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I learned that the splitting of primes in a number field $K = \mathbb{Q}(x)/p(x)$ depends on the factorization of $p(x) \pmod p$. While this is not at all obvious to me, let's use it:

$$x^3 - 2 \equiv (x-7)(x^2 + 7x + 6) \pmod {11}$$

and I do not think that second factor splits. What does this say about the factorization of $p = 11 \in \mathbb{Z}[\sqrt[3]{2}]$ ?


My best guess is that $p = 11$ splits as a linear and a quadratic. For some integers: $a,b,c,d,e \in \mathbb{Z}$ we have:

\begin{eqnarray*} 11 &=& (a + b\sqrt[3]{2}) (c + d \sqrt[3]{2}+ e\sqrt[3]{4}) \\ &=& (ac + 2be) + (bc + ad)\sqrt[3]{2} + (bd+ae)\sqrt[3]{4}\end{eqnarray*}

Most by tautology, we get 3 equations in 5 unknown integers and we can try to solve it:

\begin{eqnarray*} ac + 2be &=& 11 \\ bc + \,\,ad &=& 0 \\ bd + \,\,ae &=& 0 \end{eqnarray*}

This is looking pretty bad, however I noticed the three determinants. And maybe that could be useful.

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  • $\begingroup$ your second factor splits $=(x+6)(x+1)$. $\endgroup$ – Rene Schipperus Mar 21 '18 at 15:06
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    $\begingroup$ $$11=3^3-2(2^3)=(3-2\sqrt[3]{2})(9+6\sqrt[3]{2}+\sqrt[3]{4})$$ $\endgroup$ – Qurultay Mar 21 '18 at 15:08
  • $\begingroup$ @qurultay Good. now find me another one :-) $\endgroup$ – cactus314 Mar 21 '18 at 16:53
  • $\begingroup$ @cactus314 I just solved the equation $a^3+2b^3=11$. I don't know if there is any other solution. $\endgroup$ – Qurultay Mar 21 '18 at 18:19
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The factorization $x^3 - 2$ mod $11$ tells us about the factorization of the ideal $(11)$ in $\mathbb{Z}[\sqrt[3]{2}]$. Let $K = \mathbb{Q}(\sqrt[3]{2})$ and let $\mathcal{O}_K = \mathbb{Z}[\sqrt[3]{2}]$ be its ring of integers. Then the factorization of the ideal $(11)$ can be detected from the structure of $\mathcal{O}_K/(11)$. By the Third Isomorphism Theorem, we have $$ \frac{\mathbb{Z}[\sqrt[3]{2}]}{(11)} \cong \frac{\mathbb{Z}[x]}{(11, x^3 - 2)} \cong \frac{(\mathbb{Z}/11\mathbb{Z})[x]}{(x^3 - 2)} $$ and the structure of this last ring is determined by the factorization of $x^3 - 2$ mod $11$. For more on this, I recommend Keith Conrad's blurb Factoring After Dedekind.

So factoring mod $11$, we find $x^3 - 2 = (x - 7)(x^2 + 7x + 5)$. By Theorem 8 in the linked PDF, then we have the factorization of ideals $$ (11) = (11, \alpha - 7)(11, \alpha^2 + 7 \alpha + 5) = \mathfrak{p}_1 \mathfrak{p}_2 $$ where $\alpha = \sqrt[3]{2}$.

However, this hasn't yet given us a factorization of the element $11$. It turns out that $\mathbb{Z}[\sqrt[3]{2}]$ is a PID, so the prime ideals above are principal. Since $\mathfrak{p}_1, \mathfrak{p}_2$ have norms $11$ and $121$, then they must be generated by elements of norm $\pm 11$ and $\pm 121$. Finding such elements amounts to solving the Diophantine equations $$ a^3 - 6 abc + 2 b^3 + 4 c^3 = N(a + b \sqrt[3]{2} + c \sqrt[3]{2}^2) = \pm 11, \pm 121 \, . $$ I don't know of a good way to do this in general, but for this case it turns out that $\mathfrak{p}_1 = (1 - \sqrt[3]{2} - \sqrt[3]{2}^2)$ and $\mathfrak{p}_2 = (1 - 3 \sqrt[3]{2} - 2 \sqrt[3]{2}^2)$ and $$ 11 = (1 - \sqrt[3]{2} - \sqrt[3]{2}^2)(1 - 3 \sqrt[3]{2} - 2 \sqrt[3]{2}^2) \, . $$

