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Main properties of exterior powers

4.3 Main properties of exterior powers

Proposition 4.3.1. Let M be an R-module scheme over S. If �

R

n+1M = 0; the result follows immediately by induction.

LetN be an R-module scheme. By the definition, we have an isomorphism HomR(�

The latterR-module is trivial by hypothesis and thus, we have Homr(�

R

n+1M, N)∼= AltR(Mn+1, N)∼= AltR(Mn,HomR(M, N)) = 0

for all R-module schemes N, which implies that �

R

n+1M = 0.

Proposition 4.3.2. Let M1, M2 and P be R-module schemes over S and r1, r2

two positive integers. We have a natural isomorphism λ1,2 : AltR(M1r1 ×M2r2, P)∼= HomR(�

Proof. Using Proposition 2.2.17 we have a natural isomorphism AltR(M1r1 ×M2r2, P)∼= AltR(M1r1,AltR(M2r2, P)) which is again by Proposition 2.2.17 isomorphic to

AltR(�

HomR(�

r2M2 be the multilinear morphism in AltR(M1r1 ×M2r2,�

M2 by the isomorphism given in the Proposition 4.3.2. Then, one can easily see that the R-module scheme �

R following universal property: Given anyR-multilinear morphismϕ :M1r1× M2r2 → P which is alternating in M1r1 and M2r2, there exists a unique

alternating in each Miri such that the homomorphism HomR(�

4.3. MAIN PROPERTIES OF EXTERIOR POWERS 45

r1+r2M). Using the isomorphism of Proposition 4.3.2 with P replaced by �

That is, we have the following commutative diagram:

M1r1 ×M2r2

that together with the isomorphisms of Proposition 4.3.2 and Definition 4.1.1 (iii) gives the following commutative diagram:

AltR(Mr1+r2, P) schemes over S andr1, r2 two positive natural numbers. Then the morphism (cf.

Remark 4.3.3 3)) is an epimorphism. In particular, we have a canonical epimorphism

Proof. The morphism ϕr11 ×ϕr22 : M1r2 ×M2r2 → Mr1+r2 is an epimorphism.

Therefore, for every R-module scheme P, the induced morphism AltR(Mr1+r2, P)→AltR(M1r2 ×M2r2, P) is injective.

Using the diagram of Remark 4.3.3 4), we conclude that the induced morphism HomR(�

R r1+r2

M, P)→HomR(�

R r1

M1R

R r2

M2, P)

is injective for all R-module schemes P. This proves the first part of the state-ment. The second part follows from the first part by replacingϕi with the identity morphism of M.

Notations. LetM, M��be sub-R-module schemes ofM andP andN R-module schemes. By AltR(M�r × M��s ×Pt, N) we mean the module of R-multilinear morphisms that are alternating in Mr, M��s and (M ∩M��)r+s (as a submodule scheme of Mr+s)) and in Pt. When we say that a multilinear morphism

Mr×M��s×Pt→N

is alternating, we mean that it belongs to the module AltR(Mr×M��s×Pt, N).

Likewise, we define the module

AltR(M1r1 ×· · ·×Mnrn×P1s1 ×· · ·×Pmsm, N)

with Mi sub-R-module schemes of M and Pj’s arbitrary R-module schemes.

Lemma 4.3.5. Let π : M � M�� be an epimorphism and let ϕ : M��r → H be an R-multilinear morphism such that the composition ϕ ◦πr : Mr → H is alternating. Then ϕ is alternating as well.

Proof. The morphism π induces a morphism ∆rπ : ∆rM → ∆rM�� between diagonals and since the morphism π is an epimorphism, the morphism ∆rπ is an epimorphism too.

Similarly, we have an induced epimorphism between ∆rijM ⊂Mr and ∆rijM�� ⊂ M��r for all 1 ≤ i < j ≤ r, which we denote by ∆rijπ. In order to show that ϕ is alternating, we must show that for any 1 ≤ i < j ≤ r the composition

rijM���→M��r −−→ϕ N is trivial. But we have a commutative diagram

rijM ∆π ��

� �

ι

��

rijM��

� �

�

��Mr πr ��M��r ϕ ��N.

4.3. MAIN PROPERTIES OF EXTERIOR POWERS 47 and since the compositeϕ◦πr is alternating, the compositionϕ◦πr◦ι is trivial, and so is the composition ϕ ◦ι��◦∆π. The morphism ∆π is an epimorphism, which implies thatϕ◦ι�� is trivial.

