0

in $a\in\mathbb{N}$ we know that $x>1$. How can I prove the question above? I see the proof from this question, $x+1$ is always invertible in $\Bbb Z_{x^3}$.

But is the answer the same when the sign is just flipped?

L F
  • 3,644
  • 1
    You downvoters, could be more constructive if you leave this new member of MSE a comment about why are you voting him negatively, since he is new he will think that this site is full of selfish ****. – L F Dec 15 '20 at 00:19
  • Would int be enough to prove that $x^3$ and $x-1$ are always relatively prime? – fleablood Dec 15 '20 at 00:30

1 Answers1

1

$x^3-1 = (x-1)(x^2+x+1)$ , we know that $(x^2+x+1)$ is never $0$ since $0<1<(1+x+x^2)$.

Later, $x^3 = 1+ (x-1)(x^2+x+1)$, so $$1+ (x-1)(x^2+x+1) \equiv 0 ~~{ mod}(x^3) $$ $$(x-1)(x^2+x+1) \equiv -1 ~~{ mod}(x^3) $$ $$(x-1)(-x^2-x-1) \equiv 1 ~~{ mod}(x^3) $$ so $x-1$ is always invertible.

A Better one: Since $x^3 = 1+ (x-1)(x^2+x+1)$, then $$x^3 + (x-1)(-x^2-x-1) = 1$$ so, for Bezout's theorem, $x^3$ and $(x-1)$ are always coprime, and hence, $(x-1)$ is invertible .

Note: This observation was taken from one answer of the link attached to question.

L F
  • 3,644
  • Also, $x$ and $x-1$ are always relatively prime and $x^3$ has no prime factors other than the prime factors of $x$ so $x-1$ and $x^k$ are are always relatively prime. – fleablood Dec 15 '20 at 00:34
  • nice intuition on factors. – L F Dec 15 '20 at 00:36
  • @LuisFelipe thank you. and thanks for respecting a new member! –  Dec 15 '20 at 03:06
  • @original-tourist25 you can select this answers as correct by clicking the left “check” empty image :) – L F Dec 15 '20 at 03:08