Let $k$ be an algebraically closed field of characteristic $p>0$. All the examples of non-smooth algebraic group schemes over $k$ that I have seen (apart from "artificial" examples; see below) have been given by presentations with at least one defining relation of degree a positive power of $p$. Here are the examples I have looked at:

  • Frobenius kernels.
  • Some automorphism schemes of algebras. The paper "Non-reduced automorphism schemes" by Geiss and Voigt has an interesting example in Section 2, which is given by a presentation including some relations of degree 2. The authors state that this group scheme is not reduced if and only if $p=2$. They also mention that if $G$ is a finite $p$-group, then the group scheme $\mathrm{Aut}(k[G])$ is not reduced.
  • Results by Sopkina, "Classification of all connected subgroup schemes of a reductive group containing a split maximal torus", imply, if I have understood things correctly, that every non-smooth subgroup scheme of a reductive group over $k$ (or at least of $\mathrm{GL}_{n}$) containing a maximal torus, has a presentation including a $p$-power relation.
  • Non-smooth centralisers in classical groups in not very good characteristic.

On the other hand, it is easy to produce "artificial" examples of a non-smooth group scheme in, for example, char 3 with a presentation involving only quadratic relations. Namely, take $k$ of char 3 and $\alpha_{3}=\mathrm{Spec}\, k[x]/(x^{3})$. Since $k[x]/(x^{3})$ is isomorphic to $k[x,y]/(xy,x-y^{2})$ as $k$-algebras, we can transport the Hopf algebra structure from the former to the latter.

Question. Is there a non-smooth algebraic group scheme $G$ over $k$, and an embedding $G\rightarrow\mathbb{A}^{n}$ of $G$ as a closed subscheme of affine $n$-space, with $n$ minimal, such that every defining relation of $G$ in this embedding is of degree strictly less than $p$? (Even the case $p=3$ would be interesting.)

Using explicit equations and a minimal embedding into affine space may seem a bit unnatural, but results of Kollár and Jelonek (see this previous MO question) - which boil down to estimating degrees and Bézout's theorem - imply that if we take $p>d^{n}$, where $d$ is the maximal degree of a relation, then $p$ does not divide the nilpotency index of any element in the coordinate algebra of $G$, so a proof of Cartier's theorem can be carried through in this case. Hence an example as in the above question must have $d<p<d^n$.


1 Answer 1


No. Let $f$ be a relation of minimal degree $2 \leq d <p$. Apply the comultiplication. This must be zero in $R \otimes R$, where $R$ is the ring of functions. So if $x_1, \dots, x_n$ are the variables, then it is zero in $k[x_1, \dots, x_n] \otimes k[x_1,\dots x_n]$ modulo the various relations.

Write $f$ as a sum of monomials in the $x_i$. When we apply the comultiplication to a monomial, the leading term does not depend on which monomial we pick. It's just the sum over the ways of splitting that monomial into a product of two monomials. So each pair of monomials in $k[x_1, \dots, x_n] \otimes k[x_1,\dots x_n]$ comes from a unique monomial of $f$, so we may ignore cancellation among different monomials. Moreover, because $d \geq 2$, we may split it into two monomials, each of degree $<d$. Because $d$ was the minimal degree of relations, we may ignore the relations. So we end up with something nonzero unless the number of ways of splitting is nonzero. But the number of ways of splitting is a product of binomial coefficients. All numbers involved in the binomial coefficients are less than $p$, so the binomial coefficients are prime to $p$ and hence nonzer.

  • $\begingroup$ In general, before you get to the $p$th powers, group schemes in characteristic $p$ almost always behave exactly like group schemes in characteristic $0$ - and, in particular, all relations show up in degree $1$. $\endgroup$
    – Will Sawin
    Aug 20, 2015 at 20:43
  • $\begingroup$ Could you clarify what you mean by: 1) "the leading term" of the comultiplication applied to a monomial (my first guess was $x\otimes1+1\otimes x$, but this doesn't seem to be what you mean); 2) "the leading term does not depend on which monomial we pick"; 3) "each pair of monomials in ... comes from a unique monomial of $f$". Also, where does the minimality of $n$ enter in your argument? It seems that the argument aims to derive a contradiction, but if so, where does the non-smoothness enter (supposedly we want to pick a non-zero nilpotent element at some point)? $\endgroup$ Aug 21, 2015 at 9:32
  • $\begingroup$ @AStasinski 1) I mean you write it as a sum of tensors of monomials and take the tensors corresponding to pairs of monomials of minimal degree. Because the comultiplication applied to $x_i$ gives $x_i \otimes 1 + 1 \otimes x_i$ plus higher-order terms, we can compute the comultiplication of any monomial up to higher-order terms. 2) I meant rather that the leading term doesn't depend on the group law. $\endgroup$
    – Will Sawin
    Aug 25, 2015 at 2:33
  • $\begingroup$ @AStasinski 3) I mean that comultiplicatoin applied to a monomial like $x_1 x_2^2 x_3^3$ will be a sum of terms like $x_1 x_2 x_3 \otimes x_2 x_3^2$ where the product of the two sides of the $\otimes$ is the original monomial. The minimality of $n$ enters because I am taking a relation of minimal degree $d$ and using $d \geq 2$. If $n$ is not minimal, then there is a relation of degree $1$. Yes, the argument is attempting to derive a contradiction. $\endgroup$
    – Will Sawin
    Aug 25, 2015 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.