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I was thinking about the following problem:

Let $F$ be a field with $5^{12}$ elements.Then how can I find the total number of proper subfields of $F$?

Can someone point me in the right direction? Thanks in advance for your time.

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    Did you mean proper *subfields*? If not, what is *proper field of $F$*?2012-12-19
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    I am sorry,sir.It is just typo.I have edited my post.2012-12-19

2 Answers 2

4

I take it you mean, proper subfield.

Can you show that any subfield of $F$ contains the field of $5$ elements?

Can you show that any subfield must contain $5^r$ elements, for some $r$?

Can you show that the degree of such a subfield (over the field of $5$ elements) must be $r$? and must be a divisor of the degree of the field of $5^{12}$ elements?

Can you show that a finite field has at most one subfield of any given number of elements?

If you can do all those, you have your answer.

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    I have got my answer,sir.Thanks a lot.2012-12-19
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    The answer, btw, must be 4...2012-12-19
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    @Don, that's not the answer I got.2012-12-19
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    Well, the number of proper divisors of $\,12\,$ is $\,4\,$, isn't it? I mean, $\,2,3,4,6\,$...unless I missed some other.2012-12-19
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    @Don, I'm counting $1$ as a proper divisor. $1$ is trivial-but-proper, $12$ is nontrivial-but-improper.2012-12-20
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    I guess that's a matter of definition or of wording, and I'm really not sure.2012-12-20
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    @user120386, 12 has 6 divisors, 5 of them proper. Why do you think the answer is 4?2014-03-29
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    Gerry Myerson : I am sorry, I am editting my post.2014-03-29
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    We know that if a finite field F has characteristic p (prime), then F has cardinality pr where r=[F:Fp]. So r is a divisor of 12. I think proper subfield of $F$ is atmost 5.2014-03-29