# Strong Cardinals and Supercompact Cardinals

This is exercise 20.5 out of Jech:

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 $(V _{\lambda}, \in)$

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?

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To get curly brackets, put a backtick, `, on the outside of your dollar signs. I've edited to fix it up. –  David Roberts Oct 30 '10 at 22:47
Thanks David. Is there every any reason to not put back ticks (aside from a couple extra keystrokes), i.e. is there anything you can do without back ticks that you cannot do with them? –  Amit Kumar Gupta Oct 31 '10 at 0:19

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.