User manoj kummini - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T10:40:49Z http://mathoverflow.net/feeds/user/14895 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80957/hilbert-series-and-resolution-of-a-surface-singularity/81066#81066 Answer by Manoj Kummini for Hilbert series and resolution of a surface singularity Manoj Kummini 2011-11-16T13:47:04Z 2011-11-16T13:47:04Z <p>Assume, for now, that $e=1$. Then there exists an $(m+p)$-dimensional polynomial ring $S$ over $k$ with generators of degree $1$ and an $S$-ideal $I$ with $\mathrm{codim}(I) = p$ such that $B = S/I$. Let $0 \rightarrow M_h \rightarrow \cdots \rightarrow M_0 \rightarrow B \rightarrow 0$ be a minimal graded free resolution of $B$ as an $S$-module. Since $B$ is Cohen-Macaulay, it follows from the Auslander-Buchsbaum formula that $h=p$. The Castelnuovo-Mumford regularity $\mathrm{reg}(B)$ of $B$ is the maximum of $j-i$ such that $R(-j)$ is a free summand of $M_i$. (Eisenbud and Goto (J. Algebra, 1984) showed that this definition agrees with the definition using cohomology.) Since $B$ is Cohen-Macaulay, there exist linear forms, $\mathbf{l} = l_1, \ldots, l_m$ in $S$ that form a regular sequence on $B$. (We need that $k$ is an infinite field for this, but base change does not affect the hypotheses or conclusions, so we may replace $k$ by an algebraic closure, for instance.) Hence $\mathrm{reg}(B/\mathbf{l}B) = \mathrm{reg}(B)$ and such that $F(B/\mathbf{l}B, \lambda) = (1+p\lambda)$. </p> <p>Since $B/\mathbf{l}B$ is Artinian, its regularity is the highest degree in which it has a nonzero component, which in this case is one. Hence $\mathrm{reg}(B) = 1$, which means that $I$ is generated by quadrics and has a linear resolution, i.e, the maps in the minimal free resolution are matrices of linear forms. Then all the minimal homogeneous generators of $M_i$ have degree $(i+1)$. Moreover, $\mathrm{rank}(M_i) = i\binom{p+1}{i+1}$. I know this from a paper of Herzog and K\"uhl (Commun. Algebra, 1984), which, in turn, cites Wahl's paper. I'd guess that Wahl first shows that rational surface singularities have a linear resolution and then shows that the ranks are as above.</p> <p>Finally, let me point out that the assumption that $e=1$ is not really a restriction; the general case reduces to this, by changing the degrees of the homogeneous generators. The assertions about regularity would appear, in some way or the other, in Eisenbud's commutative algebra book or syzygies book.</p> http://mathoverflow.net/questions/79120/completion-of-local-rings-in-the-exceptional-divisor-of-a-blow-up/79580#79580 Answer by Manoj Kummini for Completion of local rings in the exceptional divisor of a blow-up Manoj Kummini 2011-10-31T02:56:29Z 2011-10-31T02:56:29Z <p>We can write $A = K[X_1, \ldots, X_n]/\mathfrak a$. Suppose that $I = (a_1, \ldots, a_m)$; then $A[It] \cong A[T_1, \ldots, T_m]/\mathfrak b$ for some ideal $\mathfrak b$. A description for $\mathfrak b$ is given in Section 1.1 of W. Vasconcelos, \textit{Integral Closure, Rees algebras, multiplicities and Algorithms}, (Springer).</p> <p>Let us suppose, for simplicity, that $P := \beta(Q)$ is defined by $(X_1, \ldots, X_n)$ and that $Q$ is defined by $(X_1, \ldots, X_n, T_1, \ldots, T_m)$. Let us write $R$ and $S$ for the local rings at $P$ and $Q$, respectively. Since we are interested in completions, we may consider the affine open subset of $Y$ defined by $T_1 \neq 0$. This can be thought of as the spectrum of $A[\frac{a_2}{a_1}, \ldots, \frac{a_m}{a_1}] \cong A[Y_2, \ldots, Y_m]/\mathfrak c$ for some ideal $\mathfrak c$. Then $\mathfrak c$ can be obtained from $\mathfrak b$ by dehomogenizing' with respect to $T_1$; see, \textit{e.g.}, Section 5.5 of \textit{Integral Closure of Ideals, Rings, and Modules} by I. Swanson and C. Huneke, (LMS Lecture Note Series 336). It contains the relations $a_1Y_i = a_i$ for all $1 \leq i \leq m$, but, in general, could have more. Hence $\widehat R \cong K[[X_1, \ldots, X_n]]/\mathfrak a$ and $\widehat S \cong\widehat R[[Y_1, \ldots, Y_m]]/\mathfrak c$. (By abuse of notation, we write $\mathfrak a$ for an ideal of $K[X_1, \ldots, X_n]$ and the ideal of $K[[X_1, \ldots, X_n]]$ generated by it.)</p> <p>Therefore, in the non-singular situation, with $A = K[X_1, \ldots, X_n]$ and $I = (X_1, \ldots, X_n)$, we have $\widehat R \simeq K[[X_1, \ldots, X_n]]$ and $\widehat S \simeq K[[X_1, Y_2, \ldots, Y_n]]$. (We use the fact that since $X_1, \ldots, X_n$ is a regular sequence in $A$, the ideal $\mathfrak b$ is generated by their Koszul syzygies', i.e., by $X_iT_j - X_jT_i, 1 \leq i &lt; j \leq n$. Hence $\mathfrak c$ is generated by $X_1Y_j - X_j, 2 \leq j \leq n$.) They are (abstractly) isomorphic as formal power series rings, but the structure morphism is not an isomorphism.</p> http://mathoverflow.net/questions/78250/can-taking-the-projective-closure-of-an-affine-variety-increase-the-degrees-of-it/78308#78308 Answer by Manoj Kummini for Can taking the projective closure of an affine variety increase the degrees of its ideal generators? Manoj Kummini 2011-10-17T04:12:01Z 2011-10-17T04:12:01Z <p>Consider $R = \Bbbk[x_1, \ldots, x_5]$ and $I = (x_1x_2^2-x_3^2, x_1x_4^2-x_5^2)$. When we homogenize w.r.t $z$, we get $(x_1x_2^2-zx_3^2, x_1x_4^2-zx_5^2)$ whose saturation w.r.t $z$ contains the binomial $x_3^2x_4^2-x_2^2x_5^2$ of degree $4$. The original generators are of degree at most $3$. </p> <p>Look at Section 15.10 of Eisenbud's commutative algebra book; one should homogenize a Groebner basis of $I$. Therefore, bound for $d'$ in terms of $d$ might turn out to be quite large. (This is just my feeling!)</p> <p>PS: in the `Simple Example', do you mean that $\hat{f_2} = x^2-yz$? As stated the ideal $\hat{I}$ is already saturated w.r.t. $z$ and does not have a component at $[0:1:0]$.</p>