Let $R$ be a Unique Factorization Domain, and also a Bezout Domain. Given an ideal of $R$, and an element $a \in I$ with a minimal number of irreducible factors, show that $I=(a)$.
I have quite a lot of difficulty understanding what this problem means. If someone could give a simple example, that could help me understand what this exercise is asking, it would be really helpful for me to try to generalize it. Thanks in advance.
Here's a sketch: Suppose $I \neq (a)$. Then, there exists $x \in I \setminus (a)$.
Thus, we have $$(a) \subsetneq (a, x) \subset I.$$ Since $R$ is Bezout, write $(a, x) = (b)$ for $b \in I$. But then, $b \mid a$ and so, $a = b$ by minimality of irreducible factors. Conclude.
Edit: This shows that $R$ must be a PID.
Indeed, given any ideal $I \neq 0$, $I$ necessarily has an element with the minimal number of irreducible factors (by well-ordering of $\Bbb N$).