Why $\mathfrak p_2\cdots\mathfrak p_r\not\subset (a)\mathcal O$

Question

If $(a)\mathcal O\subset\mathfrak p_1$ and $r$ is the minimal number such that $\mathfrak p_1\cdot\mathfrak p_2\cdots\mathfrak p_r\subset (a)\mathcal O$ then $\mathfrak p_2\cdots\mathfrak p_r\not\subset (a)\mathcal O$

I don't see, why this holds, (if it's relevant $\mathfrak p_i$'s are prime ideals in the ring of integers $\mathcal O$ and $a\in\mathfrak p_1$)

It appears in the proof of the lemma $2.4$ at the bottom of the page $21$ here

http://www1.spms.ntu.edu.sg/~frederique/antchap2.pdf#page=7

Answer 1

You are misordering the assumptions. The assumptions is that $\mathfrak p_1,\mathfrak p_2,\dots,\mathfrak p_r$ is minimal such that $\mathfrak p_1\dots\cdot\mathfrak p_r\subset (a)\mathcal O$. Then another property is used to show that for some $i$, $\mathfrak p_i=\mathfrak p$, and that $(a)\mathcal O\subseteq \mathfrak p$.

$\mathfrak p_2\cdot\dots\cdot\mathfrak p_r$ can't be a subset of $(a)\mathcal O$ because we assumed $\mathfrak p_1,\cdots,\mathfrak p_k$ was minimal.

So basically, you are asked to deduce two things from the assumption about $\mathfrak p_1,\dots,\mathfrak p_r$:

1. That $\mathfrak p_i=\mathfrak p$ for some $i$, and $(a)\mathcal O\subseteq \mathfrak p$.
2. No matter what the order, $\mathfrak p_2\cdot\cdots \cdot \mathfrak p_r\not\subset (a)\mathcal O$, simply by the minimality assumption.
3. Then reorder if necessary so that $\mathfrak p_1=\mathfrak p$.