# Being a PID is not a local property

I am proving that the property of all ideals being principal is not a local property.

It suffices to find an example. And I have a hint that we can take $\Bbb Z[x]/\langle x^2+5\rangle$ to be such an example.

I know that $\Bbb Z[x]/\langle x^2+5\rangle$ is not a PID because it is not a UFD. It can be seen by factorizing $6$.

Which left to show is $\Bbb Z[x]/\langle x^2+5\rangle$ is a PID locally, to do this we need an open cover of it. That is, we can find $f_1,...,f_n$, such that every ideal in each $(\Bbb Z[x]/\langle x^2+5\rangle)[1/f_i]$ is principal.

• Localization at $f$ corresponds geometrically to "cutting away" the zero locus of $f$. Algebraically you are making $f$ a unit. What you might want to try is finding elements that, when localized at, make the factorization of $6$ unique. – Tabes Bridges Oct 29 '17 at 4:06
• @TabesBridges So what about $6$ and $\sqrt{-5}$? Then we have $\langle 6,\sqrt{-5}\rangle=\langle 1\rangle$ so it is an open cover. But how to see that if we invert $6$ or $\sqrt{-5}$ then we get a PID? – Y.X. Oct 29 '17 at 4:46
• Consider the cover (2,3). Then $A_{2}$(that is localized wrt the set $\lbrace 1,2,2^2 ... \rbrace$) has class number one as the ideal mentioned in the comment above is invertible and hence is a PID. If we invert 3 then we are inverting $-(2-\sqrt{-5})$, also $(1+\sqrt{-5})^2 = -2(2-\sqrt{-5})$ and hence $2 \in (1+ \sqrt{-5})$ and thus the ideal $(2,1+\sqrt{-5})_3$ becomes principal and hence the class number of $A_3$ is also one and thus a PID. This seems to have resolved the problem. – random123 Nov 3 '17 at 17:35
• @User0112358 $1/((2-\sqrt{-5})) = (2+\sqrt{-5})/9$ and $-1$ is already a unit. – random123 Nov 4 '17 at 8:48