How do I compute the norm of a non-principal ideal of the ring of integers of a quadratic field without using ''large'' results

Question

I am trying to compute the norm of the ideal $I=(7, 1+\sqrt{15}) \trianglelefteq \mathbb Z[\sqrt{15}],$ the ring of integers of $\mathbb Q[\sqrt{15}].$ I knew $I^2$ would be principal, as $I\bar I=(Norm(I))$ so I computed it, $I^2=(49, 7(1+\sqrt{15}), 16+2\sqrt{15})$. But I don't know where to go from here. How does this become a principal ideal?

More generally, how do I compute the norm of a non-principal ideal of the ring of integers of a quadratic field, without using large results?

I have looked at answers to similar questions but I don't want to use big theorems like this one. In an exam, I need to be able find the norm using only the methods we were taught in lectures unless I want to prove a theorem like this which would obviously be a bad idea. I am happy to prove small results and use them, however. To clarify, the methods we were taught in lectures were the above one with conjugate ideals, and by picking a $\mathbb Z$-basis of the ideal in such a way that you can tell how big $\mathcal O_K/I$ is (but I don't understand this method as I've never seen a practical example of it - e.g. I don't know how to change $\mathbb Z$-bases as I don't know how linear algebra works over things that aren't fields).

I realise trying to answer a question like this is frustrating as I'm asking you to tie your hands behind your back and not use results that could well be useful, so I really would appreciate any help you can give.

Answer 1

I think I've managed to figure out an answer myself. Some feedback on the method would be appreciated (hence it's community wiki).

If $a \in I$ then $(a)\subseteq I$ and so $$\frac{I}{(a)}\trianglelefteq \frac{\mathcal O_K}{(a)} \implies \frac{\mathcal O_K /(a)}{I/(a)}\cong \frac{\mathcal O_K}{I} \implies \left| \frac{\mathcal O_K}{I} \right| \left| \frac{I}{(a)} \right| = \left| \frac{\mathcal O_K}{(a)} \right| \implies \left| \frac{\mathcal O_K}{I} \right| \leq \left| \frac{\mathcal O_K}{(a)} \right| $$ with equality iff $I=(a)$. Hence $N(a)$ is a proper divisor of $ Norm(I)$ if $I \supsetneq (a)$. Hence, in the above example $7 \in I \implies Norm(I)$ properly divides $49$. So it has norm $7$.

Answer 2

The best way to solve basic problems usually is to use the definitions. The norm of an ideal $I\subseteq\mathcal O_K$, where $\mathcal O_K$ is the ring of integers of a number field, is defined as the cardinality of the quotient $\mathcal O_K/I$. In your case, note that $\mathcal O_K=\mathbb Z[\sqrt{15}]$ is isomorphic to $\mathbb Z[x]/(x^2-15)$, and one possible isomorphism is the one sending $\sqrt{15}\mapsto x$. Under this isomorphism, the ideal $I$ corresponds to the ideal $(7,1+x)$. Therefore, $$\mathbb Z[\sqrt{15}]/I\simeq \mathbb Z[x]/(x^2-15,7,x+1)$$ Now note that $x^2-15=(x+1)(x-1)-7\cdot 2$, which shows that $(x^2-15,7,x+1)=(7,x+1)$. Now it is an easy exercise to show that $\mathbb Z[x]/(7,x+1)\simeq \mathbb F_7[x]/(x+1)\simeq \mathbb F_7$, proving that $I$ has norm $7$.