Lemma 4

Let $E$ be a finite dimensional vector space over a commutative field $K$ and let $u$ be a triangularisable endomorphism of $E$. Then there exists a unique diagonalisable endomorphism $v$ of $E$ which commutes with $u$ and is such that $u - v$ is nilpotent. Moreover, under these conditions the characteristic polynomials of $u$ and $v$ coincide, and there exists a polynomial $P \in K[X]$ such that $v = P(u)$.

Let $v$ be a diagonalisable endomorphism of $E$ such that $uv = vu$ and $v - u$ is nilpotent; let $\alpha$ be an eigenvalue of $v$ and let $V_\alpha$ be the corresponding eigenspace. By Lemma 3 (VII, p. 41), $V_\alpha$ is closed under $u$, and the restriction of u - α to V, is also the restriction of u - v, so is nilpotent; hence V, is contained in the subspace M, consisting of those x ∈ E which are annihilated by some power of u - a. Since E is the direct sum of the V, and also of the M, (VII, p. 34, Prop. 7), this shows that V, = M, for all a. By the corollary to Prop. 9 (VII, p. 36) it follows that χ_u = χ_v ; it also follows that v is uniquely determined by u ; its restriction to each M_α is the homothety defined by α.

Conversely, let us define v by the above condition; it is clear that v is diagonalisable and that u - v is nilpotent. By Prop. 7 of VII, p. 34 there exist polynomials q, such that, for all x ∈ E, the component of x in M, is q_α(u).x. Then v = ∑ αq_α(u); this implies that u and v commute and completes the proof.

B) Now let us return to the proof of Th. 1.

First of all, suppose that u can be written in the form s + n, where s is absolutely semi-simple and n is nilpotent, and where s and n commute. Let Ω be an algebraic closure of K; then u_{(Ω)} = s_{(Ω)} + n_{(Ω)}, where s_{(Ω)} is diagonalisable and n_{(Ω)} is nilpotent, and where s_{(Ω)} and n_{(Ω)} commute; by Lemma 4 it follows that s_{(Ω)}, and so also s, is unique, that the polynomials χ_{u_{(Ω)}} and χ_{s_{(Ω)}} in Ω[X] coincide, hence also the polynomials χ_u and χ_s, and that s can be expressed as a polynomial in u with coefficients in Ω. This shows first of all that the eigenvalues of u are the same as those of s, so are separable over K (VII, p. 42, Prop. 15); in addition, since s is an 0-linear combination of powers of u, it is also a K-linear combination of these same powers (II, p. 311, Prop. 19), and there exists a polynomial P ∈ K[X] such that s = P(u), hence n = Q(u) where Q(X) = X - P(X). Now let us show that Q (and hence P) may be chosen with no constant term. If u is invertible then its characteristic polynomial has a nonzero constant term, and the Cayley-Hamilton theorem (III, p. 541, Prop. 20) shows that 1 can be expressed as a polynomial in u with no constant term, so the assertion holds in this case. If u is not invertible, then its kernel W is nonzero and closed under n (VII, p. 41, Lemma 3); since the restriction of n to W is nilpotent, there exists a vector x ≠ 0 in W such that u(x) = n(x) = 0, which shows that Q cannot have a constant term.

Conversely, suppose that the eigenvalues of u are separable over K, and let L be a finite Galois extension of K containing these eigenvalues. By Lemma 4 we can write u_{(L)} = v + w, where v is diagonalisable and w is nilpotent, and where vw = wv. Let B be a basis of E, let B' be the corresponding basis of L ⊗_K E, and let U, V, W be the matrices of u_{(L)}, v, w with respect to B'; note that U is also the matrix of u with respect to B, so has entries in K. For every K-automorphism σ of L, and every matrix A with entries in L, let A" denote the matrix obtained by applying σ to the entries of A. Let a be a K-automorphism of L; then U = U^σ = (V + W)^σ = V^σ + W^σ, V^σW^σ = (VW)^σ = (WV)^σ = W^σV^σ; since Vu is the matrix of a diagonalisable endomorphism and W" is nilpotent, it follows from Lemma 4 that Vu = V and W^σ = W. Since this is valid for all σ, the entries of V and W are in K; if $u_s$ and $u_n$ are the endomorphisms of E with matrices V and W with respect to B, then $(u_s)_{(L)} = v$ and $(u_n)_{(L)} = w$. It follows that $u_s$ is absolutely semi-simple, that $u_n$ is nilpotent, that $u_s$ and $u_n$ commute, and that $u = u_s + u_n$. This completes the proof.

Whenever an endomorphism f admits a Jordan decomposition, we write it $(f_s, f_n)$, and the endomorphisms $f_s$ and $f_n$ are called the absolutely semi-simple component and the nilpotent component off respectively. When K is perfect, every endomorphism has a Jordan decomposition; in this case also there is no distinction between absolutely semi-simple endomorphisms and semi-simple endomorphisms, and we sometimes say « semi-simple component » for « absolutely semi-simple component ».