I doubt this can be true. I claim that:

**Lemma.** Let $k$ be a commutative ring, $A$ be a $k$-algebra, and $Q:A\to k$ be a $k$-linear map such that $Q\left(1\right)=1$. Then, the following four assertions (1), (2), (3), (4) are pairwise equivalent:

(1) The kernel $\mathrm{Ker} Q$ is a two-sided ideal of $A$.

(2) The kernel $\mathrm{Ker} Q$ is a right ideal of $A$.

(3) The kernel $\mathrm{Ker} Q$ is a left ideal of $A$.

(4) The map $Q$ is a $k$-algebra homomorphism.

*Proof of Lemma.* Clearly, (4) $\Longrightarrow$ (1) $\Longrightarrow$ (2). Now let us prove that (2) $\Longrightarrow$ (4): Assume that (2) holds. That is, we assume that $\mathrm{Ker} Q$ is a right ideal of $A$. Clearly, every $a\in A$ satisfies $Q\left(a\right)\cdot 1-a\in\mathrm{Ker} Q$ (since the $k$-linearity of $Q$ yields $Q\left(Q\left(a\right)\cdot 1-a\right)=Q\left(a\right)\cdot \underbrace{Q\left(1\right)}_{=1}-Q\left(a\right)=0$). Thus, every $a\in A$ and $b\in A$ satisfy $\underbrace{\left(Q\left(a\right)\cdot 1-a\right)} _ {\in\mathrm{Ker} Q} b \in\mathrm{Ker} Q$ (since $\mathrm{Ker} Q$ is a right ideal), so that

$0=Q\left(\left(Q\left(a\right)\cdot 1-a\right)b\right)=Q\left(Q\left(a\right)b-ab\right)=Q\left(a\right)Q\left(b\right)-Q\left(ab\right)$

(by the $k$-linearity of $Q$), so that $Q\left(a\right)Q\left(b\right)=Q\left(ab\right)$. Together with the $k$-linearity of $Q$ and $Q\left(1\right)=1$, this yields that $Q$ is a $k$-algebra homomorphism, so that assertion (4) holds. Thus we have shown that (2) $\Longrightarrow$ (4), which completes the (4) $\Longrightarrow$ (1) $\Longrightarrow$ (2) $\Longrightarrow$ (4) circle. Thus, (4) $\Longleftrightarrow$ (1) $\Longleftrightarrow$ (2). Similarly (4) $\Longleftrightarrow$ (1) $\Longleftrightarrow$ (3). This proves that all four assertions (1), (2), (3), (4) are pairwise equivalent, and the lemma is proven.

The Lemma shows that as long as you want the kernel of $r$ to be an ideal (one- or two-sided), $r$ will be forced to be a $k$-algebra homomorphism. Considering the main example of co-quasi-triangular Hopf algebras, namely the group algebra with a bicharacter, the counterexample we gave in the comments above will hold.

Maybe the "right ideal" that your references claimed refered to a different algebra structure? One of my main sources of confusion in the advanced Hopf algebra theory has always been the presence of many conflicting multiplications, comultiplications, actions etc. on one and the same set.

more or lessyield that any arbitrary bicharacter of an abelian group is a product of characters? (More or lessbecause we can conclude this only for squares of bicharacters, but I don't think this is much of a difference.) $\endgroup$ – darij grinberg Aug 24 '10 at 12:01