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I want to find the fundamental units of $K = \mathbb{Q}(\sqrt[3]{12})$. The extension degree is $n = 3$, we have $r_1 = 1$ real embedding and $2r_2 = 2$ complex embeddings. So by Dirichlet's theorem, $\mathbb{Z}_K^\times \cong \mu_K \oplus \mathbb{Z}u$ where $\mathbb{Z}_K$ is $K$'s ring of integers and $\mu_K$ is the group of roots of unity in $K$. I first proved that $B = \{1, \theta, \frac{\theta^2}{2}\}$ where $\theta = \sqrt[3]{12}$.

$\pm1$ are the only rational roots of unity. To find other roots of unity in $K$ let $\zeta_n$ be a complex primitive $n$-th root of unity in $K$. Then $\mathbb{Q} \subseteq \mathbb{Q}(\zeta_n) \subseteq \mathbb{Q}(\sqrt[3]{12})$. Then $[\mathbb{Q}(\sqrt[3]{12}): \mathbb{Q}] = 3 = [\mathbb{Q}(\sqrt[3]{12}): \mathbb{Q}(\zeta_n)]\times[\mathbb{Q}(\zeta_n):\mathbb{Q}]$. But 3 is prime and $[\mathbb{Q}(\zeta_n): \mathbb{Q}] = \phi(n)$ is even when $n \geq 2$. So $\mu_K = \{\pm1\} \cong \mathbb{Z}_2$.

Now I want to find fundamental unit $u$. This is where I'm stuck. I know $v$ is a unit $\iff$ $N_{K/\mathbb{Q}}(v) = \pm1$. So I took an element $v = a + b\theta + c\frac{\theta^2}{2}$ and calculated its norm. This gave me the following equation:

$$a^3 + 18c^3 + 12b^3 - 18abc = 1$$

I didn't follow a very formal method to find solutions for $a, b, c$. I just substituted things that satisfied the equation. For example, $a^3 = 1$ mod $3$ so I just set $a = 1$. Then $2b^3 = 0$ mod $3$. But since I don't want a real unit, I set $b = 3$ instead of $b = 0$ and so on. This gave me a unit $u = \frac{2 + 6\theta - 3\theta^2}{2}$. I found a paper that stated that $v = \frac{110 + 48\theta + 21\theta^2}{2}$ is a fundamental unit for this field. I calculated $u^{-1}$ using Euclid's algorithm and found that $u^{-1} = v$ so what I've found is a fundamental unit. I just don't know how to prove it.

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$O_K=\sum_{j=3}^3 b_j \Bbb{Z}$.

Let $\sigma_1,\sigma_2,\sigma_3$ be the 3 complex embeddings.

  • Enumerate the elements of $O_K$ until you find a unit $u\ne \pm 1$.

  • Let $m=\max(|\sigma_1(u)|,|\sigma_2(u)|,|\sigma_3(u)|,|1/\sigma_1(u)|,|1/\sigma_2(u)|,|1/\sigma_3(u)|)$.

  • Assume that you know a fundamental unit $w$.

It must be that $|\sigma_j(w)|\le m$.

Let $$B = \pmatrix{\sigma_1(b_1)&\sigma_1(b_2)&\sigma_1(b_3)\\ \sigma_2(b_1)&\sigma_2(b_2)&\sigma_2(b_3)\\ \sigma_3(b_1)&\sigma_3(b_2)&\sigma_3(b_3)}\in M_3(\Bbb{C})$$

Find some $r$ such that $|(B^{-1})_{ij}|\le r$.

Then $w=\sum_{j=1}^3 a_j b_j$ with $a_j\in [-3rm,3rm]\cap \Bbb{Z}$.

So you now have only finitely many elements of $O_K$ to enumerate, finding which one has a complex absolute value closest to $1$.

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Consider an integer $$\alpha = a + b \sqrt[3]{12} + \frac{c}{2} \sqrt[3]{12^2}= a + b \sqrt[3]{12} + c \sqrt[3]{18}$$

Denoting by $u=a$, $v=\sqrt[3]{12}$, $w= \sqrt[3]{18}$, we have $$N(\alpha) = u^3 + v^3 + w^3- 3 u v w= \frac{1}{2}(u+v+w)((u-v)^2 + (u-w)^2 + (v-w)^2)= \frac{1}{2} \alpha \cdot ((u-v)^2 + (u-w)^2 + (v-w)^2)$$

Now, we are looking for $\alpha$ units, that is $N(\alpha) = \pm 1$. Since $N(-\alpha) = - N(\alpha)$, it is enough to find $\alpha$'s of norm $1$. We conclude from the above that $\alpha>0$. Now, if $0< \alpha < 1$, then $N(\frac{1}{\alpha}) = 1$, and $\frac{1}{\alpha} >1$. So we are looking for $\alpha>1$, of norm $1$. From the above, if $\alpha>1$, and $N(\alpha) = 1$, then $(u-v)^2$, $(u-w)^2$, $(v-w)^2 < 2$. Now note that if $\ne 0$ then $|u|$, $|v|$, $|w| \ge 1$. We conclude that all of the $u$, $v$, $w$ have the same sign, so positive (like $\alpha$, their sum).

