This is exercise 20.5 out of Jech:



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> Let $\lambda \geq \kappa$ and let $U$ be a normal measure on $P\_{\kappa}(\lambda)$. The ultraproduct $\mathrm{Ult} \_U \{ (V \_{\lambda \_x},\in) : x \in P \_{\kappa}(\lambda) \}$ is isomorphic to $(V \_{\lambda}, \in)$
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Here $\lambda \_x$ simply denotes the order type of $x$. The function $x \mapsto \lambda \_x$ represents $\lambda$ in the ultrapower of $V$ by $U$. Unless I'm mistaken, the ultraproduct mentioned will be $V^M \_{\lambda} = V \_{\lambda} \cap M$ where $M$ is the ultrapower of $V$ by $U$. I don't see why this would be $V \_{\lambda}$ itself, since I don't see why $V \_{\lambda} \subset M$. Clearly $H \_{\lambda ^+} \subset M$ since $M$ is closed under $\lambda$ sequences, but I sort of doubt that $V \_{\lambda} \subset M$ -- I figure if $\lambda$-supercompactness implied $\lambda$-strongness, I would've seen that mentioned somewhere.


So did I make a mistake in computing the ultraproduct, or is there a mistake in the exercise, or does $\lambda$-supercompactness imply $\lambda$-strongness for some reason I'm not seeing?



Strong Cardinals and Supercompact Cardinals
In general, $\lambda$-supercompactness, if consistent, does
not imply $\lambda$-strongness. One can see this by
observing that the smallest cardinal $\kappa$ that is
$\kappa^+$-supercompact is never $\kappa^+$-strong, and in
fact, cannot be even $(\kappa+3)$-strong. The reason is
that $\kappa^+$-supercompactness is witnessed by a measure
on $P\_\kappa\kappa^+$, which amounts essentially to (is
coded by) a subset of $P(\kappa^+)$, and hence is witnessed inside $V\_{\kappa+3}$. Thus, if $j:V\to M$ were any embedding with
critical point $\kappa$, by minimality it follows that
$\kappa$ is not $\kappa^+$-supercompact in $M$, and hence
$M$ cannot have the true $V\_{\kappa+3}$. Thus, $\kappa$ is
not $(\kappa+3)$-strong and thus definitely not $\kappa^+$-strong.


I haven't looked at the context of the exercise, but perhaps he is merely asking you to make the observation that you did in fact make, that the ultrapower will give you $V\_\lambda^M$? 


For some kinds of $\lambda$, it does follow that $\lambda$-supercompactness implies $\lambda$-strongness. For example, if $\lambda$ is a beth-fixed point, then every $\lambda$-supercompactness embedding is also $\lambda$-strong and even $(\lambda+1)$-strong, since in this case $|V\_\lambda|=\lambda$. But if $\lambda$ is not a beth-fixed point, then a version of my argument above will still apply: if $\lambda$ is not a beth-fixed point and $j:V\to M$ is a Mitchell minimal $\lambda$-supercompactness embedding for $\kappa$, then $\kappa$ is not $\lambda$-supercompact in $M$, and this is witnessed inside $V\_\lambda$ by the assumption on $\lambda$, and so $j$ cannot be a $\lambda$-strongness embedding.


