Are algebraic groups over algebraically closed fields Cohen–Macaulay?

Question

$\DeclareMathOperator\CM{CM}\DeclareMathOperator\Spec{Spec}$Let $k$ be an algebraically closed field and let $G$ be an algebraic group over $k$.

My question: is $G$ Cohen–Macaulay? If not, are there counterexamples?

I ask this because it seems to me that the answer is yes. For every finite type algebra $A$ over a field, the CM locus is open dense in $\Spec A$. In particular, $\CM(A)$ is nonempty because otherwise $\Spec A=\emptyset$. As a nonempty open, $\CM(G)$ contains a closed point $y$ because it is locally Noetherian. So if there is a point $x\in G\backslash{\CM(G)}$, then we can translate $x$ to $y$ induced by $G(k)\rightarrow G(k)$, $g\mapsto yx^{-1}g$. Then $x\in \CM(G)$.

Answer 1

In arbitrary characteristic they are locally complete intersection, hence Cohen-Macaulay: this is in SGA3, Exposé VII$_B$, Cor. 5.5.1 or Demazure-Gabriel, Groupes algébriques, chap. III, §3, n°6. This is due to the structure theorem which says that (after extension to an algebraic closure of the base field $k$) the completed local ring of the unit element is of the form $k[[t_1,\dots,t_n]](X_1,\dots,X_r)/(X_1^{p^{n_1},\dots,X_r^{p^{n_r})$, a local complete intersection.

Answer 2

If all you want is Cohen-Macaulay, this is straightforward, and your sketch proves the result. Every locally finite type scheme over a field has a maximal open subscheme that is Cohen-Macaulay (equivalent to local equidimensionality plus flatness of one / any quasifinite dominant morphism to a smooth scheme). By homogeneity, this open subscheme equals the entire group scheme.