Prove F isomorphic to the field of rational numbers

[Shackleford]

Sorry for so many questions. This is the second-to-last problem in the last homework.

I don't think I even understand the question. I know what a group isomorphism is. How does it relate to the field of quotients of ring E of all even integers. How do I show F is isomorphic to the field of rational numbers? What's the mapping from E to the rationals?

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110806_121111.jpg?t=1312651137

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110806_121121.jpg?t=1312651148

 

[micromass]

Hi shackleford!

So, we are dealing with the space F, which is the field of fractions of the even integers. So all elements in F have the form (a,b) with a,b even and with $b\neq 0$.

The first thing we'll have to do is to find the candidate isomorphism, then we have to prove that it is an isomorphism.

Now, we need an isomorphism. So we need to map each element (a,b) in F to a rational number. Do you see an easy way to make (a,b) into a rational number?

 

[Shackleford]

Is the work I wrote down correct so far?

I wrote down S as the set of all ordered pairs (a,b) with a and b from the set of all even integers.

I then wrote down Q as the quotient field of all equivalence classes of [a,b] with (a,b) from S.

So, I need to create a mapping that is isomorphic. Could I simply do the fraction (a,b) to a/b? What operations do I consider in determining if it's isomorphic?

 

[micromass]

Is the work I wrote down correct so far?

I wrote down S as the set of all ordered pairs (a,b) with a and b from the set of all even integers.

I then wrote down Q as the quotient field of all equivalence classes of [a,b] with (a,b) from S.

Yes, so far it's all ok!

So, I need to create a mapping that is isomorphic. Could I simply do the fraction a/b?

Yes, that is the required isomorphism. So, you need to show that the map is
- well-defined (so, for example 2/4 gets sent to the same element as 4/8)
- injective
- surjective
- f(x+y)=f(x)+f(y)
- f(xy)=f(x)f(y)