Polynomials close to idempotents in quotient ring of $\Bbb R[x_1,x_2,\dots,x_n]$ Let $S=\Bbb R[x_1,x_2,\dots,x_n]/(x_1^2-x_1,x_2^2-x_2,\dots,x_n^2-x_n)$.
Given $t\in \Bbb N$, what are the polynomials $p\in S$ that satisfy the relation $$p^2=tp$$ modulo $x_i$ and $x_i-1$ for all $i$ (same as evaluation at $x_i\in\{0,1\}$)?
I think finding the result for $t=1$ suffices since:$$p^2=tp\iff\frac{1}{t}p^2=p\iff(\frac{p}{t})^2=\frac{p}{t}\iff q^2=q\mbox{ for some }q\in S.$$
What can the least degree of $p$ be?
Related question: https://mathoverflow.net/questions/138478/idempotent-polynomials However I am asking only $\mod x_i$ and $\mod (x_i-1)$?
$p=x_1$ suffices as minimal non-constant polynomial idempotent.
 A: Since $(x_i)$ and $(x_i-1)$ are coprime for $1\le i\le n$ we have by CRT
$$\frac{k[x_1,\cdots,x_n]}{(x_1^2-x_1,\cdots,x_n^2-x_n)}\cong\frac{\displaystyle\frac{k[x_1,\cdots,x_{n-1}]}{(x_1^2-x_1,\cdots,x_{n-1}^2-x_{n-1})}[x_n]}{(x_n^2-x_n)}$$
$$\cong\frac{\displaystyle\frac{k[x_1,\cdots,x_{n-1}]}{(x_1^2-x_1,\cdots,x_{n-1}^2-x_{n-1})}[x_n]}{(x_n)}\times\frac{\displaystyle\frac{k[x_1,\cdots,x_{n-1}]}{(x_1^2-x_1,\cdots,x_{n-1}^2-x_{n-1})}[x_n]}{(x_n-1)}$$
$$\cong\frac{k[x_1,\cdots,x_{n-1}]}{(x_1^2-x_1,\cdots,x_{n-1}^2-x_{n-1})}\times\frac{k[x_1,\cdots,x_{n-1}]}{(x_1^2-x_1,\cdots,x_{n-1}^2-x_{n-1})}$$
$$\cong \cdots\cdots\cdots\cong k^{2^n}$$
for any field $k$. The idempotents in $k^{2^n}$ are those with coordinates $0$ or $1$. Therefore, the idempotents in your quotient ring are obtained via subsets $A\subseteq\{1,\cdots,n\}$, coordinates $\varepsilon_A\in\{0,1\}$ (make once choice for each subset $A$) and solutions to $f\equiv\varepsilon_A\bmod (x_a,x_b-1)_{a\in A,b\not\in A}$. This gives $2^{2^n}$ idempotents.
With some combinatorial finagling via mobius functions of posets I get
$$f=\sum_{R\subseteq[n]}(-1)^{|R|}\left(\sum_{R\subseteq A\subseteq[n]}\varepsilon_A\right)\prod_{s\not\in R}x_s $$
for every choice of $\varepsilon:{\cal P}(A)\to\{0,1\}$. Maybe later I'll elaborate on this.
