What is the Characteristic of a Field with Order 2^n?

dmatador
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Homework Statement



Let F be a field with order 2^n. Prove that char (F) = 2.

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The Attempt at a Solution



My reasoning is that since a field is an integral domain, its characteristic must be either 0 or prime. After that I get confused, because would the char (F) need to somehow be related to the order of the field? Is there some reasoning that since it must divide the order of the field (just spit balling) and it must be prime, that it could just be 2? I know this is by no means a proof, but I am having difficulty finding some strong ideas to finish this.
 
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Consider the subfield generated by 1. Its order must divide the order of the field, by Lagrange's theorem, since a field is an additive group. What does this tell you?
 
If char(F)=m then doesn't that mean the field has an additive subgroup of order m?
 
TMM said:
Consider the subfield generated by 1. Its order must divide the order of the field, by Lagrange's theorem, since a field is an additive group. What does this tell you?

so the order of any element of the field must divide 2^n... so it should be a number of the form 2m (m being an integer)?
 
A field is an additive group. The additive order of any element must divide the order of the field. Period.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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