TITLE: What is difference between constant group scheme associated with cyclic group and $\mu_n$
QUESTION [7 upvotes]: Let $k$ a field. Let $\operatorname{char}(k) \not\mid n$.
Consider the two group schemes $\mu_n=spec(k[t]/(t^n-1))$ and $\underline{\mathbb{Z}/n\mathbb{Z}}$ the constant group scheme associated to the cyclic group ${\mathbb{Z}/n\mathbb{Z}}$. Are they isomorphic as group schemes?  Do they become isomorphic when $k$ is algebraically closed? 
It is false in general because $\mu_n(k)$ may not have $n$ elements. But when are they isomorphic?

REPLY [8 votes]: Here is a much lower-tech approach than Alex Wertheim's.  As you allude to, a necessary condition for $\mu_n$ and $\underline{\mathbb{Z}/n\mathbb{Z}}$ to be isomorphic is that $\mu_n(k)$ has $n$ elements, which happens iff $t^n-1$ splits over $k$.  On the other hand, if $t^n-1$ does split over $k$, this gives an isomorphism of $k$-algebras $k[t]/(t^n-1)\to k^n$ by the Chinese remainder theorem.  So as a $k$-scheme, $\mu_n$ is just a disjoint union of $n$ copies of $\operatorname{Spec} k$.  So, $\mu_n\times \mu_n$ is a disjoint union copies of $n^2$ copies of $\operatorname{Spec} k$, and the group multiplication $\mu_n\times\mu_n\to\mu_n$ is determined by what it does on $k$-points (and similarly for the inverse operation).  It follows that $\mu_n$ is isomorphic as a group scheme to the constant group scheme $\underline{\mu_n(k)}\cong\underline{\mathbb{Z}/n\mathbb{Z}}$.
This argument applies equally well to any group scheme over $k$ that is isomorphic as a scheme over $k$ to a disjoint union of copies of $\operatorname{Spec} k$.  In more abstract terms, the free functor from sets to $k$-schemes (left adjoint to the "forgetful functor" of $k$-points) is fully faithful and preserves products, so any group scheme whose underlying $k$-scheme is free must come from from a group object in sets.