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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?

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  • Closely related to http://math.stackexchange.com/questions/91087/subfields-of-finite-fields – lhf Apr 04 '12 at 12:56
  • @Kannappan, I don't think every subspace is a field. – Gerry Myerson Apr 04 '12 at 12:57
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    @KannappanSampath: Dear Kannappan, $\mathbb F_q^n$ is not a field unless $n = 1$, so perhaps you mean $\mathbb F_{p^n}$, for some prime power $p$. In this case the number of subfields is quite different to the number of sub-$\mathbb F_p$-vector spaces. Regards, – Matt E Apr 04 '12 at 12:57
  • @MattE I did mean that $\Bbb F_{q^n}$ but then I am wrong about the number of subspaces. Thank you for letting me know. –  Apr 04 '12 at 12:59
  • @GerryMyerson I am sorry about that wrong "hint". –  Apr 04 '12 at 12:59
  • @MattE Out of interest, is it not (as in the link) given by lhf that the number of subfields is $2^{15}$? Why is it that the number of sub spaces is different to the number of subfields? –  Apr 04 '12 at 13:04
  • As discussed in the related question, there is exactly 1 subfield for each divisor of the exponent. $15$'s divisors are $1,3,5,15$ so there are $4$ subfields. In general for a field of order $p^k$ there are $\tau(k)$ subfields. – Bill Cook Apr 04 '12 at 13:06
  • @lhf Thanks! I need to wake up. http://en.wikipedia.org/wiki/Divisor_function – Bill Cook Apr 04 '12 at 13:09
  • @BillCook Ah so if we look at the number of subspaces we have 15 of them, but for the subfields only $\tau(15) = 4$ of them? I assume by $\tau(k)$ you mean the arithmetic function that counts the number of divisors of an integer. –  Apr 04 '12 at 13:10
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    I believe there a many more than just 15 subspaces. The field itself is $15$-dimensional so you'll have more than $15$ one-dimensional subspaces and so on...there's a lot of subspaces. – Bill Cook Apr 04 '12 at 13:12
  • @Benjamin, the number of subspaces is much larger. Look at a little field like the one with 4 elements. It has 3 one-dimensional subspaces. – Gerry Myerson Apr 04 '12 at 13:13

2 Answers2

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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.

Arturo Magidin
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    Great explanation, but uniqueness of every subfield corresponding to r should have been mentioned. With what you have written, the number of these (nontrivial) subfields could be more than four, while it is just four – Fabian Aug 19 '15 at 12:11
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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 $x^q=x$.

lhf
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