Score:1

How is the Chinese remainder theorem used in this proof?

cn flag
Bob

enter image description here

Can you explain it in detail ?

Score:1
cn flag

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

Chris Peikert avatar
in flag
This is the right idea, but the products of ideals in your answer (starting from the first sentence) are not rings. Instead, you should refer to the *quotient rings* of the ring $R$ modulo those products of ideals.
Ievgeni avatar
cn flag
@ChrisPeikert You're right, I will modify it.
mangohost

Post an answer

Most people don’t grasp that asking a lot of questions unlocks learning and improves interpersonal bonding. In Alison’s studies, for example, though people could accurately recall how many questions had been asked in their conversations, they didn’t intuit the link between questions and liking. Across four studies, in which participants were engaged in conversations themselves or read transcripts of others’ conversations, people tended not to realize that question asking would influence—or had influenced—the level of amity between the conversationalists.