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*](http://www-users.math.umn.edu/~webb/RepBook/index.html) 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.)



Is the Cartan matrix of a finite-dimensional (Hopf) algebra invertible over the rationals?
A counterexample is given by the [Taft Hopf algebra](https://en.wikipedia.org/wiki/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).


