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Definition 3.1.1 in page 25 of this book is the definition of quasiperod and Proposition 3.1.3. shows that gcd of two quasiperiods is a quasiperiod. The whole proof is clear except for the part about CRT.

I would appreciate a simple explanation of the following claim from the proof of Proposition 3.1.3:

Now choose an integer $w_1$ such that it is from the prescribed residue class modulo $d_2/ \gcd(d_1, d_2)$, and that for any prime divisor $p$ of $q$ not dividing $d_1d_2$, we have $w_1 \not\equiv −m/d_1\pmod p$. The existence of such integers is guaranteed by the Chinese Remainder Theorem. How holds $w_1 \not\equiv −m/d_1\pmod p$ and how it comes from CRT?

PS this is an exercise in Apostol's book Ch8 and also Montgomery's book Ch9. In Apostol "quasiperiod" is named "induced modulus".

Erick Wong
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Ali
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  • In internet I found \notequiv for not congruent to, but it didn't work in the OP. Does someone knows the correct latex for that? Thanks – Ali Mar 30 '24 at 21:28
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    \not\equiv (in general \not will negate all relation symbols, usually in a visually good way) – Greg Martin Mar 30 '24 at 21:43

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$\forall i\!:\, w\not\equiv b\pmod{\!p_i}\!\iff\! \forall i\!:\, p_i\nmid w\!-\!b\!$ $\iff\! \forall i\!:\, (p_i,w\!-\!b)\!=\!1\!\iff\! \overbrace{\color{#0a0}{(P,w\!-\!b)\!=\!1}}^{\!\!\!\large P\ =\, \prod p_i}$

When $\,w\equiv a\pmod{\!d}\,$ so $\,w=a\!+\!nd,\,$ the above becomes $\,\color{#0a0}{(P,\,a\!-\!b+ n\,d)\!=\!1}.\,$ In OP we have $(P,d)\!=\!1,\,$ so it is solvable for $\,n\,$ by the Coprime Dirichlet Theorem (which - as shown there - is indeed provable by CRT, but also has a much simpler direct proof).

Bill Dubuque
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  • Thanks, I am reading your answer, for the first line p_i divides q and does not divide d_1d_2 and maybe gcd of these two not be 1 – Ali Mar 31 '24 at 02:23
  • Second comment: in OP (q,d_i) may not be 1 – Ali Mar 31 '24 at 02:25
  • So by "any prime divisor $p$ of $q$ not dividing $d_1d_2$" you mean $q\nmid d_1d_2,,$ not $,p\nmid d_1d_2?\ $ Please quote the textbook exactly as written, and you should include more of the proof for context - one should not need to follow the link - which I did not. – Bill Dubuque Mar 31 '24 at 02:28
  • No, the text means for any prime $p$ such that $p \mid q $ and $p \nmid d_1d_2$, am I right? Could you have a look at the text page 29, please? Following lines of text suggest that as well... – Ali Mar 31 '24 at 02:30
  • It is obvious from the OP. The rest of text is irrelevant that's why I didn't quote. – Ali Mar 31 '24 at 02:44
  • @Ali By definition every $,p_i,$ is coprime to $,d_1d_2$ so coprime to $,d_2,,$ so their product $,q,$ is coprime to $,d_2$ so also to its divisor $,d = d_2/(d_1,d_2),,$ i.e. $,(q,d)=1.\ \ $ – Bill Dubuque Mar 31 '24 at 02:53
  • There are two types of prime divisors of $q$. those $p_i$ coprime to $d_1d_2$ and those divide $d_1d_2$. The product of this two sets divide $q$ and may not be coprime to $d_1d_2$ if $\gcd(q,d_1d_2)>1$. In OP, I have quoted from the text for those $p_i|q$ that $p_i$ coprime to $d_1d_2$ and product of them is not $q$ – Ali Mar 31 '24 at 02:58
  • @Ali But the text requires that the incongruence $,w\not\equiv b,$ needs to hold only for those $p$ coprime to $d_1d_2\ \ $ – Bill Dubuque Mar 31 '24 at 03:05
  • So you have renamed q, ok let me think – Ali Mar 31 '24 at 03:10
  • @Ali There was a notation clash, I renamed my $,q,$ to $,P = \prod p_i\ \ $ – Bill Dubuque Mar 31 '24 at 03:13
  • @Ali If $,m/d_1\not\in\Bbb Z,$ then we can use CRT to find an integer $,b\equiv m/d_1\pmod{!p_i},$ for all $,i.,$ The issues above were caused by the omission of necessary context for the proof. – Bill Dubuque Mar 31 '24 at 03:24
  • No not relevant to CRT, since $d_1$ and $p_i$ are coprime so $w_1d_1=m$ has unique solution for $w_1$, namely $b$ - that's why I dropped my comment! – Ali Mar 31 '24 at 03:28
  • @Ali My prior comment (replying to your deleted comment), is relevant to the above proof - where $b$ is an integer (vs $,b_i := md_1^{-1}\pmod{!p_1},$ where the inverse depends on the modulus $p_i).\ $ – Bill Dubuque Mar 31 '24 at 03:34
  • I totally got it except for one thing: $(P,d)=1$ so $(P, t, d)=1$ for any $t$. So (by the link you provided) there is some $n$ such that $(P,a−b+nd)=1$ for any choice of $a$. So any $a$ or equivalently any $w$ works, a contradiction? – Ali Mar 31 '24 at 03:38
  • @Ali, Yes, glad all is clear now. – Bill Dubuque Mar 31 '24 at 03:39
  • No there is a contradiction : how any choice of w mod d is OK? – Ali Mar 31 '24 at 03:40
  • ( i didn't downvote) – Ali Mar 31 '24 at 03:43
  • @Ali Why do you think it is contradictory that any $,a,$ works for the cited excerpt in your question? (Don't worry abut the downvote - it is likely due to site politics not math). – Bill Dubuque Mar 31 '24 at 03:51
  • So we can choose any of $0,1,2,...,d-1$ for $w \equiv a$ mod $d$?! So any integer actually and one of them is congruent to $-m/d_1$ mod $p_i$, that is a contradiction. – Ali Mar 31 '24 at 03:54
  • Yes, in the notation of the linked proof, if $,(a,c)=1$ then for any $,b,$ we have that $,(ax+b,c)=1$ is solvable for $,x,,$ i.e. if an A.P. has step-size $,a,$ coprime to $,c,$ then it has a term coprime $,c.\ \ $ – Bill Dubuque Mar 31 '24 at 04:07
  • Any $a$ means any $w_1$ and that means even $w_1 \equiv -m/d_1$ mod $p_i$ but this contradicts the assumption in OP – Ali Mar 31 '24 at 04:20
  • @Ali That the system of (in)congruences has a solution $,w_a,$ for any $,a,$ does not imply that you can further choose that solution to satisfy an arbitrary additional congruence too. Indeed by construction $,w_a\not\equiv b\pmod{!p},$ so there is no solution additionally satisfying $,w_a\equiv b\pmod{!p}\ \ $ – Bill Dubuque Mar 31 '24 at 04:47
  • Thanks a lot :) {a question not relevant to OP: I have just started my PhD in Riemann zeta function and I couldn't solve this very question by myself. Should I quit? Am I too dumb for advanced maths?} – Ali Mar 31 '24 at 05:08
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    @Ali This form of CRT is not well-covered in many elementary textbooks, so it often proves a stumbling block for students. So there is no need to be discouraged by that alone. – Bill Dubuque Mar 31 '24 at 05:14