TITLE: On a dual of Kaplansky's $2^{nd}$ conjecture: admissible algebras?
QUESTION [5 upvotes]: Kaplansky's second conjecture (on Hopf algebras) deals with "admissible" coalgebras: He calls a coalgebra admissible, if there is an algebra structure making it a Hopf algebra. The conjecture states that: 

a coalgebra $C$ is admissible if and only if any finite subset of $C$ lies in a finite dimensional admissible subcoalgebra. 

As far as I know there are counterexamples refuting the conjecture. (the first one was stated by Larson if I remember correctly). 
Inspired by the above, let us lay the following definition of the notion of  admissible algebra: 

Definition: An algebra $A$ will be called admissible if there is a coalgebra structure on $A$ and a suitable linear map $S:A\to A$, such that all these data (the algebra, the coalgebra and the linear map $S$) constitute a Hopf algebra with antipode $S$. 

Now, my question is whether there is some criterion (necessary, sufficient or both) regarding to when an algebra is admissible (or non-admissible) in the sense of the above definition. In other words: 

Given an algebra $A$, under which conditions can a coalgebra structure be found on $A$ such that it will be turned into a Hopf algebra? 
When would such a coalgebra structure be unique? 
What could be examples of a non-admissible algebras, in the sense of the above definition?

REPLY [2 votes]: Note that a Hopf algebra is augmented. Let $A$ be an augmented algebra over a commutative ring $k$. A necessary condition for $A$ having a Hopf algebra structure is that the cohomology ring 
$$H^\ast(A,k) := Ext_A^\ast(k,k)$$
is graded commutative under Yoneda composition. 
In fact, the coproduct of a Hopf algebra (or, more generally, of a bialgebra) induces a graded  comutative cup product on cohomology and one can show that Yoneda composition and cup product agree on $Ext_A^\ast(k,k)$. 
When I have time, I'll look for an example there this criterion applies.