# Ideals as a product of prime ideals

Suppose we are working in $\Bbb{Q}(\sqrt{-41})$. Given a ideal, for example $(2-\sqrt{-41})$ (we especially work in $\Bbb{Z}[\omega_{-41}]$). We know that this is a Dedekind ring, thus we have unique factorization of ideals. I started to take the norm of this ideal and I get $45$ and this is equal to $5\cdot3\cdot3$. Can I say something like this:

$(2-\sqrt{-41})=(5)(3)(3)=(5,2+\sqrt{-41})(5,2-\sqrt{-41})(3,1+\sqrt{-41})^2(3,1-\sqrt{-41})^2$

The last step is deduced from the fact that the ideals $(3)$ and $(5)$ are not prime. Is this true?! Suppose this is true. Then we can do this also for $(3+\sqrt{-41})$ and we get this result:

$$(3+\sqrt{-41})=(5)(5)(2)=(5,2+\sqrt{-41})^2(5,2-\sqrt{-41})^2(2,1+\sqrt{-41})^2$$

Here we get the last step from the fact that $-41\equiv3\ \mod\ 4$. Is this all true? But how to do this for $(2-\sqrt{-41},3+\sqrt{-41})$? I hope someone can help me?!

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$(5)$ does not have norm $5$ in $\Bbb Z[\sqrt{-41}]$. However, the norm being $45$ does mean that some prime (and only one) lying over $5$ does appear in the factorization of $(2-\sqrt{-41})$. – Brandon Carter Dec 9 '12 at 17:29

First of all, we don't have $$(2-\sqrt{-41})=(5)(3)(3)=(45)$$ since $45$ does not divide $2-\sqrt{-41}$ in $\mathbb{Z}[\sqrt{-41}]$. The only thing that tells you $$N(2-\sqrt{-41})=45=3^25$$ is that the ideal $(2-\sqrt{-41})$ factorizes as $IJ$, where $I$ is a prime ideal of norm 5 and $J$ is one of norm $9$, or as $IJK$ where $I$ is a prime ideal of norm $5$ and $J$ and $K$ are prime ideals of norm $3$ not necessarily distinct.

Now, applying Kummer criterions you can see which are the possible candidates for these ideals, and determine them by divisibility. (Is it enough or do you need more?)

Finally, remember that given tow ideals $I$ and $J$ the ideal $I+J$ is the minimum one containing both, and thus given factorizations of $I$ and $J$ you can obtain the factorization of $I+J$ by taking the minimum of the exponents of the prime ideals appearing there. (Can you prove it?)

Then you only have to remember that $(a,b)=(a)+(b)$.

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First: thank you for the answer. But i don't know what that criterion is. How can i do it on a elementary way. Do i have to find a prime ideal with norm 5 and 3 resp.?! – Continued Fraction Dec 9 '12 at 17:52
Oh but i know $(5)=(5,2+\sqrt{-41})(5,2-\sqrt{-41})$ and thus both are prime ideals of norm 5. Also: $(3)=(3,1+\sqrt{-41})(3,1-\sqrt{-41})$ both with norm 3. But how to decide which of them have to been chosen? – Continued Fraction Dec 9 '12 at 18:03
@ContinuedFraction I cannot understand why you know that $(5,2+\sqrt(41))(5,2-\sqrt(41))=(5)$. Am I missing anything? Thanks. – awllower Dec 9 '12 at 19:48
@ContinuedFraction if you know the factorizations of $(3)$ and $(5)$, then you have all ideals of norm $3$ and $5$. And to see which one have to be chosen, you have to look which one contains the ideal you want to factorize. – Josué Tonelli-Cueto Dec 10 '12 at 23:15