Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$F$ 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?

share|improve this question
3  
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 at 18:19

2 Answers 2

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

share|improve this answer
1  
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 at 16:16
1  
Nice observation, but it wouldn't work for a field with $ 2^{11}$ elements because $2^{11}-1$ is not prime. –  lhf Jul 2 at 16:17
    
thus F can have only two subfields F and {0,1} –  Shreya Taneja Jul 2 at 16:18
    
@ShreyaTaneja - that's exactly right. –  Mathmo123 Jul 2 at 17:15
    
@Ihf - agreed. And your answer is definitely the more general solution! –  Mathmo123 Jul 2 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 $K$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.