Theorem 9.8.6373CA5388E72

Proof. After reducing to the case of a manifold, we blow up the image of the divisor in $X$ as long as it has codimension at least two; each step lowers the order of vanishing along the divisor of the Jacobian determinant of the model.

Step 1: reduction to a manifold. Let $\widehat X\to X$ be a resolution given by theorem 9.6.6, a finite sequence of blowups with smooth centers, since $X$ is compact. Let the prime divisor be represented by $(M_0,F_0)$, and let $M$ be a common smooth model of $M_0$ and $\widehat X$, with $E\subset M$ the strict transform of $F_0$ (proposition 9.8.3); then $(M,E)$ represents the same divisor, as we see on the common smooth model $M$. A finite sequence of blowups with smooth centers over $\widehat X$, composed with $\widehat X\to X$, is such a sequence over $X$, and smooth models of $\widehat X$ are smooth models of $X$, by corollary 9.7.3. Hence we may assume that $X$ is a compact complex manifold, and we are given a proper bimeromorphic map $f\colon M\to X$ from a compact manifold and an irreducible hypersurface $E\subset M$.

Since $E$ is irreducible, hence connected, it lies in a connected component $M_ E$ of $M$, and $f(M_ E)$ lies in a connected component $X_ E$ of $X$. These components have the same dimension $n$, because $M_ E$ contains a nonempty open set mapped isomorphically onto an open subset of $X$ (lemma 9.4.1); moreover $n\geq 1$ and $\dim E=n-1$, since $E$ is a nonempty hypersurface. If $n=1$, the center $f(E)$ considered below is a point of $X_ E$, hence a hypersurface.

Step 2: the order of the Jacobian. The Jacobian determinant $J_ f$ of $f$ in local coordinates is holomorphic and determined up to multiplication by holomorphic units. It does not vanish identically near any point, because $f$ is an isomorphism over a dense open subset of $X$ whose inverse image is dense in $M$ (lemma 9.4.1). Its order of vanishing $\operatorname {ord}_ EJ_ f$ along $E$ is the largest $k$ such that $J_ f$ is divisible by the $k$th power of a local equation of $E$ near points of $E_{\mathrm{reg}}$; definition 6.12.3 uses the same notion of order along a hypersurface. It is a finite nonnegative integer $k$, since a function divisible by all powers of a local coordinate $y$ has vanishing Taylor coefficients in $y$ and vanishes near the point. It does not depend on the point of $E_{\mathrm{reg}}$, which is a connected manifold by corollary 9.3.17: for each integer $j$, the set of points of $E_{\mathrm{reg}}$ at which the order is at least $j$ is open, and it is closed by the identity theorem on $E_{\mathrm{reg}}$, since a holomorphic function vanishing on $E$ near a regular point is divisible by a local equation of $E$.

For composable maps the chain rule gives

because orders of vanishing along $E$ add under multiplication. Replacing $M$ by a common smooth model $h\colon M'\to M$ and $E$ by its strict transform $E'$ does not change $k$: by proposition 9.8.3, $h$ is an isomorphism near generic points of $E'$, so $\operatorname {ord}_{E'}J_ h=0$ and $\operatorname {ord}_{E'}(J_ f\circ h)=k$.

Step 3: one principalization lowers the order. The center $Z=f(E)$ is an analytic subset of $X$ by theorem 9.7.2, of dimension at most $\dim E=n-1$ by lemma 9.3.15(3). It is irreducible: if $Z=Z_1\cup Z_2$ with analytic subsets $Z_1,Z_2$, then the irreducible set $E$ is the union of its analytic subsets $E\cap f^{-1}(Z_1)$ and $E\cap f^{-1}(Z_2)$, so it lies in one of them, and $Z$ equals $Z_1$ or $Z_2$. The same argument shows that images of irreducible analytic sets under proper holomorphic maps are irreducible. Suppose first that $Z$ is a hypersurface, and let $S\subset X$ be as in proposition 9.8.3 for $f$. The strict transform $Z^ M$ of $Z$ in $M$ is an irreducible hypersurface containing $f^{-1}(Z\setminus S)$. The set $E\cap f^{-1}(Z\setminus S)$ is open in $E$, and it is nonempty, because $f(E)=Z$ is not contained in $S$, which has codimension at least two. So $E\cap Z^ M$ is an analytic subset of $E$ with interior points in $E$, and $E\subset Z^ M$ by corollary 9.3.17(2); then $E=Z^ M$ by corollary 9.3.17(3). Hence $(X,Z)$ represents the given divisor, as we see on the common smooth model $M$.

