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?