Remark 4.3.6. Let M be a sub-R-module scheme of M and π : M � M�� an epimorphism. It can be shown in the same fashion that if the composition of a multilinear morphism M��r×Ms×Mt→N with the epimorphism

πr×IdMs×IdMt :Mr×Ms×Mt→M��r×Ms×Mt

is alternating (in the sense of the Notations above), then this multilinear morph-ism is also alternating.

Lemma 4.3.7. LetM1· · · , Mr be R-module schemes andψ :M1×· · ·×Mr→N an R-multilinear morphism. Assume that for some 1 ≤ i ≤ r we have an exact sequence Mi −−→ι Mi π

−−→ Mi�� → 0. If the restriction ψ|M1×···×Mi×···×Mr is zero, then there is a unique multilinear morphism

ψ :M1×· · ·×Mi��×· · ·×Mr →N

such that ψ =ψ◦(IdM1×· · ·×π×· · ·×IdMr) with π at the ith place.

Proof. By functoriality of the isomorphism in Proposition 2.2.8 we have a com-mutative diagram (we omit the superscriptRto avoid heavy notations, and so the homomorphisms and multilinear morphisms are all R-linear or R-multilinear):

Mult(M1×· · ·×Mr, N) = ��

π

��

Hom(Mi��,Mult(M1×· · ·×Mˇi×· · ·×Mr, N))

π

��Mult(M1×· · ·×Mr, N)

ι

��

= ��Hom(Mi,Mult(M1×· · ·×Mˇi×· · ·×Mr, N))

ι

��Mult(M1×· · ·×Mr, N) = ��Hom(Mi,Mult(M1×· · ·× ˇMi×· · ·×Mr, N)) where the indicated morphisms are the obvious ones and ˇMi means that this factor is omitted. The right column is exact and π is injective, because the sequence 0 → Mi −−→ι Mi π

−−→ Mi�� → 0 is exact and the functor HomR( , P) is left exact for any R-module scheme P. Therefore, the left column is exact too and π is injective. The morphism ψ is an element of MultR(M1 ×· · ·×Mr, N) which goes to zero under the morphism ι (restriction morphism). By exactness, there is a unique multilinear morphismψ ∈MultR(M1×· · ·×Mi��×· · ·×Mr, N) which is mapped to ψ underπ. This proves the lemma.

Lemma 4.3.8. Let M1 ι

−−→M2 π

−−→M3 → 0 be an exact sequence of R-module schemes over S and n a positive natural number. Then the sequence

0→AltR(M3n, N)→AltR(M2n, N)→AltR(M1×M2n1, N) is exact.

Proof. Since π:M2 →M3 is an epimorphism, the morphismπn:M2n →M3n is also an epimorphism and so the induced morphism AltR(M3n, N)→AltR(M2n, N) is injective.

Let ϕ : M2n → N be an alternating morphism and assume that the restriction ϕ|M1×M2n1 is zero. We will show that there is a multilinear morphismϕ :M3n → N such that ϕ =ϕ◦πn. The result will then follow, since by Lemma 4.3.7, ϕ is also alternating, and it is thus inside the R-module AltR(M3n, N). This will prove the exactness at the middle of the sequence (the fact that the composite

AltR(M3n, N)→AltR(M2n, N)→AltR(M1×M2n1, N)

is zero follows directly from the fact that the composition π◦ι is zero) and the proof will be achieved.

Note that sinceϕ is alternating and the restrictionϕ|M1×M2n−1 is zero, the restric-tions ϕ|M2i×M1×M2n−i−1 are zero for any 1 ≤ i≤ n−1. Set ϕ0 = ϕ. We show by induction on 0≤i≤nthat there is a multilinear morphismϕi :M3i×M2ni →N such that ϕ=ϕi◦(πi×IdMni

2 ). This is clear for i= 0, so let i >0 and assume that we haveϕi1 with the stated property. Consider the following commutative diagram

M2i1×M1×M2ni

π

��

� � ρ ��M2i1×M2×M2ni

π

��

ϕ

��

M3i−1×M1×M2n−i� � ρ ��M3i−1×M2×M2n−iϕi1 ��N where π� = πi1 ×IdM1×IdMni

2 , �π = πi1 × IdM2×IdMni

2 and ρ,ρ are the inclusion morphisms. We know that 0 = ϕ ◦ρ = ϕi−1 ◦ �π ◦ρ, which implies that ϕi1 ◦ρ◦π� = 0. The morphism �π is an epimorphism and so ϕi1◦ρ, the restriction of ϕi1 toM3i−1×M1×M2n−i, is zero. We can therefore apply Lemma 4.3.7 and obtain a multilinear morphism ϕi :M3i×M2ni →N such that

ϕii1◦(IdMi−1

3 ×π×IdMn−i 2 ).

