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For an exercise I would like to:

Factorize $(11 - \sqrt{-14})$ in $\mathbb{Z}[\sqrt{-14}]$ as a product of maximal ideals

which is possible as it is a Dedekind domain (it is the ring of integers of $\mathbb{Q}(\sqrt{-14})$).

I do not know where to start however.

I thought that the norm may be useful, so I computed the norm of an element $a + b \sqrt{-14}$ as $a^2 + 14b^2$, which gives us that the norm of $11 - \sqrt{-14} = 135 = 3^3\cdot 5$. Is this also the norm of the ideal? (I know this is the case for the ideal $\alpha \mathcal{O}_K$)

Is there an easy way to find ideals with norm $3$ and $5$? According to my computation of the norm they must be of the form $(a,b)$ (non-principal).

EDIT:

In a linked question I found some more help, which gives me:

$\mathbb{Z}[\sqrt{-14}] = \mathbb{Z}[X]/(X^2 + 14)$ and modulo $3$ we get $(X^2 + 14) = (X + 1)(X+2)$. Does this imply that there are two prime ideals with norm $3$, namely $(3, -2 + \sqrt{-14})$ and $(3, 2 + \sqrt{-14})$?

I did the same modulo $5$, which gives us $(X^2 + 14) = (X+4)(X-4)$ which gives us the ideals $(5,-4 + \sqrt{-14})$ and $(5,4 + \sqrt{-14})$. So I assume these are again ideals of norm $5$ and that $(11-\sqrt{-14})$ must be some combination of these $4$ ideals.

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I think I have solved it myself.

We know that we must find $3$ prime ideals with norm $3$ and one with norm $5$.

We start with norm $5$, and see that $3\cdot 5 + 1\cdot(-4 + \sqrt{-14}) = 11- \sqrt{-14}$ which means that $(5,-4 + \sqrt{-14}) \supset (11- \sqrt{-14})$ and therefore $(5,-4 + \sqrt{-14}) | (11- \sqrt{-14})$ which covers the norm $5$ part.

Because $3\cdot 3 + 1\cdot (2 + \sqrt{-14}) = 11- \sqrt{-14}$ we can also say that $(3,2 + \sqrt{-14}) | (11- \sqrt{-14})$.

Now we shall show that $(3) \not \supset (11- \sqrt{-14})$ which implies that $(3) \not | (11- \sqrt{-14})$ and therefore $(3,-2 + \sqrt{-14}) \not| (11- \sqrt{-14})$.

Take $a + b\sqrt{-14} \in \mathbb{Z}[\sqrt{-14}]$ then all elements of $(11- \sqrt{-14})$ are of the form $$(a + b\sqrt{-14})(11- \sqrt{-14}) = 11a + 14b + (11b-a)\sqrt{-14}$$

So certainly if $3$ was in this ideal, then $$11a + 14b = 3$$ $$11b - a = 0$$

But then $121b + 14b = 3$ for some $b \in \mathbb{Z}$, which is not possible.

Therefore $$(11- \sqrt{-14}) = (3,2 + \sqrt{-14})^3 (5,-4 + \sqrt{-14})$$

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