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I am reading The book of involutions. In page 331, (20.13) Examples (3), they said group of $n^{th}$ roots of unity $\mu_n$ is a non-connected group scheme without any explanation, so I check the case when $n=2$, i.e. consider $\mu_2$. I think $\mu_2$ is connected or not related the characterist of base field $F$. Denote the Hopf algebra of $\mu_2$ by A, then A=$F[X]/(X^2-1)$(See 20.4.5 in this book). I want to use Proposition (20.15) to check it ( let nil(A) be the set of all nilpotent elements of $A$, $A/nil(A)$ is denoted by $A_{red}$, Proposition states that $\mu_n$ is connected $\iff$ $A_{red}$ is an integral domain.

CASE1: $char(F)\neq2$. And we have $(X-1)(X+1)=0$, however $(X-1)^n\neq0$ for all $n$ i.e. $(X-1)$ is not a nilpotent element, thus $nil(A)$ is not a prime ideal which conclude $A_{red}$ is not an integral domain, thus $\mu_2$ is not connected.

CASE2: $char(F)=2$. In this case $(X-1)=(X+1)$, and $(X-1)(X+1)=0=(X-1)^2$, thus $(X-1)$ is a nilpotent element, so $(X-1)\subset nil(A)\subset M$ where $M$ is a maximal ideal of $nil(A)$. Now in $F[X]$, $(X^2-1)\subset (X-1)$, by using the third isomorphism theorem we have $(F[X]/(X^2-1))/(X-1/X^2-1)\cong F[X]/(X-1)$. LHS=$A/(X-1)$($X-1$ in A now) and $X-1$ is irreducible in $F[X]$, thus RHS is a field which means LHS is a field, so $(X-1)$ is a maximal ideal, thus $(X-1)=nil(A)=M$, so $A/nil(A)$ is a field also a integral domain. $A_{red}$ is an integral domain $\Rightarrow \mu_2$ is connected.

This is different about what book said. I wonder my proof is true or false. Any helpful answer will be grateful. Thanks for your attention!

MGIO
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1 Answers1

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You are correct. We can give a precise characterization of when $\mu_n$ is connected as follows. First recall that a commutative ring is connected if, equivalently, it has no nontrivial idempotents, or its Zariski spectrum is connected. So we are trying to understand when the ring of functions

$$K[x]/(x^n - 1)$$

on $\mu_n$ has no nontrivial idempotents. By the Chinese remainder theorem, if $x^n - 1 = \prod f_i(x)^{m_i}$ is the irreducible factorization of $x^n - 1$ over $K$, then this ring decomposes as a direct product

$$\prod K[x]/f_i(x)^{m_i}$$

which disconnects it as soon as $x^n - 1$ has more than one irreducible factor. So $\mu_n$ is connected over $K$ iff $x^n - 1$ has one irreducible factor. Since $x - 1$ is always an irreducible factor, it must be this unique irreducible factor, so $\mu_n$ is connected over $K$ iff

$$x^n - 1 = (x - 1)^n$$

which occurs iff there exists a prime $p$ such that $\text{char}(K) = p$ and $n = p^k$ for some $k$ (this is a nice exercise).

Qiaochu Yuan
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  • As a exercise, here is my solution. $x^n-1=(x-1)^n\iff p\mid (\frac{n}{i})$ for all $i$. If $n=p^k$, then $p\mid (\frac{n}{i}) $since this. Conversely, if $p\mid (\frac{n}{i})$, then $n=p^k$ since this. Thus $p\mid (\frac{n}{i})\iff n=p^k$ – MGIO Mar 26 '25 at 12:24