In [this paper](http://arxiv.org/abs/math/9605234), Gitik and Shelah make the following claim (part of Proposition 1.5):



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> **Claim (Gitik-Shelah):** Suppose $\kappa < \lambda$ are regular, $2^\lambda = \lambda^+$, and $D$ is a normal ideal on $\lambda$. If forcing with $D^+$ makes $cf(\lambda^+) < \kappa$, then it collapses all cardinals in some interval $(\alpha,\lambda^+]$ where $\alpha < \kappa$.
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I do not understand the argument for this claim. An interesting dichotomy follows:



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> **Corollary:** If GCH holds and $\eta$ is a singular cardinal, then for every normal ideal $D$ on $\eta^+$, either $D$ is presaturated or forcing with $D$ collapses $\eta$.
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Sketch of proof: Assume $D$ is not presaturated but $D^+$ preserves that $\eta$ is a cardinal. Since $D$ is not presaturated, forcing with $D^{+}$ collapses $\eta^{++}$, and since $|D^+| = \eta^{++}$, it preserves $\eta^{+3}$. By a well-known theorem of Shelah, this implies that $cf(\eta^{++}) = cf(\eta) < \eta$ in the extension by $D^+$. But by the Claim, $\eta$ is collapsed, a contradiction.


However, it seems that we should be able to force a counterexample to this dichotomy from large cardinals, but I am probably missing some subtlety. So my question is, how do you prove the Claim?



Presaturated ideals
The result follows from the following theorem:



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> **Theorem.** Suppose $\kappa$ is a regular uncountable cardinal and $|P|\leq \kappa.$ Then $\Vdash\_P cf(\kappa)=|\kappa|.$
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In your case $D^+$ has size $\lambda^+$ and it forces $cf(\lambda^+)<\kappa.$ So by above theorem it also forces $|\lambda^+|<\kappa.$


For a proof of the above theorem, see my answer given in [Singularizing forcing of "small" cardinality?](https://mathoverflow.net/questions/142086/singularizing-forcing-of-small-cardinality)


