nonstandard analysis – Legitimacy of the shadow map serving as a field homomorphism for a specific hyperfinite field formed of a union of hyperfine lattices

I’m hoping to get some comment on the legitimacy of my approach to creating a hyperfinite ring formed of a union of modular groups in order to obtain a field homomorphism from this hyperfinite space to the real numbers. As nonstandard analysis isn’t my area I feel I’m at risk of accidentally making mortal error and so I’m looking for constructive advice regarding the legitimacy of my approach.

As I said I’m looking to construct a hyperfinite space that can serve as an approximation for the reals as a field in the sense that the shadow (standard) map serves a field homomorphism between this space and $mathbb{R}$. As motivation for how I’ve tried to go about this consider taking the following set

$$
{ }^{star} mathbb{Z}_{omega}:=left{k in{ }^{star} mathbb{Z} mid-leftlceil frac{omega-1}{2} leq k leqleftlfloorfrac{omega-1}{2}rightrfloorright}right.
$$

where $omega:=omega_{mathrm{uv}} omega_{mathrm{ir}}$ for some positive $omega_{mathrm{uv}}, omega_{mathrm{ir}} in{ }^{star}mathbb{Z}$ We can define a hyperfinite abelian group with 0 as the unit with the group operation

$$
a+_{omega} b:=left{begin{array}{ll}
a+b & text { if }-leftlceilfrac{omega-1}{2}rightrceil leq a+b leqleftlfloorfrac{omega-1}{2}rightrfloor \
a+b-omega & text { if }leftlfloorfrac{omega-1}{2}rightrfloor<a+b \
a+b+omega & text { if } a+b<-leftlceilfrac{omega-1}{2}rightrceil
end{array}right.
$$

We can go further and define a ring via
$$
a cdot_{omega} b:=left{begin{array}{ll}
a cdot b & text { if }-leftlceilfrac{omega-1}{2}rightrceil leq a cdot b leqleftlfloorfrac{omega-1}{2}rightrfloor \
a cdot b-k omega & text { if }leftlfloorfrac{omega-1}{2}rightrfloor+(k-1) omega<a cdot b leqleftlfloorfrac{omega-1}{2}rightrfloor+k omega \
a cdot b+k omega & text { if }-leftlceilfrac{omega-1}{2}rightrceil-k omega leq a cdot b<-leftlceilfrac{omega-1}{2}rightrceil-(k-1) omega
end{array}right.
$$

where the ring $left({ }^{star} mathbb{Z}_{omega},+_{omega}, 0, cdot omega, 1right)$ is a field if $omega$ is prime.

Now consider the ‘scaled’ version of this structure
$$frac{1}{omega_{mathrm{uv}}} star mathbb{Z}_{omega}=left{frac{k}{omega_{mathrm{uv}}} mid k in star mathbb{Z},-left(frac{omega-1}{2}rightrceil leq k leqleftlfloorfrac{omega-1}{2}rightrfloorright}$$

Now we take the shadow of this

$$
operatorname{shd}left(left(frac{1}{omega_{mathrm{uv}}}^{star} mathbb{Z}_{omega}right)_{mathrm{fin}}right)=left{operatorname{shd}left(frac{k}{omega_{mathrm{uv}}}right) mid k in{mathrm{Z}} text { s.t. }-left(frac{omega-1}{2}right) leq k leq mid frac{omega-1}{2}rightrfloor text { and } frac{k}{omega_{mathrm{uv}}} text { is finite }} subseteq mathbb{R}
$$

Finite elements are closed under the additive group structure of shd $left(left(frac{1}{omega_{mathrm{uv}}} star mathbb{Z}_{omega}right)_{mathrm{fin}}right)$ and taking the standard part is linear with respect to said additive group structure: this means that $left(left(frac{1}{omega_{text {uv }}} star mathbb{Z}_{omega}right)_{text {fin }},+_{omega}, 0right)$ is an abelian group.

Importantly for my purposes if I choose $omega_{mathrm{uv}}$ and $omega_{mathrm{ir}}$ to be ‘infinite’ then I believe I get the following

$$text { shd }:left(left(frac{1}{omega_{mathrm{uv}}} star mathbb{Z}_{omega}right)_{text {fin }},+_{omega}, 0right) longrightarrow(mathbb{R},+, 0)$$
as the range of the modulus is now up to an infinite number as is the scaling.

Now this approach will fail for a ring because we will want to write
$$
frac{h}{omega_{mathrm{uv}}} cdot_{omega} frac{k}{omega_{mathrm{uv}}}=frac{h cdot_{omega} k}{omega_{mathrm{uv}}}
$$

but we see that what we have is
$$
frac{h}{omega_{mathrm{uv}}} cdot omega frac{k}{omega_{mathrm{uv}}}=frac{h cdot_{omega} k}{omega_{mathrm{uv}}^{2}}
$$

which isn’t in our space.

My solution is to do the following and it is the legitimacy of this which I would like to get some opinions on.

Consider this union of lattices of the type we just discussed:

$$bigcup_{n in star mathbb{Z}_{kappa}} frac{1}{omega_{mathrm{uv}}^{n}}^{star} Z_{omega omega_{mathrm{uv}}^{n-1}}$$
where $kappa$ is an ‘infinite’ hyperinteger and
$$
left.frac{1}{omega_{mathrm{uv}}^{n}}^{star} mathbb{Z}_{omega}=left{frac{k}{omega_{mathrm{uv}}^{n}}left|k in{star} mathbb{Z},-left(frac{omega omega_{mathrm{uv}}^{n-1}-1}{2}rightrceil leq k leqright| frac{omega omega_{mathrm{uv}}^{n-1}-1}{2}rightrfloorright}$$

Now this union of lattices of greater and lesser fineness means that the multiplication problem described above is dealt with though perhaps at the cost of the curious choice of the following being the case:
$$
acdot_{omega}b = frac{acdot_{omega}b}{omega^{kappa}}
$$

Note how we have division defined here as we can take the usual modular inverse $i n v_{omega_{u v}}(p)$ and for any

$y=frac{q}{omega_{u v}^{n}}$ we will have $tilde{y}=omega_{u v}^{n} cdot_{omega} operatorname{inv}_{omega_{u v}^{n}}(p)$
$$
y tilde{y}=frac{q cdot_{omega} omega_{u v}^{n} cdot_{omega} i n v_{omega_{u v}^{n}}(q)}{omega_{u v}^{n}}=1
$$

Now the meat of my question: Is this a legitimate field homomorphism?
$$
operatorname{shd}:left(left(frac{1}{omega_{mathrm{uv}}} star mathbb{Z}_{omega}right)_{fin},+_{omega}, 0, cdot omega, 1right) longrightarrow(mathbb{R},+, 0, cdot, 1)
$$

where multiplication and addition are defined as specified here.