Construction of homomorphism between $^\ast\mathbb{R}$ and $^*\mathbb{Q\cap L}$

Question

Denote by $\mathbb{I}$ the ring of infinitesimals and by $\mathbb{L}$ the ring of finite hyper-reals. Prove that $$\mathbb{R}\cong{^\ast\mathbb{Q\cap L/^\ast Q\cap I}}.$$

I thought using the first isomorphism theorem for rings by defining an homomorphism $$\phi:\mathbb{Q\cap L}\to {^\ast\mathbb{R}},\quad \phi(q)=\operatorname{st}(q),$$ where $st$ is the standard part function. Alas, that proves only that $$\mathbb{Q}\cong{^\ast\mathbb{Q\cap L/^\ast Q\cap I}}.$$ How can I build a sequence in $^\ast\mathbb{Q\cap L}$ (maybe Cauchy?) that will prove the isomorphism between $\mathbb{R}$ and the quotient ring above? Should I define another homomorphism?

Answer 1

1. $^*\mathbb{Q},\mathbb{L}$ are subrings of $^*\mathbb{R}$, then $^*\mathbb{Q}\cap\mathbb{L}$ is a subring of $^*\mathbb{R}$.
2. $\text{st}:\mathbb{L}\rightarrow\mathbb{R}$ is a ring homomorphism, then the restriction $\text{std}:{^*}\mathbb{Q}\cap\mathbb{L}\rightarrow\mathbb{R}$ is a ring homomorphism (yes, is your $\phi$, but with the * in the correct place). Therefore ${^*}\mathbb{Q}\cap\mathbb{L}/\ker(\text{std})\cong\text{im}(\text{std})$ by the first isomorphism theorem.
3. To see that $\text{im}(\text{std})=\mathbb{R}$ (i.e., that $\text{std}$ is exhaustive) you can use the density of $\mathbb{Q}$ in $\mathbb{R}$: $\forall r\in \mathbb{R}, \forall \epsilon\in \mathbb{R}_{>0},\exists q\in \mathbb{Q};r-\epsilon<q<r+\epsilon$ (proof). Therefore (by transfer, choosing $\epsilon\in\mathbb{I}_{>0}$): $\forall r\in \mathbb{R},\exists q\in {^*}\mathbb{Q};q\simeq r$. Now prove that $q\in\mathbb{L}$ and $\text{std}(q)=r$.