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I have some trouble understanding this particular proof of the Chinese Remainder Theorem.

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EDIT:

As far as I understood, this is how the proof goes:

Let $M_i$ be the integer we obtain after dividing the product of modules $M$ by one of those moduli, as follows $M_i = M/m_i $. Then we have $ \gcd(M_i, m_i) = 1$ , because every factor $m_r$ of $M$ (where $r \neq i$) is coprime to all other factors of $M$ and therefore also to $m_i$. This is equivalent to saying that $M_i \equiv_{m_i} 1$.

This means that there exists a unique integer $N_i$, which satisfies $M_i N_i \equiv_{m_i} 1$ , which means that $N_i$ is the inverse of $M_i$ modulo $m_i$. At the same time, we know that every other factor $m_r$ of $M$ (where $r \neq i$) divides $M_i$, which means $M_i \equiv_{m_r} 0$. This means that it also divides $M_i N_i$ and $a_i M_i N_i$, since if it divides $M_i$ it must also divide every multiple of it (is this true or does it follow that it must also divide the inverse of $M_i$ ?), i.e. $ a_i M_i N_i \equiv_{m_r} 0 $.

Now as far as I understood from the answers below, if we build the sum $\sum_{i = 1}^k a_i M_i N_i $ and take for every summand the remainder modulo $m_r$, where $r \neq i$, then we will get $a_i M_i N_i \equiv_{m_r} 0 \ $ for every summand except the one where $ r = i$, which will then be $a_r M_r N_r \equiv_{m_r} a_r $, because $M_r N_r \equiv_{m_r} 1 $ and if we multiply that with $a_r$ we get $a_r$ as a remainder (does this $a_r$ refer to the ones in the system of congruence equations? i.e. $x \equiv_{m_r} a_r$ ? But if that is true, why can't we just use the ones from the system of congruence equations directly?). Therefore, the entire sum $\bigg( \sum_{i = 1}^k a_i M_i N_i \bigg) \equiv_{m_r} a_r$ (and not every summand) (does this also mean that if we take a different $r$ we obtain a different $a_r$ every time?).

Therefore, the solution is $ x = R_M \bigg( \sum_{i = 1}^k a_i M_i N_i \bigg)$ (how do we get to that? Where does it follow from? And can't we just add the list $a_1, a_2, ..., a_k$ instead of computing the sum?)

And the uniqueness part I don't understand at all.

Thank you for the clarifications.

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    $\begingroup$ See here for some intuition on the CRT formula (which - as in your text - is often omitted). $\endgroup$ Jul 29, 2019 at 15:18
  • $\begingroup$ @BillDubuque Thank you for this useful post. I read it and it helped me with the intuition. However, I still can't seem to get a grasp on those questions I posted. Could you address them in an answer? $\endgroup$
    – Marwan
    Aug 5, 2019 at 13:59

2 Answers 2

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All summands but the $k$th one are $\equiv_{m_k}0 $ because for $i \neq k$ we have $M_iN_i \equiv_{m_k} 0$. Hence only the $k$th term $a_kM_kN_k$ is left and since $M_kN_k \equiv_{m_k} 1$, the whole sum is $\equiv_{m_k} a_k$.

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  • $\begingroup$ Is this $a_k$ the same one as the one in the congruence equation $x \equiv_{m_k} a_k$ ? And if yes, can't we just add them all together? $\endgroup$
    – Marwan
    Jul 31, 2019 at 13:43
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Read carefully:

He says $ M_i N_i \equiv_{m_\color{red}k} 0\ $ for $k\ne i,$ but $ M_i N_i \equiv_{m_\color{red}i} 1. $

From this it follows that $ \sum_{i=1}^r a_i M_i N_i \equiv_{m_k} a_k . $

(All terms in the sum are $\equiv0$ except the one when $i=k$.)

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