Is $\mu_n$ and $\mathbb{Z}/n\mathbb{Z}$ the same thing?

Question

As a $\mathbb{Z}/n\mathbb{Z}$-module (not a sheaf), I know that the roots of unity $\mu_n$ is noncanonically isomorphic to the $\mathbb{Z}/n\mathbb{Z}$. Also the nLab says that it is also so in the $\ell$-adic case. But I've been told that the étale sheaf $\mu_n$ defined by the Kummer sequence $$0\to\mu_n\to\mathscr{O}_X^*\overset{(-)^n}{\to}\mathscr{O}_X^*\to0$$ is not (even noncanonically) isomorphic to the constant sheaf $\underline{\mathbb{Z}/n\mathbb{Z}}_X$ without further explanation. Can you elucidate the last statement for me?

Answer 1

As explained in the comment, they might have non-isomorphic global sections. In fact, we can say a little bit more. Namely, both group schemes are representable. For simplicity, I take $X=Spec(R)$.

• $\underline{\mathbb{Z}/n\mathbb{Z}}_X$ is representable by the ring $R^n$
• $(\mu_{n})_X$ is representable by the ring $R[x]/(x^n-1)$

So for these two to be isomorphic, it is necessary (and in fact sufficient) that $R$ contains all $n$-th roots of unity (for example $k$ an algebraically closed field of characteristic coprime to $n$), which need not be the case.

It fact, here is an extreme case: Let's assume $n$ is prime and $n=0$ in $R$. Then $R[x]/(x^n-1)\cong R[x]/(x-1)^n$. So it will be actually a connected non-reduced group scheme (which is basically as far from being a constant sheaf as it can be). In terms of functors it implies that $\mu_n(Y)=1$ for any reduced scheme $Y$, whereas $\underline{\mathbb{Z}/n\mathbb{Z}}_X(Y)=\mathbb{Z}/n\mathbb{Z}^{\pi_0(Y)}$.