furthermore, since $j(\lambda)=\{[f]\mid f\colon\mathscr{P}_\lambda(\kappa)\to\lambda\}$, we have $|j(\lambda)|<|\lambda^{\mathscr{P}_\lambda(\kappa)}|^+=(2^\kappa)^+$.

This gives us that $\kappa\leq j(\lambda) < (2^\kappa)^+$, so my question is: Can we control this further? Perhaps by imposing more restrictions on $\mathcal{U}$ before implementing the ultrapower.

My hope is that either for any such $\kappa$ we can guarantee that $2^\kappa\leq j(\lambda)$; or that for any such $\kappa$ we can guarantee that $|j(\lambda)|=\kappa$.

$\aleph_1$-strong compactness is not a very strong hypothesis, so if we are unable to control $j(\lambda)$ in any meaningful way, would we be able to do so with a stronger hypothesis? I know that, e.g. strong compactness would be sufficient, so can we do better than that? Perhaps if $\lambda$ is the least $\aleph_1$-strongly compact cardinal and the least measurable cardinal?