Spring 2016 - Problem 1

ring theory

Give an example of a unique factorization domain $A$ that is not a PID. You need not show that $A$ is a UFD (assuming it is), but please show that your example is not a PID.
Let $R$ be a UFD. Let $\mathfrak{p}$ be a prime ideal such that $0 \neq \mathfrak{p}$ and there is no prime ideal strictly between $0$ and $\mathfrak{p}$ . Show that $\mathfrak{p}$ is principal.

$A = \C\br{X, Y}$ is a UFD, but not a PID. For example, the ideal $\gen{X, Y}$ is not principal; if this were not the case, write $\gen{X, Y} = \gen{f}$ for some $f \in A$ . Then $f$ must divide $X$ and $Y$ , which are prime, so because $A$ is a UFD, it follows $XY$ divide $f$ , which is impossible. Thus, $A$ is not a PID.
Let $x \in \mathfrak{p}$ , and factor $x$ into $up_1^{k_1} \cdots p_n^{k_n}$ . Without loss of generality, because $\mathfrak{p}$ is prime, we may assume that $p_1 \in \mathfrak{p}$ . But $\gen{p_1}$ is a prime subideal of $\mathfrak{p}$ , so by assumption, $\mathfrak{p} = \gen{p_1}$ , so $\mathfrak{p}$ is principal.