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I'm trying to prove that, in $\mathbb{Z}_n$, the set of $\overline{a}$ that are invertible is well-defined. That is, if $a,b \in \overline{a}$, so $a \equiv b \text{ (mod $n$)}$, then $\gcd(a,n) = \gcd(b,n) = 1$. (I'm more interested in the general result, though, that $\gcd(a,n) = \gcd(b,n)$.

My attempt Suppose $a \equiv b \text{ (mod $n$)}$. Then $a - b = kn$ for some integer $k \in \mathbb{Z}$. Then $a = b + kn$ and $b = a - kn$.

At this point, there is likely some proper of the gcd that I need to use to deduce the result from that, but I am not completely sure of what that is.

J. W. Tanner
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John P.
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  • Consider Bezout's identity or $\gcd(b + kn, n) = \gcd(b, n)$ – J. W. Tanner Sep 01 '20 at 03:59
  • I know Bezout's identity, but I can't think of a way to apply it here. – John P. Sep 01 '20 at 04:04
  • You just need to show $\gcd(b,n)=\gcd(a,n)$ right? – Anand Sep 01 '20 at 04:05
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    @JohnP. fyi: beware that the answer you accepted has various serious gaps. See the linked dupe for answers without such issues. – Bill Dubuque Sep 01 '20 at 04:41
  • Thank you. I will think more about this problem and read the linked answer. – John P. Sep 01 '20 at 04:45
  • @BillDubuque I read your answer and understand everything up to the point where you said that $a \equiv 0 \iff b \equiv 0$. Could you explain again how we know this to be true? – John P. Sep 01 '20 at 05:09
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    @JohnP. If $,a\equiv b,$ than $,b\equiv 0,\Rightarrow, a\equiv 0,$ by transitivity of 'congruence' (being an equivalence relation). The other direction is the same (by symmetry). The proof is the same as that for the analogous equality statement, i.e. if $,a=b,$ then $a=0\iff b=0$. Congruences may be viewed as generalizations of equalities that are compatible with addition and multiplication operations. – Bill Dubuque Sep 01 '20 at 05:40
  • I didn't realize it was that simple. Thank you! – John P. Sep 01 '20 at 05:42
  • @JohnP. Yes, congruences greatly simplify arguments like this. I suspect the argument intended in J.W.T's answer is similar to the Remark here, which is a very useful viewpoint to know. But be sure you understand all the details since this is not the approach that is usually first presented in most textbooks. Rather, it comes to the fore when one studies ideal theory in more advanced courses. – Bill Dubuque Sep 01 '20 at 14:11

1 Answers1

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If $\gcd(a,n)=d$, then $xa+yn$ are the multiples of $d$,

and $xb+yn=x(a-kn)+yn=xa+(y-k)n=xa+zn$,

so $xb+yn$ are the multiples of $d$, so $\gcd(b,n)=d$.

J. W. Tanner
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  • This is likely far too imprecise (for beginners) to figure out what is intended. Precisely what do you mean by "are the multiples of d" and how do you intend to use that to make the inference in the last sentence? It seems that you may be implicitly using ideal-theoretic ideas, but these are not likely familiar to most beginners. – Bill Dubuque Sep 01 '20 at 04:35
  • @BillDubuque: I mean ${xa+yn|x,y\in\mathbb Z}={md|m\in\mathbb Z}$ – J. W. Tanner Sep 01 '20 at 04:42
  • Then you need to say much more in the answer to get a proof at this level, e.g. see the Remark here (and even that should be fleshed out if if is going to be an answer vs. a remark). But this is a dupe of a FAQ so it should be deleted anyway - there is litte new that can be added on this topic. – Bill Dubuque Sep 01 '20 at 14:07