# Number of ultrafilters in an extender

Assume GCH. Suppose $$j: V\to M$$ is an elementary embedding such that $$\mathrm{crit}(j)=\kappa$$, $${}^\kappa M \subset M$$, $$M\supset V_{\kappa+2}$$. We can assume $$j$$ is defined from some extender of length $$\kappa^{++}$$. Hence, in particular, we know $$\kappa^{++}.

For each $$\delta, let $$U_\delta$$ be an ultrafilter on $$\kappa$$ be defined such that $$A\in U_\delta$$ iff $$\delta\in j(A)$$. Is it true that for any $$\delta <\delta' < j(\kappa)$$, $$U_{\delta}\neq U_{\delta'}$$? It is true if $$\delta<\kappa^{++}$$ since $$\mathrm{Ult}(V, U_\delta)$$, $$[id]_{U_\delta}=\delta$$. Of course if the length of the extender is too long then this is not true since the number of ultrafilters on $$\kappa$$ is $$\kappa^{++}$$. In particular with appropriate hypothesis, it is possible that $$j=j_E$$ where $$E$$ is some extender on $$[\delta]^{<\omega}$$ where $$\delta\in \kappa^{+++}\cap \mathrm{cf}(\kappa^{++})$$ and some $$\alpha\neq\alpha' satisfy that $$U_\alpha=U_{\alpha'}$$. But $$\delta$$ may not be $$\kappa^{++}$$.

TLDR: Yes, all the ultrafilters are different.

Suppose $$M$$ is a transitive class and $$j : V\to M$$ is an elementary embedding.

Some notation.

• For any $$x\in M$$, let $$H_x = \{j(f)(x) : f\text{ is a function}\}\prec M.$$
• Let $$j_x : V\to M_x$$ be the ultrapower by the derived ultrafilter $$U_x = \{A\subseteq X : x\in j(A)\}$$ where $$X$$ is any set such that $$x\in j(X)$$.
• Let $$k_x : M_x\to M$$ be the unique elementary embedding with $$k_x\circ j_x = j$$ and $$k_x([\text{id}]_{U_x}) = x$$.

Equivalently, $$M_x$$ is the transitive collapse of $$H_x$$, $$k_x : M_x\to M$$ is the inverse of the transitive collapse, and $$j_x = k_x^{-1}\circ j$$.

A nonstandard definition. An ordinal $$\xi$$ is a weak generator of $$j$$ if $$\xi\neq j(f)(\alpha)$$ for any function $$f$$ and any $$\alpha < \xi$$. (The difference between a generator and a weak generator is that in the definition of a generator one requires that $$\xi\neq j(f)(\vec \alpha)$$ for any $$\vec \alpha \in [\xi]^{<\omega}$$. Thus a generator is the same thing as an additively indecomposable weak generator.)

A useful fact about elementary embeddings. Suppose $$x\in M$$ and $$\xi$$ is the least ordinal such that $$x \in H_\xi$$. Then $$H_x = H_\xi$$ Thus $$j_x = j_\xi$$ and $$k_x = k_\xi$$. Moreover $$\xi$$ is a weak generator of $$j$$. We call $$\xi$$ the weak generator associated to $$x$$.

Key point: In your situation, for any distinct weak generators $$\xi,\xi'$$ of $$j$$, $$j_\xi\neq j_{\xi'}$$. (Also weak generators are the same thing as generators in your situation, but this is not really relevant.) Since $$V_{\kappa+2}\subseteq M$$, if $$\xi < (2^\kappa)^+$$ is a weak generator of $$j$$, then $$\xi < ((2^{\kappa})^+)^{M_\xi} = \min (\text{Ord}\setminus H_\xi)$$ Hence $$((2^{\kappa})^+)^{M_\xi}$$ is the least weak generator of $$j$$ above $$\xi$$. Therefore for distinct weak generators $$\xi,\xi'$$, $$((2^{\kappa})^+)^{M_\xi}\neq ((2^{\kappa})^+)^{M_{\xi'}}$$, so $$M_\xi \neq M_{\xi'}$$, and certainly $$j_\xi \neq j_{\xi'}$$.

The answer. Assume that for any distinct weak generators $$\xi,\xi'$$ of $$j$$, $$j_\xi\neq j_{\xi'}$$. Suppose $$X$$ is a set, suppose $$x,x'\in j(X)$$, and assume the derived ultrafilter $$U_x = \{A\subseteq X : x\in j(A)\}$$ is equal to $$U_{x'} = \{A\subseteq X : x'\in j(A)\}$$. We will show $$x = x'$$. Let $$\xi$$ and $$\xi'$$ be the weak generators associated to $$x$$ and $$x'$$ respectively. Since $$U_x = U_{x'}$$, the associated ultrapower embeddings $$j_x$$ and $$j_{x'}$$ are also equal. It follows that $$j_\xi = j_x = j_{x'} = j_{\xi'}$$. Thus by our assumption about weak generators, $$\xi = \xi'$$. In particular, $$k_x = k_{\xi} = k_{\xi'} = k_{x'}$$. Therefore $$x = k_x([\text{id}]_{U_x}) = k_{x'}([\text{id}]_{U_{x'}}) = x'$$ Hence $$x = x'$$, as desired.

• Thank you! I think along the same lines the following about iterations of ultrapowers is also true: if $g_0: [\kappa]^2\to \kappa$ and $g_1: [\kappa]^2\to \kappa$ are representing two different ordinals in $\mathrm{Ult}(V, E^2)$, then the ultrafilters derived from the map $V\mapsto \mathrm{Ult}(V, E^2)$ (with $[g_0], [g_1]$) will be different.
– Otto
Apr 5, 2019 at 20:21
• There are distinct ordinals in the ultrapower by $E^2$ whose derived ultrafilters are equal, namely $\kappa$ and $j_E(\kappa)$ where $\kappa = \text{crt}(j_E)$. Apr 5, 2019 at 20:33
• right of course. I meant to say if they represent ordinals in the same interval $[\kappa_k, \kappa_{k+1})$
– Otto
Apr 5, 2019 at 20:39
• This seems more subtle to me Apr 5, 2019 at 21:07