$ (x-a_1) \cap \ldots \cap (x-a_n) = (p) $ in $ \mathbb{C}[x]$ reasoning

Question

Let $ p \in \mathbb{C}[x] $ be of degree $ n $ with pairwise different roots $ a_1, \ldots, a_n \in \mathbb{C}.$

I read in a script that $ (x-a_1) \cap \ldots \cap (x-a_n) = (p) $ because $ \mathbb{C}[x] $ is a principal ideal domain.

I understand the reasoning approach, because of the principal ideal property $(x-a_1) \cap \ldots \cap (x-a_n) $ can be generated from a single $q \in \mathbb{C}[x] $.

I have two questions:

1. is the above reasoning even sufficient? Doesn't it just show that $(x-a_1) \cap \ldots \cap (x-a_n) \subseteq (p) $? Or does $ (a) \subseteq (b) \implies (a) = (b) $ apply in the main ideal ring? No, right?

2. instead, I wonder whether the equality actually applies anyway, regardless of the fact that $ \mathbb{C}[x] $ is a principal ideal domain:

$ p = (x-a_1) \cdot \ldots \cdot (x-a_n) $ holds. So $ (p) = \{ t \cdot (x-a_1) \cdot \ldots \cdot (x-a_n) \mid t \in \mathbb{C}[x] \} = (x-a_1) \cap \ldots \cap (x-a_n).$ (You could also easily show both inclusions without making use of the fact that $ \mathbb{C}[x] $ is a principal ideal domain... Or did I miss something?)

Thank you very much.

Answer 1

You didn't explain the equality $\{ t \cdot (x-a_1) \cdot \ldots \cdot (x-a_n) \mid t \in \mathbb{C}[x] \} = (x-a_1) \cap \ldots \cap (x-a_n)$. The elements of the intersection are polynomials which are multiples of each $x-a_i$, not elements that are multiples of their product.

Anyway, $\mathbb{C}[x]$ being a PID doesn't immediately imply that the ideal is generated by $p$ itself, and not by some other polynomial. The statement has to be proved. Clearly, if a polynomial in $\mathbb{C}[x]$ is divisible by $p$ then it is divisible by each $x-a_i$, and so $(p)\subseteq (x-a_1)\cap...\cap (x-a_n)$. Conversely, assume $g\in (x-a_1)\cap...\cap (x-a_n)$, i.e $g$ is divisible by each $x-a_i$. Since the polynomials $x-a_1,...,x-a_n$ are pairwise coprime, $g$ is also divisible by their product, which is $p$. Thus, $g\in (p)$ in that case.