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Let $R \subseteq \mathbb{N} \times \mathbb{N}$ such that $x, y \in \mathbb{N}$: $$xRy \iff g_1(x) = g_1(y)$$ where $g_1(x) = \left \lfloor \frac{x}{10^{\lfloor \log_{10}(x) \rfloor}} \right \rfloor$

Let $h: \mathbb{N}/R \to \mathbb{F}_9 \cong \mathbb{Z}_3[x]/(x^2+1) $ such that $$[1]_R \mapsto (x^2+1)\\ [2]_R \mapsto 1+(x^2+1)\\ [3]_R \mapsto 2+(x^2+1)\\ [4]_R \mapsto x+(x^2+1)\\ [5]_R \mapsto x+1+(x^2+1)\\ [6]_R \mapsto x+2+(x^2+1)\\ [7]_R \mapsto 2x+(x^2+1)\\ [8]_R \mapsto 2x+1+(x^2+1)\\ [9]_R \mapsto 2x+2+(x^2+1)$$ where $\mathbb{F}_9$ is the finite field of order $9$, and $\mathbb{N}/R$ as the quotient set w.r.t the equivalence relation on the set $\mathbb{N}$.

Questions:

  1. Can $h$ be shortened using an explicit formula?
  2. What is the explicit inverse of $h$?

Attempt: I tried to solve these questions, but is too difficult for me. Any hints will help.

EDIT

I changed the bijection in order to make it simpler. In this way, I found a shorthand way to have this bijection:

Let $I = (x^2+1)$ be the (two-sided) ideal of $\mathbb{Z}_3[x]$ generated by $m(x) = x^2+1$. The inverse of $h$ is the following: Take the map $h^{-1}: \mathbb{Z}_3[x]/I \to \mathbb{N}/R$ such that $$h^{-1}(f(x)+I)=[g(3)+1 \pmod{10}]_R$$ where $f(x) - g(x) \in I$ and $g(x) \in \{0, 1, 2, x, x+1, x+2, 2x, 2x+1, 2x+2\}$.

Yoyos Tutoring
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    Your question uses $g_1$ and in that case the formula simplifies to $$g_1(x)=\left\lfloor\frac{x}{10^{\lfloor\log_{10}(x)\rfloor}}\right\rfloor$$, so you could avoid posting the general formula. – jjagmath Aug 26 '22 at 00:58
  • I am actually gonna use this comment for simplicity – Yoyos Tutoring Aug 26 '22 at 01:05
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    Also, the expressions defining $h$ can be simplified and you should do it. For example $h([6]) = x^2 = -1 = 2$. – jjagmath Aug 26 '22 at 01:08
  • @jjagmath, I changed the elements for $\mathbb{F}_9$. I noticed that the elements were incorrect. Thank You! – Yoyos Tutoring Aug 26 '22 at 01:20

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