I'm trying to work out all the subfield of a finite field with $3^7$ elements. So I've said that since every subfield of that field if of the form $3^n$ where $n$ is a positive divisor of $7$. so I got the finite subfields as the finite field with $3^1$ and $3^7$ elements. just wanted to check if that was right.
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1Yes, this is correct. ${}{}{}.$ – Thomas Andrews Jun 09 '20 at 17:50
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1Yes. The field $;\Bbb F_{3^7};$ only has two subfields: it itself and the prime field $;\Bbb F_3;$ – DonAntonio Jun 09 '20 at 17:50
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1Yes, it is correct – Asv Jun 09 '20 at 17:51
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"So I've said that since every subfield of that field if of the form 3^n where n is a positive divisor of 7" You might want to explain why. – fleablood Jun 09 '20 at 17:57
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Yes, the possible sizes are $3^1$ and $3^7.$
If $m$ is the size of a subfield, then the original field must be a finite dimensional vector space over that subfield, and thus, if $d$ is the vector space dimension, the original field must have $3^7=m^d$ elements.
This means by unique factorization that $m=3^n$ for some $n$ and $nd=7.$ This means your subfields can have $3^1$ or $3^7$ elements.
Then you need to show those subfields exist, which is easy.
Thomas Andrews
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