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For a 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.

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.

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 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.

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, 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 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). )

2. 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 $I\neq R_+I$.

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.

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

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, 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 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).

2. 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 $I\neq R_+I$.

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 $\overline{R}/\overline{I}=R/I$. --- 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 $R/I$.

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.

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