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Mar 3, 2020 at 6:17 comment added Nick S A similar eample is $\mathbb Z/4 \mathbb Z[t]$. For each $n$ the element $2t^n$ is a root of $x^2=0$.
Apr 20, 2016 at 20:31 comment added Gro-Tsen @OmarAntolín-Camarena Oh right, what I wanted to write was $x^2-x$, and I got confused between "idempotent" and "one-potent"(?). But of course $x^2-1$ also works provided, as you point out, that $1\neq -1$ in $k$.
Apr 20, 2016 at 17:36 comment added Omar Antolín-Camarena Doesn't it only have one solution in $k^\mathbb{N}$ if $k$ has characteristic 2, @Gro-Tsen?
Apr 20, 2016 at 17:00 comment added Gro-Tsen To answer your last question, $x^2-1$ has an infinite number of roots in $k^{\mathbb{N}}$ for any nonzero ring $k$. So, no.
S Apr 20, 2016 at 16:30 history answered Jacob White CC BY-SA 3.0
S Apr 20, 2016 at 16:30 history made wiki Post Made Community Wiki by Jacob White