Forum for the GRE subject test in mathematics.
evelyn9293
Posts: 2
Joined: Sun Oct 18, 2015 11:28 am

"Not every field has a finite subfield."

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

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

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
Posts: 14
Joined: Wed Aug 26, 2015 6:23 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
Posts: 2
Joined: Sun Oct 18, 2015 11:28 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?

Posts: 96
Joined: Fri Mar 27, 2015 6:42 pm