Smallest ring whose field of fractions includes all the reals (subring of omnific integers?)

Question

The surreal numbers have a subring, the ring of "omnific integers" or $\mathbf{Oz}$, which have the property that every surreal number is a quotient of two omnific integers. That is, the field of fractions of the omnific integers is the entire surreal number field, which in particular includes all the reals.

A ring whose field of fractions includes all the reals seems like a useful thing. The omnific integers would seem to be much larger than necessary if that is what we want. So we can ask for simpler examples.

Of course, $\Bbb R$ is a trivial example of a ring whose field of fractions includes all of $\Bbb R$. So, to be precise, I am interested in rings which do not already have all the reals, but whose field of fractions does have all the reals.

In particular, I have the following questions:

1. Does there exist some (ordered) ring $R$, which is not a superset of $\Bbb R$ but whose field of fractions is a superset of $\Bbb R$, that is "smallest" in the sense that $R$ is isomorphic to a (ordered) subring of any other (ordered) ring with this property?
2. Does the ring of omnific integers have any smallest subring with the above property?
3. Would any ring with the above property embed into the omnific integers anyway, making these criteria all equivalent?

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To add some detail to the above:

In the omnific integers, for any real number $r$, we have that $r \omega$ is an omnific integer. So, to start, we can look at the following fragment of the omnific integers, with all elements of the form

$$z + r_1\omega + r_2\omega^2 + ... + r_n\omega^n$$

where $z$ is an integer, and the $r_n$ are all real numbers. I will notate this ring as

$$\Bbb Z \: \tilde \oplus \: \Bbb R \: \tilde \oplus \: \Bbb R \: \tilde \oplus \: ...$$

where the $\tilde \oplus$ is a kind of modified direct sum in which the polynomial coefficient multiplication is used rather than pointwise multiplication, which is always possible as long as each ring is a subring of the ring after it.

It is easy to see that the above ring has $\Bbb R$ in its field of fractions. It is also easy to see that this is true for any ring of the form

$$\Bbb Z \: \tilde \oplus \: \Bbb Z \: \tilde \oplus \: ... \tilde \oplus \: \Bbb Z \: \tilde \oplus \: \Bbb R \: \tilde \oplus \: \Bbb R \: \tilde \oplus \: ...$$

which the first ring embeds into, and which also embeds into the first ring.

So, one thing we can do is ask if there exists a countable sequence of rings $R_n \neq \Bbb R$ such that

$$ R_1 \subset R_2 \subset R_3 \subset ...$$

and

$$\Bbb R \subset \text{Quot}(R_1 \: \tilde \oplus \: R_2 \: \tilde \oplus \: R_3 \: \tilde \oplus \: ...)$$

One way this could be is if there is a countable sequence of $R_n$ such that the union of all the $R_n$ is $\Bbb R$, so that each real number appears at some point in the $R_n$.

Answer 1

Assume for contradiction that such a ring $R$ exists.

Consider the ring $R_{\omega}:=\mathbb{Z} \oplus \omega\mathbb{R}[\omega]$ which is contained in $\mathbf{Oz}$. We have $\mathbb{R} \not\subseteq R_{\omega}$ and $\operatorname{Frac}(R_{\omega})\supseteq \mathbb{R}$. Since $R_{\omega}$ is discretely ordered, so must be $R$.

Consider a ring $R_0$ given by the answer linked by Kevin Casto. Since $R_0$ is archimedean, so must be $R$.

But $\mathbb{Z}$ is the only discretely ordered archimedean ordered ring. So we must have $R=\mathbb{Z}$, which cannot be.