$(V\otimes_kk')^{\lambda}(S) = V^{\lambda}(S)\otimes_kk'$ and $(V\otimes_kk')_{\lambda}(S) = V_{\lambda}(S)\otimes_kk'$.

Let $(a_i)$ be a basis of the $k$-vector space $k'$. If $v\in V\otimes_kk',v$ can be expressed uniquely in the form $\sum v_i\otimes a_i$ where $(v_i)$ is a finitely-supported family of elements of V. For all $s\in S$,

It follows that

$v\in (V\otimes_kk')^{\lambda}(S)\Leftarrow \Rightarrow v_i\in V^{\lambda}(S)$ for all $i$,

$v\in (V\otimes_kk')_{\lambda}(S)\Leftarrow \Rightarrow v_i\in V_{\lambda}(S)$ for all $i$,

which implies the proposition.

(i) Let $f: V\rightarrow W$ be a linear map such that $q(s)f(v) =f(r(s)v)$ for $s\in S$ and $v\in V$. Then, for all $\lambda \in P,f$ maps $V^{\lambda}(S)$ (resp. $V_{\lambda}(S)$) into $W^{\lambda}(S)$ (resp. $W_{\lambda}(S)$).

(ii) Let $B : V\times V'\rightarrow W$ be a bilinear map such that

for $s\in S,v\in V,v'\in V'$. Then, for all $\lambda , \mu\in P$, B maps $V^{\lambda}(S)\times {V'}^\mu(S)$ (resp. $V_{\lambda}(S)\times V'_\mu(S)$) into $W^{\lambda+\mu}(S)$ (resp. $W_{\lambda+\mu}(S)$).

(iii) Let $B : V\times V'\rightarrow W$ be a bilinear map such that

for $s\in S,v\in V,v'\in V'$. Then, for all $\lambda , \mu\in P$, B maps $V^{\lambda}(S)\times {V'}^\mu(S)$ (resp. $V_{\lambda}(S)\times V'_\mu(S)$) into $W^{\lambda \mu}(S)$ (resp. $W_{\lambda \mu}(S)$).

In case (i), $(q(s)-\lambda (s))^nf(v) =f((r(s)-\lambda (s))^nv)$ for $s\in S$ and $v\in V$, hence the conclusion. In case (ii),

for $s\in S,v\in V,v'\in V'$, hence by induction on $n$

$(q(s)-\lambda (s)-\mu(s))^nB(v, v) =\sum B((r(s)-\lambda (s))^iv,(r(s)-\mu(s))^jv)$.

$i+j=n$

The assertions in (ii) follow immediately. In case (iii), $(q(s)-\lambda (s)\mu(s))B(v, v') = B((r(s)-\lambda (s))v, r'(s)v')+B(\lambda (s)v,(r'(s)-\mu(s))v')$ for $s\in S,v\in V,v'\in V'$, hence by induction on $n$

= $\sum(n)B(\lambda (s)^j(r(s)-\lambda (s))^iv, r'(s)^i(r'(s)-\mu(s))^jv')$.

$i+j=n$

The assertions in (iii) follow immediately.

The second assertion is a consequence of the first; hence it suffices to prove that. We distinguish several cases.

$a) S$ is empty. The assertion is trivial.

$b) S$ is reduced to a single element $s$. Let $\lambda_0, \lambda_1, . . . , \lambda_n$ be distinct elements of $k$. For $i= 0,1, . . . , n$, let $v_i\in V^{\lambda_i}(s)$ and assume that $v_0=v_1+\cdots +v_n$. It suffices to prove that $v_0$ = 0. For $i= 0, . . . , n$, there exists an integer $q_i>$ 0 such that $(r(s)-\lambda_i)^{q_i}v_i$ = 0. Consider the polynomials $P(X) =\prod_{i\geq 1}(X-\lambda_i)^{q_i}$ and $Q(X) = (X-\lambda_0)^{q_0}$. We have $Q(r(s))v_0= 0$, and

$P(r(s))v_0=\sum_{i=1}^nP(r(s))v_i= 0$. Since P and Q are relatively prime, the Bezout identity proves that $v_0= 0$.