We have thus ϕ=ϕi◦(πi×IdMni 2 ).

Now put i = n, the statement says that there is a multilinear morphism ϕn : M3n→N with ϕ =ϕn◦πn. This ϕn is the required ϕ.

For the following lemma, which is in some sense the “linearized” version of the previous one, we use notations introduced in Remark 4.3.3 3).

4.3. MAIN PROPERTIES OF EXTERIOR POWERS 49 Lemma 4.3.9. Let M1 α

−−→ M2 β

−−→M3 →0 be an exact sequence of R-module schemes and n a positive natural number. Then the sequence

M1R

Proof. The statement follows essentially from Lemma 4.3.8 and Proposition 4.3.2. Indeed, the exactness of this sequence is equivalent to the exactness of the sequence obtained by applying the contravariant functor HomR(−, N) on it for allR-module schemesN overS. So, let N be an arbitraryR-module scheme over S and consider the following diagram:

0 ��Hom(� where the first two vertical morphisms, i.e., λ23 are given by the universal property of the exterior powers and the last one by Proposition 4.3.2. The com-mutativity of the first square follows from Remark 4.1.5 3) and that of the second square from Remark 4.3.3 4). The second row is exact by Lemma 4.3.8. The ex-actness of the first row follows immediately and this achieves the proof.

Theorem 4.3.10. Assume that 0 → M1 ι

−−→M2 π

−−→M3 → 0 is a short exact sequence ofR-module schemes overS. Letn2 be a non-negative integer and write n2 =n1+n3 for non-negative integers n1 and n3. Consider the diagram

AltR(M2n2, N) ρ ��AltR(M1n1 ×M2n3, N)

AltR(M1n1��×M3n3, N)

π

��

where ρ is the restriction morphism.

(a) If �

(c) If both conditions hold, then there is a natural epimorphism ζ :�

(d) If furthermore the sequence is split, then the epimorphismζ is an isomorph-ism.

Proof. If n2 = 0 then n1 = 0 = n3 and all statements are trivially true, so assume n2 >0. We prove each point of the proposition separately.

(a) Fixn2. We show by induction on n1, with 0 ≤n1 ≤n2, that the restriction morphism gives an injective morphism

AltR(M2n2, N)�→AltR(M2n3,MultR(M1n1, N)).

Ifn1 = 0 then n3 =n2, andρ is the identity morphism, so there is nothing to show. So assume that 0 < n1 ≤ n2 and that the statement is true for n1 −1 and n3 + 1 in place of n1 and n3. Then �

R n3+1

M3 = 0 implies

R n3+2

M3 = 0 by proposition 4.3.1; so by the induction hypothesis we have an injection

AltR(M2n2, N)�→AltR(M2n3+1,MultR(M1n11, N)).

Since by hypothesis we have �

R n3+1

M3 = 0 we can use Lemma 4.3.8 (note that �

R n3+1

M3 = 0 implies that AltR(M3n3+1, P) = 0 for every R-module scheme P), and we have thus an injection

AltR(M2n3+1,MultR(M1n11, N))�→AltR(M2n3 ×M1,MultR(M1n11, N)).

The latter R-module is inside theR-module

AltR(M2n3,HomR(M1,MultR(M1n11, N))).

By proposition 2.2.10, HomR(M1,MultR(M1n11, N)) ∼= MultR(M1n1, N).

Putting these together, we conclude that there is an injection θ : AltR(M2n2, N)�→AltR(M2n3,MultR(M1n1, N)).

Following through the above isomorphisms and inclusions, one verifies that this injection is induced by the restriction morphism. Under the isomorph-ism

MultR(M2n3,MultR(M1n1, N))∼= MultR(M2n3 ×M1n1, N)

given by Proposition 2.2.8, the image of AltR(M2n2, N) by θ lies inside the submodule AltR(M2n3×M1n1, N) of MultR(M2n3×M1n1, N) and we can easily see that the injection

AltR(M2n2, N)�→AltR(M2n3 ×M1n1, N) thus obtained is given by the restriction morphism.

4.3. MAIN PROPERTIES OF EXTERIOR POWERS 51

The latter R-module is isomorphic to HomR(�

R n1+1

M1,MultR(M2n31, N)), which is zero by assumption. So, the restriction ϕ0|Mn1

1 ×M2j×M1×M2n3j1 is show, so let i < n3 and assume that we have constructed ϕi with the de-sired property and we construct ϕi+1. Consider the following commutative diagram: order to find a multilinear morphism

ϕi+1 :M3i+1×M2n3−i−1 ×M1n1 →N

Put i =n3, then the statement says that there is a multilinear morphism ϕn3 : M3n3 ×M1n1 → H such that ϕ0 = ϕn3 ◦(πn3 ×IdMn1

1 ). Since ϕ0 is alternating, by Remark 4.3.6, ϕn3 is also alternating.

