How does knowing the norm of an ideal tell us about its factors? In the ring $\mathbb{Z}[\sqrt{223}]$ we have that the ideal $\mathfrak{p}_3=(3,\sqrt{223}+1)$ is a prime ideal lying over 3, and that the ideal $\mathfrak{p}_{11}=(11,\sqrt{223}+5)$ is a prime ideal lying over 11. If we find an element $16+\sqrt{223}$ with norm 33, this is supposed to imply that $(16+\sqrt{223})=\mathfrak{p}_3\mathfrak{p}_{11}$ But I am having trouble seeing why this is the case.
 A: Let $K$ be a number field. There are a couple of facts coming in handy here:

*

*Any ideal of $\mathcal{O}_K$ has a unique factorization into positive powers of prime ideals.

*The ideal norm is multiplicative.

*If an ideal is integral, then its norm is in $\mathbb{Z}$.

*The norm of a prime $\mathfrak{p}$ over $p\in \mathbb{Z}$ is always a power of $p$.

*The norm of the ideal $x\mathcal{O}_K$ is the same as the norm of $x$.

Then we just have to put these things together... An ideal of norm $33$ has to be the product of some prime lying over $3$ and one lying over $11$. Now you just have to make sure that these ideals over $3$ and over $11$ are the right ones.
EDIT: As Jyrki Lahtonen correctly remarked, we are not done here and the computations are not as easy as I hoped:
Be careful here since both $3$ and $11$ split, so there are $4$ ideals with norm $33$ and we have to be sure that $\mathfrak{p}_3\mathfrak{p}_5$ is the ideal we are looking for and not one of the other three.
Let's calculate (by multiplying generators):
$$ \mathfrak{p}_3\mathfrak{p}_5 = (33, 11\sqrt{223}+11,3\sqrt{223}+15,228+6\sqrt{223})$$
A quick computation yields that
$$ 2\cdot (11\sqrt{223}+11)-7\cdot (3\sqrt{223}+15)+3\cdot 33 = 16+\sqrt{223}$$
Thus $(16+\sqrt{223})\subset \mathfrak{p}_3\mathfrak{p}_5$ meaning that $\mathfrak{p}_3$ and $\mathfrak{p}_5$ divide $(16+\sqrt{223})$, so by the first paragraph we have equality.
