11
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ 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
  • 3
    $\begingroup$ 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
13
$\begingroup$

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 $p=5, 11$. $$5=(\alpha^2+1)(-\alpha^2+2\alpha+1),$$ $$11=(\alpha^2+\alpha-1)(2\alpha^2+3\alpha-1).$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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 $$(3,\sqrt[3]4-\sqrt[3]2+1) = (\sqrt[3]4-\sqrt[3]2+1).$$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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
4
+150
$\begingroup$

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 $\mathcal{O}$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.