Suppose $M\_1^\#$ exists and is $\omega\_1$-iterable.


Is it consistent that we can go to a generic extension $V[G]$ where $M\_1^\#$ is no longer $\omega\_1$-iterable?


Or "worse" $M\_1^\#$ is no longer 2-iterable in Neeman's sense?


I suspect the answer is yes and that it will be relatively obvious. So more generally, I'm looking for places where problems like these might be addressed.



Destroying the iterability of $M_1^\#$
Andreas Lietz has already pointed out that $\omega\_1$-iterability can fail in a generic extension. One can also consistently get $(\omega+1)$-iterability to fail:


Suppose there is a transitive model of ZFC + "$M\_1^\#$ exists and is $(0,\omega\_1)$-iterable", and let $M$ be such with minimal ordinal height. It is easy to see we may assume that $M=L\_\alpha[X]$ for some set $X\in M$ of ordinals, where $\alpha=\mathrm{OR}^M$, and we can take $X\subseteq\lambda$ for some strong limit cardinal $\lambda$ of $M$, and $X$ coding $V\_\lambda^M$. Let $N=(M\_1^{\#})^M$.
Let $G$ be $(M,\mathrm{Coll}(\omega,\lambda))$-generic.


I claim that in $M[G]$, $N$ is not $(\omega+1)$-iterable with respect to trees based on $N|\delta^N$ (the Woodin of $N$).
For suppose otherwise. We have $M[G]=L\_\alpha[x]$ for a certain real $x$,
and we can take $x$ to code $X$ in a simple manner. So since $\mathcal{P}(\delta^N)\cap N$ is countable in $M[G]$, working in $M[G]=L\_\alpha[x]$, we can form a Neeman genericity iteration on $N$ (see Theorem 3.3 of "AN INNER MODELS PROOF OF THE KECHRIS-MARTIN THEOREM"), producing a tree $\mathcal{T}$ on $N$ of length $\omega$, such that if $b$ is any $\mathcal{T}$-cofinal wellfounded branch, then there is a $\mathrm{Coll}(\omega,\delta^{M^{\mathcal{T}}\_b})$-generic $g\in L\_\alpha[x]$
with $x\in M^{\mathcal{T}}\_b[g]$.
But letting $\kappa$ be the critical point of the active extender of $M^{\mathcal{T}}\_b$, then note that $\kappa<\alpha$ and $L\_\kappa[x]\models$ ZFC and $X\in L\_\kappa[x]$. Therefore $L\_\kappa[X]\models$ ZFC. But since $\kappa<\alpha$,
we have $L\_\kappa[X]\subseteq M$,
so $V\_\lambda^{L\_\kappa[X]}=V\_\lambda^{M}$,
so $L\_\kappa[X]\models$ ZFC + "$M\_1^{\#}$ exists and is $(0,\omega\_1)$-iterable",
contradicting the minimality of $\alpha$.


However, one should note that if $M\models$ ZFC + "$M\_1^{\#}$ exists and is $(0,\omega\_1)$-iterable" and there is a larger model $W$ of ZFC with $M\subseteq W$
and $\mathrm{OR}^M=\mathrm{OR}^W$
and $(M\_1^{\#})^M$ is fully iterable in $W$
(so $(M\_1^{\#})^M=(M\_1^{\#})^W$)
then for any set generic extension $M[G]$ with $\mathbb{R}^{M[G]}\subseteq W$, we will have that $(M\_1^{\#})^M$ is $(0,\omega+1)$-iterable in $M[G]$. This is just by $\Sigma^1\_2$ absoluteness between $M[G]$ and $W$: given a length $\omega$ tree $\mathcal{T}\in M[G]$, there is a $\mathcal{T}$-cofinal wellfounded branch $b\in W$, but this is just a $\Sigma^1\_2$ assertion about $\mathcal{T}$ which is true in $W$, hence true in $M[G]$. Of course, the "sharp" at the top of $M^{\mathcal{T}}\_b$ need not be iterable.