$c) S$ is finite and non-empty. We argue by induction on the cardinal of S. Let $s\in S$ and $S'= S$-- $\{s\}$. Let $(v_{\lambda})_{\lambda\in P}$ be a finitely-supported family of elements of V such that $\sum_{\lambda\in P}v_{\lambda}= 0$ and $v_{\lambda}\in V^{\lambda}(S)$. Let $\lambda_0\in P$. Let $P'$ be the set of $\lambda \in P$ such that $\lambda |S'=\lambda_0|S'$. By the induction hypothesis applied to $S'$, we have $\sum_{\lambda\in P'}v_{\lambda}= 0$. If $\lambda , \mu$ are distinct elements of $P',\lambda (s)\not=\mu(s)$.

Since the sum $\sum_{\alpha\in k}V^{\alpha}(s)$ is direct by b), and since $v_{\lambda}\in V^{\lambda(s)}(s),v_{\lambda}= 0$ for all $\lambda \in P'$, and in particular $v_{\lambda_0}= 0$, which we had to prove.

d) General case. Let $(v_{\lambda})_{\lambda\in P}$ be a finitely-supported family of elements of V such that $\sum_{\lambda\in P}v_{\lambda}= 0$ and $v_{\lambda}\in V^{\lambda}(S)$. Let $P'$ be the finite set of $\lambda \in P$ such that $v_{\lambda}\not= 0$, and let $S'$ be a finite subset of S such that the conditions $\lambda \in P'$, $\mu\in P',\lambda |S'=\mu|S'$ imply that $\lambda =\mu$. We have $v_{\lambda}\in V^{\lambda|S'}(S')$; applying c), we see that $v_{\lambda}= 0$ for $\lambda \in P'$, which completes the proof.

Recall that, if $x\in$ End(V), we denote by ad $x$ the map $y \rightarrow xy-yx= [x, y]$ from End(V) to itself.

(i) Assume that V is finite dimensional. Then $x$ is triangularizable if and only if $V =\sum_{a\in k}V^a(x)$.

(ii) If there exists an integer $n$ such that (ad $x$)$^ny= 0$, each $V^a(x)$ is stable under $y$.

(iii) Assume that V is finite dimensional. If $V =\sum_{a\in k}V^a(x)$ and if each

$V^a(x)$ is stable under $y$, there exists an integer $n$ such that (ad $x$)$^ny= 0$.

Part (i) follows from Algebra, Chap. VII, §5, no. 2, Prop. 3.

Let E = End(V). Let B be the bilinear map $(u, v) \rightarrow u(v)$ from $E\times V$ to V. By the definition of ad $x$,

$x(B(u, v)) =B(u, x(v)) +$ B((ad $x$)$(u), v$)

for $x\in E,u\in E,v\in V$. Let $x$ operate on E via ad $x$. By Prop. 2 (ii), $B(E^0(x),V^a(x))\subset V^a(x)$ for all $a\in k$. If (ad $x$)$^ny= 0$, then $y\in E^0(x)$, so $y(V^a(x))\subset V^a(x)$, which proves (ii).

To prove (iii), we can replace V by $V^a(x)$ and $x$ (resp. $y$) by its restriction to $V^a(x)$. Replacing $x$ by $x-a$, we can assume that $x$ is nilpotent. Then, (ad $x$)$^{2 dim V-1}= 0$ (Chap. I, §4, no. 2), which proves (iii).

(i) Condition (AC) is satisfied and, for all $s\in S,r(s)$ is triangularizable.

(ii) For all $\lambda \in P$, $V^{\lambda}(S)$ is stable under $r(S)$, and $V =\sum_{\lambda\in P}V^{\lambda}(S)$.

