Let A be a UFD and $A_{P}$ be the localisation of $A$ at prime ideal $P$. Then prove that $A_{P}$ is principal

Question

I have proved that $A_{P}$ is a local ring that is, it has unique maximal ideal containing elements of type $\frac{a}{r}$ where $a\in P$ and $r\in A\backslash P$. But, I have no idea how to prove that $A_{P}$ is principal.

Answer 1

This is not true. Recall that the prime ideals of $S^{-1} A$ are in a one to one order preserving correspondence with the prime ideals of $A$ that do not intersect $S$. In the case $S=A-\mathfrak p$, this means that the primes of $A_{\mathfrak p}$ correspond to the primes contained in $\mathfrak p$. Now let $A=k[x, y]$ for $k$ a field. This is a UFD and we can then consider the chain of primes $0 \subsetneq (x) \subsetneq (x, y) = \mathfrak p$. By correspondence, this yields a chain of primes in $A_{\mathfrak p}$: $0 \subsetneq (x) A_{\mathfrak p} \subsetneq (x, y) A_{\mathfrak p}$. In a PID all nonzero prime ideals are maximal, but we see here that $(x) A_{\mathfrak p}$ is a prime ideal which is strictly contained in $(x, y) A_{\mathfrak p}$, another prime ideal. Hence, $A_{\mathfrak p}$ is not a PID.

We can phrase this proof a bit more succinctly with dimension theory (although it's really the exact same proof). The chain of primes I listed above proves that $\mathrm{ht}(\mathfrak p) \geq 2$. Thus, $\dim(A_{\mathfrak p}) = \mathrm{ht}(\mathfrak p) \geq 2$ but a PID must have dimension 1 (or 0 if you allow fields to be PIDs).