Let \(X\) be a noetherian scheme, and let \(Y, Z\) be two closed subschemes, neither one containing the other. Let \(\tilde X\) be obtained by blowing up \(Y \intersect Z\), defined by the ideal sheaf \(\mci_Y + \mci_Z\). Show that the strict transforms \(\tilde Y\) and \(\tilde Z\) of \(Y\) and \(Z\) in \(\tilde X\) do not meet.
Solution
Put \(W=Y\cap Z\) and let \(b:\tilde X\to X\) be its blowup, with exceptional divisor \(E=b^{-1}(W)\). The strict transform of a closed subscheme is the scheme-theoretic closure of its inverse image outside \(W\); equivalently, remove from its total inverse image the sections supported on \(E\).
1. Locally over \(X\), the blowup is covered by charts indexed by generators of the ideal of \(Y\) and generators of the ideal of \(Z\).
On an affine open \(\Spec A\subseteq X\), let \(I=(a_1,\ldots,a_r)\) and \(J=(b_1,\ldots,b_s)\) define \(Y\) and \(Z\) there. Then \(K=I+J\) defines \(W\). In the Rees algebra \(R=\bigoplus_{q\ge0}K^q\), the degree-one elements corresponding to the \(a_i\) and \(b_j\) generate the algebra over \(A\). Their distinguished opens therefore cover \(\operatorname{Proj}R\).
For any one of these generators \(c\in K\), denote its degree-one copy by \(c^{(1)}\) and put \[ C_c=(R[1/c^{(1)}])_0. \] The corresponding chart is \(\Spec C_c\). The identity \(h=c(h/c)\) for \(h\in K\) shows that \(KC_c=cC_c\), so the exceptional divisor on this chart is cut out by \(c\) and its complement is \(D(c)\). These statements use homogeneous localization and remain valid when \(A\) has zero divisors; a chart may be empty.
2. On a chart indexed by \(a_i\in I\), the strict transform \(\tilde Y\) is empty; on a chart indexed by \(b_j\in J\), the strict transform \(\tilde Z\) is empty.
On \(\Spec C_c\), the total inverse image of \(Y\) has coordinate ring \(C_c/IC_c\). Its part outside the exceptional divisor has coordinate ring \((C_c/IC_c)[1/c]\). The scheme-theoretic closure in this affine chart is therefore defined by the kernel of \[ C_c\longrightarrow(C_c/IC_c)[1/c]. \] This kernel consists of the elements whose image modulo \(IC_c\) is killed by a power of \(c\), which is precisely the removal of exceptional torsion in the strict-transform definition.
When \(c=a_i\in I\), its image in \(C_c/IC_c\) is zero. Inverting it gives the zero ring, so the kernel is all of \(C_c\) and the strict transform is empty on the chart. Interchanging \(I\) and \(J\) proves the assertion for the \(b_j\) charts and \(\tilde Z\).
3. Q.E.D.
The charts in step 1 cover the blowup over every affine open of \(X\), hence cover \(\tilde X\). By step 2, at least one of \(\tilde Y\) and \(\tilde Z\) is empty on each chart. Their scheme-theoretic intersection is consequently empty, and in particular their underlying sets do not meet.