Is the Cartan matrix of a finite-dimensional (Hopf) algebra invertible over the rationals? This is probably well-known to representation theorists, but this doesn't imply being well-known to me.
Let $k$ be a field, and let $A$ be a $k$-algebra that is finite-dimensional as a $k$-vector space. Let $S_1, S_2, \ldots, S_m$ be a complete list of pairwise non-isomorphic simple $A$-modules (up to isomorphism). Let $P_1, P_2, \ldots, P_m$ be the projective covers of $S_1, S_2, \ldots, S_m$, respectively. (Thus, $P_1, P_2, \ldots, P_m$ is a complete list of pairwise non-isomorphic indecomposable projective $A$-modules.) The Cartan matrix of $A$ is defined to be the $m\times m$-matrix in $\mathbb{Z}^{m\times m}$ whose $\left(i,j\right)$-th entry (for all $i$ and $j$) is the number of composition factors of $P_j$ isomorphic to $S_i$. (In other words, its $\left(i,j\right)$-th entry is $\dfrac{\dim \operatorname{Hom}_A \left(P_i, P_j\right)}{\dim \operatorname{End}_A \left(S_i\right)}$.) See §7.4 of Peter Webb's A Course in Finite Group Representation Theory for proofs and some background.
1. Is it true that the Cartan matrix of $A$ is invertible as a matrix in $\mathbb{Q}^{m\times m}$ ? (I think the answer is "no", but I don't know of a counterexample.)
2. Does this change if we require $k$ to be algebraically closed (or at least to be a splitting field for $A$) ?
3. Does this change if we furthermore require $A$ to be a Hopf algebra?
What I know is that the Cartan matrix of $A$ is invertible if $A$ is the group algebra of a finite group. This is proven in Corollary 10.2.4 of Peter Webb's above-mentioned book, but the proof uses the theory of Brauer characters, which as far as I know is particular to group algebras. (Or not? Is there a notion of Brauer characters for Hopf algebras? I certainly wouldn't find it strange, at least compared with the strangeness of classical Brauer character theory, but I have never seen such a notion.)
 A: A counterexample is given by the Taft Hopf algebra. (I checked this at least for n=2, where this Hopf algebra also goes by a different name that I forget today).
A: Perhaps it's worthwhile to mention one very large and natural class of finite dimensional Hopf algebras (over fields of prime characteristic $p$) for which the Cartan matrix is certainly not invertible: the restricted universal enveloping algebras of the Lie algebras of semisimple algebraic groups.   Recall that if $G$ and hence its Lie algebra $\mathfrak{g}$ has dimension $d$, the restricted enveloping algebra has dimension $p^d$; it has for each ordered basis of $\mathfrak{g}$ an ordered basis of PBW type obtained by truncating at $p$th powers according to the $[p]$-operation on $\mathfrak{g}$.      See for example the remarks at the end of section 4 in my 1971 paper here.  
Take the simplest case, with $\mathfrak{g} = \mathfrak{sl}_2$. Here the $p$ simple modules have dimensions $1,2, \dots, p$.   One block (indecomposable 2-sided ideal) involves just the simple/projective Steinberg module of dimension $p$, with Cartan invariant 1. Each other block involves just two simple modules of "linked" highest weights.     If weights are identified with non-negative integers $0,1, \dots, p-1$ (with $p-1$ the highest weight of the Steinberg module), such blocks correspond to pairs of weights adding up to $p-2$.    Then the Cartan invariants of the block are all 2.      
Larger ranks lead to much more complicated (and not easily computed) Cartan invariants, but the theory developed in older work such as my paper yields the same result on the vanishing determinant of the matrix of these invariants.