If $V =\sum_{\lambda\in P}V^{\lambda}$(S), then $V =\sum_{a\in k}V^a(s)$ for all $s\in S$, and it follows from

Lemma 1 that (ii) implies (i). Assume that condition (i) is satisfied. Lemma 1 and formula (1) imply that each $V^{\lambda}(S)$ is stable under $r(S)$. It remains to prove that $V =\sum_{\lambda\in P}V^{\lambda}(S)$. We argue by induction on dim V. We distinguish two cases.

a) For all $s\in S,r(s)$ has a single eigenvalue $\lambda (s)$. Then $V = V^{\lambda}(S)$.

b) There exists $s\in S$ such that $r(s)$ has at least two distinct eigenvalues. Then V is the direct sum of the $V^a(s)$ for $a\in k$, and dim $V^a(s)<$ dim V for all $a$. Each $V^a(s)$ is stable under $r$(S), and it suffices to apply the induction hypothesis.

Let $r': S\rightarrow$ End(V$\otimes_kk'$) be the map defined by $r$. If $s_1, s_2\in S$, there exists an integer $n$ such that (ad $r(s_1)$)$^nr(s_2) = 0$, hence (ad $r'(s_1)$)$^nr'(s_2) = 0$. It now suffices to apply Th. 1.

(i) $V^0(S)$ and $V^+(S)$ are stable under $r(S)$;

(ii) $V = V^0(S)\oplus V^+(S)$;

(iii) every vector subspace W of V, stable under $r(S)$ and such that $W^0(S) = 0$, is contained in $V^+(S)$;

(iv) $\sum_{s\in S}r(s)V^+(S) = V^+(S)$.

Moreover, $V^+(S)$ is the only vector subspace of V with properties (i) and (ii). For any extension $k'$ of $k,(V\otimes_kk')^+(S) = V^+(S)\otimes_kk'$.

The last assertion is immediate. Thus, taking Prop. 1 into account, in proving the others we can assume that $k$ is algebraically closed. By Th. 1, V = $\sum_{\lambda\in P}V^{\lambda}$(S), and the $V^{\lambda}(S)$ are stable under $r(S)$. If $s\in S$, the characteristic

polynomial of $r(s)|V^{\lambda}(S)$ is $(X-\lambda (s))^{dim V^{\lambda}(S)}$; it follows that $\bigcap_{i\geq 1}r(s)^iV^{\lambda}(s)$

is zero if $\lambda (s) = 0$ and is equal to $V^{\lambda}(S)$ if $\lambda (s)\not= 0$; hence,

which proves (i), (ii) and (iv). If W is a vector subspace of V stable under $r$(S), then $W =\sum_{\lambda\in P}W^{\lambda}(S)$ and $W^{\lambda}(S) = W\cap V^{\lambda}(S)$. If $W^0(S) = 0$, we see that $W\subset V^+$(S), which proves (iii).

Let $V'$ be a vector subspace of V stable under $r(S)$ and such that $V'\cap V^0(S) = 0$. Then ${V'}^0(S) = 0$, so $V'\subset V^+(S)$ by (iii). If, in addition, $V = V^0(S) + V'$, we see that $V'= V^+(S)$. Q.E.D.

We sometimes call $(V^0(S),V^+(S))$ the Fitting decomposition of V, or of the map $r: S\rightarrow$ End(V). If S reduces to a single element $s$, we write $V^+(s)$ or $V^+(r(s))$ instead of $V^+(\{s\})$. We have that $V = V^0(s)\oplus V^+(s), V^0(s)$ and $V^+(s)$ are stable under $r(s),r(s)|V^0(s)$ is nilpotent and $r(s)|V^+(s)$ is bijective.

In view of Prop. 1, we are reduced to the case in which $k$ is algebraically closed. We have $V =\bigoplus_{\lambda\in P}V^{\lambda}$(S), $V'=\bigoplus_{\lambda\in P}{V'}^{\lambda}(S)$ by Th. 1, and $V'=f(V) =$ $\sum_{\lambda\in P}f(V^{\lambda}(S))$. Finally, $f(V^{\lambda}(S))\subset {V'}^{\lambda}(S)$ by Prop. 2 (i), hence the corollary.

(i) For all $\lambda \in k,V^{\lambda}(u) = V^{\lambda}(u_s) = V_{\lambda}(u_s)$.

(ii) If V has an algebra structure and if $u$ is a derivation of V$,u_s$ and $u_n$ are derivations of V.

