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. 

 1. 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][1], but one can use matrix coefficient trick to prove it directly. (Here we use the assumption that $R$ is noetherian. )

 2. 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. )

 3. 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. )

 4. 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.  

 1. Note that $I$ admits a finite finitely generated free (twisted) resolution due to Hilbert’s syzygy theorem and classification of twisted projective modules over polynomial ring. 

 2. As the annihilator of $R/I$, it consists some non zero divisor. So is $I\setminus R_+ I$ by [(strong) prime avoidance][2] and the fact $I\neq R_+I$. 

 3. Pick such nonzero divisor $x\in I\setminus R_+ I$, then consider $\overline{R}=R/xR$, and $\overline{I}$ the image of $I$. 

 4. It is clear, now $\overline{I}/\overline{I}^2=I/(I^2+xR)$ is free of less rank than $\overline{R}/\overline{I}=R/I$. 

 5. $\overline{I}$ 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][3]. 

But it is also not clear what will happen for general graded ring. 


  [1]: https://stacks.math.columbia.edu/tag/090U
  [2]: https://stacks.math.columbia.edu/tag/00DS
  [3]: https://core.ac.uk/download/pdf/82227314.pdf