Isolated lc-centre SNC vanishing and polynomial nonvanishing

Use the given log resolution and $D=\mathrm{ceil}(-B)$. The lc condition says that every coefficient of B is at most one; negative coefficients occur only on exceptional divisors because G is effective. Therefore $D=P-E$, with P effective exceptional and E the reduced sum of the coefficient-one divisors. For a regular function f downstairs, its pullback is regular on Y. Membership in ${\mu }_{\ast }{O}_Y(D)$ imposes exactly $or{d}_F(f)\ge 1$ for each component F of E and no further condition: the positive exceptional allowance P does not enlarge the ring of regular functions on the normal X. The valuation inequality $or{d}_F(f)\ge 1$ is equivalent to f vanishing identically on its centre. Since the union of these centres is precisely the reduced Z, this proves

${\mu }_{\ast }{O}_Y(D)=I_Z$.

This is the ideal of the reduced centre, not an unexplained multiplier-ideal identification.

Here is the needed vanishing from the actual smooth SNC adjoint supplier. Put $F=B+\mathrm{ceil}(-B)$, so $0\le \mathrm{coeff}(F)<1$ and F is SNC, and observe

$D+{\mu }^{\ast }(tA)=K_Y+F+{\mu }^{\ast }(tA-K_X-G)$, whose last term is numerically equivalent to (t+r-gamma)mu*A.

If $t+r-\gamma >0$, the last pullback has the class of a positive multiple of an ample divisor. Use the assumed effective exceptional rational divisor T with -T relatively ample. For the specified blowup models it is constructed inductively: a blowup has relatively ample negative tautological exceptional divisor, and at each later blowup a sufficiently large multiple of the previous choice plus the new exceptional divisor retains relative ampleness. This construction only requires effective exceptional Cartier tautological divisors, not smoothness of intermediate bases. For small rational $ϵ>0$, ${\mu }^{\ast }(tA-K_X-G)-ϵ$ T is ample on Y. Increase the SNC boundary from F to F+epsilon T; all its coefficients remain below one. The line bundle on the left has not changed. The SNC adjoint vanishing supplier therefore gives

$H^q(Y,O_Y(D+{\mu }^{\ast }(tA)))=0$, $q>0$.

The same argument with t replaced by arbitrarily large integers proves relative vanishing $R^q$ ${\mu }_{\ast }{O}_Y(D)=0$ for $q>0$. Indeed twist the Leray spectral sequence by $O_X(mA)$; for large m Serre vanishing annihilates every positive cohomology of its finitely many coherent higher direct images. The spectral sequence degenerates to $H^q(Y,O(D+{\mu }^{\ast }mA))=H^0(X,R^q{\mu }_{\ast }O(D)\otimes O(mA))$. The left side vanishes and a nonzero coherent sheaf has nonzero H0 after a sufficiently high ample twist, so each higher image is zero. Consequently

$H^q(X,I_Z(tA))=0$ whenever $q>0$ and $t+r-\gamma >0$.

For the empty boundary put $D0=K_Y-{\mu }^{\ast }{K}_X$. It is effective exceptional: in the smooth case by the blowup Jacobians, and in the terminal Gorenstein case by the actual positive integral discrepancies. The same normal-function calculation gives ${\mu }_{\ast }{O}_Y(D0)=O_X$, and the same large-twist relative vanishing gives $R^q$ ${\mu }_{\ast }{O}_Y(D0)=0$ for $q>0$. Since $D0+{\mu }^{\ast }(tA)=K_Y+{\mu }^{\ast }(tA-K_X)$, whose adjoint difference is numerically equivalent to (t+r)mu*A, smooth SNC vanishing on Y with the exceptional perturbation gives $H^q(X,O_X(tA))=0$ for $q>0$ and $t+r>0$. No vanishing theorem on a singular X is inferred without this calculation. For the remaining nonvanishing assertion assume $\gamma <c-1$, where $z=\dim$ Z and $c=n-z$. Effectivity gives $\gamma \ge 0$, by intersecting G with a general complete-intersection curve from a very ample multiple of A. Thus for every integer $t\ge 1-z$, both vanishings apply: $t+r-\gamma \ge c-1-\gamma >0$ and $t+r\ge t+r-\gamma >0$. The restriction sequence gives $H^{>0}(Z,O_Z(tA))=0$ for that whole range, and $H^0(X,A)\to H^0(Z,A)$ is onto.

