Is $I=\langle7, 3+\sqrt{19}\rangle$ a principal ideal of $\Bbb Z[\sqrt{19}]$? 
Is $I=\langle7, 3+\sqrt{19}\rangle$ a principal ideal of $\Bbb Z[\sqrt{19}]$?

I defined the norm :
$$N(a+b\sqrt{19})=(a+b\sqrt{19})(a-b\sqrt{19})=a^2-19b^2$$
Then we can see the multiplicative property :
$$N((a+b\sqrt{19})(c+d\sqrt{19}))=N(a+b\sqrt{19})N(c+d\sqrt{19})$$
Since $N(7)=49$, $N(3+\sqrt{19})=-10$, if $z | 7$ and $z | 3+\sqrt{19}$ then $N(z)=1$. But it means that
$$u^2-19v^2=1$$
I know that this is called Pell's equation and it has a solution. In fact, $u=170, v=39$ satisfies equation. 
I need your help. Thanks in advance.

You know that any common divisor $z$ of $7$ and $3+\sqrt{19}$ has norm $1$, so it is a unit. Then any principal ideal $J\subseteq\mathbb{Z}[\sqrt{19}]$ containing $7$ and $3+\sqrt{19}$ also contains $1$ and is therefore equal to $\mathbb{Z}[\sqrt{19}]$. You now have to check whether $I=\mathbb{Z}[\sqrt{19}]$: If yes, then $I=\langle 1\rangle$, else $I$ is not a principal ideal in $\mathbb{Z}[\sqrt{19}]$.
