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Working within the field $K=\mathbb{Q}(\sqrt[3]{n})$, for any cube root of $n$, how does one factor the unramified rational prime ideals $(p)$?

For starters, I'm relatively new to this and not too sure I completely understand factoring in these extensions, and so I'll instead talk about factoring $x^3-n$ over $\mathbb{F}_{p}$.

Assuming for simplicity that $\mathcal{O}^{K}=\mathbb{Z}(\sqrt[3]{n})$, it's obvious that if $p \equiv 2 \pmod{3}$, then $x^3-n=0$ has exactly one solution in $\mathbb{F}_{p}$. For $p \equiv 1 \pmod{3}$, I have no clue what is supposed to be done.

I know that there must be three solutions or no solutions, but getting there is the problem. All I've been able to think of that might be key to figuring this out is assigning the Frobenius a conjugacy clas s of $S_{3}$ due to the fact that $\mid\rho(p)\mid=\chi_{3}(p)$, where $\rho(p)$ is some representation of an element within the conjugacy class and $\chi_{3}(p)$ is the non-trivial Dirichlet Character of modulus 3.

This seems nice since it alludes to working within $\mathbb{Z}(\frac{-1+\sqrt{-3}}{2})$ for the answer, and for one case, $\mathbb{Q}(\sqrt[3]{2})$, the solution does indeed involve $\mathbb{Z}(\frac{-1+\sqrt{-3}}{2})$ (understanding how this connection is made is difficult).

user26857
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Dave huff
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  • I want to add that I don't know how to manipulate said approach to yield anything fruitful. – Dave huff Mar 19 '17 at 06:47
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    If $p \equiv 1 \bmod 3$ there are three solutions if and only if $n$ is a cubic residue modulo $p$. If this does not satisfy you, you will have to use cubic reciprocity to turn this into a statement about the representation of $p$ in the form $4p = L^2 + 27M^2$ or something equivalent. –  Mar 20 '17 at 13:19
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    The problems on ramification/decomposition of primes in an extension of number fields are generally not trivial, and they are less even so when you ignore the technical tools which were forged precisely to deal with them. I can only suggest that you take a good textbook on ANT, such as D. Marcus' "Number Fields", go straight tio the chapter on ramification and try to study it, even if this means going back sometimes to prerequisites if needed. In Marcus' chapters 3 and 4, you will find a complete exposure of Dedekind's (almost) general theorem on the factorization of primes (thm.27 of chap.3) – nguyen quang do Mar 20 '17 at 14:02
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    ... as well as detailed examples of different types of decomposition in cubic fields (ex. 12, 13, 26 of chap. 3, and p. 103 of chap. 4). NB: a complete description of the ring of integers of a pure cubic field is given at the end of chap. 2. – nguyen quang do Mar 20 '17 at 14:12
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    One of the difficulties that was explained to me when I asked a more specific question http://math.stackexchange.com/questions/1924378/how-to-factorize-5-in-mathbbz-root-3-of-2 is that an irrational number can have a norm of $p^2$ and be prime nevertheless, whereas in a quadratic integer ring a norm of $p^2$ and the number being prime means it's rational. – Mr. Brooks Mar 20 '17 at 21:23
  • @nguyenquangdo I pulled an old print of your reference. I found it much more intuitive to understand cubic reciprocity within the context of the Gauss Sum of the cubic character. I managed to figure out enough for criteria when $n \equiv 2 \pmod{3}$ and determined how a rational prime ideal $(p)$ splits for $p \equiv \pmod{3}$ in $K=\mathbb{Q}(\sqrt[3]{2})$, which resorted to using franz' suggestion on the representation of $4p=L^{2}+3(3M)^{2}$. Thank you for your input. – Dave huff Mar 27 '17 at 23:00
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    @Mr.Brooks Yes this was very difficult to understand for me as well. If I'm not mistaken, I think it is because $K$ is not Galois over $\mathbb{Q}$. A way for me to avoid this dilemma is to interpret how $x^3-2$ factors over $\mathbb{F}{p}$ for $p \neq 2, 3$. I wanted to know the euler product for $\zeta{K}$, and so all that mattered to me was the degree of each irreducible factor of $x^3-2$ over $\mathbb{F}_{p}$ to determine the norms of each $(\mathfrak{b})$ lying over $(p)$. – Dave huff Mar 27 '17 at 23:06
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    @Dave huff Unless I missed something, your original question was about the factorization of a rational prime $p$ in a cubic ring of integers. This problem can be divided in two parts : 1) Factorize mod $p$ the irreducible polynomial giving rise to the cubic field or ring ; 2) "Lift" that factorization mod $p$ to a factorization of $p$ in the cubic ring. The Dedekind theorem which I cited does this job in an "almost" general way, provided you know precisely the ring of integers of the number field. All the references which I gave (in Marcus' book) relate exclusively to this second step. – nguyen quang do Mar 28 '17 at 07:32
  • @nguyenquangdo I'm sure you didn't miss anything. I'm relatively new to algebraic number theory, and so it is most likely the case I have phrased something incorrectly. My goal to gain from this question was to learn 2 things: how many prime ideals $(\mathfrak{p})$ are over $(p)$ for $p \equiv 1 \pmod{3}$, and what is $N_{K}(\mathfrak{p})$ for each. All within a pure cubic extension over $\mathbb{Q}$. I am currently working through the Chapter 3 in Marcus and intend to nail the second part (Lifting the factorization) but in the setting of an abelian extension over $\mathbb{Q}$. Thanks again. – Dave huff Mar 30 '17 at 03:14

