Show that $\left< \mathbb{Q}\left[\sqrt 2\right], < \right> \simeq \left<\mathbb{Q}, < \right>$

Question

Show that $\left< \mathbb{Q}\left[\sqrt 2\right], < \right> \simeq \left<\mathbb{Q}, < \right>$,

where $\mathbb Q$[$\sqrt 2$] is the smallest subfield of $\mathbb R$ containing $\sqrt 2$.

I do not really know where to go from here, i.e. start the proof.

Answer 1

Any two "countable, dense, totally ordered sets without endpoints" are isomorphic (with respect to the order).

"Dense" refers to that for any $a < b$, there exists a $c$ inbetween; so there is a $c$ such that $a < c < b$. And "without endpoints" means that for any $a$, there is at least one larger and at least one smaller element (there exists $b > a$ and there exists $c < a$).

Your problem is an instance of this because $\mathbb{Q}$ and $\mathbb{Q}[\sqrt{2}]$ are both countable (prove this), dense (prove this also), and without endpoints (prove this finally).

Now to prove that any two countable, dense, totally ordered sets without endpoints are isomorphic: this is a classic result that we approach as follows. Let the sets be $A$ and $B$. Since they are countable, we can number them: \begin{align*} A &= \{a_1, a_2, a_3, \ldots \} \\ B &= \{b_1, b_2, b_3, \ldots \} \end{align*} (But be careful to note that these numberings are not in order, so it's not true that $a_1 < a_2 < a_3 < \cdots$. They could be in a completely random order.)

Anyway, the idea is then to match up $A$ and $B$ one element at a time: we build an isomorphism between the two:

• First pick the element $a_1$ of $A$. Match it with the first unmatched element of $B$, which is $b_1$.

• Now pick the first element of $B$ on the list that is not yet matched, that is, $b_2$. If $b_2 > b_1$, then pick some element of $A$ that is larger than $a_1$, say, $a_{i_1}$; otherwise pick some element of $A$ that is smaller than $a_1$. Match this element with $b_2$.

• Now pick the first element of $A$ on the list that is not matched. Maybe it's $a_2$, or $a_3$, for example. Consider how it compares to all the previous $a$'s in the ordering ($<$), and pick an element of $B$ that compares the same way. This is always possible due to "denseness" and "without endpoints".

Basically, we keep alternately picking the first element on the $A$ list that is not matched, and the first element on the $B$ list that is not matched; we compare the element with everything matched up so far, and we pick a corresponding element of $B$ (or $A$) to match it to which is consistent with the total orderings $<$ on $A$ and $B$.

This always preserves the ordering $<$ by construction -- and it actually eventually matches all elements of $A$ and all elements of $B$ , because we always match up the first element on the lists that is not matched, so we eventually get to every element of $A$ and every element of $B$. So it is an isomorphism.