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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.

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  • $\begingroup$ Is that ideal generated by $x_i^2-x_i$ for a specific $i$ or for $i=1,\cdots,n$? Also by "when $x_i\in\{0,1\}$" do you mean the equality is true when pushed forward to the image of $S$ mod $x_i$ and mod $x_i-1$? $\endgroup$
    – anon
    Nov 12, 2014 at 4:38
  • $\begingroup$ Yes, evaluation at a point is an algebra homomorphism. $\endgroup$
    – anon
    Nov 12, 2014 at 4:42
  • $\begingroup$ If $x_i\mid(p^2-tp)$ and $(x_i-1)\mid(p^2-tp)$ then $x_i(x_i-1)\mid(p^2-tp)$, so you don't need to write the "where $x_i\in\{0,1\}$" condition. $\endgroup$
    – anon
    Nov 12, 2014 at 4:50

1 Answer 1

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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.

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  • $\begingroup$ @AsafF you just need me to elaborate on the last formula, right? $\endgroup$
    – anon
    Nov 12, 2014 at 6:25
  • $\begingroup$ @Asaf, exercise: for any rings $R$ and $S$, the idempotents of $R\times S$ are of the form $(r,s)$ for idempotents $r\in R$ and $s\in S$. Now generalize. $\endgroup$
    – anon
    Nov 12, 2014 at 6:33
  • $\begingroup$ No it's an element of $R\times S$ ..... $\endgroup$
    – anon
    Nov 12, 2014 at 6:36
  • $\begingroup$ @AsafF the factors of $k^{2^n}$ are more precisely of the form $$ \frac{k[x_1,\cdots,x_n]}{(x_{a_1},\cdots,x_{a_k},x_{b_1}-1,\cdots,x_{b_l}-1)} $$ for subsets $A\subseteq[n]:=\{1,\cdots,n\},$ so we want $f$ to be $0$ or $1$ mod each of these possible denominators. Every element of your ring is of the form $f=\sum_{A\subseteq[n]}a_A\prod_{a\in A}x_a$ for various coefficients $a_A$. Take this mod one of the denominators and get $\sum_{A\subseteq S}a_A=\varepsilon_A$. From there I used the theory of mobius functions to solve for the $a_A$s. $\endgroup$
    – anon
    Nov 12, 2014 at 6:43
  • $\begingroup$ I will elaborate on the combinatorics part later. It is late here. $\endgroup$
    – anon
    Nov 12, 2014 at 6:43

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