Talk:Idempotents in group rings (Ex)

Recall that by the representation theory of finite groups we have

$\displaystyle \C\lbrack \Z/p\Z \rbrack \cong \prod_p \C$ $$\displaystyle \C\lbrack \Z/p\Z \rbrack \cong \prod_p \C$$

since all irreducibles are $1$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}1$-dimensional as the group is abelian. Hence the idempotents of $\C\lbrack \Z/p\Z \rbrack$$\C\lbrack \Z/p\Z \rbrack$ are precisely $p$$p$-tupels with entries either $0$$0$ or $1$$1$, since $x \in \C^p$$x \in \C^p$ fulfills $x^2 = x$$x^2 = x$ if and only if each $x_i \in \C$$x_i \in \C$ fulfills $x_i^2 = x_i$$x_i^2 = x_i$ which implies the claim.

Let us turn to the other two rings and let us focus on the case

Tex syntax error

$\F_p\lbrack \Z/p\Z \rbrack$ first. It is a standard fact that this ring is isomorphic to

Tex syntax error

$$\displaystyle \F_p\lbrack x \rbrack/(x^p-1) \cong \F_p\lbrack x \rbrack/(x-1)^p$$

where the last equality holds since we are in characteristic $p$$p$.

Now it is a fact from commutative algebra that the ring

Tex syntax error

$\F_p\lbrack x \rbrack/(x-1)^p$ is local. This can be shown as follows. Suppose

Tex syntax error

$\mathfrak{m} \subset \F_p\lbrack x \rbrack/(x-1)^p$ is a maximal ideal. We let

Tex syntax error

$\mathfrak{m}^\prime \subset \F_p\lbrack x \rbrack$ be its preimage under the canonical projection. It is a prime ideal that contains the ideal $(x-1)^p$$(x-1)^p$ and hence must contain the ideal $(x-1)$$(x-1)$. But since $(x-1)$$(x-1)$ is maximal in

Tex syntax error

$\F_p\lbrack x \rbrack$ we must have equality, i.e., $\mathfrak{m}^\prime = (x-1)$$\mathfrak{m}^\prime = (x-1)$ and hence $\mathfrak{m}$$\mathfrak{m}$ is the image of the ideal generated by $(x-1)$$(x-1)$ which determines $\mathfrak{m}$$\mathfrak{m}$ uniquely. Thus

Tex syntax error

$\F_p\lbrack x \rbrack/(x-1)^p$ has a unique maximal ideal. So we have shown that

Tex syntax error

$\F_p\lbrack \Z/p\Z \rbrack$ is a local ring. In particular every element fulfills that either $x$$x$ or $1-x$$1-x$ is invertible (in a local ring the complement of the maximal ideal consists of the units). Thus if $x$$x$ is an idempotent it must be $0$$0$ or $1$$1$. We can put this in more geometric terms as follows. The ring

Tex syntax error

$\F_p\lbrack \Z/p\Z \rbrack$ is a local, artinian ring (this ring is not only finitely generated as an

Tex syntax error

$\F_p$-algebra but also as an

Tex syntax error

$\F_p$-module and is thus artinian) and hence its spectrum consists of precisely one point (every prime ideal is maximal). But a non-trivial idempotent $e \in R$$e \in R$ for any ring decomposes the ring into the product $eR \times (1-e)R$$eR \times (1-e)R$ and hence the spectrum of $R$$R$ is a disjoint union of two non-empty subsets. Of course, this cannot be the case if the spectrum consists only of one point.

We now claim that also in the ring $\Z\lbrack \Z/p\Z\rbrack$$\Z\lbrack \Z/p\Z\rbrack$ for general prime numbers $p$$p$ there are only the trivial idempotents. This follows from the previous considerations as follows. Suppose we have an idempotent $x \in \Z\lbrack \Z/p\Z \rbrack$$x \in \Z\lbrack \Z/p\Z \rbrack$. Then we have just shown that the reduction mod $p$$p$ of $x$$x$ is either $0$$0$ or $1$$1$. Without loss of generality we may assume that it is zero. But then it follows that $x = p\cdot y$$x = p\cdot y$ for some $y \in \Z\lbrack \Z/p\Z \rbrack$$y \in \Z\lbrack \Z/p\Z \rbrack$. But since $x$$x$ is an idempotent it fulfills $x^k = x$$x^k = x$ for all $k \geq 2$$k \geq 2$. Calculating $x^k = (py)^k = p^k\cdot y^k$$x^k = (py)^k = p^k\cdot y^k$ shows that $x$$x$ is divisible by $p^k$$p^k$ for all $k \geq 2$$k \geq 2$ which implies that $x = 0$$x = 0$.