I don't know much about the ring $\mathbb{Z}[\sqrt[3]{2}]$. According to this note it is a Euclidean domain.

I wanted to know how primes split in $\mathbb{Q}[\sqrt[3]{2}]$. There's an obvious case: $2 = (\sqrt[3]{2})^3$ but this already tells us that splitting occurs. Apparently $p = 3$ also splits as: \begin{eqnarray*} 3 &=& \big(\sqrt[3]{2} + 1\big)\big(\sqrt[3]{4} - \sqrt[3]{2} +1\big) \\ 5 &=& \big(\sqrt[3]{4} + 1\big)\big(1 +2 \sqrt[3]{2} - \sqrt[3]{4}\big) \end{eqnarray*} There's a small typo in the note.

If we have a prime $p \in \mathbb{Z}$ do we know when it factors in the ring $\mathbb{Z}[\sqrt[3]{2}]$ ? Can we get more formula like the two listed above?

  • $\begingroup$ The very first theorem there tells you all the history: since the extension's discriminant is $\;-3\;$ , only the prime $\;3\;$ is ramified , and all other primes split according to how $\;x^2+x+1\pmod p\;$ behaves (This is just Kronecker's Theorem) . $\endgroup$ – DonAntonio Mar 19 '18 at 11:44
  • $\begingroup$ @DonAntonio OK. which Kronecker theorem? Let's try $x^2 + x + 1 \equiv (x-2)(x-4) \pmod 7 $, what did this tell us about $7 \in \mathbb{Z}[\sqrt[3]{2}]$ ? $\endgroup$ – cactus314 Mar 19 '18 at 11:53
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    $\begingroup$ @DonAntonio Are you thinking of $\mathbb Q(\sqrt {-3})=\mathbb Q(\zeta_3)$ instead of $\mathbb Q(\sqrt[3]2)$? $\endgroup$ – Mathmo123 Mar 19 '18 at 12:42
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    $\begingroup$ I might be under a misapprehension here, but isn't $2 = (\root 3 \of 2)^3$ an example of ramification rather than splitting? $\endgroup$ – Robert Soupe Mar 19 '18 at 19:53
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    $\begingroup$ In a quadratic ring, like say $\mathbb Z[\sqrt 2]$, 2 ramifies since $(\sqrt 2)^2 = 2$, 3 and 5 are inert and 7 splits since $(3 - \sqrt 2)(3 + \sqrt 2) = 7$. In terms of ideals in $\mathcal O_{\mathbb Q(\sqrt d)}$, $\langle p \rangle$ ramifies if the number $p$ is a divisor of $d$, so, for example, in $\mathbb Z[\sqrt{-6}]$, $\langle 2 \rangle = \langle 2, \sqrt{-6} \rangle^2$ even though 2 as a number is irreducible. It would seem a simple matter to carry these definitions to cubic rings, but there are probably several traps for the unwary. $\endgroup$ – Robert Soupe Mar 20 '18 at 0:51

Building on Mathmo123 and DonAntonio's answers:

Let's write $\alpha=\sqrt[3]{2}$. The kind of prime factorization of $p$ that may occurs are as follows:

(0) $p$ may ramify : it has square factors. This occurs only when $p$ divides the discriminant of the number field, here $-108$, so $p=2$ or $p=3$. For $p=2$, as already mentioned, the decomposition is: $$2=(\alpha)^3.$$ For $p=3$, it's less obvious from the expression given in the OP. So the correct decomposition is in fact $$3=(\alpha+1)^3(\alpha-1).$$ Note that $\alpha-1$ is a unit, its inverse being $\alpha^2+\alpha+1$, so $3$ is a cube, up to a unit.

(i) $p$ may completely split as a product of 3 non-associated irreducible factors $$p=xyz,$$ where $x,y,z$ have each norm $p$.

(ii) $p$ may split as a product of 2 non-associated irreducible factors $$p=xy,$$ where $x$ has norm $p$, $y$ norm $p^2$.

(iii) $p$ may stay prime (inert prime)

We assume $p>3$ from now on. Now, to describe the condition which case occur, let us consider the ring $A=\mathbb{Z}[\sqrt[3]{2}]/(p)=\mathbb{F}_p[X]/(X^3-2)$. As Mathmo123 says, $A$ is the product of $k$ fields, $k$ depending on the number of factors of $X^3-2 \bmod p$, which is also the number of prime factors of $p$.

If $p\equiv2 \bmod 3$, then $3$ is coprime to $p-1$, so $x\mapsto x^3$ has trivial kernel and is a isomorphism of $\mathbb{F}_p^\times$; in particular, $X^3-2$ has a unique root. So we are in case (ii). That is the case for $p=5$, as noticed by the OP, but also $11, 17, 23, \ldots$

If $p\equiv1 \bmod 3$, then $3 \mid p - 1$ so there are two primitive $3$-root of unity $\zeta_3$ in $\mathbb{F}_p$. Notice that if there is a root of $X^3-2$, multiplying by $\zeta_3$ and $\zeta_3^2$ gives the two other roots. So $X^3-2$ has either zero, or three roots in $\mathbb{F}_p$, so we are either in case (iii) or (i) respectively.

