Finite Morphisms and Finite Coherence Theorem

Exercise 1.4.6.Let (X, x) be a reduced complex space germ, and let (A, x)⊂(X, x) be a closed subgerm such that dim(A, x) = dim(X, x). Prove that (A, x) contains an irreducible component of (X, x).

Exercise 1.4.7.Let a1, . . . ,am∈Z^r be integer vectors. Define a complex space germ (X,0)⊂(C^m,0) by the ideal I⊂C{x}=C{x1, . . . , xm}, gener- ated by the binomialsx^k−xfor allk= (k1, . . . , km),= (1, . . . , m)∈N^m satisfyingm

i=1kiai=m i=1iai.

Prove that (X,0) is non-singular if there are 1≤i1<· · ·< in≤m such thatai1, . . . ,ain are linearly independent overZand each other vectoraj ad- mits a (unique) representationaj=u1ai1+. . .+unain withu1, . . . , un∈N. Show further that such a non-singular germ has dimensionn.

is an open neighbourhood ofyinY. As (1) and (2) are obvious, it only remains to show that the restrictionsfUi,V :Ui→V are closed maps.

For this, let A⊂U_i be closed. Then, by (1) and (2), A is also closed in f⁻¹(V), that is, there is a closed subsetA ⊂X such that A=A ∩f⁻¹(V).

Sincef is closed, the imagef(A)⊂Y is closed. Therefore,f(A) =f(A)∩V

is closed inV.

Definition 1.55.A morphism f :X →Y of complex spaces is called finite (atx∈X)if the underlying mapf :X →Y of topological spaces is finite (at x). A morphism of germs f : (X, x)→(Y, y) is called finite if it has a finite representativef :U →V (or, equivalently, if each representative off is finite atx).

Proposition 1.56.Let f :X→Y be a finite morphism of complex spaces, y∈Y andf⁻¹(y) ={x1, . . . , xs}. Further, let V ⊂Y andU1, . . . , Us⊂X be open subspaces satisfying the conditions of Lemma 1.54, and letF be anOX- module. Then there are isomorphisms

(1)f_∗F |V ∼=

%s i=1

(fU_i,V)_∗(F |U_i)of OV-modules , (2)(f_∗F)_y∼=

%s i=1

Fxi of OY,y-modules .

Proof. LetW ⊂V be an open subspace. Then

f⁻¹(W)⊂f⁻¹(V) =U₁∪. . .∪U_s.

Since theUiare pairwise disjoint, we get isomorphisms ofΓ(W,OY)-modules Γ(W, f_∗F) =Γ

f⁻¹(W),F ∼=

%s i=1

Γ

f⁻¹(W)∩Ui,F

∼=

%s i=1

Γ f_U⁻¹

i,V(W),F |Ui

=

%s i=1

Γ

W,(f_Ui,V)_∗F |Ui

(using that f_∗F is an OY-module via f:OY →f_∗OX). Since the isomor- phisms are compatible with the restriction maps to open subsets, we obtain

(1) and (2).

The proposition has the following important corollary:

Corollary 1.57.If f :X →Y is a finite morphism of complex spaces, then the direct image functorf_∗ is an exact functor.

Proof. Let 0→ F → F → F →0 be an exact sequence of OX-modules, y∈Y and f⁻¹(y) ={x1, . . . , xs}. Then, by Proposition 1.56, we obtain a commutative diagram ofOY,y-modules

0

%s i=1

Fx_i

∼=

%s i=1

Fxi

∼=

%s i=1

Fx _i

∼=

0

0 (f_∗F)y (f_∗F)y (f_∗F )y 0.

Since the upper sequence is exact, the lower is exact, too.

Next, we consider special finite maps, the so-called Weierstraß maps, and prove the finite coherence theorem for these maps. Later in this section, we will reduce the general case to this special case.

Definition 1.58.LetB ⊂Cⁿ be an open subset, and let

f(y, z) =z^b+a1(y)z^b−1+. . .+ab(y)∈Γ(B,OCⁿ)[z].

