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 $\overline{R}/\overline{I}=R/I$. --- Here we use that it is over some field, and graded, otherwise, one cannot claim like this, since $x$ may not extend to a basis of $I/I^2$ over $R/I$.
$\overline{I}$ 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.
Edit: My classmate remind me that this is also true.
Let $R$ be a connected noetherian ring, $M$ a finitely generated module. Then if the annihilator of $M$ is not trivial, then it contains a nonzero divisor in $R$.
By the same way. So we also have this
For a connected graded ring $R$ over field $k$ with $R^0=k$, only nonnegative degree, a homogenous ideal $I$ is generated by a regular sequence of homogenous elements if $I/I^2$ is free over $R/I$ and $I$ has finite projective dimension.
Besides, for any set of element presenting a basis, by our choice, our choice of regular sequence presents a set of basis of $I/I^2$ over $R/I$, so it differs by our choice a invertible matrix. Then it reduces to exchange two element. We permute them by degree reason. So in conclusion
In above case, any set of basis presenting a set of basis for $I/I^2$ over $R/I$ forms a regular sequence.