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Recall that $x$ is regular point if the local ring $\mathcal{O}_{x}$ is a regular local ring.

I ask because secretly I'm thinking $X_k=G$ is an algebraic group over $k$, so that the existence one regular point, by homogeneity, implies all points are regular (justifying the name variety). Not being well versed in modern AG, I'm having trouble interpolating between the approaches of classical varieties over algebraically closed fields and the scheme theoretic approach taken, for example, in these notes.

For a classical variety $V$ over $k$ (so $k$ is algebraically closed) one shows that the set of regular points is an open dense subset of $V$, hence if one defines an algebraic group on top of this structure it is automatically smooth.

However, in the aforementioned notes, they (sec.1 pgs 3-4) define an algebraic group as a finite type $k$-scheme with the added condition of smoothness, where $x$ is a smooth point iff $\mathcal{O}_{{X_{\overline{k}}} ,x}$ is regular, and $X_{\overline{k}}$ is base change to $\overline{k}$.

From the notes:

Let $G$ be a $k$-group scheme [of finite type]. The group $G(\overline{k})$ acts on $G_{\overline{k}}$ by translation. Now $G$ is smooth if and only if the local rings of $G_k$ are regular, and over an algebraically closed field it is enough to check smoothness at the classical points $G_{\overline{k}}(\overline{k}) = G(\overline{k})$. But by commutative algebra plus the aforementioned “homogeneity”, it is thus enough to check that the completed local ring $\mathcal{O}_{G_{\overline{k}},e}$ is regular.

So my question is: if it's enough to check regularity at classical points, why isn't smoothness immediate? In sections $6$ and $7$ of the notes, a lot of work is put into showing the equivalence between ($\mathcal{O}_{G_{\overline{k}},e}$ is regular) $\iff$ (Grothendieck's infinitesimal criterion for smoothness) which is then put to use to show smoothness of some classical groups like $G=Sp_{2n}$, which amounts to computing the tangent space at $e$ (kernel of $G(k[\epsilon]) \to G(k)$ where $\epsilon^2=0$), and showing the dimension of the tangent space equals the Krull dimension of $G$.

This seems like a lot of work for something we get for free with classical varieties, what exactly prevents us from reproducing the density of regular points on scheme theoretic varieties? If it is an issue with the ground field, could I avoid all this work if I just assumed $k$ to be a perfect field or characteristic $0$? i.e. is this lemma from Stacks enough? I'm not sure because the lemma requires varieties to be reduced, so I'm not sure how all these subtly different definitions play out.

If that lemma fails for non-perfect ground fields, I would still be interested in seeing an example of a finite type group scheme over $k$ without regular points, i.e. a non-smooth group, or a scheme $X$ in general.

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  • $\begingroup$ $\mu_p:=\text{Spec}(k[t]/(t^p-1))$ is non reduced (so nonsmooth) over any field $k$ of characteristic $p$. It has one scheme theoretic point. Of course it is a finite type scheme, and is a group scheme (it's functor is just $p$th roots of unity in a given algebra; in particular it's not trivial). There are more examples in characteristic $p$ of nonreduced group schemes of finite type. $\endgroup$ – Eoin May 5 '17 at 6:43
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If you require your algebraic group to be a variety then you surely want it to be geometrically reduced. But then your group is smooth.

By the way, Cartier proved that every group scheme of finite type over a field of characteristic $0$ is (geometrically) reduced. For an example of a group without any regular point, take $k$ to be a field of characteristic $p > 0$ and consider the infinitesimal group $\mu_p = \mathrm{Spec}\ k[t]/(t^p-1)$ of $p$th roots of $1$.

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  • $\begingroup$ Thanks for the example.Then for example $Sp_{2n}$ should be automatically smooth? Then why does Brian Conrad (in the notes I mentioned) go through the trouble of proving smoothness using infinitesimal criterion rather than just use the fact that it's geometrically reduced finite type, he makes the same point here on page 11 example 1.1.12: math.stanford.edu/~conrad/papers/luminysga3.pdf $\endgroup$ – ws898989 May 5 '17 at 6:57
  • $\begingroup$ Is the geometric reducedness trivial ? $\endgroup$ – BrL May 5 '17 at 7:09
  • $\begingroup$ Well I suppose no, if I allow the coordinate ring to have nilpotents. Now I'm confused. $\endgroup$ – ws898989 May 5 '17 at 7:44
  • $\begingroup$ The fact is that, only with the definition of the symplectic group, it is not obvious (at least to me!) that this group is geometrically reduced. But you can argue as Brian Conrad to show that it is so. $\endgroup$ – BrL May 5 '17 at 8:23
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$\operatorname{Spec} k[\epsilon]$ is an example of a scheme with no regular points. I believe it can even be made into a group $k$-scheme, induced by the cogroup

$$ e : k[\epsilon] \to k : \epsilon \mapsto 0 $$ $$ i : k[\epsilon] \to k[\epsilon] : \epsilon \mapsto -\epsilon $$ $$ \mu : k[\epsilon] \to k[\epsilon, \epsilon'] : \epsilon \mapsto \epsilon + \epsilon' $$

(note that $\epsilon \epsilon' \neq 0$ in that ring)

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