I am reading from Brodmann, Sharp Local Cohomology about Castelnuovo-Mumford (CM)-regularity, and although I think had understood some things about it I got stuck in an exercise which wants to prove that $\mathrm{reg}(R_0[X_1,...,X_n])=0$, for $R_0$ commutative, Noetherian ring. Although that fits in my intuition so far, that is, CM-regularity measures in a sense the complexity of a given module $M$ in terms of its free graded resolution I can't really understand why is $0$ based on its definition only.
Also in some other example (which apparently is related), they prove that $$H^n_{(X_1,...,X_n)}(R_0[X_1,...,X_n]) \cong R_0[X^{-}_1,...,X^{-}_n], \thinspace.$$ where the latter is the "module of inverse polynomials over $R_0$". Again although the construction of it makes sense, I have a problem understand its grading, since inside the book for some it's described quite complicate.
Can you please help me sort out the above? Thank you!
Let me answer you question when $R_0 = k$ a field. Let $S = k[x_1,\dots,x_n]$. Then $x_1, \dots, x_n$ forms an $S$-regular seqeunce. Hence $H^i_m (S) = 0$ unless $i = n$, where $m = (x_1, \dots, x_n)S$.
One can compute the local cohomology modules using the Cech complex with respect to $x_1,\dots, x_n$. Then it is easy to observe that $$ H^n_m (S) = \{ x_1^{a_1}\cdots x_n^{a_n} \mid a_i \le -1\} $$ as $k$-vector spaces by analyzing the tail of the Cech complex $$ \oplus S_{x_1\cdots \hat{x_l} \cdots x_n} \to S_{x_1\cdots x_n} \to 0. $$ Here $\hat{x_l}$ means omitting $x_l$. Hence $\sup \{ i \mid [H^n_m(S)]_i \neq 0\} = -n$ as everything respects grading. This answers your first question.
I am not familiar with the concept of the module of inverse polynomials over $k$. However, the right hand side of $$ H^n_m (S) = \{ x_1^{a_1}\cdots x_n^{a_n} \mid a_i \le -1\} $$ is isomorphic to $\frac{1}{x_1\cdots x_n} k[x_1^{-1}, \dots, x_n^{-1}]$ as $k[x_1^{-1}, \dots, x_n^{-1}]$-modules.