Regulare sequences and tensor product of modules Let $(R,\mathfrak m)$ be a Noetherian local ring of Krull dimension $d$ and $M$,$N$ two maximal Cohen-Macaulay $R$-modules,that is Cohen-Macaulay with their dimensions equal to dimension of $R$. Suppose that ${\bf x}=x_1,\cdots ,x_d \in \mathfrak m $ is an $R$-regular sequence.I refer the reader to "Bruns. Herzog, Cohen-Macaulay Rings" for the definition of regular sequences and maximal Cohen-Macaulay modules. It is well-known that ${\bf x}$ is also an $M$- and $N$-regular sequence.
Let $H^d_{\mathfrak m}(M)$ denotes the $d$-th local cohomology module of $M$ with respect to $\mathfrak m$ and assume that $Tor^R_i(N,H^d_{\mathfrak m}(M))=0$, for all $i>0$.
With these assumptions, I want to show that ${\bf x}$ is an $M\otimes_RN$-regular. Does anyone have an idea to prove it? I don't know if these assumptions are enough?

 A: Yes, it is true. It suffices to show that $M \otimes_R N$ is maximal Cohen-Macaulay. We may pass to the completion to suppose $R$ is complete and thus has a canonical module $\omega_R$. We let $(-)^{\vee}:=\operatorname{Hom}_R(-,\omega_R)$.
By examining free resolutions, or via a standard spectral sequence argument (see Corollary 10.63 in Rotman's Homological Algebra text) we have
$$\operatorname{Hom}_R(\operatorname{Tor}^R_i(N,H_{\mathfrak{m}}^d(M)),E_R(k)) \cong \operatorname{Ext}^i_R(N,\operatorname{Hom}_R(H_{\mathfrak{m}}^d(M),E_R(k)))$$.
By Grothendieck local duality, the latter term is isomorphic to $\operatorname{Ext}^i_R(N,M^{\vee})$. Thus, supposing $\operatorname{Tor}^R_i(N,H_{\mathfrak{m}}^d(M))=0$ is equivalent to supposing $\operatorname{Ext}^i_R(N,M^{\vee})=0$. A result of Lyle-Montaño (Lemma 3.4 (1) here cf. Lemma 5.3 here) then shows that $M \otimes_R N$ is maximal Cohen-Macaulay with the supposed vanishing. In fact, one only needs to assume that $\operatorname{Ext}^i_R(N,M^{\vee})=0$ for all $1 \le i \le d$, equivalently that $\operatorname{Tor}^R_i(N,H^d_{\mathfrak{m}}(M))=0$ for all $1 \le i \le d$.
