By Jürgen Müller

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N}, implying n − rk(M k ) = ki=1 λi . Hence Nλ is uniquely determined by the rank sequence [n − ki=1 λi ∈ N0 ; k ∈ {0, . . , n}]; note that we have rk(M 0 ) = n and rk(M n ) = 0 anyway. Moreover, for µ n and N ∈ Nµ we have µ λ if and only if λ µ , which holds if and only if rk(M k ) ≥ rk(N k ), for all k ∈ {0, . . , n}. Thus we have k N ∈ N λ := µ λ Nµ ⊆ N if and only if rk(N k ) ≤ n − i=1 λi , that is II Summation N k ∈ Rn− k λ , for all k ∈ {0, . . , n}. This implies that N i=1 i subset, containing Nλ , thus we have N λ ⊆ N λ .

R , . . , µt−1 , µt + 1, µt+1 , . . , µs−1 , µs − 1, µs+1 , . . , µn ] ν = λ, a contradiction. Let conversely λ be as asserted, and let ν = [ν1 , . . , νn ] n such that µ ν λ. Hence for i ∈ {r, . . , s} we have νi = µi . Thus if s = r+1 we conclude νr = µr +1 and νr+1 = µr+1 − 1, thus ν = λ. If s > r + 1 and hence µr = µs , then there are r ≤ r < s ≤ s such that νi = µi for i ∈ {r , s } as well as νr = µr + 1 and νs = µs − 1. Since µr = νr − 1 ≤ νr −1 − 1 = µr −1 − 1 < µr −1 , whenever r > 1, and µs = νs + 1 ≥ νs +1 + 1 = µs +1 + 1 > µs +1 , this implies r = r and s = s, hence ν = λ in this case as well.

I) Let X additionally have a zero element 0. Then we get an injective Q-linear map F (X) → A(X) : f → f ◦ , where f ◦ (x, y) := δ0,x f (y), for all x, y ∈ X. Then F (X)◦ ≤ A(X) is a right ideal: For f ∈ F (X) and g ∈ A(X) we have (f ◦ g)(x, y) = x≤z≤y f ◦ (x, z)g(z, y), where x ≤ y ∈ X, hence (f ◦ g)(x, y) = 0 if x = 0, and (f ◦ g)(0, y) = z≤y f (z)g(z, y), thus f ◦ g ∈ F (X)◦ . II Summation 36 This shows that F (X) becomes a right A(X)-module by letting f ∗ g ∈ F (X) be deﬁned as (f ∗g)(x) := z≤x f (z)g(z, x), for x ∈ X.