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Every feild has a subset isomorphic to rational numbers?

  1. Dec 20, 2012 #1
    I am reading linear algebra by Georgi Shilov. It is my first encounter with linear algebra. After defining what a field is and what isomorphism means he says that it follows that every field has a subset isomorphic to rational numbers. I don't see the connection.
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  3. Dec 20, 2012 #2


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    Either you're misinterpreting the statement or it's very wrong. I can't see how a field with a finite number of elements could be isomorphic to Q. Could directly quote the section?
  4. Dec 20, 2012 #3


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    That is a good book. As a recall that statement does not apply to the most general field, there is some qualification. With a qualification (perhaps the feild must be of characteristic 0 so that Ʃ1=0 only when the sum is empty) it is obviously true a/b is just a 1's divided by b 1's.
  5. Dec 20, 2012 #4


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    What is true is that every field has a subfield which is isomorphic to either ##Q## or to the field ##Z_p## (integers modulo ##p##) for some prime ##p##.
  6. Dec 21, 2012 #5
    I've got a copy of Shilov in front of me, and on page 2 while defining a field (or number field, as he calls it) he writes

    "The numbers 1, 1+1=2, 2+1=3, etc. are said to be natural; it is assumed that none of these numbers is zero."

    That is, he is only working with fields of characteristic zero. In this case, it is immediate that every such field contains a subfield isomorphic to the field of rational numbers, i.e. the rationals can be isomorphically embedded in any field of characteristic zero.
  7. Dec 21, 2012 #6


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    I wouldn't say that it's "immediate", but it's fairly easy to prove. Denote the field by ##\mathbb F##. For each positive integer n, define n1=1+...+1, where 1 is the multiplicative identity of ##\mathbb F##, and there are n copies of 1 on the right. Also define (-n)1=(-1)+...+(-1), and 01=0, where the 0 on the left is the additive identity in the field of integers, and the 0 on the right is the additive identity of ##\mathbb F##.

    Now you can define a function ##f:\mathbb Q\to\mathbb F## by
    $$f\left(\frac p q\right)=(p1)(q1)^{-1}.$$ This only makes sense if we can prove that the right-hand side depends only on the quotient p/q, so you would have to do that. Then you would of course also have to prove that this f is a field isomorphism.
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