Number of subfields of finite field 
Possible Duplicate:
Subfields of finite fields 

Field with $2^{15}$ elements
How many subfields? Including trivial one and the whole thing I guess $15$. Right or not my guess? I think every subfield gives a subspace of dimension dividing $2^{15}$. Hence 15 of them? 
 A: 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.
In particular, $\mathbb{F}_{2^{15}}$ contains a field with $2^r$ elements if and only if $r|15$.
While every subfield is a subspace, not every subspace is a subfield. For example, the field with four elements can be realized as
$$\{a+b\alpha\mid a,b\in\mathbb{F}_2,\ \alpha^2=\alpha+1\}.$$
Then $\{b\alpha\mid b\in\mathbb{F}_2\}$ is a subspace, but not a subfield.
A: Your given field has dimension 15 over $\mathbb F_2$ and so every subfield has dimension a divisor of 15. The key point is that there is exactly one subfield for each divisor, and so the lattice of subfields is isomorphic to the lattice of divisors of 15. Uniqueness is proved by showing that a subfield with $q$ elements is the set of solutions of $x^q=x$.