Set A:=V(f)⊂B×Cand OA:=OB×C/f. We refer to the mapA−→^π B induced by the projection B×C→B as aWeierstraß mapof degreeb.

Note that, for eachy∈B, the fibre π⁻¹(y) = y

× z∈Cf(y, z) = 0 is finite. Indeed, for y fixed, f(y, z) is a polynomial in z of degree b and has, therefore, at mostb roots (see Figure 1.3).

y B

z

Fig. 1.3. The (local) zero-set of a Weierstraß polynomial.

Lemma 1.59.Each Weierstraß mapπ:A→B is a finite holomorphic map.

Proof. It suffices to show that πis a closed map. LetM ⊂Abe closed, lety be a point in the closure of the image π(M) inB ⊂Cⁿ, and let (y_ν)ν∈Nbe a sequence in π(M) which converges toy. For each ν∈N choose zν∈C such that (y_ν, zν)∈M ⊂A, that is, such that

−z^b_ν=a1(y_ν)z_ν^b−1+. . .+ab(y_ν). Either|z_ν|<1 or, by the above expression,

|zν| = |zν|^b

|z_ν|^b−1 ≤ b i=1

|ai(y_ν)|.

Since (y_ν)ν∈N is convergent, and since the ai are continuous functions, the sequences

|ai(y_ν)|

ν∈N, i= 1, . . . , b, are bounded. Thus, (|zν|)ν∈N is also bounded. It follows that there is a subsequence (z_νk)_k∈Nwhich converges to somez∈C. SinceM is closed, the limit (y, z) of the sequence (y_ν

k, z_νk)_k∈N is a point ofM. Asπis continuous,π(y, z) = limk→∞π(y_ν

k, zν_k) =y. Hence, y∈π(M), and we conclude thatπ(M) is closed.

Remark 1.59.1.Let (y_ν)_ν∈N⊂B be a sequence converging toyand consider the sequence of polynomials

f(y_ν, z)

ν∈N⊂C[z]. If we choosez_ν to be any root off(y_ν, z),ν ∈N, then the proof of Lemma 1.59 shows that there exists a subsequence of (zν)ν∈N converging to a root of f(y, z)∈C[z]. This fact is sometimes referred to as thecontinuity of the roots of a Weierstraß polynomial (see also Exercise 1.5.5).

Theorem 1.60 (General Weierstraß division theorem). Let f(y, z) = z^b+a1(y)z^b−1+. . .+ab(y)∈Γ(B,O_Cⁿ)[z]

be a Weierstraß polynomial, withB⊂Cⁿ open, and letπ:A=V(f)→B be the corresponding Weierstraß map. Fixy∈B, and letπ⁻¹(y) ={x1, . . . ,xs}. Then, for each gi∈ OB×C,x_i,i= 1, . . . , s, there exist unique r∈ OB,y[z]

andhi∈ OB×C,xi such that g1 =h1f+r ,

...

gs=hsf+r ,

deg_z(r)≤b−1.

The theorem says that we cansimultaneously divide the germsgi∈ OB×C,x_i

by the germ defined byf inOB×C,x_i, with acommon remainderr∈ OB,y[z].

Proof. Without loss of generality, we can assume y=0.

The cases= 1 is just the usual Weierstraß division Theorem 1.8. Lets≥2, and letxi= (0, zi),i= 1, . . . , s. Then

f(0, z) = (z−z1)^b1·. . .·(z−zs)^bs, bi>0, s i=1

bi=b . By Hensel’s lemma, there are monic polynomials fi∈ OB,0[z] of degree bi, i= 1, . . . , s, such thatf =f1·. . .·fs andfi(0, z) = (z−zi)^bi. In particular, eachfi is (z−zi)-general of orderbi.

We setei:=f1·. . .·fi−1·fi+1·. . .·fs∈ OB,0[z], which has degreeb−bi

in z, and which is a unit in the local ring OB×C,xi. Applying the Weier- straß division theorem toe⁻_i¹g_i∈ OB×C,xi gives the existence ofr_i∈ OB,0[z], h_i∈ OB×C,xi such thate⁻_i ¹g_i=h_if_i+r_i with deg_z(r_i)< b_i.

