The question is as follows.

Let $D= \{m+n\sqrt 2\text{ s.t. }(m,n) \in \mathbb Z^2\}$ which is an integral domain. Let $Q$ be its field of fractions and $\phi: D \to Q$ the usual map.

Find an isomorphism

$$\alpha : \{a+b\sqrt 2\text{ s.t. }(a,b) \in \Bbb Q^2\} \to Q$$

whose restriction to $D$ is $\phi$.

You should write the map explicitly, prove that it's injective and surjective.


So i have gone over what is written below and some other things in my textbook. I have no idea what the usual map $\phi: D \to Q$ is i understand that this is an integral domain going to a field of fractions Q and i believe that the below post has defined what Q must look like as a field to have this integral domain embedded in it. my confusion arises in constructing the map from the above integral domain to the below Q and proving its a bijection i know that all the pieces are here i just can't quite fit them together.

i have the operation so to speak on either side of $\phi$ defined by above and below just that the map is eluding me.

  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Martin Sleziak Mar 16 '13 at 7:08
  • $\begingroup$ Hmm that makes a lot more sense actually then the gibberish in my text book ty. $\endgroup$ – Faust Mar 17 '13 at 17:44

Using the trick $$\frac1{a+b\sqrt2}=\frac{a-b\sqrt2}{a^2-2b^2}\,,$$ we can see that the field of fractions $Q$ is $\{r+s\sqrt2\mid r,s\in\Bbb Q\}=\Bbb Q(\sqrt2)$.

Basically this was asked. However, the field of fractions is also defined in an abstract way, as equivalence classes of pairs $(u,v):\ u,v\in D,\,v\ne 0$, and define the operations as $(u,v)$ would stand for '$u/v$'. In this way, $\alpha$ can be explicitly given on these $(u,v)$ pairs, using the above equality.

  • $\begingroup$ Yeah i couldn't figure out how to write root 2 so i did that haha $\endgroup$ – Faust Mar 16 '13 at 4:08

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