# Fundamental units of $\mathbb{Q}(\sqrt{12})$

I want to find the fundamental units of $$K = \mathbb{Q}(\sqrt{12})$$. 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 $$\theta = \sqrt{12}$$.

$$\pm1$$ 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 $$\mathbb{Q} \subseteq \mathbb{Q}(\zeta_n) \subseteq \mathbb{Q}(\sqrt{12})$$. Then $$[\mathbb{Q}(\sqrt{12}): \mathbb{Q}] = 3 = [\mathbb{Q}(\sqrt{12}): \mathbb{Q}(\zeta_n)]\times[\mathbb{Q}(\zeta_n):\mathbb{Q}]$$. 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.

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

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 $$1$$.

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

Denoting by $$u=a$$, $$v=\sqrt{12}$$, $$w= \sqrt{18}$$, 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 $$\alpha = a + b\sqrt{12} + c\sqrt{18}>1$$ 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 $$\alpha = 55 + 24 \sqrt{12} + 21\sqrt{18}=164.98\ldots$$

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 $$\alpha^{\frac{1}{3}} =5.48\ldots < 1+\sqrt{12} + \sqrt{18}$$, 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 $$\alpha = a + b \sqrt{12} + c\sqrt{18}$$ is a unit then the numbers $$a$$, $$b \sqrt{12}$$, $$c\sqrt{18}$$ are approximately equal. For instance in our case $$\alpha = 55 + 24 \sqrt{12} + 21 \sqrt{18}$$ we have $$(55 , 24 \sqrt{12}, 21 \sqrt{18}) = (55, 54.9463\ldots, 55.0356\ldots)$$

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

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

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

• 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 $\sqrt{12} \equiv 1 \bmod \mathfrak p$. Then $\sqrt{18} = 6/\sqrt{12} \equiv 6 \bmod \mathfrak p$ 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!
– KCd
Jun 26, 2022 at 20:21
• 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 $\sqrt{12} \equiv 3 \bmod \mathfrak p'$ 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.
– KCd
Jun 26, 2022 at 20:27