Defining r:=s

i=1e_ir_i∈ OB,0[z], we get deg_z(r)< b. Moreover, as f_j, j =i, is a unit inOB×C,xi, we obtain in this ring

gi = h_ieifi+r−

j=i

rj

f_jfjej =

h_i−

j=i

rj

f_j

& '( )

=:h_i

f +r ,

and the existence part is proven. The uniqueness is left as Exercise 1.5.6.

Now, we are well-prepared to prove an isomorphism of sheaves which will be the basis for the proof of Oka’s coherence theorem.

Theorem 1.61 (Weierstraß isomorphism). Let π:A→B be a Weier- straß map of degreeb. Thenπ_∗OA is a locally freeOB-module of rankb. More precisely, the map π⁰:O^b_B→π_∗OA defined by

Γ(U,O^bB) =Γ U,OB

b

−→Γ

π⁻¹(U),OA

(r1, . . . , rb) −→

r1y^b−1+. . .+rbmodf is an isomorphism ofOB-modules.

Proof. Sinceπ⁰is anOB-linear morphism of sheaves, we have to show that for eachy∈B the morphism of stalksπ_y⁰ :O^bB,y→(π_∗OA)y is an isomorphism.

Ifπ⁻¹(y) ={x1, . . . ,xs}, Proposition 1.56 gives an isomorphism ofOB,y- modules (π_∗OA)y ∼=OA,x₁⊕. . .⊕ OA,x_s. Given gi ∈ OA,x_i=OB×C,x_i/f, i= 1, . . . , s, we deduce from the general Weierstraß division theorem that there is a uniquely determined polynomialr∈ OB,y[z] of degree at mostb−1 such thatgi= (rximodf) for eachi= 1, . . . , s.

Writingr=r₁z^b−1+. . .+r_b, we conclude that (g₁, . . . , g_s) has the unique preimage (r₁, . . . , r_b)∈ O^bB,y underπ⁰_y. Lemma 1.62.Let π:A→B be a Weierstraß map, and let F be an OA- module such that π_∗F is a finite (resp. coherent) OB-module. Then F is a finite (resp. coherent) OA-module.

Proof. Step 1. Since π_∗F is a finite OB-module, B can be covered by open sets V ⊂B such that, locally onV, the direct image sheaff_∗F is generated byg1, . . . , gk∈Γ(V, π_∗F) =Γ(π⁻¹(V),F). We claim thatg1, . . . , gk generate alsoF|π⁻¹(V)asOπ⁻¹(V)-module (which yields, in particular, thatFis locally a finite OA-module). Indeed, for each y∈V, the stalk (π_∗F)y is generated by the germs of g1, . . . , gk as OB,y-module. Thus, Proposition 1.56 (2) yields

that for each point x∈π⁻¹(y) the stalk Fx is generated by g1, . . . , gk as OB,y-module, hence also asOA,x-module.

Step 2. Letπ_∗F be a coherentOB-module. We have to show that, for each open subset U ⊂A and for each surjection ϕ:O^q_U → F|U, the kernel is an OU-module of finite type. Letx∈U andy=π(x). Then Lemma 1.54 yields an open neighbourhoodV ⊂B ofysuch thatπ⁻¹(V) is the disjoint union of U1, . . . , Us, where eachUicontains precisely one point of the fibreπ⁻¹(y). We may assume thatx∈U1=U and extendϕto a mapϕ:O_π^q−1(V)→ F|π⁻¹(V)

by settingϕ_Ui = 0 for alli= 2, . . . , s. Since the direct image functorπ_∗ for the restrictionπ=π|π⁻¹(V)is exact (Lemma 1.57), we get an exact sequence ofOV-modules 0→π_∗Ker(ϕ) →(π_∗OA|V)^q →π_∗F|V →0.

The Weierstraß isomorphism Theorem 1.61 yields that π_∗OA|V is a free OV-module. Thus, the coherence of π_∗F implies that π_∗Ker(ϕ) is an OV- module of finite type. Sinceπis a Weierstraß map, Step 1 applies, showing that Ker(ϕ) is an Oπ⁻¹(V)-sheaf of finite type. In particular, Ker(ϕ) |U =Ker(ϕ)

is anOU-module of finite type.

