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Let $k$ be an algebraically closed field of characteristic $p > 0 $ and let $A$ be an algebra over $k$, which is a local ring. There is an isomorphism of algebras $\prod_{i=1}^p A \cong A \otimes k[z]/(z^p - z)$, the left hand side denoting the $p$-fold direct product of algebras.

Recall there is up to isomorphism only one Hopf algebra structure on $k[z]/(z^p - z)$, which we can call $\widetilde \Delta : z \mapsto z \otimes 1 + 1 \otimes z$ (a universal enveloping algebra for a 1-dimensional $p$-Lie algebra, with counit $z \mapsto 0$ and antipode $z \mapsto -z$).

Is it possible for there to be a Hopf algebra structure on $A \otimes k[z]/(z^p - z)$ which is not isomorphic to the tensor product $\Delta \otimes \widetilde\Delta$ for some Hopf algebra structure $\Delta$ on $A$? I suspect this is impossible after studying examples in small dimension. We could assume $A$ is finite dimensional over $k$ and commutative if it helps, but infinite or noncommutative counterexamples would be interesting too.

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    $\begingroup$ If I understand the question correctly, the answer is no. It seems you are fixing the algebra structure and varying the coalgebra structure, right? Then the question can be rephrased as: if $S = \{0,\ldots,p-1\}$, then is there a unique algebraic group structure on $S$? (Yes: $\mathbf Z/p\mathbf Z$). And: if $X$ is any other affine $k$-scheme, does every algebraic group structure on $X \times S$ come from one on $X$? (No: if $X = S$ then $X \times S \cong \mathbf Z/p^2\mathbf Z$ as sets or $0$-dimensional $k$-schemes, giving a group structure that is not a product.) $\endgroup$
    – R. van Dobben de Bruyn
    Commented 21 hours ago
  • $\begingroup$ @R.vanDobbendeBruyn Ah, yes, this is a trivial example I forgot to consider. I actually want that the algebra $A$ is a local ring. I think this becomes a much more interesting question now. The small dimensional examples I have been working with were all local rings. I have edited my question to add this hypothesis. $\endgroup$
    – Justin Bloom
    Commented 19 hours ago

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$\def\Spec{\text{Spec}}\def\GG{\mathbb{G}}\def\ZZ{\mathbb{Z}}$Yes, there are other Hopf structures. First I'll give a non-local (but connected) example, and then I'll modify it to be local.

Geometrically, I'm going to form the semidirect product $k^2 \rtimes (\ZZ/p \ZZ)$ with $z \in \ZZ/p \ZZ$ acting by $(x,y) \mapsto (x+zy, y)$. In other words, geometrically, I'm going to consider the group structure on $k \times k \times \ZZ/p \ZZ$ by $$(x_1, y_1, z_1) \ast (x_2, y_2, z_2) = (x_1+x_2, y_1+y_2+z_1 x_2, z_1 + z_2).$$

Switching to Hopf algebra language, my ring is $k[x,y,z]/\langle z^p-z \rangle$ and my comultiplication is $$\Delta(z) = x \otimes 1 + 1 \otimes x,\ \Delta(y) = y \otimes 1 + 1 \otimes y + z \otimes x,\ \Delta(z) = z \otimes 1 + 1 \otimes z.$$ So, as a ring, this is $k[x,y][z]/\langle z^p-z \rangle$, but the Hopf structure product isn't a product.

Now, this has $A = k[x,y]$, which isn't local. But try the same formulas with $k[x,y,z]/\langle x^p, y^p, z^p-z \rangle$ and everything still works, and $k[x,y]/\langle x^p, y^p \rangle$ is local (and even finite dimensional over $k$).

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    $\begingroup$ This isn't exactly the Frobenius kernel; that would have the ideal $\langle x^p, y^p, z^p \rangle = \langle x^p, y^p, z \rangle$. The connected component of the identity in the local example is the Frobenius kernel of the connected component of the identity in the first example, and then I observed that this also worked on the other components. I don't know a general theorem which allows this; I just checked. $\endgroup$
    – David E Speyer
    Commented 8 hours ago
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    $\begingroup$ For co-commutative Hopf structures, I think there should only be product structures. Here is how I want the argument to go: Let $A$ be a finite dimensional local $k$ algebra, so $A/\sqrt{A} = k$. Suppose we have a Hopf structure on $B:=A[z]/\langle z^p-z \rangle$. Then I believe the Hopf structure should descend to the reduced algebra $B/\sqrt{B} = k[z]/\langle z^p-z \rangle$. This gives a closed subgroup isomorphic to the cyclic group of order $p$. $\endgroup$
    – David E Speyer
    Commented 8 hours ago
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    $\begingroup$ On the other hand, we can always take the group of connected components of a group scheme (Chapter 5.i in Milne's online notes jmilne.org/math/CourseNotes/iAG200.pdf ). This gives a quotient map to the cyclic group of order $p$. $\endgroup$
    – David E Speyer
    Commented 8 hours ago
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    $\begingroup$ So the group scheme $\text{Spec}(B)$ should be the semidirect product of its identity component, $\text{Spec}(A)$, and the cyclic group of order $p$. If the group scheme is commutative, the semidirect product should be direct. $\endgroup$
    – David E Speyer
    Commented 8 hours ago
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    $\begingroup$ But I remember that there are some subtleties about taking the reduction of a group scheme, and I don't remember what they are. Maybe we should require that $k$ is perfect. $\endgroup$
    – David E Speyer
    Commented 8 hours ago

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