(iii) If V has an algebra structure and if $u$ is an automorphism of V, then $u_s$ and $1 +u^{-1}_su_n$ are automorphisms of V.

In view of Prop. 1, we can assume that $k$ is algebraically closed, so

The semi-simple component of $u|V^{\lambda}(u)$ is the homothety with ratio $\lambda$ in $V^{\lambda}(u)$. This proves (i).

Assume from now on that V has an algebra structure. Let $x\in V^{\lambda}(u)$, $y\in V^\mu(u)$.

If $u$ is a derivation of V, then $xy\in V^{\lambda+\mu}(u)$ (Prop. 2 (ii)), so

This proves that $u_s$ is a derivation of V. Then $u_n=u-u_s$ is a derivation of V.

If $u$ is an automorphism of V, Ker($u_s$) $= V^0(u) = 0$, so $u_s$ is bijective. On the other hand, $xy\in V^{\lambda \mu}(u)$ (Prop. 2 (iii)), so

This proves that $u_s$ is an automorphism of V; but then so is

Since $V^{\lambda}(S)$ is stable under $r(S)$ (Lemma 1 and formula (1) of no. 1), we can assume that $V = V^{\lambda}(S)$. Let $n=$ dim V. Thus, for $s\in S$,

det(X $-r(s)$) $= (X-\lambda (s))^n$.

On the other hand, the expansion of the determinant shows that

det(X $-r(s)$) $= X^n+a_1(s)X^{n-1}+\cdots +a_i(s)X^{n-i}+\cdots$

where $a_i: S\rightarrow k$ is a homogeneous polynomial function of degree $i$. Write $n=qm$ where $q$ is a power of the characteristic exponent of $k$ and $(q, m) = 1$. Then $(X-\lambda (s))^n= (X^q-\lambda (s)^q)^m$; hence $-m\lambda (s)^q=a_q(s)$, which implies the result.

The first equality follows from Prop. 1. To prove the second, we can assume that $V = V^0(S)$ and so $V'{= V'}^0(S)$. Let $(s_1, . . . , s_m)$ be a basis of S and $(e_1, . . . , e_n)$ a basis of V. There exist polynomials $P_{ij}(X_1, . . . ,X_m)$ such that

for $1\leq j\leq n$ and $a_1, . . . , a_m\in k'$. By hypothesis, $r'(s)^n= 0$ for all $s\in S$, in other words $P_{ij}(a_1, . . . , a_m) = 0$ for $1\leq i, j\leq n$ and $a_1, . . . , a_m\in k$. Since $k$ is infinite, $P_{ij}= 0$. Consequently, every element of $r'(S')$ is nilpotent and $V'{= V'}^0(S')$.

(i) The function $s \rightarrow P(s)$ is polynomial on S. We have $\widetilde{S} =\{s\in S|P(s)\not= 0\}$; this is an open subset of S in the Zariski topology (App. 1).

(ii) $\widetilde{S}$ is non-empty, and $V^+(s) = V^+(S)$ for all $s\in \widetilde{S}$.

The fact that $s \rightarrow P(s)$ is polynomial follows from the linearity of $r$. If $s\in S, V^0(s)\supset V^0$(S), with equality if and only if $r(s)$ defines an automorphism of $V/V^0$(S), hence (i).

Now let $k'$ be an algebraic closure of $k$, and introduce $V',S', r'$ as in Prop. 6. We remark that $S'$ satisfies condition (AC) by continuation of the polynomial identity (ad $r(s_1)$)$^{2 dim V-1}r(s_2) = 0$ valid for $s_1, s_2\in S$ (no. 1, Remark ). Applying Th.1, we deduce a decomposition

with $\lambda_i\not= 0$ for $1\leq i\leq m$. For $1\leq i\leq m$, there exists a polynomial function $P_i$ non-zero on $S'$ and an integer $q_i$ such that $\lambda^{q_i}_i= P_i$ (Prop. 5). Since $k$ is infinite, there exists $s\in S$ such that $(P_1. . .P_m)(s)\not= 0$, cf. Algebra, Chap. IV, §2, no. 3, Cor. 2 of Prop. 9. Then $\lambda_i(s)\not= 0$ for all $i$, so ${V'}^0(S'{) = V'}^0(s)$ and consequently $V^0(S) = V^0(s)$ (Prop. 6), which shows that $\widetilde{S}\not=\emptyset$. If $s\in \widetilde{S}$, the fact that $V^+(S)$ is stable under $r(s)$ and is a complement of $V^0(s)$ in V implies that $V^+(S) = V^+(s)$ (Cor. 2 of Th. 1).