Based on the results for Weierstraß maps obtained so far, we can give a proof of Oka’s coherence theorem [Oka]:

Theorem 1.63 (Coherence of the structure sheaf ). The structure sheaf OX of a complex spaceX is coherent.

Proof. Coherence being a local property, we may suppose thatX is a complex model space defined by an ideal sheafJ ⊂ OD of finite type,D⊂Cⁿ open (see also A.7, Fact 6). By A.7, Facts 2 and 6, OD/J is coherent if OD is coherent. Therefore, we can assume (X,OX) = (D,OD).

We use induction onn, the casen= 0 being trivial. To show the coherence ofOD, we have to show that for each morphism

ϕ:O^kD−→ OD, (a₁, . . . , a_k) −→a₁f₁+. . .+a_kf_k,

Ker(ϕ) is of finite type. SinceO^kDis of finite type, we may assumeϕ= 0 and, without loss of generality,f :=f1= 0.

We claim that the sheaf OD/fOD is a coherent sheaf of rings at any point x∈D. If f(x)= 0 then f has no zero in some neighbourhood of x andOD/fOD is locally the zero sheaf, hence coherent. Thus, we may assume f(x) = 0. We may also assume that x=0and thatf is xn-general of order b. By the Weierstraß preparation theorem, there exists a Weierstraß polyno- mialg0∈ O_Cⁿ⁻¹,0[xn] such thatg0OD=f0OD. We choose a neighbourhood B⊂Cⁿ⁻¹of0such that the germg0has a representativeg∈Γ(B,O_Cⁿ⁻¹)[xn] withgOU =fOU for a sufficiently small neighbourhoodU ⊂Dof0. We con- sider the Weierstraß map

π: A= (y, z)∈B×Cg(y, z) = 0

−→B .

The Weierstraß isomorphism Theorem 1.61 yieldsπ_∗OA∼=OB^b withOBbeing coherent by the induction hypothesis. Hence, π_∗OA is OB-coherent. Now, Lemma 1.62 applies, showing that OA= (OB×C/gOB×C)|A is OA-coherent.

Then the trivial extension to B×C, i_∗OA is also a coherent sheaf of rings.

Since the sheafOD/fODlocally coincides withi_∗OAnear0, we get the claim.

During the following construction, letO=OD|U for a sufficiently small neighbourhoodU ⊂Dof0which we allow to shrink. We consider the following commutative diagram of sheaf morphisms with exact bottom row

O^q⊕ O^k

φ

O^{k ϕ}

f

O

f

O^q _ψ O^k

π

ϕ O

π

(O/fO)^q

ψ (O/fO)^k

ϕ O/fO.

Sincef0= 0, the multiplication mapf is injective.πis the canonical projec- tion, ϕis theO/fO-linear map induced byϕ, andψexists since O/fO is a coherent O/fO-module.ψ is an O-linear lift of ψ, which exists since O^q is free and generated byΓ(U,O)^q. We define

φ: O^q⊕ O^k−→ O^k, (a,b) −→ψ(a) +fb.

By diagram chasing, we see that φ surjects onto K:=Ker(π◦ϕ)⊂ O^k. In particular, Kis finitely generated.

Sincef is injective, for eacha∈ K, there is a unique h(a)∈ Osuch that f ◦h(a) =ϕ(a). This obviously defines a splittingϕ|_K=f◦hofϕ|_Kthrough anO-linear maph:K → Owithh|_Ker(ϕ)= 0. Define

χ:K −→ O^k, a −→a−

h(a),0

Then ϕ◦χ(a) = 0, that is, χ(K)⊂ Ker(ϕ). Since Ker(ϕ)⊂ K and since χ|_Ker(ϕ)= id_Ker(ϕ), we get thatχ surjects ontoKer(ϕ). Therefore,χ◦φde- fines a surjection O^q⊕ O^kKer(ϕ), proving that Ker(ϕ) is of finite type.

