Lemma 40.10.9. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $U = \mathop{\mathrm{Spec}}(k)$ with $k$ a field. Assume $s, t$ are locally of finite type. Then

$R$ is equidimensional,

$\dim (R) = \dim _ r(R)$ for all $r \in R$,

for any $r \in R$ we have $\text{trdeg}_{s(k)}(\kappa (r)) = \text{trdeg}_{t(k)}(\kappa (r))$, and

for any closed point $r \in R$ we have $\dim (R) = \dim (\mathcal{O}_{R, r})$.

**Proof.**
Let $r, r' \in R$. Then $\dim _ r(R) = \dim _{r'}(R)$ by Lemma 40.10.3 and Morphisms, Lemma 29.28.3. By Morphisms, Lemma 29.28.1 we have

\[ \dim _ r(R) = \dim (\mathcal{O}_{R, r}) + \text{trdeg}_{s(k)}(\kappa (r)) = \dim (\mathcal{O}_{R, r}) + \text{trdeg}_{t(k)}(\kappa (r)). \]

On the other hand, the dimension of $R$ (or any open subset of $R$) is the supremum of the dimensions of the local rings of $R$, see Properties, Lemma 28.10.3. Clearly this is maximal for closed points $r$ in which case $\text{trdeg}_ k(\kappa (r)) = 0$ (by the Hilbert Nullstellensatz, see Morphisms, Section 29.16). Hence the lemma follows.
$\square$

## Comments (0)