Bourbaki

Exercise 14, § 2

Tag 0733, permanent. Cite it as [Bourbaki, Tag 0733].

Algebra, VII, p. 58

Bourbaki marks this one as harder than the rest.

Let M be a reduced τ-primary module. In the notation of Ex. 13, for each ordinal a < τ we can consider (M(π) ∩ M^{(α)}/(M(π) ∩ M^{(α + 1)}) as a vector space over the field F_π = A/Aπ; let c(α) denote the dimension of this vector space, and call c(α) the multiplicity of the ordinal a in the module M (cf. VII, p. 24). a) Show that if M and N are two τ-primary modules each of which is a direct sum of cyclic submodules (which implies that τ ≤ ω, the first infinite ordinal), then M and N are isomorphic if and only if they have the same terminal ordinal τ and the multiplicity of every integer n < τ is the same in M and in N. b) Give an example of two τ-primary modules with no elements of infinite height, which are not isomorphic, but in which each integer n has the same multiplicity (cf. Ex. 10, f)). c) Let M and N be two reduced τ-primary modules, each having a countably infinite system of generators. Show that if M and N have the same terminal ordinal τ, and if each ordinal a < τ has the same multiplicity in M and in N, then M and N are isomorphic

(« Ulm-Zippin Theorem »). (Given two submodules S ⊂ M and T ⊂ N, an isomorphism h from S onto T is said to be indicial if, for all x ∈ S, the height of x in M is equal to the height of h(x)h(x) in N. Define an indicial isomorphism from M onto N by induction, and by alternating the rôles played by M and N, in such a way as to reduce the problem to the following: given two finitely generated submodules S ⊂ M and T ⊂ N, an indicial isomorphism h from S onto T, and an element x∈M∩CSx \in M \cap C S, extend h to an indicial isomorphism from S + Ax onto a submodule of N containing T. Show that this can be reduced to the case where τx∉S\tau x \notin S; using Ex. 5, choose an element w, in (S+Ax)∩CS(S + Ax) \cap C S with greatest possible height aa in M, then choose an element w2w_2 in S∩M(α)S \cap M^{(\alpha)} such that the height of π(w1+w2)\pi(w_1 + w_2) in M is greatest possible; show that w=w1+w2w = w_1 + w_2 is proper with respect to S (Ex. 13, b)). Let z=h(τw)∈Tz = h(\tau w) \in T; show that, if there exists an element t of N, of height aa in N, proper with respect to T, and such that πt=z\pi t = z, then an extension h‾\overline{h} of h of the desired form can be obtained by putting h‾(w)=t\overline{h}(w) = t. It remains to prove the existence of the element t. If

γ(z)=γ(πw)=α+1<τ, \gamma(z) = \gamma(\pi w) = \alpha + 1 < \tau,

show that there exist elements uu in N(α)∩CTN^{(\alpha)} \cap C T such that z=πuz = \pi u, and that t may be taken to be any such element. If on the other hand γ(z)=γ(πw)>α+1\gamma(z) = \gamma(\pi w) > \alpha + 1 or else if γ(z)=γ(πw)=α+1=τ\gamma(z) = \gamma(\pi w) = \alpha + 1 = \tau, use Ex. 13, d) to prove the existence of the element t, noticing that (in the notation of Ex. 13) the image under ψ\psi of S∩M~(α)S \cap \tilde{M}^{(\alpha)} in (M(τ)∩M(α))/(M(τ)∩M(α+1))(M(\tau) \cap M^{(\alpha)})/(M(\tau) \cap M^{(\alpha + 1)}) is finite dimensional over the field FσF_{\sigma}.

*d) Obtain from c) a new proof of Ex. 12, d) (use also Th. 1 of VII, p. 14).

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