Prove: Every Subfield of C Contains Rational Numbers

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Every subfield of the complex numbers (C) must contain the rational numbers (Q) due to the requirements of field properties, including closure under addition, subtraction, multiplication, and the existence of multiplicative inverses. The discussion highlights that while a set like {0,1} can form a field, it does not qualify as a subfield of C because it has a characteristic of 2, which contradicts the characteristic of C. Fields of characteristic zero, which include C, must contain Q, as they require distinct elements 0 and 1. The proof relies on the fact that all operations in the subfield must yield consistent results with those in C. Ultimately, any valid subfield of C must include Q as a subset.
poolwin2001
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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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