Why are the $Q_i$ unique?

Question

I am studying the theorem:

Let $R$ be a Noetherian domain of dimension 1. Then every non-zero ideal $I$ of $R$ has a unique expression as a product of primary ideals with distinct radicals.

Proof. Let $I=⋂_{i=1}^nQ_i$ such that $Q_i$ are primary and each $P_i=rad (Q_i )$ is maximal since $R$ is of dimension 1. Since no $P_i⊆P_j$ for $i≠j$ then $P_i$ are embedded primes and hence the $Q_i$ are unique. The $P_i$ are pairwise comaximal ($P_i+P_j=R$ for all $i≠j$) whence so are the $Q_i$. It follows that $Q_1∩…∩Q_n=Q_1…Q_n$.

Conversely if $I=Q_1'…Q_m'$ where $Q_i'$ are primary with distinct radicals $P_i'$. As before the $Q_i'$ are comaximal, so $I=∏_{i=1}^mQ_i' =⋂_{i=1}^mQ_i'$ . By uniqueness of decomposition, $m=n$ and $Q_i=Q_i'$ after permuting.

I am not understanding why $P_i$ are maximal? Moreover where it is used the fact that $R$ is of dimension 1? Would you help me, please? Thank you in advance.

Answer 1

... and each $P_i=nilrad(Q_i)$ is maximal since $R$ [is $1$-dimensional]

Is that what you mean? It kind of looks like you didn't finish typing it. You would need $1$-dimensionality to conclude that these nonzero primes were maximal...