Suppose now that $Z$ has codimension at least two. Its ideal $\mathcal I_ Z$ is coherent by theorem 9.3.10, and it is not identically zero on any component of $X$, since $Z$ is nowhere dense. Let $\sigma =\sigma _1\circ \cdots \circ \sigma _ t\colon X'\to X$ be the finite sequence of blowups with centers $C_ i\subset X_{i-1}$ and transformed ideals $\mathfrak a_ i$, $\mathfrak a_0=\mathcal I_ Z$, given by theorem 9.6.7; it is proper and bimeromorphic by corollary 9.7.3, so $X'$ is a smooth model of $X$. Let $h\colon M'\to M$ and $f'\colon M'\to X'$ be the maps from a common smooth model of $M$ and $X'$, so that $f\circ h=\sigma \circ f'$, let $E'$ be the strict transform of $E$, let $\rho _ i\colon M'\to X_ i$ be the induced maps, and let $Z_ i=\rho _ i(E')$; thus $Z_0=Z$.

Let $r\geq 2$ be the codimension of $Z$, which is pure by corollary 9.3.17(1). We claim that for some $i$ the center $C_ i$ contains $Z_{i-1}$, and that the connected component of $C_ i$ containing $Z_{i-1}$ has codimension $r$. The sets $Z_ j$ are irreducible, as images of the irreducible set $E'$, and $\sigma _{j+1}(Z_{j+1})=Z_ j$. Put $\tau _ j=\sigma _1\circ \cdots \circ \sigma _ j$. Suppose that $Z_{l-1}\not\subset C_ l$ for all $l\leq j$. We show by induction on $j$ that there are a nowhere dense analytic subset $A_ j\subset Z$ and an open set $N_ j\subset X\setminus A_ j$ containing $Z^\circ _ j=Z\setminus A_ j$ such that $\tau _ j$ is an isomorphism over $N_ j$ and, on $\tau _ j^{-1}(N_ j)$,

For $j=0$ take $A_0=\varnothing $ and $N_0=X$.

Assume the assertion for some $j<t$, and suppose that $Z_ j\not\subset C_{j+1}$. On $\tau _ j^{-1}(N_ j)$ we have $C_{j+1}\subset V(\mathfrak a_ j)=\tau _ j^{-1}(Z^\circ _ j)\subset Z_ j$. By corollary 9.3.17(2), $C_{j+1}\cap Z_ j$ is nowhere dense in $Z_ j$. Its image $\tau _ j(C_{j+1}\cap Z_ j)\subset Z$ is analytic by theorem 9.7.2, and it is nowhere dense in $Z$: its part in $Z^\circ _ j$ corresponds under the isomorphism $\tau _ j$ to the nowhere dense part of $C_{j+1}\cap Z_ j$ in the open subset $\tau _ j^{-1}(Z^\circ _ j)$ of $Z_ j$, and $A_ j$ is nowhere dense. Let $A_{j+1}=A_ j\cup \tau _ j(C_{j+1}\cap Z_ j)$ and $N_{j+1}=N_ j\setminus A_{j+1}$. Then $\tau _ j^{-1}(N_{j+1})$ does not meet $C_{j+1}$, so $\sigma _{j+1}$ is an isomorphism over it and does not change the ideal there: $\mathfrak a_{j+1}=\mathcal O_{X_{j+1}}(E_{j+1})\mathfrak a_ j\mathcal O_{X_{j+1}}=\mathcal I_ Z\mathcal O_{X_{j+1}}$ on $\tau _{j+1}^{-1}(N_{j+1})$. Moreover $Z_{j+1}\cap \tau _{j+1}^{-1}(N_{j+1})=\sigma _{j+1}^{-1}(Z_ j\cap \tau _ j^{-1}(N_{j+1}))=\tau _{j+1}^{-1}(Z^\circ _{j+1})$, because $\sigma _{j+1}(Z_{j+1})=Z_ j$ and $\sigma _{j+1}$ is injective there. This completes the induction.

