$R$ is a P.I.D. Show that every proper ideal can be written as a product $P_{1}P_{2}...P_{k}$ of maximal ideals and this writing is unique. (Don't take my question as a dublicate because I ask about something different)
$R$ PID $\Rightarrow R$ UFD.
Every proper ideal in PID has the form $(a)$ for $a\in R$. I am in UFD so $a=ux_{1}x_{2}...x_{n}$, $u \in U(R) , x_{i}$ irreducible. I want to show that $(a) = (x_{1})(x_{2})...(x_{n})$.
$(\subseteq)$: I have $c \in (a)$ then $c=ya=yux_{1}x_{2}...x_{n}$ for $y \in R$ so $c \in (x_{1})(x_{2})...(x_{n})$
$(\supseteq )$: I have $c \in (x_{1})(x_{2})...(x_{n})$ then $c=wq_{1}q_{2}...q_{n}$, $w \in R, q_{i} \in (x_{i})$.
and $q_{i} \in (x_{i})$ so there is $k_{i}$ such that $q_{i}=k_{i}x_{i}$
Finally $c=w(k_{1}x_{1})(k_{2}x_{2})...(k_{n}x_{n}) = w(k_{1}k_{2}...k_{n})(x_{1}x_{2}...x_{n}) = w(k_{1}k_{2}...k_{n})a \in (a)$
so $(a) = (x_{1})(x_{2})...(x_{n})$.
The problem is that I think I am missing something. Is everything ok? A friend of mine told me that I should also write that the product $P_{1}P_{2}...P_{n}$ is in the intersection $P_{1} \cap P_{2} \cap ... \cap P_{n}$ but I can't see why and I don't know how
Generally, for any finitely generated ideals $I = (a_1, \dots, a_m)$ and $J = (b_1, \dots, b_m)$ of a commutative ring, we have that $IJ = (a_i b_j \,|\, 1 \leq i \leq m \text{ and } 1 \leq j \leq n).$ (Check out the first answer provided in this post, for instance.) Considering that your ring $R$ is a PID, it is by definition a commutative ring, hence we have that $(a) = (ux_1 \cdots x_n) = (x_1 \cdots x_n) = (x_1) \cdots (x_n),$ as desired.