# Bounding Fundamental Units

Let $$N$$ be a positive integer. Denote by $$\varepsilon(N)$$ to be the number of fundamental units which appear for some real quadratic field. The goal is to prove that $$\varepsilon(N)$$ is asymptotic to $$2N$$.

Here is an incomplete argument that I have thus far.

If $$\varepsilon$$ is a fundamental unit of $$L = \mathbb{Q}(\sqrt{d})$$ where $$d>1$$ and square-free then $$\varepsilon = a + b \sqrt{d}$$ where $$a,b$$ are positive integers or positive half-integers (depending on $$d$$). If also require that $$1 \leq \varepsilon \leq N$$ then we see that $$a\leq N$$. It follows from here $$\text{tr}_{L/\mathbb{Q}}(\varepsilon) = 2a$$ and $$N_{L/\mathbb{Q}}(\varepsilon) = \pm 1$$. Thus, the minimal polynomial for $$\varepsilon$$ is given by $$X^2 - cX \pm 1$$ where $$1\leq c \leq 2N$$.

The larger root of this minimal polynomial is given by $$c + \sqrt{c^2 \pm 4}$$. If $$c > N+1$$ then this larger root is $$>N$$ and falls outside the allowable interval. Thus, we conclude that the fundamental units in the interval $$[1,N]$$ have minimal polynomial given by $$X^2 - cX \pm 1$$ where $$1\leq c\leq N+1$$.

We have a total of $$2(N+1)$$ such polynomials. The fundamental unit will be the larger root (by definition of fundamental unit in a real quadratic field). Thus, we have at most $$2(N+1)$$ such roots in the interval $$[1,N]$$. Of course, some of those roots might be units which are not fundamental, however, this is just an upper bound, and so we have, $$\varepsilon(N) \leq 2(N+1)$$

Now we need some lower bound estimate. I have been trying to do several attempts but they are not working out so far. My intuition tells me the lower bound estimate should look something like $$2N - \sqrt{N}$$. But I cannot find any argument for that as I do not see how one can tell if such a polynomial has a fundamental unit as its root.

I was trying to estimate those $$c$$ for which $$c^2 \pm 4$$ is square-free. Those such $$c$$ will definitely lead to polynomials which have a root as a fundamental unit. But it is not clear to me how to estimate that quantity.

• @reuns For me, "fundamental unit" refers to the positive unit larger than 1. In my problem $K$ is not fixed! The question says, "count number of fundamental units of some real quadratic number field". Emphasize on the word "some". Commented Feb 2, 2019 at 1:24
• @reuns My definition of "fundamental unit" is exactly what is written in the above comment. No, I do not mean the number of elements in $\bigcup_{d>1} \mathbb{Z}( L_d)^{\times} \cap [1,N]$ (where $L = Q(\sqrt{d})$). Read my original question again. That is not what I am asking. Commented Feb 2, 2019 at 2:22
• @reuns You have not shown this is a lower bound. Some of those polynomials have roots which are not fundamental units. Take for example, the golden ratio $\varphi$. This is the fundamental unit for $\mathbb{Z}(L_5)$. The minimal polynomial for $\varphi$ is $X^2 - X - 1$. However, $\varphi^2$ is also a unit in $\mathbb{Z}(L_5)$ and its minimal polynomial is equal to $X^2 - 3X + 1$. Therefore, the polynomial $X^2 - 3X + 1$ must be rejected, since its roots are $\varphi^2$ and $\varphi^{-2}$, which are not fundamental units in $\mathbb{Z}(L_5)$. Commented Feb 2, 2019 at 2:58
• From your question it is impossible to understand that you meant counting the number $\varepsilon(N)$ of $u \in [1,N]$ that are generators of $O_{K_u}^\times/\{1,-1\}$ for some real quadratic field $K_u$. To improve the $2N+O(1)$ then use exclusion-inclusion $\varepsilon(N) \approx \sum_{d \le N} \mu(d)f(N^{1/d})$ where $f(N) = \# \{ X^2-cX\pm 1, c \in \{1\ldots N\}\} = 2N$ to avoid counting $u^2,u^3,...$. Commented Feb 2, 2019 at 3:15
• I got stuck on your first paragraph. I thought real quadratic rings only have one fundamental unit, the smallest unit greater than 1, and infinitely many units greater than that. e.g., $1 + \sqrt 2$, $3 + 2 \sqrt 2$, $7 + 5 \sqrt 2$, etc. Commented Feb 3, 2019 at 16:43

With $$U$$ the units of every ring of integers of real quadratic fields let $$r(y) = \sum_{u \in U, u \in (1,y]} 1 = \sum_{X^2-cX+1, c \ge 3,u = \frac{c+\sqrt{c^2-4}}{2} \le y} 1+\sum_{X^2-cX-1, c \ge 1, u = \frac{c+\sqrt{c^2+4}}{2} \le y} 1 \\= 2y+O(1)$$ with $$F$$ the fundamental units of every ring of integers of real quadratic fields let $$\varepsilon(y) =\sum_{u \in F, u \in (1,y]} 1$$ since $$O_{\mathbb{Q}(\sqrt{d})}^\times = \{1,-1\} \times u_d^\mathbb{Z}, \qquad O_{\mathbb{Q}(\sqrt{d})}^\times \cap (1,y] = \{ u_d^k, k \in 1 \ldots\frac{\log y}{\log u_d}\}$$ we have $$r(y) = \sum_{k \ge 1}\sum_{u \in F, u^k \in (1,y]} 1 = \sum_{k \ge 1}\varepsilon(y^{1/k})$$ and by Möbius inversion
$$\varepsilon(y)= \sum_{k \ge 1} \mu(k) r(y^{1/k}) = \sum_{k \ge 1} (\mu(k)2y^{1/k}+O(1))= 2y- 2y^{1/2}+O(2y^{1/3}\log_{3/2} y)$$