For real numbers, we know that $x^{0}=1$ except when $x=0$. Does it apply to any ring $R$ as well (ie, $\forall x\in R,$ does $x^{0}=1_{R}$ except when $x=0_{R}$)?
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$x^0$ makes sense for units $x^\in R^\times$ (as $x^n / x^n$) for non-units it depends on what you want to mean with $x^0$. – reuns Oct 30 '22 at 20:07
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@reuns I should have clarified this...I saw the definition of nilpotent in a textbook: An element $x\in R$ is a nilpotent iff $x^{k}=0_{R}$ for some $k$. Then I started to wonder what happens if $k=0$. Is it possible for a nilpotent $x$ to satisfy $x^{0}=0_{R}$? This brings out the question I posted. I don't know if $x^{0}$ makes sense for ring multiplication... – Irene Oct 30 '22 at 20:17
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No really $x^0$ doesn't have any natural meaning in algebra, unless $x$ is a unit, because for $x\in R^\times$ the map $n\mapsto x^n, \Bbb{Z}\to R^\times$ is a natural group homomorphism (with $x^0=1$). You are still free to define $x^0$ for non-units $x$, but don't expect any canonical (unique and natural) definition. – reuns Oct 30 '22 at 20:29
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@reuns thanks! this helps a lot. – Irene Oct 30 '22 at 20:32
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3Any "good" definition will satisfy $x^n\cdot x^0=x^n$ for any $n\ge0$, and one good way of insuring this holds is defining $x^0:=1$. – Kenta S Oct 30 '22 at 21:38
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See here for one elementary exposition that works in domains (or for cancellable elements). – Bill Dubuque Oct 31 '22 at 03:14