Addendum: Here’s how we can use lattice reduction algorithms to find generators for $\mathfrak{p}_1$ and $\mathfrak{p}_2$. Using the Minkowski embedding we embed $\mathfrak{p}_1$ and $\mathfrak{p}_2$, obtaining lattices $L_1$ and $L_2$ in $\mathbb{R} \times \mathbb{C} \cong \mathbb{R}^3$. The LLL algorithm allows us to find short vectors \begin{align*} v_1 &= (1.8473221018630726395, -3.4275743288228073350, -0.40107945302261334667)\\ v_2 &= (5.9545652536210184438, -6.3318338134128589259, 0.74092293817182559026) \end{align*} for $L_1$ and $L_2$, respectively, which should correspond to generators for $\mathfrak{p}_1$ and $\mathfrak{p}_2$. The first entries $\alpha_1$ and $\alpha_2$ of each of these vectors corresponds to the real embedding. To recognize these real numbers as algebraic numbers, we embed $1, \sqrt[3]{2}, \sqrt[3]{2}^2$ in $\mathbb{R}$, and then use numerical linear algebra to compute (approximate) linear relations among $\alpha_1, 1, \sqrt[3]{2}, \sqrt[3]{2}^2$ and among $\alpha_2, 1, \sqrt[3]{2}, \sqrt[3]{2}^2$. Using the following code in Magma

QQ := Rationals();
QQx<x> := PolynomialRing(QQ);
f := x^3 - 2;
K<nu> := NumberField(f);
OK := Integers(K);
places := InfinitePlaces(K);
P1 := ideal< OK | 11, nu - 7>;
P2 := ideal< OK | 11, nu^2 + 7*nu + 5>;
L1 := Lattice(P1);
L2 := Lattice(P2);
v1 := ShortestVectors(L1)[1];
v2 := ShortestVectors(L2)[1];
nu_RR := Evaluate(nu, places[1]);
LinearRelation([ComplexField() | v1[1], 1, nu_RR, nu_RR^2]);
LinearRelation([ComplexField() | v2[1], 1, nu_RR, nu_RR^2]);

we find relations $$ [ 1, 1, -1, -1 ], [ 1, 1, -3, -2 ] $$ which yield exactly the generators we eyeballed before. You can try the code out on the online Magma calculator here: http://magma.maths.usyd.edu.au/calc/ .

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  • $\begingroup$ Thanks for the bounty! I think there is a probably a nice way to find generators for the ideals using lattice reduction methods. Hopefully I'll have time to add something about that in the near future. $\endgroup$ – André 3000 Apr 2 '18 at 21:48
  • $\begingroup$ Broken link.... $\endgroup$ – lhf Dec 25 '18 at 11:16
  • $\begingroup$ @lhf Thanks for pointing that out. Unfortunately, it looks like the link on Keith's website is broken, too: if you go to math.uconn.edu/~kconrad/blurbs and click on the link for this article, you get the same error. Hopefully he'll fix it soon... $\endgroup$ – André 3000 Dec 25 '18 at 18:50
  • $\begingroup$ Here's a backup link in the meantime: semanticscholar.org/paper/Factoring-after-Dedekind-Conrad/… $\endgroup$ – André 3000 Dec 25 '18 at 19:10
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I'm not a bounty sniper, I swear. I'm just thinking out loud. I'm wondering how much of the naïve methods that work for quadratic rings can be carried over to cubic rings.

For example, we find that 11 splits in $\mathbb Z[\sqrt{-2}]$ since $-2$ is a square modulo 11, e.g., $3^2 = 11 - 2$. Then $3 + \sqrt{-2}$ is a divisor of 11, and we find that $$\frac{11}{3 + \sqrt{-2}} = 3 - \sqrt{-2}.$$ Since those are both conjugates with norms of 11 each, we're done.

Similarly in $\mathbb Z[\root 3 \of 2]$ with find that 2 is a cube modulo 11, e.g., $7^3 \equiv 2 \pmod{11}$. But $N(7 + \root 3 \of 2) = 345$... damn it. Gotta do $N(7 - \root 3 \of 2) = 341$ instead, but then we must figure out how to find $x \in \mathbb Z[\root 3 \of 2]$ such that $\langle 11, 7 - \root 3 \of 2 \rangle = \langle x \rangle$. I guess that can't be done by naïve methods.

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  • $\begingroup$ i think this is a good question. this is why i ask the same question over and over. not because i am confused $\endgroup$ – cactus314 Apr 2 '18 at 22:48
  • $\begingroup$ I agree, this is a good, interesting question. Sometimes it helps to rephrase a question, or to look at it from a broader or narrower perspective. You might occasionally draw the ire of pedants and DRYers, but hopefully others are more appreciative. $\endgroup$ – Robert Soupe Apr 3 '18 at 4:22

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