Put $L={\mathcal{O}}_X(A){\mathrm{\mid }}_Z$ and $z=\dim Z$. If $z=0$, the integral complex projective $Z$ is a point and $h^0(Z,L)=1$. Suppose $z\ge 1$. We first construct the Euler polynomial for the ample $L$ without assuming it very ample. Fix a very ample line bundle $H$ on the projective $Z$. By F2, $H^{-1}\otimes L^d$ is globally generated for all sufficiently large integers $d$. Its generating sections define a morphism $j_d:Z\to {\mathbb{P}}^{N_d}$; the graph with the embedding $i_H:Z\hookrightarrow {\mathbb{P}}^N$ is a closed immersion into ${\mathbb{P}}^N\times {\mathbb{P}}^{N_d}$. The Segre map, whose coordinates are all products of the two coordinate systems, is a closed immersion: its image is cut out by the two-by-two minors, and the coordinate ratios recover both factors on each standard chart. The composite pulls back $\mathcal{O}(1)$ to $H\otimes (H^{-1}\otimes L^d)=L^d$. Hence choose $m\ge 1$ with both $L^m$ and $L^{m+1}$ very ample. For $d=m,m+1$ and $0\le a<d$, F3 applied with this actual embedding and coherent sheaf $L^a$ gives a polynomial $q_{d,a}(k)=\chi (Z,L^{a+dk})$ for every $k\in \mathbb{Z}$, of degree $z$. Its leading coefficient is positive: Serre vanishing identifies its large values with nonnegative $h^0$, and its degree is exactly $z$.

Define $f:\mathbb{Z}\to \mathbb{Z}$ by $f(t)=\chi (Z,L^t)$ and let $T$ be the shift operator $(Tf)(t)=f(t+1)$. Each residue-class polynomial from step 2.1 has degree $z$, so $(T^m-1{)}^{z+1}f=0$ and $(T^{m+1}-1{)}^{z+1}f=0$ on all integers. The polynomials $u^m-1$ and $u^{m+1}-1$ have greatest common divisor $u-1$ (subtract $u(u^m-1)$ from $u^{m+1}-1$), and all their roots are simple in characteristic zero. Their $(z+1)$st powers therefore have greatest common divisor $(u-1{)}^{z+1}$. The Euclidean algorithm yields $U,V\in \mathbb{Q}[u]$ with $U(u)(u^m-1{)}^{z+1}+V(u)(u^{m+1}-1{)}^{z+1}=(u-1{)}^{z+1}$. Substitution $u=T$ proves ${\mathrm{\Delta }}^{z+1}f=0$ for $\mathrm{\Delta }=T-1$. Thus $f(t)$ equals the polynomial $p(t)={\sum }_{j=0}^z({\mathrm{\Delta }}^{j}f)(0)\left(\genfrac{}{}{0px}{}{t}{j}\right)$ at every integer: this polynomial has the same first $z+1$ values, and the recurrence ${\mathrm{\Delta }}^{z+1}f=0$ propagates equality both forward and backward. Since $p(mk)=q_{m,0}(k)$, its degree is exactly $z$ and its leading coefficient is positive. This proves the ample-twist extension required here, including negative twists, using only the stated fixed-embedding suppliers and algebra.

A nonzero section of $L^{-j}$, $j>0$, would give an injection $L^t\hookrightarrow L^{t-j}$ because $Z$ is integral. For large $t$, Serre vanishing would then give $p(t)=h^0(Z,L^t)\le h^0(Z,L^{t-j})=p(t-j)$, contradicting the positive leading coefficient and $z\ge 1$. Hence negative powers have no sections. By step 1.10 their higher cohomology also vanishes for $t=-1,\dots ,-(z-1)$, so these integers are roots of $p$; moreover $p(0)=h^0(Z,{\mathcal{O}}_Z)\ge 0$. Consequently $p(t)={\prod }_{j=1}^{z-1}(t+j)(at+b)$, with $a>0$ and $b=p(0)\mathrm{/}(z-1)!\ge 0$; for $z=1$ the product is $1$. Step 1.10 also gives $h^0(Z,L)=p(1)=z!(a+b)>0$. Its restriction surjectivity lifts a nonzero section to $X$, including the zero-dimensional case already handled.

The strict bound $\gamma <c-1$ yields a section of A not identically zero on Z; it does not exclude the existence of Z. In particular, if one additionally assumes Z is contained in the base locus Bs|A| (the common zero set of all global sections of A), the conclusion is a contradiction. Under that additional base-locus assumption, $\gamma \le 1$ excludes $c\ge 3$, and $\gamma <1$ excludes $c=2$. Without it such centres may exist. The equality case $\gamma =1,c=2$ requires the separate curve/surface subadjunction–nonvanishing argument. Normality of Z was not required for this polynomial lemma.