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

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

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

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

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

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

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

Are there any mistakes?

Bye,
Alexander