Since $A_ j$ is nowhere dense in $Z$, the set $Z^\circ _ j$ is nonempty, and so is $\tau _ j^{-1}(Z^\circ _ j)$. If $Z_{l-1}\not\subset C_ l$ held for all $l\leq t$, the induction would reach $j=t$, and $\mathfrak a_ t$ would be the ideal of the nonempty set $\tau _ t^{-1}(Z^\circ _ t)$ on $\tau _ t^{-1}(N_ t)$, whereas $\mathfrak a_ t$ is the unit ideal. So there is a least index $i$ with $Z_{i-1}\subset C_ i$, and the assertion holds for $j=i-1$. On $\tau _{i-1}^{-1}(N_{i-1})$ we have $Z_{i-1}\subset C_ i\subset V(\mathfrak a_{i-1})=\tau _{i-1}^{-1}(Z^\circ _{i-1})\subset Z_{i-1}$, so $C_ i$ coincides there with a copy of the open subset $Z^\circ _{i-1}$ of $Z$, which is pure of codimension $r$. The set $Z_{i-1}$ is connected, being the closure of the connected manifold $(Z_{i-1})_{\mathrm{reg}}$, so it lies in one connected component of the manifold $C_ i$, and this component has codimension $r$.

By (9.6.4) and proposition 9.6.3, near points over this component the Jacobian determinant of $\sigma _ i$ is, in any local coordinates, a unit times the $(r-1)$st power of a local equation of the exceptional divisor $E_ i=\sigma _ i^{-1}(C_ i)$. Since $Z_ i=\rho _ i(E')$ lies over $Z_{i-1}\subset C_ i$, the function $J_{\sigma _ i}\circ \rho _ i$ vanishes on $E'$; it does not vanish identically near any point, because $J_{\sigma _ i}$ vanishes only on $E_ i$, and $\rho _ i^{-1}(E_ i)$ is nowhere dense, $\rho _ i$ being proper and bimeromorphic as the composite of $f'$ with $\sigma _{i+1}\circ \cdots \circ \sigma _ t$ (corollary 9.7.3). Hence it has positive order along $E'$. The Jacobians of the other blowups composed with the corresponding maps $\rho _ j$ are holomorphic, so they have nonnegative order, and by the chain rule $J_\sigma \circ f'=\prod _ j(J_{\sigma _ j}\circ \rho _ j)$. Thus

Applying (9.8.7) to $f\circ h=\sigma \circ f'$ and using Step 2, we obtain

Step 4: conclusion. The pair $(M',E')$ represents the given prime divisor, as we see on the common smooth model $M'$, and $X'$ is a compact manifold. We repeat Step 3 with $X'$, $f'$ and $E'$ in place of $X$, $f$ and $E$. Each repetition in which the center has codimension at least two lowers the nonnegative integer $k$ by at least one, so after at most $k$ repetitions the center is a hypersurface on the current model, which is obtained from $X$ by a finite sequence of blowups with smooth centers, and it represents the given prime divisor. Finally, by lemma 9.8.5(2), principalizing the ideal of this hypersurface by theorem 9.6.7 gives a further sequence of blowups with smooth centers, finite because the model is compact, on which the strict transform of the hypersurface is smooth and represents the same prime divisor. $\square$