# How to factorize $5$ in $\mathbb{Z}[\root 3 \of 2]$?

Since $5$ has a norm of $125$ in this domain, and $N(1 + (\root 3 \of 2)^2) = 5$, it seems like a sensible proposition that $5 = (1 + (\root 3 \of 2)^2) \pi_2 \pi_3$, where $\pi_2, \pi_3$ are two other numbers in this domain having norms of $5$ or $-5$. This is supposed to be a unique factorization domain, right?

I am encouraged by the fact that $$N\left(\frac{5}{1 + (\root 3 \of 2)^2}\right) = N(1 + 2 \root 3 \of 2 - (\root 3 \of 2)^2) = 25.$$

But I am discouraged by the fact that $$\frac{5}{(-1 - (\root 3 \of 2)^2)(1 + (\root 3 \of 2)^2)} = \frac{7 - 6 \root 3 \of 2 - 2 (\root 3 \of 2)^2}{5}$$ is an algebraic number but not an algebraic integer. I have found a couple of other numbers with norms of $5$ or $-5$ but can't get them to multiply to $5$ in any of the combinations of three of them that I have tried. I feel like I'm going around in circles.

• did you try in $\mathbb{Z}[\sqrt[3]{2},e^{2i \pi / 3}\sqrt[3]{2}]$ ? – reuns Sep 12 '16 at 21:27
• I assumed that this domain is purely real. Did I make an incorrect assumption? – Mr. Brooks Sep 12 '16 at 21:29
• you used the field norm of $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{2}e^{2i\pi/3})$, right ? – reuns Sep 12 '16 at 21:38
• It looks like you are viewing $\mathbb{Z}[\sqrt[3]{2}]$ as a ring of integers in $\mathbb{Q}[\sqrt[3]{2}]$. So, I don't think there is any need of including nontrivial cube root of unity. – Sungjin Kim Sep 12 '16 at 21:38
• Here's a great set of notes by Keith Conrad on factoring in rings of integers of number fields: math.uconn.edu/~kconrad/blurbs/gradnumthy/dedekindf.pdf – Richard D. James Sep 12 '16 at 22:18

We have

$$\mathbf Z[\sqrt[3]{2}]/(5) \cong \mathbf Z[x]/(x^3 - 2, 5) \cong \mathbf Z_5[x]/(x^3 - 2) \cong \mathbf Z_5[x]/(x+2) \times \mathbf Z_5[x]/(x^2 + 3x + 4)$$

so that the ideal $(5)$ factors as $(5) = \mathfrak p_1 \mathfrak p_2$. To find the ideals $\mathfrak p_1$ and $\mathfrak p_2$, note that they correspond to the maximal ideals in the ring $\mathbf Z_5[x]/(x^3 - 2)$. We therefore have the following factorization:

$$(5) = (\sqrt[3]{2} + 2, 5)(\sqrt[3]{4} + 3 \sqrt[3]{2} + 4, 5)$$

Since the ring $\mathbf Z[\sqrt[3]{2}]$ is a principal ideal domain, these ideals are principally generated by prime elements of norm $5$ and $25$, respectively. Some easy trial and error by hand with the norm form yields that the first ideal is generated by $2\sqrt[3]{4} - 3$, and dividing $5$ by this number finally gives the factorization

$$5 = (2\sqrt[3]{4} - 3)(6 \sqrt[3]{4} + 8 \sqrt[3]{2} + 9)$$

Here's some more detail on the trial and error part: we know that $\mathfrak p_1$ is a prime ideal of norm $5$, and we know that $(5) = \mathfrak p_1 \mathfrak p_2$. Since any number with norm $5$ generates a prime ideal, we can conclude by unique factorization of ideals that any such element would have to generate $\mathfrak p_1$. After that, we just look for solutions to the equation $x^3 + 2y^3 + 4z^3 - 6xyz = 5$ (the norm form), and it is easily seen that $x = -3$ and $z = 2$ do the trick.

Note that the factorization you are asking for is impossible: the prime $5$ does not factor as the product of three prime ideals, but as two.

• Very good answer. I would only add that the student of quadratic rings who goes on to study cubic rings will be properly aware that just as rational primes have square norms in quadratic rings they have cubic norms in cubic rings, but may fail to realize that now some irrational primes may have square norms. – Bob Happ Sep 13 '16 at 21:39
• Very good point. – Ege Erdil Sep 13 '16 at 22:27
• So how do you distinguish between a prime with square norm like $9 + 8 \root 3 \of 2 + 6 \root 3 \of 4$ and a composite with square norm like $1 + 2 \root 3 \of 2 + 2 \root 3 \of 4$? – David R. Sep 14 '16 at 21:40
• I would try taking the square root to see if the result is also in the ring, e.g., $\sqrt{1 + 2 \root 3 \of 2 + 2 \root 3 \of 4} = 1 + \root 3 \of 4$. But if that's not the case, then I'm in the same boat as you. – Robert Soupe Sep 15 '16 at 2:18
• @DavidR. If you identified one generator of $\mathfrak p_2$, any other generator of that ideal must be associate to that. So, calculate $(6 \sqrt[3]{4} + 8 \sqrt[3]{2} + 9)/(1 + 2\sqrt[3]{2} + 2 \sqrt[3]{4}) = 1/25 \cdot (41+22 \sqrt[3]{2} +24 \sqrt[3]{4})$, which is not an integer of $\mathbf Q(\sqrt[3]{2})$, implying that the two elements are not associated. – Ege Erdil Sep 15 '16 at 6:49

This answer is a try to use the number $1+\sqrt[3]{4}$ as OP did.

As proved by @Starfall, the prime $5$ factors as a product of two primes. With a help from SAGE, the prime factors as $$5 = (1+\sqrt[3]{4})(1+2\sqrt[3]{2} - \sqrt[3]{4}).$$ One might wonder why the factors look different from @Starfall's answer. In fact, we have $$2\sqrt[3]{4} -3 = (1+\sqrt[3]{4})/(1+\sqrt[3]{2}+\sqrt[3]{4})^2,$$ with $1+\sqrt[3]{2}+\sqrt[3]{4}$ being a fundamental unit (i.e. a generator of the free part of unit group) in $\mathbb{Z}[\sqrt[3]{2}]$. As $\mathbb{Z}[\sqrt[3]{2}]$ is a PID, it is also a UFD. This means each prime ideal is generated by a prime element in the ring. So, the factors $1+\sqrt[3]{4}$ and $1+2\sqrt[3]{2} - \sqrt[3]{4}$ are prime elements in $\mathbb{Z}[\sqrt[3]{2}]$.