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Let $A$ be a ring. The order of $1_A$ in (A,+) is p (prime). For $k\in \mathbb Z,\ (k,p)=1$ the element $k\cdot1_A$ is invertible. I tried to prove this.

$(k,p)=1 \to \exists m,n\in \mathbb Z\ s.t \ \ mk+np=1$. I don't know what can I do now.

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  • $\begingroup$ What is $(k\cdot 1_A)(m\cdot 1_A)?$ $\endgroup$ – J. W. Tanner Mar 10 at 18:01
  • $\begingroup$ It is $km$ .... $\endgroup$ – Gaboru Mar 10 at 18:06
  • $\begingroup$ It's $km\cdot 1_A$, and use $mk=1-np$ $\endgroup$ – J. W. Tanner Mar 10 at 18:07
  • $\begingroup$ But the + and . are not the same from "$mk+np=1$"(these are the standard + and . from Z) $\endgroup$ – Gaboru Mar 10 at 18:10
  • $\begingroup$ That's why I'm confused $\endgroup$ – Gaboru Mar 10 at 18:11
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You're on the right track.

Since $(k\cdot1_A)(m\cdot1_A)=km\cdot1_A=(1-np)\cdot1_A=1\cdot1_A-np\cdot1_A=1_A-n0_A=1_A,$

$(k\cdot1_A)$ has an inverse in $A$, namely, $m\cdot1_A.$

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  • $\begingroup$ Thank you very much! $\endgroup$ – Gaboru Mar 10 at 18:14

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