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Let $k$ be an algebraically closed field of characteristic $p>0$ and let $C$ be a $k$-algebra of finite dimension over $k$ such that $k[C^p]=k$. Under these hypothesis it is known by results of L. Harper and S. Yuan that the following conditions are equivalent.

  1. $C$ is differentiably simple, i.e., it has no proper ideals $I\subset C$ such that $D(I)\subseteq I$ for all $k$-derivation $D:C\to C$.
  2. $\Omega_{C/k}$ is a free $C$-module.
  3. $\operatorname{Hom}_k(C,C)$ is generated as $C$-algebra by $k$-derivations $D:C\to C$.
  4. $C$ is of the form $k[X_1,\ldots,X_n]/\langle X_1^p,\ldots,X_n^p\rangle$ for some $n$.

Note that condition 4 implies that $C$ is a complete intersection of length a power of $p$. I would like to know if there are others. In other words, given an algebraically closed field $k$ of characteristic $p$, I would like to know if there is a $k$-algebra $C$ satisfying a) $k[C^p]=k$, b) $C$ is a complete intersection ring, c) the dimension of $C$ over $k$ is a power of $p$, d) $C$ is not of the form given in 4.

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1 Answer 1

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Let $k$ be perfect of characteristic $p$, and let $C=k[X_1,\dots,X_n]/(f_1,\dots,f_n)$ be a complete intersection of dimension a power of $p$. I'm assuming that when you write $k[C^p]=k$, you mean that every $p$th power of an element of $C$ is in $k$.

If $X_i^p$ is in $k$, say $X_i^p=\lambda_i$, then letting $\mu_i=\lambda_i^{1/p}$, we can rewrite this relation as $(X_i-\mu_i)^p=0$. So $C$ is a quotient of $k[Y_1,\dots,Y_n]/(Y_1^p,\dots,Y_n^p)$, where $Y_i=X_i-\mu_i$. Thus $C$ has dimension at most $p^n$. But the dimension of $C$ is the product of the degrees of the $f_i$, so the degree of each $f_i$ is a power of $p$. If the product of $n$ positive powers of $p$ is at most $p^n$ then each is exactly $p$. Thus $C$ is equal to $k[Y_1,\dots,Y_n]/(Y_1^p,\dots,Y_n^p)$.

On the other hand, there are non-commutative complete intersections of the form you describe. For example, $k\langle X_1,X_2\rangle /(X_1^p,X_2^p,X_1X_2+X_2X_1)$. These interesting algebras occur as the basic algebras of non-principal blocks of finite groups. This one, for example, comes from $(\mathbb{Z}/p\times\mathbb{Z}/p)\rtimes Q_8$ for $p$ odd, and has dimension $p^2$.

Edit: I apologise, I guess I've made an assumption: that once you've written everything in terms of the $Y_i$ then the highest degree terms of your regular sequence still form a regular sequence. This implies that the intersection of the hypersurfaces contains no points at infinity when you homogenise, so you can count the multiplicity at the origin (which is the length) using Bézout's theorem. I'm afraid I don't know what happens without this assumption, but my guess is that $C$ is still equal to $k[Y_1,\dots,Y_n]/(Y^p_1,\dots,Y^p_n)$.

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  • $\begingroup$ Nice proof and remark. I am not used to complete intersections. Do you have a reference for the expression of the length as the product of the degrees of the polynomials? $\endgroup$ Commented Sep 21 at 15:47

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