TITLE: When is the exterior algebra a Hopf algebra?
QUESTION [6 upvotes]: 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 $n$).

REPLY [3 votes]: 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.