# Every feild has a subset isomorphic to rational numbers?

1. Dec 20, 2012

### Tyler314

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. Dec 20, 2012

### pwsnafu

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?

3. Dec 20, 2012

### lurflurf

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. Dec 20, 2012

### Erland

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. Dec 21, 2012

### A. Bahat

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. Dec 21, 2012

### Fredrik

Staff Emeritus
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.