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Every feild has a subset isomorphic to rational numbers?

by Tyler314
Tags: feilds, isomorphism, linear algebra
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Tyler314
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Dec20-12, 02:49 AM
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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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pwsnafu
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Dec20-12, 04:06 AM
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Quote Quote by Tyler314 View Post
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.
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?
lurflurf
#3
Dec20-12, 07:07 AM
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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.

Erland
#4
Dec20-12, 09:26 AM
P: 345
Every feild has a subset isomorphic to rational numbers?

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##.
A. Bahat
#5
Dec21-12, 10:58 AM
P: 150
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.
Fredrik
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Dec21-12, 06:23 PM
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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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