1 Answers1

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The whole process is given in Şaban Alaca and Kenneth Williams introductory book Introductory Algebraic Number Theory $([A1])$ that contains many examples, particularly in cubic rings. In chapter $10$ this process is explained. But there are many other good books that explain this process.

Let $\varphi\in\mathcal{O}_{K}$ be an algebraic integer that generates the number field $K=\mathbb Q(\varphi)$. The discriminant of this algebraic integer $\varphi$ (definition $6.4.2, [A1]$) divided by the discriminant of the number field $K$ (definition $7.1.2, [A1]$) is the square of the index of this algebraic integer $\varphi$ (definition $7.1.4, [A1]$). If the prime p does not divide this index Dedekind has given a procedure of factorizing this prime to prime ideals (theorem $10.3.1$ and $10.5.1, [A1]$). You simply factorize the minimal monic polynomial $f(\varphi)\in\mathbb Z[\varphi]$ of the rational integer $\varphi$ over the field $\mathbb F_p$

$$f(\varphi)\equiv g_1(\varphi)^{e_1}\cdot g_2(\varphi)^{e_2}\cdots g_k(\varphi)^{e_k}\;\;\;\;(mod\;p)$$

to irreducible polynomials $g_k(\varphi)$ over $\mathbb F_p$. Then the prime $p$ factorizes to the prime ideals

$$\langle p\rangle=\bigl\langle p,g_1(\varphi)\bigr\rangle ^{e_1}\,\bigl\langle p,g_2(\varphi)\bigl\rangle^{e_2}\cdots\bigl\langle p,g_k(\varphi)\bigl\rangle^{e_k}.$$

If the prime p divides the index of the algebraic integer $\varphi$ you can try to find another algebraic integer $\vartheta$ generating the number field $K$ with an index not divisible by $p$. This does not always help because there are algebraic rings of integers in which the index of every algebraic integer is divisible by this prime (see example $7.4.3, [A1]$). Those primes will normally be small so that you can determine the possible ideals by simply running through the ideals

$$\langle p, \vartheta\rangle$$

(in a brute force attack) with an integral basis $(\omega_k)$ of $\mathcal{O}_{K}$, the algebraic integer $\vartheta=\sum_k a_k\omega_k$ and the coeffcients $0\le a_k\lt p$. An ideal with norm $p$ will certainly be prime ($10.1.6, [A1]$).

If the polynomial $f(\varphi)$ factorizes in a cubic field over the field $\mathbb F_p$ it must have a linear factor $\varphi-z_0$. The linear factor is quickly found because $z_0$ is a zero of the polynomial $f(\varphi)$ over the field $\mathbb F_p$. The factorization of the remaining quadratic polynomial is just as easy. If the polynomial is irreducible over the field $\mathbb F_p$, the ideal $\langle p\rangle$ is prime. You cannot rely on a factorization to only one prime ideal $\langle p\rangle$ or to three prime ideals in a cubic ring. In example $10.4.1, [A1],$ the cubic ring of integers of the field $\mathbb Q(\sqrt[3]{2})$ factorizes the prime $5$ to only $2$ prime ideals.

Matemáticos Chibchas
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Reimer Brüchmann
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