Exercise 1, § 3

For an integral domain to be factorial, it is necessary and sufficient that there exist a mapping $x \mapsto s(x)$ of $\mathbf{A} - \{0\}$ to $\mathbf{N}$ such that $s(xy) = s(x) + s(y)$, the relation $s(x) = 0$ implies that $x$ is invertible in $\mathbf{A}$ and finally, for any two elements $x, y$ of $\mathbf{A} - \{0\}$, neither of which divides the other, there exist elements $a, b, z, t$ of $\mathbf{A} - \{0\}$ such that $ax + by = zt, s(z) < s(x)$, $t$ being prime to $x$ and toy. (If this condition is fulfilled, show first that every non-zero element of $\mathbf{A}$ is a product of extremal elements and then that, for every extremal element $p, Ap$ is a prime ideal; conclude with the aid of Theorem 1 (d)).