By Jürgen Müller

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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.

We show that M ∈ Nλ if = 0, while lim →0 M = M0 ∈ Nµ : If = 0 then, since a > b, the unit vector e1 ∈ Cn×1 has minimum polynomial X a ∈ C[X] with respect to M . Moreover, the unit vector ea+1 ∈ Cn×1 has minimum polynomial X b ∈ C[X]; here for b = 0 we let ea+1 := 0 ∈ Cn×1 . From ea+1 M = ea+1 , . . , en C and e1 M ∩ ea+1 , . . , en C = {0} we conclude Cn×1 = e1 M ⊕ ea+1 M , hence M has Jordan normal form Ja ⊕ Jb . If = 0 then e2 ∈ Cn×1 and e1 ∈ Cn×1 have minimum polynomials X a−1 ∈ C[X] and X b+1 ∈ C[X], respectively, with respect to M0 .

Moreover, we have ∂ 1 1 n−1 = n≥0 n! X n = exp ∈ Q[[X]]. n≥1 (n−1)! X ∂X exp = If K is a ﬁeld of characteristic 0, then for f ∈ XK[[X]] we have exp(f ) := 1 n ∈ 1 + XK[[X]] ⊆ K[[X]], fulﬁlling the identity exp(f + g) = n≥0 n! (n−k)! 1 g j ) = n≥0 n! exp(f ) · exp(g) ∈ K[[X]], for all f, g ∈ XK[[X]], hence exp(f )−1 = exp(−f ) ∈ K[[X]]. (n−k)! = δ0,n , that is k=0 (−1)n−k nk = δ0,n , for n ∈ N0 . Note that going over to the associated Taylor series yields the identity exp(x) · exp(−x) = 1 ∈ C, for all x ∈ C; but conversely the identity 1 1 m exp(x + 1) = e · exp(x) ∈ C, yielding n≥0 n!

### Algebraic combinatorics by Jürgen Müller

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