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I am trying to factor the ideals (7) and (35) into products of prime ideals in the ring of integers of $K:= \mathbb{Q}(\sqrt[3]{2}).$

Well, I know, that the ring of integers of $K$ is $ O_K =\mathbb{Z}[\sqrt[3]{2}]$ (see: Easy way to show that $\mathbb{Z}[\sqrt[3]{2}]$ is the ring of integers of $\mathbb{Q}[\sqrt[3]{2}]$).

I am lacking a lot of theory and I do not know how to start.

$(35) = (7) (5)$, so I have to find factorizations of (7) and (5).

I thought that maybe I can do this: $$ O_K \cong \mathbb{Z}[X] / (X^3 -2), $$ so $$O_K / (7) \cong \mathbb{F}_7[X] / (X^3 -2), ~ O_K / (5) \cong \mathbb{F}_5[X] / (X^3 -2).$$

But I do not know what to do next. Can you help me?

user26857
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Maria
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1 Answers1

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I'm assuming you have to follow a somewhat elementary approach; otherwise the link by Sebastian Monnet to Dedekind's Factorization Theorem will help.

You're taking the right approach (and you're indeed going in the direction of that theorem). Those two rings $O_K/(7)$ and $O_K/(5)$, are they integral domains? (And hence even fields?) If they are, the ideals are prime; if they're not, then they have zero divisors. Lifting those zero divisors back to $O_K$ will tell you something about the factorization in $O_K$.

user26857
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Magdiragdag
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  • Thank you! In $\mathbb{F}_5[X]:$ $(-2)^3 = -8 \equiv 2 ~\text{mod} ~5 $ and $$X^3 + 8 = (X +2 )(X^2 -2X+4) = (X +2) (X^2 -2X-1) ~\text{mod} ~5 ,$$ since $-2$ is not a root of $(X^2 -2X-1)$ it is irreducible in $\mathbb{F}_5[X]$ – Maria Feb 01 '22 at 22:03
  • How do I lift it back for $O_K/ (5)?$ – Maria Feb 01 '22 at 22:19
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    @Maria The lift would be that in $O_K$, you have $(\sqrt[3]{2} + 2) (\sqrt[3]{4} - 2\sqrt[3]{2} - 1) \in (5)$. So, from there, I think the trick would be to hypothesize that $(5) = (5, \sqrt[3]{2} + 2) (5, \sqrt[3]{4} - 2 \sqrt[3]{2} - 1)$ and the two ideals on the right are prime (and in particular neither is the unit ideal). – Daniel Schepler Feb 01 '22 at 22:34
  • Thanks, that was really helpful. $X^3−2$ is irreducible in $O_K/(7)$, but how do I do it here $(7) = (7, \sqrt[3]{8} -2) = (7,0)$ so this feels wrong ? – Maria Feb 03 '22 at 21:41
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    Note that $X^3 - 2$ is irreducible over ${\mathbb F}_7$ (not in $O_K/(7) = {\mathbb F}_7[X]/(X^3-2)$, because it is simply $0$ in that ring). But, anyway, that means that $O_K/(7)$ is an integral domain (even a field), so $(7)$ is a prime ideal (even maximal), so $7$ is prime in $O_K$. – Magdiragdag Feb 03 '22 at 22:34