Prove that $\forall n \geq 1$ there exists a field containing $n$ subfields

Question

Then Find a minimal finite field containing exactly 8 subfields. I assume that the last part of this question will be more obvious once I have the proof down.

I know that a finite field $\mathbb{F}_q$ is contained in the finite field $\mathbb{F}_r$ if and only if there exists a prime $p$ and integers $n$ and $m$ such that $q=p^m$, $r=p^n$, and $m|n$. This follows because $\mathbb{F}_r$ is a vector space over $\mathbb{F}_q$, and so will have $q^d$ elements for some $d$; and because finite fields must have order a power of a prime. How does this mean that there is always a field that contains n elsements?

Answer 1

Let $r=p^m$. Then $\mathbb{F}_r$ contains a (unique) subfield with $s$ elements iff $s=p^k$ where $k$ is a divisors of $m$. Thus $\mathbb{F}_r$ contains exactly $n$ subfield iff the number $d(m)$ of divisors of $m$ equals $n$. So, for example, for $m=2^{n-1}$ the corresponding field $\mathbb{F}_r$ contains exactly $n$ subfields.

Answer 2

Assuming that you don’t mean “exactly eight proper subfields”, I suppose you want the smallest number with precisely eight divisors, and that’s $30$. So $\Bbb F_{2^{30}}$ should fit the bill.