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Suppose you have a Deligne Mumford stack which is a quotient $[X/G]$ of a scheme $X$ by an algebraic group $G$ .

What is the normalization of that? Is it true that its normalization is a quotient stack of the form $[X'/G]$ , where $X'\rightarrow X $ is the normalization map of $X$ ?(where $X$ is not necessarily reduced or irreducible)

Same question for $X$ an algebraic space.

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I assume $X$ is meant to be of finite type over a field (or excellent ring) and that $G$ is a smooth group scheme of finite type over the base ring. The formation of normalization (for underlying reduced structure) commutes with smooth base change, and $X \mapsto Y:=[X/G]$ is a smooth surjection, even a $G$-torsor, so if the DM stack $Y'$ is the normalization of $Y$ then $X':=X\times_Y Y'$ is the normalization of $X$ and the map $X' \rightarrow Y'$ from a scheme to a DM stack is $G$-invariant, identifying $Y'$ with $[X'/G]$. So the answer is "yes"; it works the same way for Artin stacks. –  user36938 Jul 20 '13 at 0:15
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Since $[X/G]_{\rm{red}} = [X_{\rm{red}}/G]$ for $X$ of finite type over a reduced ring and $G$ smooth of finite type over that ring, most likely in your question it makes no real difference to assume $X$ is reduced. –  user36938 Jul 20 '13 at 0:17

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