Determine all idempotents, nilpotents, and units in $F[x]/\langle h\rangle$, where $F$ is a field, $h=x^2-x$.

Question

Problem: Determine all idempotents, nilpotents, and units in $F[x]/\langle h \rangle$, where $F$ is a field, $h=x^2-x$.

I know $F[x]/\langle h \rangle = \{a_0 + a_1t \mid a_i \in F, t^2-t=0 \}$, So first starting with idempotents, if $z \in F[x]/\langle h \rangle $ is an idempotent, then $z=a_0+a_1t$ satisfies $z^2=z \rightarrow a_0^2+2a_0a_1t+a_1^2t^2=a_0+a_1t$ which implies that $a_0=0$ or $1$, $a_1=0$.

For the nilpotents, if there is some $k \geq 0$ such that $z^k=0$, I'm not sure what can be said about $z$. Hints appreciated.

Answer 1

CRT is wonderful, but sometimes I prefer to manipulate directly the polynomials: $z^n = 0 \Rightarrow (a_0+a_1t)^n = 0 \Rightarrow \sum_{i=0}^n \binom{n}{i} a_0^i a_1^{n-i} t^{n-i} = 0$. Now $t = t^2 = \ldots = t^n$, so $a_0^n + \sum_{i=0}^{n-1} a_0^i a_1^{n-i} t = 0 \Rightarrow a_0^n = 0 \Rightarrow a_0 = 0$ and $\sum_{i=0}^{n-1} a_0^i a_1^{n-i} = 0 \Rightarrow a_1^n = 0 \Rightarrow a_1 = 0$.

Be careful, because in $F[x]/(h)$, $0 \neq x = x^2$! $z = z^2 \Rightarrow a_0 + a_1 t = a_0^2 + 2 a_0 a_1 t + a_1^2 t^2 = a_0^2 + (2 a_0 a_1 + a_1^2 )t \Rightarrow a_0 = 0,1; a_1 = 0$ or $a_1 = 2 a_0 a_1 + a_1^2 \Rightarrow 1 = 2a_0 + a_1$.

Answer 2

Hint:

For the nilpotent elements, using the Chinese remainder theorem, you can prove that $F [x]/(h)$ is a product of fields.

Edit: the isomorphism of the Chinese remainder theorem is, explicitly, the following: \begin{align} \varphi:F[x]/(x^2-x)&\xrightarrow{\enspace\sim\enspace}F[x]/(x)\times F[x]/(x-1)\simeq F\times F\\ p(x)&\longmapsto \bigl(p(0),p(1)\bigr) \end{align}