Conclusion: if $\alpha = a + b\sqrt[3]{12} + c\sqrt[3]{18}>1$ is a unit, then $a$, $b$, $c>0$. This is important. Now, if $1< \alpha< \alpha'$ are both units, then $\frac{\alpha'}{\alpha}>1$ is a unit, so all of its "components" are positive. We conclude that $\alpha' = \frac{\alpha'}{\alpha} \cdot \alpha$, and so the components of $\alpha'$ are larger than the components of $\alpha$. Therefore: there exists a smallest unit $\alpha_1>1$. Moreover, every other unit $\alpha>1$ is a power of $\alpha_1$ ( a general fact, reproved).

Now consider the unit $$\alpha = 55 + 24 \sqrt[3]{12} + 21\sqrt[3]{18}=164.98\ldots$$

Assume that $\alpha \ne \alpha_1$, the fundamental unit $>1$. Then $\alpha = \alpha_1^n$ for some $n \ge 2$, and so $\alpha_1 = \alpha^{\frac{1}{n}}$.

Now, we have $\alpha^{\frac{1}{3}} =5.48\ldots < 1+\sqrt[3]{12} + \sqrt[3]{18}$, so $\alpha$ cannot be a power $\alpha_1^{n}$, with $n\ge 3$. Now one should just check that $\alpha$ is not a square. If it were then $\alpha_1 = \alpha^{\frac{1}{2}} = 12.84\ldots$. Write

$$1=N(\alpha_1) = \frac{1}{2}\alpha_1 \cdot ((u-v)^2 + (v-w)^2 + (u-w)^2)$$

Now note that $$(u-v)^2 + (v-w)^2 + (u-w)^2 = 3 ((u-m)^2 + (v-m)^2 + (w-m)^2)$$ where $m = \frac{u+v+w}{3}$. Therefore we get $$(u-\frac{\alpha_1}{3})^2 + (v-\frac{\alpha_1}{3})^2+ (w-\frac{\alpha_1}{3})^2= \frac{2}{3 \alpha_1}= \frac{2}{3 \cdot 12.84\ldots}$$

and so $$|u - \frac{12.84\ldots}{3}| < (\frac{2}{3 \cdot 12.84\ldots})^{\frac{1}{2}} = 0.22\ldots$$ while $\frac{12.84\ldots}{3} = 4.28\ldots$, and this is a contradiction with $u$ being a (positive) integer.

Conclusion: $\alpha$ is not the power of another unit, and so is a fundamental unit.

$\bf{Added:}$ If $\alpha = a + b \sqrt[3]{12} + c\sqrt[3]{18}$ is a unit then the numbers $a$, $b \sqrt[3]{12}$, $c\sqrt[3]{18}$ are approximately equal. For instance in our case $\alpha = 55 + 24 \sqrt[3]{12} + 21 \sqrt[3]{18}$ we have $$(55 , 24 \sqrt[3]{12}, 21 \sqrt[3]{18}) = (55, 54.9463\ldots, 55.0356\ldots)$$

$\bf{Added:}$ We can also check that $55 +24 \sqrt[3]{12},+21 \sqrt[3]{18}$ is not a square by using congruences $\mod 4$, since

$$(p + q \sqrt[3]{12}+ r \sqrt[3]{18} )^2 = (p^2 + 12 q r) + \textrm{ irrat. part}$$

and note that $p^2 + 12 qr \equiv 1 \mod 4$, while $55\equiv 3 \mod 4$.

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    $\begingroup$ That's an ugly way to show $\alpha$ isn't a square in $\mathcal O_K$. A nicer way: get a prime ideal $\mathfrak p$ s.t. $\alpha \not\equiv \Box \bmod \mathfrak p$. Since ${\rm disc}(x^3-12) = -2^43^5$ isn't divisible by $11$ and $x^3- 12 \equiv (x-1)(x^2+x+1) \bmod 11$, $11\mathcal O_K = \mathfrak p\mathfrak q$ where $\mathfrak p$ has norm $11$ and $\sqrt[3]{12} \equiv 1 \bmod \mathfrak p$. Then $\sqrt[3]{18} = 6/\sqrt[3]{12} \equiv 6 \bmod \mathfrak p$ and $\alpha \equiv 0 + 2(1) + (-1)6 \equiv 7 \bmod \mathfrak p$. Since $7$ isn't a square mod $11$, $\alpha \bmod \mathfrak p$ isn't a square! $\endgroup$
    – KCd
    Commented Jun 26, 2022 at 20:21
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    $\begingroup$ I used the prime $11$ in my previous comment since I wanted a prime number bigger than $3$ with a prime ideal factor of prime norm, and the first such prime $5$ did not help: $5 = \mathfrak p'\mathfrak q'$ where $\mathfrak p'$ has norm $5$ and $\sqrt[3]{12} \equiv 3 \bmod \mathfrak p'$ and it turns out that $\alpha \equiv 4 \bmod \mathfrak p'$, so we don't get a contradiction. The nice thing about this prime ideal method is that a nonsquare in $\mathcal O_K$ will be a nonsquare modulo half of all prime ideals, and in practice it doesn't take long to find such a prime ideal. We only need one. $\endgroup$
    – KCd
    Commented Jun 26, 2022 at 20:27

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