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Let $G$ be a finite flat non-smooth group scheme over an algebraically closed field $k$, for example, $G$ can be $\operatorname{Spec}(\overline{\mathbb{F}}_p[t]/(t^p))$. Then the classifying stack $\mathcal{X}:=BG$ is a smooth algebraic stack over $k$, so the diagonal of this stack is also smooth. However, by some computation, the stabilizer group scheme of a point $x: \operatorname{Spec}(k)\to \mathcal{X}$ is $G$, which is a base change of the smooth morphism $\triangle:\mathcal{X}\to \mathcal{X}\times_k \mathcal{X}$ along $x: \operatorname{Spec}(k)\to \mathcal{X}$, but $G$ is not smooth.

Question: which statement(s) in the above paragraph is wrong?

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    $\begingroup$ The stack is not smooth. $\endgroup$
    – Ariyan Javanpeykar
    Commented Mar 21 at 22:57
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    $\begingroup$ @AriyanJavanpeykar actually it is! See [Tag 0DLS]. But I don't see why the diagonal of a smooth stack must be smooth — this already fails for smooth schemes, doesn't it? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 22 at 0:03
  • $\begingroup$ @R.vanDobbendeBruyn I think you are right. For a geometrically connected variety $X$, the diagonal is a closed embedding. If it is smooth, it is in particular open. Then it violates the connectedness of $X\times X$. It is just a little counter-intuitive that closed immersion of smooth subscheme into a smooth scheme isn't always smooth. $\endgroup$
    – mhahthhh
    Commented Mar 22 at 1:21
  • $\begingroup$ There exist smooth varieties over fields of positive characteristic with non-reduced automorphism group schemes. $\endgroup$
    – Daniel Loughran
    Commented Mar 22 at 10:16
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    $\begingroup$ The diagonal of a morphism of schemes $f \colon X \to Y$ is a locally closed immersion, so it is smooth if and only if it is an open immersion, i.e. $f$ is unramified. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 22 at 15:53

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