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$$Recall that by the representation theory of finite groups we have <wikitex> $$ \C\lbrack \Z/p\Z \rbrack \cong \prod_p \C $$ </wikitex> since all irreducibles are <wikitex>1$-dimensional as the group is abelian. Hence the idempotents of $\C\lbrack \Z/p\Z \rbrack$$\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

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$\mathfrak{m} \subset \F_p\lbrack x \rbrack/(x-1)^p$ 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

Tex syntax error

$x = 0$.

$-dimensional as the group is abelian. Hence the idempotents of $\C\lbrack \Z/p\Z \rbrack$ are precisely $p$-tupels with entries either $1$-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

Tex syntax error

$x = 0$.

$ or 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

Tex syntax error

$x = 0$.

$, since $x \in \C^p$ fulfills $x^2 = x$ if and only if each $x_i \in \C$ fulfills $x_i^2 = x_i$ which implies the claim. Let us turn to the other two rings and let us focus on the case $\F_p\lbrack \Z/p\Z \rbrack$ first. It is a standard fact that this ring is isomorphic to $$ \F_p\lbrack x \rbrack/(x^p-1) \cong \F_p\lbrack x \rbrack/(x-1)^p $$ where the last equality holds since we are in characteristic $p$. Now it is a fact from commutative algebra that the ring $\F_p\lbrack x \rbrack/(x-1)^p$ is local. This can be shown as follows. Suppose $\mathfrak{m} \subset \F_p\lbrack x \rbrack/(x-1)^p$ is a maximal ideal. We let $\mathfrak{m}^\prime \subset \F_p\lbrack x \rbrack$ be its preimage under the canonical projection. It is a prime ideal that contains the ideal $(x-1)^p$ and hence must contain the ideal $(x-1)$. But since $(x-1)$ is maximal in $\F_p\lbrack x \rbrack$ we must have equality, i.e., $\mathfrak{m}^\prime = (x-1)$ and hence $\mathfrak{m}$ is the image of the ideal generated by $(x-1)$ which determines $\mathfrak{m}$ uniquely. Thus $\F_p\lbrack x \rbrack/(x-1)^p$ has a unique maximal ideal. So we have shown that $\F_p\lbrack \Z/p\Z \rbrack$ is a local ring. In particular every element fulfills that either $x$ or 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

Tex syntax error

$x = 0$.

-x$ is invertible (in a local ring the complement of the maximal ideal consists of the units). Thus if $x$ is an idempotent it must be $1$-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

Tex syntax error

$x = 0$.

$ or 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

Tex syntax error

$x = 0$.

$. We can put this in more geometric terms as follows. The ring $\F_p\lbrack \Z/p\Z \rbrack$ is a local, artinian ring (this ring is not only finitely generated as an $\F_p$-algebra but also as an $\F_p$-module and is thus artinian) and hence its spectrum consists of precisely one point (every prime ideal is maximal). But a non-trivial idempotent $e \in R$ for any ring decomposes the ring into the product $eR \times (1-e)R$ and hence the spectrum of $R$ 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 $\Z\lbrack \Z/p\Z\rbrack$ for general prime numbers $p$ there are only the trivial idempotents. This follows from the previous considerations as follows. Suppose we have an idempotent $x \in \Z\lbrack \Z/p\Z \rbrack$. Then we have just shown that the reduction mod $p$ of $x$ is either $1$-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

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

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

Now it is a fact from commutative algebra that the ring

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

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

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$\mathfrak{m}^\prime \subset \F_p\lbrack x \rbrack$ 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

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$\F_p\lbrack x \rbrack$ 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

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

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$\F_p\lbrack \Z/p\Z \rbrack$ 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

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

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$\F_p$-algebra but also as an

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$\F_p$-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

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$x = 0$.

$ or 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

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

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

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

Now it is a fact from commutative algebra that the ring

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

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

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$\mathfrak{m}^\prime \subset \F_p\lbrack x \rbrack$ 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

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$\F_p\lbrack x \rbrack$ 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

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

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$\F_p\lbrack \Z/p\Z \rbrack$ 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

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$x = 0$.

$. Without loss of generality we may assume that it is zero. But then it follows that $x = p\cdot y$ for some $y \in \Z\lbrack \Z/p\Z \rbrack$. But since $x$ is an idempotent it fulfills $x^k = x$ for all $k \geq 2$. Calculating $x^k = (py)^k = p^k\cdot y^k$ shows that $x$ is divisible by $p^k$ for all $k \geq 2$ which implies that $x = 0$.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

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

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

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

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$\mathfrak{m} \subset \F_p\lbrack x \rbrack/(x-1)^p$ 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

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

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$x = 0$.