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Suppose $G$ is a group scheme over a field $k$, i.e., $G$ is a functor from the category $\text{Alg}_k$ of unital commutative, associative $k$-algebras to the category of $\text{Groups}$. Suppose that $k[G]$ be the corresponding Hofp algebra, i.e., for $R\in\text{Alg}_k$, $G(R)={\rm{Hom}_{\text{Alg}_k}}(k[G],R)$.

Question: What is the definition of rational group scheme? (I looked into Waterhouse's book but did not find the definition). In particular, how to conclude that the group scheme ${\rm PGL}_1(A):={\rm GL}_1(A)/G_{m,k}$ embeds as an open subset in the projective space $\mathbb{P}(A)$? Here $A$ is a central simple algebra over $k$ and $GL_1(A)(R):=(A\otimes_k R)^*$.

Thanks!

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  • $\begingroup$ I don't know what is a rational group scheme, but $\mathrm{PGL}_1(A)$ embeds into $\mathbb{P}(A)$ in an obvious way : $\mathrm{GL}_1(A)$ is open in $A$ and stable under homotheties, hence $\mathrm{PGL}_1(A)$ is open in $(A\smallsetminus\{0\})/ \mathbb{G}_m=\mathbb{P}(A)$. $\endgroup$
    – abx
    Commented Mar 9, 2015 at 17:41
  • $\begingroup$ Perhaps rational group scheme just means that the underlying scheme is a rational variety (over $k$). $\endgroup$
    – naf
    Commented Mar 10, 2015 at 6:14
  • $\begingroup$ @abx, I think in above notation you mean $GL_1(A)(k)$ In that case argument is clear. ulrich, what would be the underlying scheme in the above example? $\endgroup$
    – user69002
    Commented Mar 11, 2015 at 3:51

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