help: about finite subfield

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evelyn9293
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Joined: Sun Oct 18, 2015 11:28 am

help: about finite subfield

Post by evelyn9293 » Sun Oct 18, 2015 11:31 am

"Not every field has a finite subfield."

I don't understand why. Can you give me a counterexample?

Ivanjam
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Joined: Tue Mar 17, 2015 2:29 am

Re: help: about finite subfield

Post by Ivanjam » Sun Oct 18, 2015 12:17 pm

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.

zhangvict
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Joined: Wed Aug 26, 2015 6:23 pm

Re: help: about finite subfield

Post by zhangvict » Sun Oct 18, 2015 2:46 pm

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.

evelyn9293
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Joined: Sun Oct 18, 2015 11:28 am

Re: help: about finite subfield

Post by evelyn9293 » Mon Oct 19, 2015 11:13 am

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.
Thank you very much! Still, I have one more question. In the example of the rationals, can {0} form a subfield?

p-adic
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Joined: Fri Mar 27, 2015 6:42 pm

Re: help: about finite subfield

Post by p-adic » Mon Oct 19, 2015 1:44 pm

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.

Ivanjam
Posts: 60
Joined: Tue Mar 17, 2015 2:29 am

Re: help: about finite subfield

Post by Ivanjam » Mon Oct 19, 2015 3:53 pm

evelyn9293 wrote:Thank you very much! Still, I have one more question. In the example of the rationals, can {0} form a 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}.



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