When is the exterior algebra a Hopf algebra?

Question

I have several questions on the exterior algebra of a vector space:

Q1:When has the exterior algebra A (viewed just as an algebra, not considered as a graded algebra) of an $n$-dimensional vector space over a field $K$ the structure of a Hopf algebra? (depending on n and K)

Note that it is not always a Hopf algebra, for example in the easiest case the exterior algebra is $K[x]/(x²)$ and this should be a Hopf algebra iff the characteristic of the field is 2.

Q2: Is there a finite dimensional, nonprojective module M over A with $Ext_A^{1}(M,M)=0$?

This question has the answer no when the exterior algebra is a Hopf algebra and thus this question is related to Q1.

(Q2 is also open in the graded case and has a positive solution in a special case, see the last chapter of https://arxiv.org/pdf/1701.01149.pdf )

Q3:Can one classify all periodic modules over this algebra?

In general the exterior is a wild algebra for more than 2 variables and it is hopeless to give a classification of all indecomposable modules, but maybe there is an interesting classification of special modules such as indecomposable periodic modules (a module is periodic in case $\Omega^n(M) \cong M$ for some $n$).

Answer 1

I will show that the exterior algebra over a nonzero vector space can only support a co-algebra structure in characteristic $2$. Over a field of characteristic $2$, the map $v \mapsto v \otimes 1 + 1 \otimes v$ is a coproduct which, with the identity map as antipode, makes $\bigwedge^{\bullet} V$ into a Hopf algebra, so the answer is "if and only if the ground field has characteristic $2$.

Notation Let $\Lambda$ be the exterior algebra on $V$, and let $M$ be its radical, $\bigoplus_{j \geq 1} \bigwedge^j V$. By a coproduct, we mean an algebra map $\Delta: \Lambda \to \Lambda \otimes \Lambda$ with the standard ring structure $(u_1 \otimes v_1) (u_2 \otimes v_2) = (u_1 u_2) \otimes (v_1 v_2)$ on $\Lambda \otimes \Lambda$. In the comments, people discuss more exotic ring structures, but the OP makes it clear that is not what they are looking for. Suppose that $\Delta$ is a coproduct and $\epsilon: \Lambda \to k$ is a counit, making a coalgebra structure.

Since $\epsilon$ is a map of $k$-algebras, its kernel must be $M$ and it must be the obvious map $\Lambda \to \Lambda/M \cong k$. Let $x \in V$. The counit axiom is equivalent to saying that $\Delta(x) \equiv x \otimes 1 \bmod \Lambda \otimes M$, and $\Delta(x) \equiv 1 \otimes x \bmod M \otimes \Lambda$, so $$\Delta(x) \equiv x \otimes 1 + 1 \otimes x \bmod M \otimes M.$$

We deduce that $$\Delta(x^2) \equiv x^2 \otimes 1 + 2 x \otimes x + 1 \otimes x^2 \bmod M^2 \otimes M + M \otimes M^2 .$$ But $x^2=0$, so this says $$2 x \otimes x \equiv 0 \bmod M^2 \otimes M + M \otimes M^2.$$ If $2 \neq 0$ and $x \neq 0$, this is a contradiction. So there are no solutions over nonzero vector spaces in characteristic not $2$.

Answer 2

Another proof: The exterior algebra is Koszul, it's Koszul dual is the symmetric algebra, that is commutative, but NOT super commutative, unless Char=2. Assuming $char \neq 2$, the exterior algebra cannot be Hopf because for $H$ a Hopf algebra, $Ext_H(k,k)$ is a subalgebra of Hochschild cohomology of H, hence, it should be super commutative.