I am correct in thinking that the only subfields of are and itself? How do we prove that these are the only ones?
Is it fair to say that there are no infinite fields 'smaller' than since is the field of fractions for so that no 'smaller' field can contain , which we know it must?
So combining the reasoning I included in my post and your hint, do we know that contains no subfields, and contains no subfields but because they could not include and thus must be the rationals (but in this case, how do we know this)?
NonCommAlg: how do we know that every subfield must contain ? Your method is beyond what I know, but it does make me look forward to Galois theory!
ok, well, if you don't know galois theory, then you have to solve the problem directly. let be a subfield of . then and thus since is a subfield, it contains the inverse of all non-zero elements and so .
now suppose that then there exists thus let then ...
try to finish the proof yourself and if you couldn't, see the rest:
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can you finish it now? if not, see below
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