$\kappa$-complete, $\lambda$-saturated ideal equivalence More or less the follow-up to this question, I asked a few days ago. 
Now I need to show the equivalence to the seemingly weaker version of $S(\kappa,\lambda,\mathbb{I})$ - the version where there is no family $\{X_\alpha : \alpha < \lambda\} \subset P(\kappa)$ such that each $X_\alpha \notin \mathbb{I}$ but $\alpha \neq \beta \rightarrow (X_\alpha \cap X_\beta) = 0$.
Can someone give me a hint on the less obvious direction of the equivalence? Somehow, I'm not being able to find a path that would lead me anywhere.
Thanks.
 A: You omitted one hypothesis: you’re trying to prove that if $\lambda\le\kappa$ and $\Bbb I$ is a $\kappa$-complete ideal on $\kappa$ containing all of the singletons, and if there is no pairwise disjoint family $\{X_\alpha:\alpha<\lambda\}\subseteq\wp(\kappa)\setminus\Bbb I$, then $\Bbb I$ is $\lambda$-saturated.
Suppose, on the contrary, that there is a family $\{X_\alpha:\alpha<\lambda\}\subseteq\wp(\kappa)\setminus\Bbb I$ such that $X_\alpha\cap X_\beta\in\Bbb I$ whenever $\alpha<\beta<\lambda$. Suppose first that $\lambda<\kappa$. Let $I=\bigcup\{X_\alpha\cap X_\beta:\alpha<\beta<\lambda\}$; $I$ is the union of $\lambda$ elements of $\Bbb I$, so $I\in\Bbb I$. For $\alpha<\lambda$ let $Y_\alpha=X_\alpha\setminus I$; clearly $Y_\alpha\notin\Bbb I$, and $Y_\alpha\cap Y_\beta=0$ if $\alpha<\beta<\lambda$. By hypothesis no such family $\{Y_\alpha:\alpha<\lambda\}$ exists, so we must have $\lambda=\kappa$. 
For this case we need a slightly more sophisticated version of the same basic idea. For each $\alpha<\kappa$ let $$Y_\alpha=X_\alpha\setminus\bigcup\{X_\alpha\cap X_\beta:\beta<\alpha\}\;;$$ $|\alpha|<\kappa$, so $\bigcup\{X_\alpha\cap X_\beta:\beta<\alpha\}\in\Bbb I$, and therefore $Y_\alpha\notin\Bbb I$. But clearly $Y_\alpha\cap Y_\beta=0$ if $\alpha<\beta<\kappa$, since $Y_\alpha\cap Y_\beta\subseteq X_\alpha\cap \left(X_\beta\setminus\big(X_\beta\cap X_\alpha\big)\right)$, and again we contradict the hypothesis that no such family exists. This complete the proof.
Note that I could actually have done it in a single case using the more sophisticated version even when $\lambda<\kappa$; I just thought that it would be a little clearer if I started with the simple version of the idea.
