"Not every field has a finite subfield."
I don't understand why. Can you give me a counterexample?
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.
evelyn9293 wrote:Thank you very much! Still, I have one more question. In the example of the rationals, can {0} form a subfield?
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