(c) If both conditions hold, then by (a), ρ is injective and therefore the

which is natural, in other words we have a natural injection of functors τ : HomR(�

It is a known fact that any natural transformation between such functors is induced by a unique morphism

ζ :�

in fact, this morphism is the image of the identity morphism of �

R n2M2

under this transformation. This means that for any R-module scheme N, τN : HomR(� of τ implies thatζ is an epimorphism.

(d) Let s:M3 →M2 be a section of π, i.e., π◦s= IdM3 and r :M2 →M1 the corresponding retraction of ι, that is,r◦ι = Id and that the short sequence

0→M3 s

−−→ M2 r

−−→M1 →0 is exact. Then we show that the morphism

µ: AltR(M2n2, N)→AltR(M1n1×M3n3, N)

whose composition with π isρ (given by (b)) is induced by the inclusion j :=ιn1 ×sn3 :M1n1 ×M3n3 �→M2n2.

Indeed, given a morphism f ∈AltR(M2n2, N), we have ρ(f) =π(µ(f)), or in other words, µ(f)◦(IdnM11×πn3) =f◦(ιn1 ×IdnM32). Hence the following

4.3. MAIN PROPERTIES OF EXTERIOR POWERS 53 induced by j as we claimed.

Now define a morphismω:M2n2 →M1n1×M3n3 as follows: for anyk-algebra

and it is straightforward to see that in fact the image lies inside the sub-module AltR(M2n2, N). We also denote by ω the homomorphism

AltR(M1n1 ×M3n3, H)→AltR(M2n2, N)

obtained by restricting the codomain of ω. Since the composites r◦s and π◦ι are trivial andr◦ι and π◦s are the identity morphisms, we see that the composition ω◦j is the identity morphism of M1n1 ×M3n3. Therefore the composite µ◦ω is the identity homomorphism. Consequently, the homomorphism µ: AltR(M2n2, N)→AltR(M1n1 ×M3n3, N) is an epimorph-ism. We know from (c) that it is a monomorphism, and hence it is an isomorphism. We obtain thus

As we know, this homomorphism is induced by the morphism

Since it is an isomorphism, the morphism ζ must be an isomorphism as well.

Lemma 4.3.11. Let M be an R-module scheme over S and r ∈ R. Then for every positive natural number n we have a commutative diagram

M⊗R

Proof. Remark 4.1.6 3) gives us the following commutative diagram:

M ×· · ·×M (r.)×IdM×···×IdM ��

Using Remark 4.3.3 1), we can split this diagram into two squares:

M ×· · ·×M

The commutativity of the bottom square follows from the commutativity of the top diagram, diagram 3.12 and the universal property of the morphism

λ1,2 :M×· · ·×M →M ⊗R

R n−1M

4.3. MAIN PROPERTIES OF EXTERIOR POWERS 55 given in Remark 4.3.3 1); more precisely, we have:

(Id∧�n1

Id)◦((r.)⊗RId)◦λ1,2 = (Id∧�n1

Id)◦λ1,2◦(r.)×Idn1 = (r.)◦(Id∧�n1

Id)◦λ1,2

and the universal property of λ1,2 implies that (r.)◦(Id∧�n1

Id) = (Id∧�n1

Id)◦((r.)⊗RId).

The fact that the vertical arrows in the diagram of the lemma are epimorphisms has been proved in Lemma 4.3.4.

Proposition 4.3.13. Let M be anR-module scheme over S, and Q the cokernel of multiplication by an element r∈R, i.e., we have an exact sequence

M −−→r. M −−→p Q → 0. Then, for any positive natural number n the following

Proof. It follows from Remarks 4.1.6 and 4.3.3 that the composition (Id∧�n−1

Id)◦((r.)⊗RId) as in previous lemma is equal to (r.)∧�n−1

Id. So, the diagram of the previous lemma can be rewritten as:

M⊗R� to obtain the following exact sequence:

M ⊗R

Using diagram 3.14, we can factorize the first morphism of the sequence, so that the following diagram is commutative with exact row

M ⊗R

Since the morphism Id∧�n1

nM is an epimorphism, we conclude that the sequence

is exact as well, and the proof is achieved.