Let F be a field of order 32. Show that the only subfields of F are F itself and {0,1}.

Question

$F$ is a field of order $32$. $F$ and {$0,1$} are trivial subfields of $F$.

But how can we show that these are the only subfields of $F$? Can someone give me a direction to this question?

Answer 1

Hint: if $K$ is a subfield of $F$, then in particular, $K^*$, the multiplicative group of $K$ must be a subgroup of $F^*$

Answer 2

More generally, a field with $p^n$ elements contains a subfield with $p^m$ elements iff $m$ divides $n$.

In your case, we have $p=2$ and $n=5$, which has no nontrivial divisors.

Here is a proof of one direction, the one that concerns the question:

If a field $F$ has $p^n$ elements and contains a subfield $K$ with $p^m$ elements, then $F$ is a finite dimensional vector space over $K$ and so $p^n=(p^m)^d=p^{md}$, where $d$ is the dimension of $F$ over $K$.

Answer 3

If there is a field $F$ of order $p^n$, $p$ is a prime & $n$ is a positive integer, then for each divisor $m$ of $n$ there is a unique subfield of order $p^m$. These are the only subfields of $F$. Now, we have $32=2^5$. The divisors of $5$ are $1,5$. So there are only two subfields, one is of order $2^1$ and another is of order $2^5$. And hence the only subfields are {0,1} and the field itself.