## Ideal a is irreducible <--> a=p^n, p is prime ideal

Hello.

I proofed the following proposition and I am not sure whether there aren't any mistakes:

Let $\displaystyle A$ be a principal ideal domain. Then the ideal $\displaystyle \mathfrak a$ is irreducible if and only if $\displaystyle \mathfrak a=\mathfrak p^n$ for a prime ideal $\displaystyle \mathfrak p$ and $\displaystyle n \in \mathbb N$.

($\displaystyle \Leftarrow$): If $\displaystyle \mathfrak a=\mathfrak p^n$, then $\displaystyle \mathfrak a=(x)$ for some $\displaystyle x \in A$ and $\displaystyle (x)=(p^n)$ for a prime element $\displaystyle p \in A$.
Let $\displaystyle (x)=(p^n)=(b) \cap (c) = \mathrm{lcm}(b,c)$, $\displaystyle b,c \in A$.
$\displaystyle \Rightarrow p^n=\mathrm{lcm}(b,c)$. It follow: $\displaystyle b=p^n$ or $\displaystyle c=p^n$ and therefore $\displaystyle \mathfrak a$ is irreducible.

($\displaystyle \Rightarrow$): Let $\displaystyle \mathfrak a=(x)$ be a irreducible ideal of $\displaystyle A$ and $\displaystyle x=p_1 ^{\varepsilon_{p_1}} \cdot ... \cdot p_n ^{\varepsilon_{p_n}}$ be the prime factorisation of $\displaystyle x$ with $\displaystyle p_i \neq p_j$ for $\displaystyle i\neq j$. Then $\displaystyle p_1 ^{\varepsilon_{p_1}},...,p_1 ^{\varepsilon_{p_1}}$ are coprime. It follows:

$\displaystyle (x)=(p_1 ^{\varepsilon_{p_1}} \cdot ... \cdot p_n ^{\varepsilon_{p_n}} ) = (p_1 ^{\varepsilon_{p_1}}) \cap ... \cap (p_n ^{\varepsilon_{p_n}})$.

Let $\displaystyle (q):= (p_2 ^{\varepsilon_{p_2}}) \cap ... \cap (p_n ^{\varepsilon_{p_n}})$

Because $\displaystyle (x)=(p_1 ^{\varepsilon_{p_1}}) \cap (q)$ is irreducible, one can assume without loss of generality: $\displaystyle (x)=(p_1 ^{\varepsilon_{p_1}})$
$\displaystyle \Rightarrow p_1 ^{\varepsilon_{p_1}} = r x, r \in A^{\times}$.

$\displaystyle \Rightarrow (x)=(x) \cap (q)$, therefore $\displaystyle q=sx, s \in A^{\times}$ because of $\displaystyle (q)=(x)$.
Now, $\displaystyle q$ and $\displaystyle p_1 ^{\varepsilon_{p_1}}$ both have the factor $\displaystyle x$ and therefore $\displaystyle p_1 ^{\varepsilon_{p_1}},...,p_n ^{\varepsilon_{p_n}}$ are not coprime. This is a contradiction and we get $\displaystyle p_1 = p_2 =...=p_n$, thus $\displaystyle (x)=(p^n)$.

Are there any mistakes?

Bye,
Alexander