On the paper "On the asymptotic linearity of Castelnuovo-Mumford regularity"

Question

I have posted this question on MSE, however it seems not to be interested by member there, so I decided to post it here. I am sorry if you feel it is not appropriate for MO.

I am now reading the paper on Castelnuovo-Mumford regularity of Ngo Viet Trung and Hsin-Ju Wang, which can be found here

In this paper, the authors gave the concept of flat extension of ring, and in my opinion, they gave an example of it (in the lemma 1.2):

Let $A$ be a Noetherian ring, $R$ be standard graded finitely generated $A$-algebra, $R_{+}$ be its irrelevant ideal, $M$ be a finitely generated $R$-module. Let $Q$ be a $M$-reduction of $R_+$, generated by linear form $x_{1},...,x_{s}$. For $i=1,...,s$, put $z_{i}=\sum_{j=1}^{s}u_{ij}x_{j}$ where $U=\lbrace u_{ij}|i,j=1,...,s\rbrace$ is a matrix of indeterminates. Put $A'=A[U,\text{det}(U)^{-1}], R'=R\otimes_{A}A', M'=M\otimes_{A}A'$ then $A'$ is a flat extension of $A$.

As I understand $A'$ is a flat extension of $A$ if regarded as $A$ module, tensoring a exact sequence with $A'$ will give us an exact sequence.My questions are

1. How can we prove that the ring $A'$ constructed above satisfies the above property ?
2. From a commutative ring $A$, how can we construct its flat extension ?

I also have questions in the proof of the proposition 2.1 in that paper

Let $S=A[X_{1},...,X_{S}, Y_{1},...,Y_{v}]$ be a polynomial ring over a commutative Noetherian ring $A$. This ring can be viewed as a bigraded ring if we associate the degree for $X_{i}$ and $Y_{j}$ as $\text{deg}(X_{i})=(1,0), i=1,...,s; \text{deg}(Y_{j})=(d_{j}, 1), j=1,...,v$

Now, let $\mathcal{M}$ be a finitely generated graded module over $A[X_{1},...,X_{s}, Y_{1},]$. For a fixed number $n$, put $\mathcal{M}_{n}=\oplus_{a\ge0}\mathcal{M}_{(a,n)}$. Then $\mathcal{M_n}$ is a finitely generated graded module over the naturally graded polynomial ring $A[X_1,...,X_s]$. If $s=0$ then $\text{reg}(\mathcal(M_{n})$ is the invariant : $a(\mathcal{M_{n}})=\text{max}\lbrace a|\mathcal{M_{(a,n)}}\neq 0\rbrace$

3-Question : Why $\text{reg}\mathcal{M_n})$ is equal to $a(\mathcal{M_n})$ ?

As I understand , in $\mathcal{M_{(a,n)}}$, the index $a$ is graded for the variables $X_i$ and the index $n$ is graded for the variables $Y_{J}$ and then $\mathcal{M_{n}}$ is graded module over $A[X_1,…,X_s]$ but it is not a graded module over $A[Y_1,…,Y_v]$. So if $s=0$, and $n$ is fixed then we get nothing on $\mathcal{M_n}$. So, what is the point here ?

Proposition 2.1 in that paper says that

Let $\mathcal{M_n}$ be a finitely generated bigraded module over the bigraded ring $A[Y_1,...,Y_v]$. Then $a(\mathcal{M_n})$ is asymptotically a linear function with slope $\leq d_v$

Here is the argument in the proof :

The case $v=0$ is trivial.Consider the exact sequence : $0\rightarrow [0_{\mathcal{M}}:Y_{v}]_{(a,n)}\rightarrow \mathcal{M}_{(a,n)}\xrightarrow{Y_{v}}\mathcal{M}_{(a+d_{v},n+1)}\rightarrow \mathcal{M/Y_{v}M}_{(a+d_{v},n+1)}\rightarrow 0$

Since $0_{\mathcal{M}}:Y_{v}$ and $\mathcal{M}/Y_{v}\mathcal{M}$ can be viewed as bigraded modules over $A[Y_1,...,Y_{v-1}]$, using induction we may assume that $a([0_{\mathcal{M}}:Y_{v}]_{n})$ and $a([\mathcal{M}/Y_{v}\mathcal{M}]_{n})$ are asymptotically linear functions with slope $\leq d_v$. As consequence : $$a([0_{\mathcal{M}}:Y_{v}]_{n})+d_v\ge a([0_{\mathcal{M}}:Y_{v}]_{n+1})$$ $$a([\mathcal{M}/Y_{v}\mathcal{M}]_{n})+d_v\ge a([\mathcal{M}/Y_{v}\mathcal{M}]_{n+1})$$ for large $n$.

My question for this part is :

4- How did the author use the induction ? If they use the induction on $v$ then why can they get the above two inequalities, since I think we can get the inequalities if we use induction on $n$.

Since $a(\mathcal{M}_{n})\ge a([0_{\mathcal{M}}:Y_{v}]_{n})$ for all $n$, we only need to consider the case that $a(\mathcal{M}_{m})> a([0_{\mathcal{M}}:Y_{v}]_{m})$ for infinitely many $m$. Putting this condition into the above exact sequence we get : $$a(\mathcal{M}_{m+1})=\text{max}\lbrace a(\mathcal{M}_{m})+d_{v},a([\mathcal{M}/Y_{v}\mathcal{M}]_{n+1})\rbrace $$

My question for this part is :

5-How can the author get the above information from the exact sequence ?

Here is my argument : We have : $Y_{v}\mathcal{M}_{n}=\oplus_{a\ge 0}Y_{v}\mathcal{M}_{(a,n)}:=\oplus_{a\ge 0}\mathcal{N}_{(a+d_{v},n+1)}$. Then : $$a(Y_{v}\mathcal{M}_{n})=a(\oplus_{a\ge 0}\mathcal{N}_{(a+d_{v},n+1)})$$ $$=\text{max}\lbrace a+d_{v}|\mathcal{N}_{(a+d_{v},n+1)}\neq 0\rbrace$$ $$=\text{max}\lbrace a+d_{v}|Y_{v}\mathcal{M}_{(a,n)}\neq 0\rbrace$$ $$=d_{v}+\text{max}\lbrace a|\mathcal{M}_{(a,n)}\neq 0\rbrace$$ $$=d_{v}+a(\mathcal{M_{n}})$$ Since $Y_{v}\mathcal{M}_{n}\subseteq\mathcal{M}_{n+1}$ then $a(Y_{v}\mathcal{M}_{n})\le a(\mathcal{M}_{n+1})$ or $a(\mathcal{M}_{n+1})\ge d_{v}+a(\mathcal{M}_{n})$.

6-Am I wrong anywhere ?

7-What is the aim of the authors in the rest of the proof ?

I know that my questions are very long and may be hard to follow. If you have any question on this or this kind of questions are not appropriate with MO please let me know.

Thank for reading!

Answer 1

The flatness has nothing to do with the module or even the gradedness of the ring. If you take any ring and adjoin a matrix of indeterminates and the inverse of their determinant, you get a flat extension. This is because adjoining indeterminates is flat, adjoining inverses is flat, and a composition of flat morphisms is flat. All these facts are obvious from the definition.