To know in which case we are, we just want to know if $2$ is a cube in $\mathbb{F}_p$. For this, refer to Ireland and Rosen, A classical introduction to modern number theory. Proposition 9.6.2 of that book tells us that $x^3=2$ has a solution if and only if $p$ can be written $$p=x^2+27y^2,$$ for some integers $x,y$.

To summarize, $p=2,3$ are associated to cubes, and for $p>3$

  • If $p\equiv 1 \bmod 3$, and $p=x^2+27y^2$, then $p=q_1q_2q_3$ is the product of three distinct primes. This is the case for example of $p=31$. $$31=(\alpha^2+3)(2\alpha^2-1)(\alpha^2+2\alpha-1).$$

  • If $p\equiv 1 \bmod 3$, but $p\neq x^2+27y^2$, then $p$ is prime (inert). This is the case for example of $p=7, 13, 19$.

  • If $p\equiv 2 \bmod 3$, then $p=q_1q_2$ is the product of two distinct primes. This is the case for example of $p=5, 11$. $$5=(\alpha^2+1)(-\alpha^2+2\alpha+1),$$ $$11=(\alpha^2+\alpha-1)(2\alpha^2+3\alpha-1).$$

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  • $\begingroup$ For each of your examples $p= 5, 7, 31$ could you find an example of a factorization. $\endgroup$ – cactus314 Mar 23 '18 at 21:03
  • $\begingroup$ Edited to include the examples of 5,11, 31. 5 is as you said in your question; 7 does not admit non-obvious factors as it is an inert prime. To find the factors of 31, for example, I found the cubic root of 2 mod 31, namely 4, 7,20, then try to compute gcd(\alpha-4,31) and so on; but since I don't have an euclidean algorithm at hand (is there one ?), I did this mostly by trial and error. $\endgroup$ – user120527 Mar 24 '18 at 7:50
  • $\begingroup$ I'm still reading, this looks terrific. Factoring into primes was never easy in the first place beyond what we learned in grade school over $\mathbb{Z}$. $\endgroup$ – cactus314 Mar 24 '18 at 9:31

The Kummer-Dedekind theorem states in this case that the splitting of a prime $p$ in $\mathbb Z[\sqrt[3]2]$ depends entirely on the splitting of the polynomial $X^3-2\pmod p$.

Explicitly, if $X^3-2$ factors as $$f_1(X)\cdots f_r(X)\pmod p$$ with $1\le r\le 3$, then $(p)$ factors as $$(p) = \big(p,f_1(\sqrt[3]2)\big)\cdots \big(p,f_r(\sqrt[3]2)\big).$$

For example, $X^3-2 \equiv (X+1)(X^2-X+1)\pmod 3,$ and one can check that $$(3, \sqrt[3]2+1)=(\sqrt[3]2+1),$$ and $$(3,\sqrt[3]4-\sqrt[3]2+1) = (\sqrt[3]4-\sqrt[3]2+1).$$

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    $\begingroup$ How about $p = 11$ then $x^3 - 2 \equiv (x-7)(x^2 + 7x + 6) \pmod 11$ ? Then how does $11 \in \sqrt[3]{2}$ split ? $\endgroup$ – cactus314 Mar 19 '18 at 12:56
  • $\begingroup$ how about $p=11$ then $x^3 - 2 \equiv (x-7)(x^2 + 7x + 6) \pmod {11}$ then how does $11 \in \mathbb{Z}[\sqrt[3]{2}]$ split? $\endgroup$ – cactus314 Mar 20 '18 at 1:12

Note $\mathcal{O} = \mathbf{Z}\left[\sqrt[3]{2}\right]$ and $k = \textrm{Frac}(\mathcal{O}) = \mathbf{Q}\left(\sqrt[3]{2}\right)$. How a prime $p\in\mathbf{Z}$ factors (or not) in $\mathcal{O}$ is given by how the polynomial $T^3 - 2$ factors in $\mathbf{F}_p [T]$. (This is a result of Dedekind.) To know the latter we need to know when $2$ is cube in $\mathbf{F}_p$ and when it is and when $p\not=2$ whether or not there is a primitive cubic root of $1$ in $\mathbf{F}_p$. Previous condition for $p\not=3$ is the same as $-3$ being a square in $\mathbf{F}_p$ which is the same (by the law of quadratic reciprocity) as $p \equiv 1[3]$. Take now a prime $p\geq 5$. If $p \equiv 2[3]$ then $2$ is a cube in $\mathbf{F}_p$ and there's no primitive cubic root of $1$ in $\mathbf{F}_p$ so that $(p) = (x)(y)$ where the ideals $(x)$ and $(y)$ have respective norms $p$ and $p^2$. As $-1$ is a norm in $k$, modulo action from $\mathcal{O}^{\times}$, you write $p=xy$ with $N_{k/\mathbf{Q}}(x)=p$ and $N_{k/\mathbf{Q}}(y)=p^2$. If If $p \equiv 1[3]$ then if $2^{\frac{p-1}{3}}\equiv 1[p]$ the polynomial $T^3-2$ splits completely in $\mathbf{F}_p[T]$ and $p$ will be (in $\mathcal{O}$) the product of three elements of norm over $\mathbf{Q}$ equal to $p$. If $2^{\frac{p-1}{3}}\not\equiv 1[p]$ then the polynomial $T^3-2$ is irreducible in $\mathbf{F}_p[T]$ and $p$ is prime in $\mathcal{O}$.

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