I found the right reference, and read them and carry them properly to graded case.
The main lemma is Auslander-Buchsbaum's argument
Let $R$ be a noetherian ring, $M$ a finitely generated module. Assume $M$ admit a finite finitely generated free resolution. Then if the annihilator of $M$ is not trivial, then it contains a nonzero divisor in $R$.
The sketch of the proof is as the following.
Firstly, show that $M_{\mathfrak{p}}$ is free for any associaed prime $\mathfrak{p}$ of $R$. --- This is essentially the main process of Auslander-Buchsbaum equality, but one can use matrix coefficient trick to prove it directly. (Here we use the assumption that $R$ is noetherian. )
But the annihilator $\mathfrak{a}$ kills $M_{\mathfrak{p}}$, so $\mathfrak{a}_{\mathfrak{p}}=0$ or $M_{\mathfrak{p}}=0$. Since the rank to get the rank of $M_{\mathfrak{p}}$ does not depend on $\mathfrak{p}$. (Here we use the assumption of finiteness of free resolution. )
But if $\mathfrak{a}_{\mathfrak{p}}=0$, then the annihilator of $\mathfrak{a}$ is not contained in any prime associated to $R$, so their union, the zero divisor. Thus $\mathfrak{a}=0$. (Here we use the assumption that $R$ is noetherian. )
So $M_{\mathfrak{p}}=0$, then then the annihilator of $\mathfrak{a}$ is not contained in any prime associated to $R$, so their union, the zero divisor. (Here we use the assumption that $M$ is finitely generated. )
So we are done.
For a polynomial ring $R$ over field, the homogenous ideal $I$ is generated by a regular sequence of homogenous elements if $I/I^2$ is free over $R/I$.
The sketch of the proof is. the following.
Note that $I$ admits a finite finitely generated free (twisted) resolution due to Hilbert’s syzygy theorem.
As the annihilator of $R/I$, it consists some non zero divisor. So is $I\setminus R_+ I$ by (strong) prime avoidance and the fact $I\neq R_+I$.
Pick such nonzero divisor $x\in I\setminus R_+ I$, then consider $\overline{R}=R/xR$, and $\overline{I}$ the image of $I$.
It is clear, now $\overline{I}/\overline{I}^2=I/(I^2+xR)$ is free of less rank than $\overline{R}/\overline{I}=R/I$.
$\overline{I}$ admits a finite finitely generated free (twisted) resolution. --- remind the prove $pd_R M=pd_{R/xR} M/xM$ for non zero divisor $x$ for both $R$ and $M$.
Then it follows from induction.
The above process follows for neotherian local ring, as done in Ideals generated by R-sequences.
But it is also not clear what will happen for general graded ring.