When is a Unique Factorization Domain a Principal Ideal domain "Let $R$ be a Unique Factorization Domain and $(a,b)=(c)$ for $a,b,c \in R$. Show that $R$ is a Principal Ideal domain."
To be honest I found this very hard, here is my naive try:
Lets assume $R$ is a Unique Factorization Domain and $(a,b)=(c)$ $\Rightarrow c$ is gct of $a$ and $b$ then $\Rightarrow (b)\subseteq (c)$ and $(a)\subseteq (c)$ . Since this is true for every pair of Elements of $R$ we can construct a series $(a_1) \subseteq (a_2)\subseteq (a_3) ... \subseteq(a_n)$ in $R$. Because $R$ is a Unique Factorization Domain there is a $N$ with $a_N \in R$ and $(a_i)=(a_N)$ for $i \geq N$.
Let $I$ be an Ideal but not a Principal Ideal, so $I=(d,e,f,...z)$ with $d,e,f,...z \in R$. Then $I= r_1 \cdot d + r_2 \cdot e + r_3 \cdot f ... r_j \cdot z$ for all $r_i \in R$. But because $d,e,f,...z \in R$ and $(d),(e),(f),...,(z) \in (a_N)$, so $d,e,f,...z \in (a_N)$ so there are $\alpha, \beta, ..., \omega $ so that $d=\alpha \cdot a_N$, $e=\beta \cdot a_N$,..., $z=\omega \cdot a_N$. This means:
$$ \begin{align*} I &= r_1 \cdot d + r_2 \cdot e + r_3 \cdot f \ldots r_j \cdot z \\ &= r_1 \cdot \alpha \cdot a_N + r_2 \cdot  \beta \cdot a_N + \ldots r_j \cdot \omega \cdot a_N \\ &= a_N(r_1 \cdot \alpha + r_2 \cdot  \beta \cdot + \ldots r_j \cdot \omega) \in (a_N) \end{align*}
$$
I know that I am wrong somewhere, but I don't really know how to show it. Maybe there is a way to show that $R$ is an Euclidean Ring?

Here are some hints hints: every finitely generated ideal is principal (why?), so it's enough to show there is no ideal which is not finitely generated. If $I$ is such an ideal, take elements $a_1,a_2,\dots\in I$ such that $(a_1)\subsetneq(a_1,a_2)\subsetneq(a_1,a_2,a_3)\subsetneq\dots$. Each of those ideals is generated by some element, say $(a_1,\dots,a_n)=(b_n)$. What relations are there between $b_n$? Writing them in terms of their unique factorizations, can you see why we reach a contradiction?
