Castelnuovo-Mumford regularity and postulation numbers

Question

I have a problem about Castelnuovo-Mumford regularity. This is a proposition from Castelnuovo-Mumford regularity, relation types and postulation numbers by M. Brodmann and C. H. Linh.

My questions are:

1. What's $x^*G_{\mathfrak{q}}(A)$?
2. How do we have the isomorphism: $G_{\mathfrak{q}}/x^*G_{\mathfrak{q}}(A)\cong G_{\mathfrak{q}(A/xA)}(A/xA)$?
3. Please explain with more details this: "By induction, we have $\operatorname{reg}(G_{\mathfrak{q}(A/xA)}(A/xA))=p(G_{\mathfrak{q}(A/xA)}(A/xA))+d-2$..."? I really understand that we make induction base on: $\operatorname{dim}(A)$ or $\operatorname{depth}(A)$ or $\operatorname{depth}(A/xA)$...? and how to make it as an induction: base case, induction step?

Thanks for your regarding!

Answer 1

1. $x^*G_{\mathfrak{q}}(A)$ is the ideal generated by $x^*$ in $G_{\mathfrak{q}}(A)$.

2. $G_{\mathfrak{q}}(A)/x^*G_{\mathfrak{q}}(A)\cong G_{\mathfrak{q}(A/xA)}(A/xA)$ because they have isomorphic graded pieces: the homogeneous degree $n$ piece of $G_{\mathfrak{q}}(A)/x^*G_{\mathfrak{q}}(A)$ is $\displaystyle\frac{q^n/q^{n+1}}{x^*(q^{n-1}/q^n)}=\frac{q^n}{xq^{n-1}+q^{n+1}}$ and of $G_{\mathfrak{q}(A/xA)}(A/xA)$ is $\displaystyle\frac{q^n+xA}{q^{n+1}+xA}$. The last quotient is isomorphic to $\displaystyle\frac{q^n}{q^n\cap(q^{n+1}+xA)}$ and then it remains to show that $q^n\cap(q^{n+1}+xA)=xq^{n-1}+q^{n+1}$. The inclusion "$\supseteq$" is clear. For "$\subseteq$" let $y\in q^n\cap(q^{n+1}+xA)$. Then $y\in q^{n+1}+xA$, so we can write $y=z+xa$ with $z\in q^{n+1}$ and $a\in A$. Since $z\in q^n$ we get $xa\in q^n$ and all we have to do is to show that $a\in q^{n-1}$. This follows easily by induction on $n$ using that $x^*$ is regular. (This can be written as $q^n:x=q^{n-1}$ for all $n\ge 1$.)

3. You can assume that $A/m$ is infinite and apply the homogeneous prime avoidance in order to choose $x$ a non-zero divisor on $A$, too.

The induction is on $d=\dim A$.
For $d=1$ (the base case), if $\operatorname{depth}A=0$ and $\operatorname{depth}G_q(A)=0$ then we have to show that $\operatorname{reg}(G_q(A))=p(G_q(A))$. But this follows from Lemma 3.3.
For $d\ge 2$, note that since $x$ is chosen regular on $A$, then $\dim A/xA=d-1$. Moreover, $\operatorname{depth}A/xA=\operatorname{depth}A-1$ and since $x^*$ is regular on $G_q(A)$ we also have $\operatorname{depth}G_{q(A/xA)}(A/xA)=\operatorname{depth}G_q(A)-1$.