# Radin generics from iterated ultrapowers

Let $$u$$ be a measure sequence derived from some embedding $$j:V\rightarrow M$$. I heard that from iterating $$j$$ is possible to define a generic filter for some Radin forcing $$\mathbb{R}_w$$. Specifically, $$\mathbb{R}_w$$ is the Radin forcing defined using the measure sequence $$w=j_{\alpha}(u)$$, where $$j_\alpha$$ is the $$\alpha^{th}$$-iterated of $$j$$.

My question is the following: Which are the measures that we use to carry out this process? Are the second coordinates (i.e. $$u_\gamma(1)$$) of the corresponding measure sequence $$u_\gamma$$? Are the whole measure sequences $$u_\gamma$$? In this latter case, how looks like the definition of a ultrapower given by a measure sequence $$u_\gamma$$?

More details: The iteration is the usual one, using $$j$$ (see my comment below). Now the point is that is $$u$$ is the measure sequence in $$V$$, then there is $$\sigma: ORD \to V$$ so that $$\sigma(\alpha)=j_{0, \alpha}(u)\restriction \beta_\alpha,$$ for some suitable $$\beta_\alpha$$ and such that for any $$\alpha, \sigma \restriction \alpha+1$$ is $$\mathbb{R}_{\sigma(\alpha)}$$-generic over $$M_\alpha$$.
• Thanks for the reference! Anyway I'm still having some doubts on it. I may understand that if $j:V\rightarrow M$ is an elementary embedding then what we are iterating not the measures $u_\gamma(1)$ but the extender generating $j$. More precisely, if $j:V\rightarrow M_1$ is generated by $E$ then $V$ thinks $u$ is generated by $E$, hence $M$ thinks that $j(u)$ is generated by $j(E)$. Now as $M_2$ we take the ultrapower of $M_1$ by the extender $j(E)$ so that $M_2$ thinks $j_{2}(u)$ is generated by $j_2(E)$. Does it make sense? Thank you in advance! – Cesare Apr 4 '19 at 8:05
• The iteration by j may be considered by applying j at successor steps $(j_{0,\alpha+1}=j(j_{0,\alpha}) \circ j)$, and taking direct limits at limit ordinals – Mohammad Golshani Apr 7 '19 at 1:55