(i) The sum $\sum_{\lambda\in P}L^{\lambda}(P)$ is direct.

(ii) If $f: L\rightarrow M$ is a homomorphism of $\mathfrak{h}$-modules, $f(L^{\lambda}(\mathfrak{h}))\subset M^{\lambda}(\mathfrak{h})$ for all $\lambda \in P$.

(iii) If $f: L\times M\rightarrow N$ is a bilinear $\mathfrak{h}$-invariant map,

for all $\lambda , \mu\in P$.

This follows from Props. 2 and 3.

(i) Each $M^{\lambda}(\mathfrak{h})$ is an $\mathfrak{h}$-submodule of M. If $x_M$ is triangularizable for all $x\in \mathfrak{h}$, then $M =\sum_{\lambda\in P}M^{\lambda}(\mathfrak{h})$.

(ii) If $k$ is infinite, there exists $x\in \mathfrak{h}$ such that $M^0(x) = M^0(\mathfrak{h})$.

(iii) If $k$ is of characteristic 0, and if $\lambda \in P$ is such that $M^{\lambda}(\mathfrak{h})\not= 0$, then $\lambda$ is a linear form on $\mathfrak{h}$ vanishing on $[\mathfrak{h},\mathfrak{h}]$, and $M_{\lambda}(\mathfrak{h})\not= 0$.

(iv) If $f: M\rightarrow N$ is a surjective homomorphism of finite dimensional $\mathfrak{h}$-modules, then $f(M^{\lambda}(\mathfrak{h})) = N^{\lambda}(\mathfrak{h})$ for all $\lambda \in P$.

(v) If N is a finite dimensional $\mathfrak{h}$-module, and B a bilinear form on $M\times N$ invariant under $\mathfrak{h}$, then $M^{\lambda}(\mathfrak{h})$ and $N^\mu(\mathfrak{h})$ are orthogonal relative to B if $\lambda +\mu\not= 0$. Moreover, if B is non-degenerate then so is its restriction to $M^{\lambda}(\mathfrak{h})\times N^{-\lambda}(\mathfrak{h})$ for all $\lambda \in P$.

Part (i) follows from no. 1, Lemma 1 and Th. 1. Part (ii) follows from no. 2, Prop. 7. Part (iv) follows from no. 1, Cor. 3 of Th. 1. We prove (iii). We can assume that $M = M^{\lambda}(\mathfrak{h})$. Then, for all $x\in \mathfrak{h},\lambda (x) =$ (dim $M$)$^{-1}$Tr($x_M$); this proves that $\lambda$ is linear (which also follows from Prop. 5) and that $\lambda$ vanishes on $[\mathfrak{h},\mathfrak{h}]$. Consider the map $\rho :\mathfrak{h}\rightarrow$ End$_k(M)$ defined by

from the above, this is a representation of $\mathfrak{h}$ on M, and $\rho (x)$ is nilpotent for all $x\in \mathfrak{h}$. By Engel’s theorem (Chap. I, §4, no. 2, Th. 1), there exists $m\not= 0$ in M such that $\rho (x)m= 0$ for all $x\in \mathfrak{h}$, so $m\in M_{\lambda}(\mathfrak{h})$.

The first assertion of (v) follows from no. 1, Prop. 2 (ii). To prove the second, we can assume that $k$ is algebraically closed in view of Prop. 1 of no. 1; it then follows from the first and the fact that $M =\sum_{\lambda}M^{\lambda}(\mathfrak{h}), N =\sum_{\lambda}N^{\lambda}(\mathfrak{h})$, cf. (i).