as [@Shly2000 Lemma 3.2] but only on a purely algebraic level. In fact, we used the surjective homomorphism ${\operatorname{ev}}_X: {\mathbb{C}}\langle Z_1,\dots,Z_n\rangle \rightarrow {\mathbb{C}}\langle X_1,\dots,X_n\rangle$ in order to pass from $({\mathbb{C}}\langle X_1,\dots,X_n\rangle,\tau)$ to the non-commutative probability space $({\mathbb{C}}\langle Z_1,\dots,Z_n\rangle,\tau_X)$, where $\tau_X := \tau\circ{\operatorname{ev}}_X$. Due to this lifting, the algebraic relations between the generators disappear whereas the relevant information about their joint distribution remains unchanged.

In this section, our aim is to show that the existence of a conjugate system guarantees that $X_1,\dots,X_n$ do not satisfy any algebraic relations. This will be the content of Theorem \[main1\] below. Its proof proceeds in two steps, which are performed in the following two propositions.

\[key-prop-1\] Let $(M,\tau)$ be a tracial $W^\ast$-probability space and let self-adjoint $X_1,\dots,X_n\in M$ be given. Assume that there are elements $\xi_1,\dots,\xi_n\in L^2(M,\tau)$, such that $(\xi_1,\dots,\xi_n)$ satisfies the conjugate relations for $X=(X_1,\dots,X_n)$.

Then the following implication holds true for any non-commutative polynomial $P\in {\mathbb{C}}\langle Z_1,\dots,Z_n\rangle$: $$\label{reduction}
P(X) = 0 \quad\Longrightarrow\quad \forall j=1,\dots,n:\ (\partial_j P)(X) = 0$$

Before beginning with the proof, let us introduce a binary operation $\sharp$ on the algebraic tensor product $M \otimes M$ by bilinear extension of $$(a_1 \otimes a_2) \sharp (b_1 \otimes b_2) := (a_1 b_1) \otimes 