We introduce the concept of finite field density and explore some of its properties.
Let $F$ be a finite field of order $p^n$. Then $F \cong GF(p^n)$ with $p^n$ elements and has a unique subfield of order $p^m$ for each postive divisor m of n. Thus, the number of subfields of $F$ is $\tau(n)$, where $\tau$ is the divisor function. We define the Finite Field Density of $GF(p^n)$ as:
Given a finite field $F$, with $F \cong GF(p^n)$ and $|F| = p^n$, we define the finite field density as the ratio of the number of subfields and the number of elements in the field. $$D(GF(p^n)) = \frac{\tau(n)}{p^n}$$
So the density, $D$, of a finite field is the ratio of the number of subfields and the number of elements in the field. It can be thought of as a measure of how "field-rich" a given finite field is. Let's take a look at some examples.
Let $F$ be $GF(3^1)$. So $n = 1$, where the only positive divisor of $1$ is $1$, and $\tau(1)=1$. Them $|F| = 3^1 = 3$, so $D(F) = \frac{1}{3}$.
Let $F$ be $GF(7^4)$. So $n = 4$, the positive divisors of $4$ are $\{1,2,4\}$, so $\tau(4)=3$. The $|F| = 7^4 = 2401$, and $D(F) = \frac{3}{2401}$.
Let $F$ be $GF(2^1)$. So again, $n = 1$, where the only positive divisor is $1$ and $\tau(1)=1$. The $|F| = 2^1 = 2$, so $D(F) = \frac{1}{2}$.
Let $F$ be $GF(2^6)$. So $n = 6$, the positive divisors of $6$ are $\{1,2,3,6\}$, and $\tau(6)=4.$ The $|F| = 2^6 = 64$, so $D(F) = \frac{4}{64} = \frac{1}{16}$.
Let $F$ be $GF(2^{10})$. So $n = 10$, and the positive divisors of $10$ are $\{1,2,5,10\}$, and $\tau(10)=4.$ The $|F| = 2^{10} = 1024$, so $D(F) = \frac{4}{1024} = \frac{1}{256}$.
Let $F$ be $GF(2^{12})$. So $n = 12$, the positive divisors of $12$ are $\{1,2,3,4,6,12\}$, and $\tau(12)=6.$ So $|F| = 2^{12} = 4096$, so $D(F) = \frac{6}{4096} = \frac{3}{2048}$.
Observe that when $n=1$, the density is just $\frac{1}{p}$, where $p$ is the order of the field. Also notice that when fixing the base prime, the density decreases for larger powers. These two observations lead to the following theorems.
Let $p$ be a fixed prime number. Then the sequence of finite field densities $D(GF(p^n))_{n=1}^{\infty}$ converges to $0$. $$\lim_{n \to \infty} D(GF(p^n)) = 0$$
We use known bounds on the divisor function $\tau(n)$, and compare it to the exponential growth of $p^n$. From number theory [1], the divisor function satisfies: $$ \tau(n) = O(n^\delta) \quad \text{for any } \delta > 0 $$ More precisely, for every $\delta > 0$, there exists a constant $C_\delta > 0$ such that: $$ \tau(n) \leq C_\delta \cdot n^\delta \quad \text{for all } n \geq 1 $$ Let's fix any $\delta > 0$. Then: $$ D(GF(p^n)) = \frac{\tau(n)}{p^n} \leq \frac{C_\delta \cdot n^\delta}{p^n} $$ Since $p^n$ grows exponentially and $n^\delta$ grows subexponentially, we know: $$ \lim_{n \to \infty} \frac{C_\delta \cdot n^\delta}{p^n} = C_\delta\lim_{n \to \infty} \frac{n^\delta}{p^n} = C_\delta \cdot 0 = 0 $$ Also, $D(GF(p^n)) \ge 0$, Therefore, by the Order Limit Theorem [2], the entire expression tends to zero: $$ \lim_{n \to \infty} D(GF(p^n)) \leq \lim_{n \to \infty} \frac{C_\delta \cdot n^\delta}{p^n} = 0 $$ Thus, $$ \lim_{n \to \infty} D(GF(p^n)) = 0 \qquad \blacksquare $$
For all finite fields $F$, the finite field density satisfies $D(F) \leq \tfrac{1}{2}$, with equality if and only if $F \cong GF(2)$ or $F \cong GF(2^2)$.
We need to show $D(GF(p^n)) \leq \frac{1}{2}, \forall p,n$ with equality only at $(p,n)=(2,1)$ or $(2,2)$. We proceed by cases.
Case A: $n=1$
Then $D(p^1) = \frac{1}{p} \leq \frac{1}{2}$, with equality only when $p=2$.
Case B: $n=2$
Then $D(p^2) = \frac{\tau(2)}{p^2} = \frac{2}{p^2} \leq \frac{2}{4} = \frac{1}{2}$, with equality only when $p=2$.
Case C: $n \geq 3$
From number theory [1], it's been show that $\tau(n) \leq 2\sqrt{n}$. Thus, $$ D(p^n) = \frac{\tau(n)}{p^n} \leq \frac{2\sqrt{n}}{p^n}. $$ So, it is enough to show that $\frac{\sqrt{n}}{p^n} \lt \frac{1}{4}$. The inequality $\frac{\sqrt{n}}{p^{n}}\lt \frac{1}{4}$ for integers $p\ge 2$ and $n\ge 3$ can be proven by induction on $n$. Base Case $(n=3)$: For $n=3$, we need to prove $\frac{\sqrt{3}}{p^{3}}\lt \frac{1}{4}$ for $p\ge 2$. This is equivalent to $4\sqrt{3}\lt p^{3}$. Since $4\sqrt{3}\approx 6.928$ and the minimum value of $p^{3}$ is $2^{3}=8$, the inequality holds. (Inductive Step) Assuming the inequality holds for some integer $k\ge 3$, i.e., $\frac{\sqrt{k}}{p^{k}}\lt \frac{1}{4}$, we need to prove it for $n=k+1$. We have $\frac{\sqrt{k+1}}{p^{k+1}}=\frac{\sqrt{k+1}}{p\sqrt{k}}\cdot \frac{\sqrt{k}}{p^{k}}$. From the inductive hypothesis, $\frac{\sqrt{k}}{p^{k}}\lt \frac{1}{4}$. Since $k\ge 3$ and $p\ge 2$, we have $k+1\le 4k$, which implies $\sqrt{k+1}\le 2\sqrt{k}\le p\sqrt{k}$. Therefore, $\frac{\sqrt{k+1}}{p\sqrt{k}}\le 1$. Combining these results gives $\frac{\sqrt{k+1}}{p^{k+1}}\lt 1\cdot \frac{1}{4}=\frac{1}{4}$.
Thus in all cases, $D(GF(p^n)) \le \frac{1}{2}$, and for $n \ge 3$ the inequality is strict. Equality only holds when $n=1$ and $p=2$, or $n=2$ and $p=2$. $ \qquad \blacksquare $
We've introdued the concept of Finite Field Density and proven some theorems about it. Some potential properties are still open to further exploration. For instance, under what conditions will two finite fields will have the same density? Could this concept be useful to areas like cryptography or coding theory? Could the concept be extended to other algebraic structures?
[1] Hardy, G. H., & Wright, E. M. (1979). An Introduction to the Theory of Numbers (5th ed.). Oxford University Press.
[2] Abbott, S. (2015). Understanding analysis (2nd ed.). Springer.
[3] Gallian, J. (2021). Contemporary Abstract Algebra (10th ed.). Chapman and Hall/CRC