Let A be an r x (r – 1) matrix. Set \u03b4_i = (-1)^i times the determinant of the (r – 1) x (r – 1) matrix you get by deleting the ith row. Then we have the equality<\/p>\n
(\u25ca) (\u03b4_1, …, \u03b4_r) A = 0.<\/p>\n
Let I be an ideal in a Noetherian regular local ring R with dim(R) = dim(R\/I) + 2 = depth(R\/I) + 2. Then I has a minimal resolution of the form<\/p>\n
0 —> R^{r – 1} —> R^r —> I —> 0<\/p>\n
where r is the minimal number of generators for I. Denote A the matrix defining the map R^{r – 1} —> R^r. In this situation Burch’s theorem tells us that I is generated by the \u03b4_i, and in fact the map R^r —> I is (up to a unit) given by the row vector (\u03b4_1, …, \u03b4_r).<\/p>\n
Why is this useful? Well, suppose you want to deform R\/I (see this post<\/a>). It is often easy to see that there are lots of deformations, but what isn’t so easy is to prove that there are any unobstructed deformations. But in the situation above we can just choose a family of matrices A(t) formally depending on an auxiliary parameter t. Then the minors \u03b4_1(t), …, \u03b4_r(t) of A(t) generate an ideal I(t) in R[[t]]. Then R[[t]]\/I(t) is a flat deformation of R\/I by the criterion from the post on deformation theory: all the relations lift to R[[t]] because the equation (\u25ca) is universal and hence holds also for our matrix A(t).<\/p>\n As an example consider a fat point in C<\/strong>^2, for example given by the ideal I = (x^n, x^{n – 1}y, x^{n – 2}y^2, …, y^n). The matrix A is the matrix whose ith column look is (0, …, 0, x, -y, 0, …, 0) with x in the ith spot. We can deform this by picking (0, …, 0, x – ta_i, -y + tb_i, 0, …, 0) with a_i, b_i 2n pairwise distinct complex numbers. The deformed scheme for t = 1 has n(n + 1)\/2 reduced points, namely the points (a_i, b_j) with j\u00a0\u2265 i.<\/p>\n In fact you can show that any deformation is given by deforming the matrix (by applying the Burch’s theorem which is more general than what I said above to the resolution of the deformed ideal), and hence all deformations are unobstructed and the deformation space of the singularity defined by I is smooth. This in particular shows that the Hilbert scheme of points of a smooth surface is smooth.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Let A be an r x (r – 1) matrix. Set \u03b4_i = (-1)^i times the determinant of the (r – 1) x (r – 1) matrix you get by deleting the ith row. Then we have the equality (\u25ca) … Continue reading