V-filtration of D-modules associated to a monomial

Hi

In Mixed Hodge modules Saito computes the Verdier specialisation of a D-modules with respect to a monomial $g = x_1^{m_1}ldots x_n^{m_n}$. This is a very nice result as I find such explicit computations are quite rare in the litterature but I'm having problems understanding his proof.

Let $X$ be a polydisc, $D_i = {x_i = 0}$, $D_I = bigcap_{iin I} D_i$. Consider $M$ a regular holonomic algebraic quasi-unipotent right D-module with characteristic variety contained in the union of the conormal bundles to the $D_I$. For $nu = (nu_1,ldots,nu_n)$, set
$$
M^nu := langle uin M ~|~ u(x_ipartial_i - nu_i)^p = 0~~ textrm{for $pgg 0$} rangle
$$
We have $bigoplus_{nu in mathbb{Q}^n} M^nu = M$.

Let $g:Xto S$, $S$ an open disk, given by $g=x_1^{m_1}ldots x_n^{m_n} = x^m$ and $i_g:X to Xtimes S$ its graph. The D-module $tilde{M} = (i_g)_* M$ is $M[partial_t]$ with the action of $D_{Xtimes S}$ given by:
$$
(uotimes partial_t^j)x_i = ux_iotimes partial_t^j, quad
(uotimes partial_t^j)partial_i = upartial_iotimes partial_t^j - u(partial_ig)otimes otimes partial_t^{j+1}
$$
$$
(uotimes partial_t^j)t = ugotimes partial_t^j + ju otimes partial_t^{j-1} , quad
(uotimes partial_t^j)partial_t = ux_iotimes partial_t^{j+1}
$$

Theorem (3.4 p. 280): The V-filtration of $tilde{M}$ along $t = 0$ is generated over $D_X$ by
$$
M^nu otimes 1 ~~textrm{with $nu_i leq m_ialpha$} quad textrm{if $alpha< 0$}
$$
$$
M^nu otimes partial_t^j ~~textrm{with $nu_i leq m_i(alpha-j)$} quad textrm{in general}
$$

It is enough to check that the filtration defined in the theorem satifies the properties of the V-filtration. Saito says:

"We have
$$
(uotimes partial_t^j)x_ipartial_i = (uotimes partial_t^j)(N_i + nu_i - m_i(s-j) )
qquad forall u in M^nu
$$
where $s = tpartial_t$ and $(uotimes partial_t^j)N_i = u(x_ipartial_i-nu_i)otimes partial_t^j$ if $uin M^nu$. Thus we get that $s-alpha$ is nilpotent on $Gr^V_alpha tilde{M}$ and $V_alpha tilde{M}$ ar $V_0D_{Xtimes S}$-submodules."

I don't understand this last statement. $Gr^V_alpha tilde{M}$ is generated over $D_X$ by the $u otimes partial_t^j in V_alpha tilde{M}$ with at least one $i$ so that $nu_i = m_i(alpha-j)$. For such $nu$, the above equation reads
$$
(uotimes partial_t^j)x_ipartial_i = (uotimes partial_t^j)(N_i - m_i(s-alpha))
$$
Why would this imply that $s-alpha$ is nilpotent?

Edit: I think I have found the answer. Consider $uotimes partial_t^j$ with $nu_i leq m_i(alpha-j)$ forall $i$. Set $I := langle i ~|~ nu_i = m_i(alpha-j),~ m_ineq 0rangle$. Then
$$
(uotimes partial_t^j)(prod_{iin I} x_ipartial_i) = (u prod_{iin I}x_i otimes partial_t^j) (prod_{iin I}partial_i)
$$
is in $V_{<alpha} tilde{M}$. So, by applying the relation above, we have
$$
(uotimes partial_t^j)prod_{iin I} (N_i - m_i(s-alpha)) =
(uotimes partial_t^j)(prod_{iin I} x_ipartial_i) = (u prod_{iin I}x_i otimes partial_t^j) (prod_{iin I}partial_i) = 0
$$
in $Gr^V_alpha tilde{M}$. So
$$
(uotimes partial_t^j)(s-alpha)^{|I|} = (uotimes partial_t^j)(sum_{k=0}^{k=|I|-1} sigma_k(N_i/m_i)(s-alpha)^k) )
$$
The operator on the right hand side is a polynomial in the $N_i$'s with coefficient in $Q[s-alpha]$ and no constant term. So if $p_i$ is the nilpotency order of $N_i$ operatating on $uotimes partial_t^j$ and we elevate the relation to the power $sum p_i$ we find that $(uotimes partial_t^j)(s-alpha)^{|I|(sum p_i)} = 0$. So $s-alpha$ is indeed nilpotent.