$\mu_n : \mathbf{alg_k} \to \mathbf{Group}$ given by $A \mapsto \{a \in A : a^n=1\}$ is casually stated to be an affine algebraic group scheme because $\mu_n(A) \cong Hom(k[x]/(x^n - 1), A)$. I want to know how to go about showing this isomorphism?
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1A $k$-algebra homomorphism $k[x] \longrightarrow A$ is determined by the image of $x$. If $x \mapsto a$ and $a^n=1$, then it factors through $k[x]/(x^n-1)$. On the other hand, given any $k$-algebra homomorphism $k[x]/(x^n -1) \longrightarrow A$, the image of $x$ under the composite map $f: k[x] \longrightarrow k[x]/(x^n-1) \longrightarrow A$ must satisfy $f(x)^n=1$. – Krish Oct 29 '17 at 13:47
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A $k$-algebra homomorphism from $k[x]/(x^n-1)$ to $A$ is a $k$-algebra homomorphism from $k[x]$ to $A$ with $\phi(x^n-1)=0$. The $k$-algebra homomorphisms from $k[x]$ to $A$ are the $\phi_a:f(x)\mapsto f(a)$ for $a\in A$. Such a map vanishes on $x^n-1$ iff $a^n=1$.
Angina Seng
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