How is the Chinese remainder theorem used in this proof？

Question

Can you explain it in detail ?

Answer 1

You have to apply the chinese remainder theorem to the ring $\prod_{1\leq i\leq r} \mathfrak{p}_i^{e_{i}+1}$.

This ring is isomorphic to the ring $\times_{1\leq i\leq r} \mathfrak{p}_i^{e_{i}+1}$. Let call $\phi$ this isomorphism.

Let recall that for all $x \in \prod_{1\leq i\leq r} \mathfrak{p}_i^{e_{i}+1}$, $x \mod \mathfrak{p}_i^{e_{i}+1}$ is the $i$th coordinate of $\phi(x)$.

Then we compute $t :=\phi^{-1}(t_1 \mod \mathfrak{p}_1^{e_{1}+1}, \dots, t_r \mod \mathfrak{p}_r^{e_{r}+1})$.

Thus $t\in \prod_{1\leq i\leq r} \mathfrak{p}_i^{e_{i}+1} $ (because $\phi^{-1} : \times_{1\leq i\leq r} \mathfrak{p}_i^{e_{i}+1} \rightarrow \prod_{1\leq i\leq r} \mathfrak{p}_i^{e_{i}+1}$).

And $t\mod \mathfrak{p}_i^{e_{i}+1} = \phi^{-1}(t_1 \mod \mathfrak{p}_1^{e_{1}+1}, \dots, t_r \mod \mathfrak{p}_r^{e_{r}+1}) \mod \mathfrak{p}_i^{e_{i}+1}$ $$ = t_i \mod \mathfrak{p}_i^{e_{i}+1} \mod \mathfrak{p}_i^{e_{i}+1} =t_i \mod \mathfrak{p}_i^{e_{i}+1}$$