Corollary 1.64.LetX be a complex space.

(1) If Y is a complex subspace ofX, given by the ideal sheafJY ⊂ OX, then JY andOY = (OX/JY)|Y are coherent.

(2) A closed subsetA⊂X is analytic iff there exists a coherent sheaf F such thatA= supp(F).

Proof. A subsheaf of a coherent sheaf is coherent iff it is of finite type. Hence, (1) follows from Oka’s Theorem 1.63. For (2), note that ifA⊂Xis an analytic

set, thenA= supp(OX/J) for some ideal sheaf J ⊂ OX of finite type. By Oka’s theorem,OX and J are coherent, thus OX/J is coherent, too (A.7, Fact 2). Conversely, if A= supp(F) then A= supp

OX/Ann(F)

, and F being coherent implies thatAnn(F) is coherent, too (A.7, Fact 5).

Corollary 1.65.Letπ:A→B be a Weierstraß map. IfFis a coherentOA- module thenπ_∗F is a coherentOB-module.

Proof. Lety∈B,π⁻¹(y) ={x1, . . . ,xs}. By Lemma 1.54, there are an open neighbourhoodV ⊂Bofyand pairwise disjoint open neighbourhoodsUi⊂A ofxi,i= 1, . . . , s, such thatπ⁻¹(V) =U1∪. . .∪Usand such that the restric- tionsπU_i,V :Ui→V are finite maps. SinceFis coherent, we may assume that, for eachi= 1, . . . , s, there is an exact sequence

O^qUⁱ_i−→ O^kUⁱ_i−→ F|U_i −→0 (1.5.1) (shrinkingV if necessary). By adding direct summands, we may assume that q_i=q,k_i=kfor eachi.

We set U =π⁻¹(V)⊂A. As U is the disjoint union of U₁, . . . , U_s, the exact sequences (1.5.1) yield an exact sequence O^q_U → O^k_U → F |U →0. As π:U →V is finite, the direct image functor π_∗ is exact (Corollary 1.57).

Thus, the induced sequence

(π_∗O^q_A)|V −→(π_∗OA^k)|V −→π_∗F|V −→0,

is exact, too. Applying the Weierstraß isomorphism Theorem 1.61, we get an exact sequence ofOV-modules O^qbV → O^kbV →π_∗F |V →0, where b is the degree ofπ. Finally, asOBis coherent, the existence of such an exact sequence (for eachy∈B) implies thatπ_∗F is a coherentOB-module (see A.7).

The following theorem appears to be the main result about finite holomorphic maps. It has numerous applications, in particular in singularity theory. Its main advantage is that the assumption is purely topologically and very easy to verify.

Theorem 1.66 (Local finiteness theorem). Let f :X→Y be a mor- phism of complex spaces, lety∈Y, and letxbe an isolated point of the fibre f⁻¹(y). Then there exist open neighbourhoods U ⊂X of x and V ⊂Y of y such thatf(U)⊂V and

(1)fU,V :U →V is finite.

(2) For each coherent OU-module F the direct image(fU,V)_∗F is a coherent OV-module.

Proof. All statements being local, it suffices to consider the case that X and Y are complex model spaces. Further, it suffices to consider the case that f is a projection: consider the graph off,

X ^∼=_ϕ

f

Γ(f) X×Y

pr

Y .

Then an OX-module F is coherent iff the direct image ϕ_∗F is a coherent OΓ(f)-module.

Thus, altogether, assume that there are (sufficiently small) open subsets B ⊂Cⁿ andD⊂C^k such thatX is a closed subspace ofB×D, given by the coherent ideal sheafI ⊂ OB×D, andf is the projection

f = pr_X,Y :B×D⊃X →Y ⊂B .

Further, assume that y=0∈B and that x= (0,0). Since x is an isolated point of the fibre f⁻¹(0) =X∩({0}×D), after shrinking D⊂C^k, we have X∩({0}×D) ={(0,0)}.

