Prove: Every Subfield of C Contains Rational Numbers

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I am self studying linear algebra from `Linear Algebra' by Hoffman and Kunze.
One of exercise Q is:
Prove that Every subfield F of C contains all rational numbers.

But doesn't the set {0,1}(with the usual +,-,.) satisfy all conditions to be a field?
 
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what's 1+1?
 
As Matt implies, closure under the operations is a requirement.
 
EEK!I forgot about 1+1 :(
to have closure under addn & subtr you need to have Z.
to have closure under multiplication and division(or existence of x^-1 for all x) you need Q.Therefore All subfields of C should have atleast Q in them.
Is my proof correct?
 
Wait a minute!
My set can be a field with characteristic 2 (1+1=0).(or is it characteristic 1)
Which brings me to the next Question.
P.T. All zero characteristic fields contain Q.
Any hints how to begin?
Thanks is advance
 
Yes that's basically correct:

1 and 0 must be elements (actually I am a little unclera on this is the trivial field technically a subfield of C?) thus any 1+1+1...+1 is also an element so all the natural nunmbers must be elements and by additve inverse all integers must be elements. Any number in Q can be given by n*1/m where n and m are integers (m not equal to zero), by muplicative inverse 1/m must be in the any subfield of C, therefore any subfield of C has Q as a subfield.
 
Thanks .but is the charactesitic 1 or 2?
 
Please don't give away the whole ans.Just gimme a hint.Thanks anyway
 
poolwin2001 said:
Wait a minute!
My set can be a field with characteristic 2 (1+1=0).(or is it characteristic 1)
Which brings me to the next Question.
P.T. All zero characteristic fields contain Q.
Any hints how to begin?
Thanks is advance

Yes, but it's not a subfield of C though is it.

Just look at the definition of a field with charestic 0.
 
  • #10
True
C is a zero charecteristic field so are its subfields thereof
Why?
 
  • #11
because of the prefix sub. If it is a subfield then adding two elements in the subfield must give the same answer as adding them in the field, so if 1+1..+1=0 in the subfield, it equals zero in the field and hence the field has characteristic p for soem prime.


All fields must contain 0 and 1 and these are distinct (so the set {0} with addition and multiplication isn't a field, jcsd), so all fields of char 0 contain a copy of Q. The proof is the same as for the large field being C. You didn't actually use anything other than it was a field of characteristic zero did you?
 
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  • #12
So {0,1,+,.} is a field with charecteristic 2.But it is not a subfield of C.
Thanks.
 
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