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I was reading a description of the CRT which went like this

$x = a \mod p$
$x = b \mod q$ has a unique solution for $x \mod pq$

It also said - The reverse direction is trivial. Give $x \mod pq$ we can reduce it to $\mod p$ & $\mod q$

What are the steps to do the reverse. If I have an equation for $\mod pq$, what do I to obtain equations in $\mod p$ & $\mod q$?

user93353
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    if $x \equiv n \pmod{pq}$ then $x \equiv n \pmod{p}$ and $x \equiv n \pmod{q}$. – Oussema Apr 01 '21 at 16:44
  • Consider a simple example, eg $x\equiv 6\pmod{10}$. Then $x\equiv 6\pmod{2}$ and $x\equiv 6\pmod{5}$ – PM 2Ring Apr 01 '21 at 16:45
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    If you have $x=a$ mod$pq$ than $x=a$ mod $p$ and $x=a$ mod $q$, indeed the map of the isomorphism $\frac{\mathbb{Z}}{(p)}\oplus\frac{\mathbb{Z}}{(q)}\longleftarrow\frac{\mathbb{Z}}{(pq)}$ is easy to verify it is injective, the problem is that is surjective. – yo yo Apr 01 '21 at 16:47
  • But it is not very clear for me what you were looking for – yo yo Apr 01 '21 at 16:48
  • @yoyo I think it's very unlikely the person who asked this question knows what an isomorphism is. There is no need to complicate something very simple. – Oussema Apr 01 '21 at 16:49
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    $x \equiv n \pmod{pq} \iff x = n + kpq$ for some integer $k$ $\iff x = n + (kp) \cdot q \implies x \equiv n \pmod{q}$. – Oussema Apr 01 '21 at 16:50
  • Maybe, but what does reverse direction means? This is why i write the isomorphism – yo yo Apr 01 '21 at 16:52
  • @yoyo I think they just asked whether we could know the residue of $x$ mod $p$ and $q$ given its residue mod $pq$. As opposed to the CRT which gives you the residue mod $pq$ if you know the residue mod $p$ and mod $q$ and they are relatively prime. – Oussema Apr 01 '21 at 16:55
  • I confess, i never seen CRT in that form, if i complicate it was not my intention. – yo yo Apr 01 '21 at 17:31
  • Congruences always persist mod factors of the modulus - see the linked dupe. – Bill Dubuque Apr 01 '21 at 18:29
  • @yoyo - I am looking for the steps. – user93353 Apr 01 '21 at 22:44

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I think it's a lot easier to understand with an example. Suppose we are told $x\equiv 7 \bmod 15$, then we have $x\equiv 7 \bmod 3$ and $x\equiv 7\bmod 5$. Of course we can "reduce" this to $x\equiv 1 \bmod 3$ and $x\equiv 2\bmod 5$.

Hopefully this clears it up.

Asinomás
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  • I know we can reduce. My question is about the steps to do it. – user93353 Apr 01 '21 at 22:41
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    If the original equation is $x\equiv a \bmod pq$ you just write $x\equiv a \bmod p$ and $x\equiv a \bmod q$. – Asinomás Apr 01 '21 at 23:14
  • What if the original equation is more complex - can the same be done? For e.g. $c \equiv (ax +by) \mod pq$ - can it still be split as $c \equiv (ax +by) \mod p$ and $c \equiv (ax +by) \mod q$ – user93353 Apr 02 '21 at 00:53
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    Yes indeed ! (although this is really the direction which should be easy. Notice that if $x\equiv a \bmod n$ then we have $x\equiv a \bmod d$ for every divisor $d$ of $n$. This is because if $x-a$ is a multiple of $n$ it is also of $d$. – Asinomás Apr 02 '21 at 00:53
  • Ok, thank you ..... – user93353 Apr 02 '21 at 00:55
  • Just to make sure I am adressing what you asked: If you have any equivalence of the form $expression_1 \equiv expression_2 \bmod n$ then you can always break it into a bunch of equivalences mod the prime powers of $n$. – Asinomás Apr 02 '21 at 00:56