If F is a field, prove that the field of quotients, Q, is isomorphic to F
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. $\displaystyle \frac{a}{b}=ab^{-1}\in F$, so I am not entirely sure what this question is asking for you to show
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 $\displaystyle \phi_s:F \rightarrow S^{-1}F$ given by $\displaystyle f \mapsto fs/s$ for any s in S, where S is the set of all non-zero elements of a field F, and $\displaystyle \phi_s(s)$ is a unit in $\displaystyle S^{-1}F$ for any $\displaystyle s \in S$. It is easy to check that $\displaystyle \phi_s$ is an injective homomorphism. To see $\displaystyle \phi_s$ is surjective, for any $\displaystyle f/s \in S^{-1}F$, we have $\displaystyle f/s = \phi_s(fs^{-1})$.
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'
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