Now, we prove the theorem by induction onk, starting with k= 1. Since X∩({0}×D) ={(0,0)}, there exists a germg∈ I(0,0) such thatg(0,0) = 0 and z →g(0, z) is not the zero map on D. By the Weierstraß preparation theorem, there exists a Weierstraß polynomial g and a unit u∈ OD×B,(0,0)

such that

ug = g = z^b+a1z^b−1+. . .+ab∈ O_Cⁿ,0[z], ai(0) = 0,

i= 1, . . . b. After shrinking B and D, we may assume that ai∈Γ(B,OCⁿ), and we may consider the Weierstraß map defined by the projection on B,

A= (y, z)∈B×Dg(y, z) = 0 p

−→B⊂Cⁿ.

Due to Lemma 1.59, p is finite. Sincei:X →A is a closed embedding, the restrictionpX,Y =f :X→Y is finite, too. Moreover, ifF is a coherentOX- module, then the trivial extension i_∗F of F to A is a coherent OA-module (A.7, Fact 6). Thus, Corollary 1.65 yields that p_∗i_∗F ∼=f_∗F is a coherent OB-module.

For the induction step, let k >1 and assume that D=D ×D, where D ⊂C^k−1 and D ⊂C are open neighbourhoods of the origin. Then f = pr_X,Y is induced by the composition of two projections:

B×D=B×(D ×D) ^p

−→ B×D ^p

−→ B .

As the statement holds fork= 1, we may assume that (after shrinkingB,D and D) the restrictionp|X :X →B×D is finite and that for each coher- entOX-moduleF the direct image (p|X)_∗Fis a coherentOB×D-module. In particular, by Lemma 1.44, the image X1:=p(X) is a closed complex sub- space ofB×D , endowed with one of the structure sheaves of Definition 1.45.

Note that, in each case, the restrictionπ =p_X,X1 :X →X1is finite andπ_∗F is a coherentOX1-module.

SinceX∩({0}×D) ={(0,0)}, we also haveX1∩({0}×D ) ={(0,0)}. Thus, the induction hypothesis implies that (after shrinkingB and D ) we may assume that the restrictionπ =p _X

1,Y is finite and that the direct image π_∗ G of a coherent OX1-moduleG is a coherentOY-sheaf. Together with the above, we get thatf =π ◦π is finite and that the direct imagef_∗F ∼=π _∗π_∗F of a coherentOX-moduleF is a coherentOY-module.

Taking into account the considerations on finite maps at the beginning of this section, the local finiteness theorem implies the finite coherence theorem which succinctly says that for a finite morphismf the direct image functor f_∗ preserves coherence:

Theorem 1.67 (Finite coherence theorem, FCT). Let f :X →Y be a finite morphism of complex spaces, and letF be a coherent OX-module. Then f_∗F is a coherent OY-module.

Proof. Lety∈Y andf⁻¹(y) ={x1, . . . , xs}. By Lemma 1.54 and Proposition 1.56, there are open neighbourhoodsV ⊂Y ofyandUi ⊂Xofxi,i= 1, . . . , s, such that the restrictionsfU_i,V :Ui→V are finite and

f_∗F|V ∼=

%s i=1

(fU_i,V)_∗(F|U_i).

The local finiteness theorem implies that (after shrinkingV andUi) we may assume that (fU_i,V)_∗(F|U_i) is a coherent OV-module. Since direct sums of coherent sheaves are coherent (A.7, Fact 2) and since coherence is a local property, we deduce thatf_∗F is coherent, as claimed in the theorem.

Together with Corollary 1.64 (2), the finite coherence theorem shows that the image of a finite morphism of complex spaces is analytically closed. More precisely, we obtain:

Corollary 1.68 (Finite mapping theorem). Let f :X→Y be a finite morphism of complex spaces andZ⊂X a closed complex subspace ofX. Then f(Z)⊂Y is an analytic subset of Y (which can be endowed with one of the structure sheaves of Definition 1.45).

The finite coherence theorem and the local finiteness theorem are due to Grauert and Remmert (cf. [GrR2]). We emphasize again that, in both theo- rems, the assumptions are of a purely topological nature (thus, independent of the structure sheaf).

