Show that in any field holds $((x^r-1),(x^s-1)) = x^d-1 \iff d= \gcd(r,s)$

Asked Dec 18 '18 at 09:33

Active Dec 21 '18 at 17:42

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Show that in any field holds $((x^r-1),(x^s-1)) = x^d-1 \iff d= \gcd(r,s)$.

I define $r=ad$ and $s=bd$, then we have $((x^{ad}-1),(x^{bd}-1)) = x^d-1$, but the annoying thing for me to proceed are those "$-1$".

From literature:

For non-negative integers a and b, where a and b are not both zero, provable by considering the Euclidean algorithm in base n we know $\gcd(n^a − 1, n^b − 1) = n^{gcd(a,b)} − 1$.

Some hints for a proof that doesn't use the Euclidean Algorithm?

edited Dec 21 '18 at 17:42

Shaun

44,997

asked Dec 18 '18 at 09:33

Alessar

1

https://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 – lab bhattacharjee Dec 18 '18 at 09:39
I can't understand this passage; if $n > m$, then
$$\gcd(a^n - 1, a^m - 1) =\color{red}{ \gcd(a^n - 1, a^n - a^{n-m}) = \gcd(a^{n-m} - 1, a^m - 1)}$$
– Alessar Dec 18 '18 at 10:02
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If $d$ divide both, $d$ must divide $a^n-a^m=a^m(a^{n-m}-1)$ But $d\nmid a^m$ – lab bhattacharjee Dec 18 '18 at 10:08
In my case, your $d$ is my $x^d-1$, right? Sorry I want to understand clearly – Alessar Dec 18 '18 at 10:23
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I've stated common divisor, ultimately this leads to the greatest common divisor – lab bhattacharjee Dec 18 '18 at 10:25
Thanks for the answer and for the link to the other tread – Alessar Dec 18 '18 at 10:36

Show that in any field holds $((x^r-1),(x^s-1)) = x^d-1 \iff d= \gcd(r,s)$

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