(b_2 a_2).$$ Note that, since $M \otimes M$ is naturally a $M$-bimodule, this corresponds exactly to our earlier definition of $\sharp$ and will therefore not lead to any confusion.

Assume that $P\in {\mathbb{C}}\langle Z_1,\dots,Z_n\rangle$ satisfies $P(X)=0$ for $X=(X_1,\dots,X_n)$ and choose any $j=1,\dots,n$. If we take arbitrary $P_1,P_2\in {\mathbb{C}}\langle Z_1,\dots,Z_n\rangle$, we have by iterating the product rule twice that $$\partial_j (P_1 P P_2) = (\partial_j P_1) P P_2 + P_1 P (\partial_j P_2) + P_1 (\partial_j P) P_2$$ and therefore, by evaluating this identity at $X$ and applying $\tau\otimes\tau$ subsequently, $$(\tau \otimes \tau)\big((\partial_j (P_1 P P_2))(X)\big) = (\tau \otimes \tau)(P_1(X) (\partial_j P)(X) P_2(X)).$$ Furthermore, according to , we may deduce that $$(\tau \otimes \tau)\big((\partial_j (P_1 P P_2))(X)\big) = \tau(\xi_j (P_1 P P_2)(X)) = 0.$$ Thus, we observe that $$(\tau \otimes \tau)((P_1 \otimes P_2)(X) \sharp (\partial_j P)(X)) = (\tau \otimes \tau)(P_1(X) (\partial_j P)(X) P_2(X)) = 0$$ for all $P_1,P_2 \in {\mathbb{C}}\langle Z_1,\dots,Z_n\rangle$ and hence by linearity $$(\tau \otimes \tau)(Q(X) \sharp (\partial_j P)(X)) = 0$$ for all $Q\in {\mathbb{C}}\langle Z_1,\dots,Z_n\rangle^{\otimes 2}$. If we apply this observation to $Q = (\partial_j P)^\ast$, the faithfulness of $\tau\otimes\tau$ (recall that $\tau$ was assumed to be faithful) implies $(\partial_j P)(X) = 0$, as claimed.

The second proposition 