Remarks and Exercises

(A) Proper Maps and the Proper Coherence Theorem. Recall that a contin- uous map is calledproper if the preimage of any compact set is compact.

Much deeper than the finite coherence theorem (and much more difficult to prove) is the coherence theorem for proper maps due to Grauert [Gra], which says that for a proper morphism of complex spaces the direct image functor preserves coherence. A proof can be found in [FoK], respectively in [GrR2, Ch. 10]. As for finite morphisms, one can deduce as a corollary that the image of an analytic set under a proper morphism of complex spaces is analytically closed. This statement is also referred to as the proper mapping theorem. It was proved first by Remmert [Rem].

LetX, Y be complex spaces withX being compact. Then the projection X×Y →Y is a proper morphism. In particular, the projectionPⁿ×Y →Y is proper.

Exercise 1.5.1.Show that finite maps between complex spaces are proper.

Exercise 1.5.2.LetX be a complex space which is compact and connected, and letf :X →Cbe a holomorphic map. Prove thatf is constant.

Hint. Apply the proper mapping theorem.

(B) Computing the Image by Elimination. Let X =V(g₁, . . . , g_k)⊂P^m(C) be defined by homogeneous polynomials g1, . . . , gk∈C[x] =C[x0, . . . , xn], and let f0, . . . , fn∈C[x] be homogeneous polynomials of the same de- gree d with V(f0, . . . , fn)∩X=∅. Then we get a (proper) morphism f :X→Pⁿ(C), x →(f0(x) :. . .:fn(x)). The annihilator structure on the image of f can be computed effectively in a computer algebra system like Singular by eliminatingxfrom the ideal

J =y0−f0, . . . , yn−fn, g1, . . . , gk_C[x,y]

that is, by computing the elimination ideal J∩C[y] (see [GrP, App. A.7]

for a much broader discussion of the geometric meaning of elimination). For instance, the followingSingularsession computes the image of the morphism f :P¹→P², (x0:x1) →(x³₀:x²₁x0:x³₁):

ring r=0,(x(0),x(1),y(0),y(1),y(2)),dp;

poly f(0),f(1),f(2) = x(0)^3,x(1)^2*x(0),x(1)^3;

ideal J=y(0)-f(0),y(1)-f(1),y(2)-f(2);

eliminate(J,x(0)*x(1));

//-> _[1]=y(1)^3-y(0)*y(2)^2 Hence,f(P¹) =V(y₁³−y₀y₂²)⊂P².

For details on how to compute elimination ideals usingSingular(and on the implemented algorithms), we refer to [GrP, Sect. 1.8.2], resp. [DeL, Sects.

3.6.2 and 9.5].

Exercise 1.5.3.Let X =T =C, and Y =C², each equipped with the re- duced structure. Moreover, letf :X →Y be given byt →(t², t³). Show that f is a finite morphism and that the Fitting, annihilator, and reduced structure of the imagef(X) coincide (see Exercise 1.6.4 for a more general statement).

Moreover, show that the annihilator structure and the reduced structure are not compatible with the base changeg:T →Y, x →(x,0) (see Remark 1.45.1 (2)).

Exercise 1.5.4.Let f :X →Y be a finite morphism, let A⊂X be an an- alytic set, and let y∈Y. Moreover, let π⁻¹(y) ={x1, . . . , xs} and assume that the germ (A, xi) decomposes intoriirreducible components,i= 1, . . . , s.

Prove that the germ of the imagef(A) aty decomposes into at mosts i=1ri

irreducible components.

Exercise 1.5.5.Let π:A→B be a Weierstraß map and (y, z)∈A. Prove the following statements:

(1)πis anopen map, that is, it maps open sets inA to open sets inB.

(2) To every sequence (y_ν)ν∈N⊂B converging to y there exists a sequence (zν)ν∈N⊂Csuch that (y_ν, zν)∈A and (zν)ν∈N converges toz.

Hint for (1). Use Hensel’s lemma to reduce the statement to the case that π⁻¹(y) ={(y, z)}.

Exercise 1.5.6.Prove the uniqueness statement in the general Weierstraß division Theorem 1.60.

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