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 Quillen–Suslin theorem. (Since we only need an existence, maybe the one who do not want to use such big theorem can use only Scheja-Stroch's computational proof of Hilbert’s syzygy theorem (for example, Weibel page 114). )
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.