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David Loeffler
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If I'm not mistaken, the kernel of $\alpha$ is necessarily flat, and indeed either equal to $\mu_p$ or $0$ if $S$ is connected. The key is that $\mu_p$ is of multiplicative type, and we have the following result (Corollary B.3.3 in Conrad's notes on reductive group schemes):

Let $S$ be a scheme and $H$ a subgroupan $S$-group scheme of multiplicative type. Then any fppf closed subgroup of $H$ is also of multiplicative type (hence flat).

If I'm not mistaken, the kernel of $\alpha$ is necessarily flat, and indeed either equal to $\mu_p$ or $0$ if $S$ is connected. The key is that $\mu_p$ is of multiplicative type, and we have the following result (Corollary B.3.3 in Conrad's notes on reductive group schemes):

Let $S$ be a scheme and $H$ a subgroup of multiplicative type. Then any fppf closed subgroup of $H$ is also of multiplicative type (hence flat).

If I'm not mistaken, the kernel of $\alpha$ is necessarily flat, and indeed either equal to $\mu_p$ or $0$ if $S$ is connected. The key is that $\mu_p$ is of multiplicative type, and we have the following result (Corollary B.3.3 in Conrad's notes on reductive group schemes):

Let $S$ be a scheme and $H$ an $S$-group scheme of multiplicative type. Then any fppf closed subgroup of $H$ is also of multiplicative type (hence flat).

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If I'm not mistaken, the kernel of $\alpha$ is necessarily flat, and indeed either equal to $\mu_p$ or $0$ if $S$ is connected. The key is that $\mu_p$ is of multiplicative type, and we have the following result (Corollary B.3.3 in Conrad's notes on reductive group schemes):

Let $S$ be a scheme and $H$ a subgroup of multiplicative type. Then any fppf closed subgroup of $H$ is also of multiplicative type (hence flat).