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If $K=\mathbb{Q}[\sqrt[3]{19}]$, what are the prime ideals in the factorization of $3\mathcal{O}_K = \mathfrak{p}_1^2\mathfrak{p}_2$.

I'm not able to determine $\mathfrak{p}_1$ and $\mathfrak{p}_2$ using the methods suggested here: $\mathbb{Q}(\sqrt[3]{17})$ has class number $1$

I can factorize $2\mathcal{O}_K, 5\mathcal{O}_K$ and $7\mathcal{O}_K$ using standard theorem (factorizing $x^3-19 \mod p$).

I can factorize it using SageMath but don't know the algorithm it is using to calculate factors (since it can compute factors for any number field).

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rationalbeing
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  • So when you we using the methods from http://math.stackexchange.com/questions/1607361/mathbbq-sqrt317-has-class-number-1 we you considering norms that were multiples of 3, or just norms precisely 3? – John M Jul 29 '16 at 16:35
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    You can use the minimal polynomial $m(x)$ of $\alpha=(1+\root3\of{19}+\root3\of{19^2})/3$, when we have $$m(x)=x^3-x^2-6x-12.$$ You see it has a multiple factor modulo $3$, but that cannot be helped when there is ramification. Anyway, $3$ is a simple factor of $-4332=-2^2\cdot3\cdot19^2=\operatorname{disc}(m(x))$, so powers of $\alpha$ form a local integral basis, and therefore factors of $m(x)$ modulo $3$ can be used. – Jyrki Lahtonen Jul 29 '16 at 16:43
  • @JohnM I couldn't find any element of norm 2,3 or 5. I found many elements of norm 8,12,18,27,30,.. – rationalbeing Jul 29 '16 at 17:18
  • @JyrkiLahtonen I'm not aware of the terminology "local integral basis". Sorry I couldn't follow your argument. (which theorem being used?). – rationalbeing Jul 29 '16 at 17:22
  • Basically what I am saying is that ${1,\alpha,\alpha^2}$ is not (at least not necessarily) an integral basis. But the discriminant proves that the ring $R$ they span is of index $n$ in $\mathcal{O}$ such that $(3,n)=1$. This is enough for factors of $m(x)$ modulo $3$ to give you the prime ideals lying above $3$ by the usual recipe. Local integral basis is defined using valuation rings. – Jyrki Lahtonen Jul 29 '16 at 17:36
  • @JyrkiLahtonen I'm sorry, but your hints doesn't ring any bells with me. My knowledge of ideal factorization is limited to the one given up to chapter 5 of Marcus' Number Fields (this question arises while solving an exercise of ch 5). I don't understand the general Galois theory arguments given in wikipedia. – rationalbeing Jul 29 '16 at 17:53
  • Sorry, I can't point you at another source. The idea is that modular factorization of the polynomial $x^{19}-1$ gives too coarse a picture of the splitting of some primes, and we look for another polynomial that could be used in the exceptional cases (if one can be found). $x^{19}-1$ can be used for primes other than $3$ and $19$. The polynomial $m(x)$ can be used for primes other than $2$ and $19$. The fully ramified $19$ is a separate case. – Jyrki Lahtonen Jul 29 '16 at 18:14
  • @JyrkiLahtonen Ok, I think I'm able to get a broad picture of the arguments needed. But, now I believe that I should spend time reading more literature to fully understand this. Can you suggest some book/notes where I can hope to find similar examples. Also a bit curious: "How computer algebra systems are able to factorize and multiply all the ideals I give them within seconds?" – rationalbeing Jul 29 '16 at 18:25
  • So it shouldn't be surprising that you couldn't find elements of norm 2, 3 or 5 because the class number is not 1, so no longer a PID. Anyway Jyrki's method is the best approach. @JyrkiLahtonen - Might I suggest that you write up your hints as an answer? I don't think there is an example on MSE of finding a different polynomial $m(x)$. – John M Jul 29 '16 at 23:09
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    As far as your question about how computer systems factor ideals, the standard reference is Henri Cohen's book on computational number theory. For material online, see section 4.3.2 here: http://wstein.org/books/ant/ant.pdf – John M Jul 30 '16 at 00:25
  • @JohnM Thanks for the reference of Stein's book. – rationalbeing Jul 30 '16 at 09:48

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EDIT: The following is a complete rewrite of my original posting here. Not that I told any (or very many) lies, it’s just that what I wrote was woefully incomplete. I originally said that doing this computation shouldn’t be hard once you knew the integral basis, and I find myself rather humbled to discover that the task was not at all easy.

The field is $\Bbb Q(z)$, where $z^3=19$, and a good integral basis is $\{1,z,\frac{1+z+z^2}3\}$. For convenience, I’ll call $(1+z+z^2)/3=b$.

To shortcut the whole argument below, let me say that $\mathfrak p_1=(3,b)$ and $\mathfrak p_2=(3,b-1)$. If you want to see why, read on.

The first thing to notice is that $(z-1)b=6$, which allows you to compute $(3,z-1)(3,b)=(9,3b,3(z-1),6)=(3,3b,3(z-1))=(3)$. If we call $I=(3,z-1)$ and $J=(3,b)$, then $IJ=(3)$, which is as far as I got in my original answer to this question. Now, the norm of $(3)$ is $27$, and so one of $I,J$ has norm $3$, the other has norm $9$. But which is which? The one of norm $3$ is maximal, so prime.

But notice that $b^2=\frac{13 + 7z+z^2}3=4+2z+b$, so that $4+2z\in(3,b)=J$, so $-2+2z\in J$, and consequently $z-1\in J$. It follows that $I\subset J$, properly, so that $J$ is the maximal ideal, and since $IJ=(3)=\mathfrak p_1^2\mathfrak p_2$, it follows that $J=\mathfrak p_1$ and $I=\mathfrak p_1\mathfrak p_2$.

How to find $\mathfrak p_2$? I looked at the minimal polynomial for $b$, which is $X^3-X^2-6X-12$, and modulo $3$, this factors as $X^2(X-1)$. The $X^2$-factor has something to do with the ideal that $b$ is in, namely $J=\mathfrak p_1$, and the factor $X-1$ should have to do with the other prime. Anyway, I tried the ideal $(3,b-1)=(3,\frac{-2+z+z^2}3)=\mathfrak a$, and by fairly laborious hand computation, I saw, first, that $\mathfrak a\mathfrak p_1=(3,z-1)=I$, so that $\mathfrak a=\mathfrak p_2$, and to be sure, I calculated $\mathfrak p_1^2\mathfrak p_2$ and sure enough got $(3)$. You do see at a glance that $\mathfrak p_1+\mathfrak p_2=(1)$, the full ring of integers. I haven’t included here any of the ideal-multiplication computations beyond the one I showed at the beginning for $IJ=(3)$, but they all go like that, they’re just more complicated.

Notice that I didn’t use any systematic method, I just thrashed around till I got the right combination of ingredients.

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