"Not every field has a finite subfield."
I don't understand why. Can you give me a counterexample?
help: about finite subfield
Re: help: about finite subfield
Not to be pedantic  but I take it that you meant example, not counterexample. Here is one: the rationals (under ordinary addition and multiplication) form an ordered field that has no finite subfield. Proof: Assume that there exists such a finite subfield, choose the largest element  add it to itself and you get an element that is not in the subfield  that is a contradiction. I am sure that by now you can come up with at least two other similar examples.
Re: help: about finite subfield
In fact, it is not hard to see a field has finite subfields iff it has non zero characteristic, without using any properties of ordering. To do this, look at the field generated by 1, i.e. the prime subfield. Every subfield contains the prime subfield, and the prime subfield is infinite iff the field is characteristic 0.

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Re: help: about finite subfield
Thank you very much! Still, I have one more question. In the example of the rationals, can {0} form a subfield?Ivanjam wrote:Not to be pedantic  but I take it that you meant example, not counterexample. Here is one: the rationals (under ordinary addition and multiplication) form an ordered field that has no finite subfield. Proof: Assume that there exists such a finite subfield, choose the largest element  add it to itself and you get an element that is not in the subfield  that is a contradiction. I am sure that by now you can come up with at least two other similar examples.
Re: help: about finite subfield
Pretty sure field definition says 0 != 1 (sometimes even in ring definitions). For some silly stuff, look here: https://en.wikipedia.org/wiki/Field_with_one_element
Zhangvict is spot on. Finite fields have characteristic p for some prime p. If E is a subfield of F with char p, then 1 + ... + 1 (p times) = 0. That equation also holds in F, so F has char p as well.
Zhangvict is spot on. Finite fields have characteristic p for some prime p. If E is a subfield of F with char p, then 1 + ... + 1 (p times) = 0. That equation also holds in F, so F has char p as well.
Re: help: about finite subfield
No, because {0} would then be a finite subfield of the rationals. Alternatively, {0} is not a subfield of the rationals because the multiplicative identity 1 != 0 is not an element of {0}.evelyn9293 wrote:Thank you very much! Still, I have one more question. In the example of the rationals, can {0} form a subfield?