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

Question

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.

Answer 1

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 $\mathbb{Z}[\sqrt{19}]$.

Answer 2

Actually $\Bbb Z[\sqrt{19}]$ is a PID, that is, all its ideals are principal.

In this case $I=\Bbb Z[\sqrt{19}]$ because $1=3\cdot 7+(6-2\sqrt{19})(3+\sqrt{19})\in I$.