How do I prove that $\mathbb{Z}[\sqrt{19}]$ is not a UFD? Moreover, how do I prove that $(7,3+\sqrt{19})$ is not a principal ideal?
This is the first time I'm dealing with a quadratic integer ring where $\omega>0$, so I'm not sure how to approach this problem. Thank you in advance.
The question is incorrect; $\mathbb{Z}[\sqrt{19}]$ is a UFD. The Minkowski bound for the field $\mathbb{Q}(\sqrt{19})$ is $\sqrt{19}$, and the ring of integers of $\mathbb{Q}(\sqrt{19})$ is $\mathbb{Z}[\sqrt{19}]$. It is a theorem that the ideal class group of a number field is generated by the ideals of norm less than the Minkowski bound.
Hence, to prove that $\mathbb{Z}[\sqrt{19}]$ is a UFD, it is sufficient to check that the $\mathbb{Z}$-primes 2 and 3 (the only primes less than $\sqrt{19}$) factor into principal ideals in $\mathbb{Z}[\sqrt{19}]$.
This is equivalent to checking that one of $\pm 2$ is a norm, and one of $\pm 3$ is a norm. (In the real quadratic case, we must not forget about the sign of the norm!) Since $\left(\frac{-2}{19}\right) = \left(\frac{-3}{19}\right) = 1$, we should choose the minus sign in both cases, and we can find these elements: $N(13 - 3 \sqrt{19}) = -2$ and $N(4 - \sqrt{19}) = -3$.
Therefore $\mathbb{Z}[\sqrt{19}]$ is a UFD.