Ideals generated by regular sequences

Question

In Vasconcelos' paper (Ideals generated by R-sequences), he proved

If $R$ is a local ring, $I$ an ideal of finite projective dimension, and $I/I^2$ is a free $R/I$ module, then $I$ can be generated by a regular sequence.

This is a theorem for local ring.

In Kac's paper, (Torsion in cohomology of compact Lie groups and Chow rings of reductive algebraic groups), he referred this result (in appendix, Proof of Theorem 1), but used for non-local ring.

More precisely, he constructed a map for a compact lie group $K$, and a field $k$, $S(M)\stackrel{\psi}\to H^\bullet(K/T;k)$, where $M=L\otimes k$ with $L$ the weight lattice, and $T$ the maximal torus. He claim $\ker \psi$ is generated by a homogenous regular sequence.

Furthermore, what I believe to be right, is the following

For polynomial ring $R$ over field, and a graded ideal $I$ such that $I/I^2$ is free over $R/I$ (as graded module), then $I$ is generated by a homogenous regular sequences.

My question is, how to prove this if it is true? If not, is the $\ker \psi$ in the paper generated by a regular sequence?

Maybe some useful remarks,

1. This is not true for example $k[x]/x^2$, and $I=(x)$. Since $k[x]/x=k$ never admits a finite projective(=free since local) $k[x]/x^2$ resolution by dimension argument.
2. When the ring is local and $I$ is the maximal ideal, this is exactly the theorem of regular local ring. I tried to move the proof, but fails, because of the above example.
3. The main step of Vasconcelos' paper, is to a result due to auslander and buchbaum. It discussed local ring specifically.
4. Generally, there is a concept called regular ideal, but it is local.
5. I do not even know whether we can pick the sequence to be arbitary choice of representative of basis.
6. I also wounder whether it is true for all graded ring with $I$ of finite projective dimension.
7. For $\psi$, it is more crutial when the field of positive characteristic. When it is of characteristic zero, the $\psi$ is nothing but the classical thing, the projection to coinvariant algebra.

Answer 1

I am sure even your (6) is correct, but am a bit lazy today to check things carefully, so let me answer your question for polynomial rings.

If $I\subset R$, a graded ideal, it is immediate that one can pick a minimal set of generators for $I$ which are homogeneous. With your hypothesis, these become a regular sequence after localizing at the `irrelevant' maximal ideal by Vasconcelos.

Let $x_1,\ldots, x_k$ be the homogeneous generators of $I$. If they were not a regular sequence, say $x_1,\ldots x_l$ are, but $x_{l+1}$ is a zero divisor modulo $x_1,\ldots,x_l$. Then, $x_{l+1}$ is contained in an associated prime of $(x_1,\ldots,x_l)$, but this ideal is graded and thus so are all its associated primes. In particular, the associated prime containing $x_{l+1}$ is graded. This prime is contained in the irrelevant maximal ideal and thus survives when you localize. But, this says that $x_1,\ldots, x_{l+1}$ do not form a regular sequence after localizing, which is a contradiction.

Answer 2

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.

---

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.