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The inclusion $(m\mathbb{Z} + k) \cap (s \mathbb{Z} + t) \supset c + lcm(m,s)\mathbb{Z}$ is trivial, but I've been stuck with the other one for some time now. I thought about the chinese remainder theorem but couldn't actually apply it. Help?

OBS: Here we assume that the intersection is non empty.

Bill Dubuque
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Matheus Andrade
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    @Did My original title actually had words, but someone else edited it (and I like it better now actually). Imho, excellent communication requires only that the intended meaning is conveyed clearly, correctly and without ambiguity, and I think this applies here too... – Matheus Andrade Feb 08 '19 at 13:35
  • Words and clarity are not antagonistic... Many student passes by a phase when they think that symbols are more precise, then they grow up (and/or they study the masters). – Did Feb 08 '19 at 17:26
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    @Did I agree with you in general, but in this case I think it doesn't matter. I'm quite aware symbols can often be antagonistic to clarity and then it's better to use words, but for this specific situation I think there is no loss of clarity. I admit I've already passed that phase you mention. – Matheus Andrade Feb 08 '19 at 17:51
  • Hmmm... you are the one who linked words to lack of clarity, right? And I disagreed. But the link of symbols to lack of clarity was not made, so there is no need to debunk it. If you want to know, I would rather invoke ugliness here... – Did Feb 08 '19 at 18:10
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    What I meant was that words are sufficient for clarity, but not absolutely always necessary for it. I agree it's ugly alright and I wouldn't let that title up if it was anything any more advanced, but in this case it's not. I think this discussion is meaningless since imo we seem to agree on almost everything. You can rest easy knowing I try to use as many words as I can as often as possible. – Matheus Andrade Feb 08 '19 at 18:17
  • Then why you let the title in its sorry present state is something I cannot fathom (but that you have every right to do, of course). – Did Feb 08 '19 at 18:23
  • @Did I tried to think of a reason too and couldn't... so there, problem solved. Thanks for the patience. – Matheus Andrade Feb 08 '19 at 21:27

1 Answers1

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Hint $\,c',c\in (k\!+\!m\Bbb{Z}) \cap (t\!+\!s \Bbb{Z})\,\Rightarrow\,c'\!-c\in m\Bbb Z,s\Bbb Z\,\Rightarrow\,c'\!-c\in m\Bbb Z\cap n\Bbb Z = {\rm lcm}(m,n)\Bbb Z$

Remark $ $ This is equivalent to the uniqueness of a solution $\,x = c\,$ of the following congruences

$$\begin{align} x&\equiv k\!\!\pmod{\!m}\\ x&\equiv t\!\!\pmod{\!s}\end{align}$$

If $\,x = c'$ is another solution then $\, c'\equiv x\equiv c\pmod{\! m}\,$ so $\ m\mid c'-c.\,$ Similarly $\,s\mid c'-c\,$ therefore $\,\ell := {\rm lcm}(m,s)\mid c'-c,\,$ thus $\,c'\equiv c\pmod{\!\ell},\,$ i.e. any solution is unique $\!\bmod \ell.\,$ Therefore if you know that form of CRT then it follows immediately from the uniqueness part.

As explained here, the existence part of CRT (solvability criterion) can be expressed similarly, i.e.

$$ {k\! +\! m\Bbb Z\,\cap\, t\! +\! s\Bbb Z \neq \phi \iff k-t \in m\Bbb Z+s\Bbb Z = \gcd(m,s)\Bbb Z}\qquad $$

Bill Dubuque
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