Confused about a solution: proving every prime ideal is maximal

Question

I'm looking at this solution to this problem:

I'm getting thrown off by the special case where $n = 2$. If $n = 2$, why must it be that $x = 1$? All that we then know is that $x^2 = 1$ or that $x = x^{-1}$. However, I don't see how this implies that $x = 1$. Could someone enlighten me?

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Resolved: If $n = 2$, in $R/P$ we have $x^2 = x$ or $x(x-1) = 0$. Since $R/P$ is an integral domain, there are no zero divisors and so $x - 1 = 0$ or $x = 1$.

Answer 1

Probably what is throwing you off is that $x\in R/P$ and not $x\in R$.

Of course, $x^2=x$ in $R$ without $x$ being $1$. But in the quotient, there are no zero divisors (because you're quotienting by a prime ideal.) So you must examine $x^2-x=0$.

Answer 2

Because as $n=2$ we have that $x^2 = x$ then $x(x-1) = 0$ and $x$ is not zero. In a integral Domain what happens to $x$?

Answer 3

You're plugging in wrong: $n=2$ is not the case $x^2 = 1$, but rather the case $x^1 = 1$.

But the special case isn't part of the proof, just an illustration. In general, the point is that if some power—any power—of $x$ equals $1$, then $x$ is invertible.