Given a ring A for which $a=1+1+\cdots+1$ ($n$ times) is invertible, I'm supposed to find all elements $x$ for which $(x-1)^3=0$ and $x^n=1$. I've yet to encounter such exercise, so I dont have a clear image of what I'm supposed to be "looking for". What I've noticed, and I'm not quite sure of, is that $x^n=1$ means $x$ is invertible, and I'm now trying to figure out how that would be of any help, since it usually is.
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1$\begingroup$ Assuming $n\geq 2$, if $x^n=1$ then $x\cdot x^{n-1} = 1$ and $x^{n-1}\cdot x = 1$, so $x$ has a left and a right inverse, both of which are equal, so $x$ is invertible. $\endgroup$– MPWCommented Mar 21 at 14:40
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1$\begingroup$ In fact for $n>1$ , $x^n=1$ implies that $x$ is invertible (a unit). If $x-1$ is invertible (a unit) , then the only solution is $x=1$. So you can concentrate on the case that $x-1$ is a zero-divisor (not invertible). $\endgroup$– PeterCommented Mar 21 at 14:40
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1$\begingroup$ @Peter if $x-1$ was invertible, then wouldn't the solution $x=1$ be a contradiction, since 0 cannot be invertible? Then that leaves us with $x-1$ nilpotent and $1-(x-1)$ invertible. Im guessing this is the direction im supposed to be heading perhaps? $\endgroup$– RemWheelCommented Mar 21 at 15:46
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1$\begingroup$ This type of question can indeed be daunting when you're meeting it for the first time! It will turn out in this case there is a very nice answer. (You can do some "meta" second-guessing to make a pretty good guess of what the answer will be.) A natural strategy is to observe there is an obvious answer $x = 1$, and first ask "is it ever possible that $x \ne 1$?". In general when you see an $n$th power and you know something about $n$, you should think about the binomial theorem. To this end, try rephrasing as: "find all $y$ such that $y^3 = 0$ and $(y + 1)^n = 1$" and see how far you get... $\endgroup$– Izaak van DongenCommented Mar 21 at 16:10
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1$\begingroup$ Exactly! Nice work :). If you like, you can type up a short summary of this answer in the answers section below. $\endgroup$– Izaak van DongenCommented Mar 21 at 16:30
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With the help of everyone in the comments I managed to end up with this: Instead of working with the formulas we were given, let's write $x-1$ as $y$. Now we have $y^3=0$ and $(y+1)^n=1$. Expand the binomial and knowing $y^3=0$ (1) you get ${n \choose n-2}y^2+{n \choose n-1}y+1=1$ (2), then you subtract 1 and multiply by $y$, leaving you with $ny^2=0$ (1), and since $n=a=1+1+...+1$ invertible, $y^2=0$. Heading back to (2) you now have $ny=ay=0$, then $y=0$. So, the only possible element is $x-1=0 => x=1$. Nice!
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1$\begingroup$ Very nice. And would work had we been given that any $y^k=0$, since the argument implies that then $y^{k-1}=0$. $\endgroup$ Commented Mar 22 at 7:45