1. ## Field of Quotients

If F is a field, prove that the field of quotients, Q, is isomorphic to F

2. ## Can you clear it up a bit?

What do you mean by the field of quotients? I am only familiar with the field of fractions when dealing with rings. In a field it doesnt really make sense to talk about a field of fractions because your object is already a field. $\frac{a}{b}=ab^{-1}\in F$, so I am not entirely sure what this question is asking for you to show

3. I checked this problem in wiki, "The field of fractions of a field is isomorphic to the field itself" (link).

You only need a well-defined isomorphism $\phi_s:F \rightarrow S^{-1}F$ given by $f \mapsto fs/s$ for any s in S, where S is the set of all non-zero elements of a field F, and $\phi_s(s)$ is a unit in $S^{-1}F$ for any $s \in S$. It is easy to check that $\phi_s$ is an injective homomorphism. To see $\phi_s$ is surjective, for any $f/s \in S^{-1}F$, we have $f/s = \phi_s(fs^{-1})$.

4. I'm a little confused on how you're showing the isomorphism, phi, is well-defined. What is the set S^-1 F defined to be?

5. Originally Posted by Gamma
What do you mean by the field of quotients? I am only familiar with the field of fractions when dealing with rings. In a field it doesnt really make sense to talk about a field of fractions because your object is already a field. $\frac{a}{b}=ab^{-1}\in F$, so I am not entirely sure what this question is asking for you to show
Yes, it does make sense (albeit trivial) to speak of the field of fractions of a field.

6. Originally Posted by Coda202
If F is a field, prove that the field of quotients, Q, is isomorphic to F
If you consider the definition and construct of Field of quotients, Q as a field of all equivalence classes (equivalence class of (a,b), i.e. [a,b] are all (x,y) where ay=bx and b,y<>0. Think of [a,b] as a/b), then

Consider phi(x)=[x,1] for all x in F.

It is easy to show that above is an isomorphism from F onto Q.

Note if F were an Integral Domain, we can define an isomorphism from F into Q. In field the fact that every non-zero element has an multiplicative inverse turns this isomorphism from 'into' -> 'onto'

7. I'm having trouble showing that phi is an isomorphism, especially the well-defined part.

8. Originally Posted by Coda202
I'm having trouble showing that phi is an isomorphism, especially the well-defined part.
Can you be more specific plz? Where exactly are you having trouble.

9. I'm trying to show that phi: D --> Q defined by phi(a) = [a,1] is an isomorphism.
I don't have a problem showing that phi is onto. I'm having a problem showing that phi is 1-1 and preserves the operations + and *

10. Originally Posted by Coda202
I'm trying to show that phi: D --> Q defined by phi(a) = [a,1] is an isomorphism.
I don't have a problem showing that phi is onto. I'm having a problem showing that phi is 1-1 and preserves the operations + and *
For onto
Consider [a,b] in Q where a,b belong to F.
[a,b]=[ab^-1,bb^-1]=[ab^-1,1]
F being a field and b<>0 ensure there exists b^-1.
So phi(ab^-1)=[a,b]
Hence onto

For 1-1
Let phi(a) = phi(b)
[a,1]=[b,1]
By definition of equality in Q a.1=1.b => a=b
Hence 1-1
For 1-1 you can also show phi(a)=[0,1] => a=0

For +,* well defined
Note phi(a+b) = [a+b,1] = [a,1]+[b,1] = phi(a)+phi(b)
Similarly for *
These result follow directly from the definition of +,* in Q.

Hope this clarifies

11. Very much so, thanks.