# Idempotent polynomials

Let $R$ be a commutative ring with identity and let $f \in R[x]$. There are well known characterizations for $f$ to be a nilpotent element of $R[x]$ or to have a multiplicative inverse in $R[x]$. Is there any characterization for idempotent elements in $R[x]$ ?

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Let $f = a_0 + a_1x + ... + a_nx^n$ be idempotent. Then $a_0^2 = a_0$. Also $a_0a_1 + a_1a_0 = a_1$. Multiply by $a_0$ to get $a_0a_1 = 0$ which means that $a_1 = 0$ and by induction it is easy to show that $a_2 = ... = a_n = 0$ Therefore $f$ is idempotent iff its constant term is idempotent and other coefficients are zero.

Note that this is not true if we drop the commutativity condition. For example consider the polynomial $f(x) = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}x$ in $M_2(\Bbb{R})[x]$ which is clearly an idempotent polynomial.

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Maybe I'm just not seeing the obvious here, but why does $a_0 a_1 \implies a_1 = 0$ hold? Why can't $a_1$ be a nonzero element of $(1-a_0) R$ ? –  Johannes Hahn Aug 3 '13 at 19:46
You have $a_0a_1 = 0$, now plug this back into $a_0a_1 + a_1a_0 = a_1$ (as long as the ring is commutative). This confused the heck out of me, too! –  darij grinberg Aug 3 '13 at 20:22
Here's a geometric argument. Idempotents in a (commutative) ring $R$ are naturally in bijection with clopen subsets $C\subseteq \operatorname{Spec} R$ (given a clopen subset, take the element of $T$ that is $1$ on $C$ and $0$ on its complement; every idempotent is of this form). Now $\mathbb{A}^1_k$ is connected for any field $k$, so for any scheme $X$, every fiber of the projection $X\times \mathbb{A}^1\to X$ is connected. Thus any clopen subset of $X\times \mathbb{A}^1$ is a union of fibers, and it follows easily that every clopen subset is of the form $C\times\mathbb{A}^1$ for $C\subseteq X$ clopen. If $X=\operatorname{Spec}(R)$, this says exactly that any idempotent in $R[x]$ must be a constant.