Register to reply 
Every feild has a subset isomorphic to rational numbers? 
Share this thread: 
#1
Dec2012, 02:49 AM

P: 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.



#2
Dec2012, 04:06 AM

Sci Advisor
P: 820




#3
Dec2012, 07:07 AM

HW Helper
P: 2,263

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.



#4
Dec2012, 09:26 AM

P: 339

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##.



#5
Dec2112, 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. 


#6
Dec2112, 06:23 PM

Emeritus
Sci Advisor
PF Gold
P: 9,268

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 righthand 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. 


Register to reply 
Related Discussions  
Nxmod(1) is a discrete subset of [0,1] iff x is a rational?  Calculus & Beyond Homework  3  
Rational numbers  bounded subset with no least upper bound  Calculus & Beyond Homework  2  
Rational numbers with denoms not divisible by a prime p mod I is isomorphic to Z_p  Calculus & Beyond Homework  2  
Prove F isomorphic to the field of rational numbers  Calculus & Beyond Homework  3  
Is this correct? (sum of two rational numbers is rational)  Precalculus Mathematics Homework  8 