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

$F$ 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?

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Do you remember what happens to degrees of field extensions in a tower? Say if $K\subset L\subset F$ are fields, then $[F:K]$ is the _______ of $[F:L]$ and $[L:K]$. –  Jyrki Lahtonen Jul 2 '14 at 18:19

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

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so that means order of F* is 31 (prime number) and it can have just two subgroups with respect to multiplication - F* and {1}. –  Shreya Taneja Jul 2 '14 at 16:16
Nice observation, but it wouldn't work for a field with $2^{11}$ elements because $2^{11}-1$ is not prime. –  lhf Jul 2 '14 at 16:17
thus F can have only two subfields F and {0,1} –  Shreya Taneja Jul 2 '14 at 16:18
@ShreyaTaneja - that's exactly right. –  Mathmo123 Jul 2 '14 at 17:15
@Ihf - agreed. And your answer is definitely the more general solution! –  Mathmo123 Jul 2 '14 at 17:17

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 $K$.

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