Locally finite endomorphisms.
Andrzej Nowicki, Toru´n, February 17, 1994
1 Invariant submodules and integral elements.
Let k be a ring, M a k-module, and let ϕ : M −→ M be a k-endomorphism of M . We say that M is finite if M is finitely generated over k. A k-submodule M0 of M is said to be ϕ-invariant if ϕ(M0) ⊆ M0.
The module M becomes an k[t]-module, where k[t] is the ring of polynomials in the variable t if, for a polynomial f = a0 + a1t + · · · + antn and an element x ∈ M , we set f x = a0x + a1ϕ(x) + · · · + anϕn(x). Every k[t]-module can thus be obtained from the k-endomorphism given by the formula ϕ(x) = tx. A submodule M0 of M is ϕ-invariant iff M0 is a k[t]-submodule of the k[t]-module M .
Let x ∈ M . We shall say that x is ϕ-integral if there exists a monic polynomial f ∈ k[t]
such that f x = 0, that is, x is ϕ-integral if and only if
ϕn(x) = a0x + a1ϕ(x) + · · · + an−1ϕn−1(x),
for some n ∈ N and a0, . . . , an−1 ∈ k. We shall denote by Mxϕ the smallest ϕ-invariant submodule of M containing x, that is, Mxϕ is the k-submodule of M generated by the set {x, ϕ(x), ϕ2(x), . . . }.
Proposition 1.1. Let x ∈ M . The following conditions are equivalent:
(1) Mxϕ is finite.
(2) There exists a finite ϕ-invariant submodule of M containing x.
(3) x is ϕ-integral.
(4) Mxϕ = kx + kϕ(x) + · · · + kϕn−1(x), for some n ∈ N.
If k is a field then the above proposition is evident. For a proof in a general case we need the following well known lemma (see, for example, [1] p. 21).
Lemma 1.2. Let k be a ring, M0 be a finite k-module and let ψ be a k-endomorphism of M0. Then ψ satisfies an equation of the form
ψn+ an−1ψn−1+ · · · + a0 = 0 where a0, . . . , an−1 ∈ k.
Proof of Proposition 1.1. The implications (1) ⇒ (2) and (4) ⇒ (1) are clear.
(2) ⇒ (3). If M0is a finite ϕ-invariant submodule of M containing x then put ψ = ϕ|M0 and use Lemma 1.2.
(3) ⇒ (4). Denote M0 = kx + kϕ(x) + · · · + kϕn−1(x). Then M0 is a k-submodule of M and x ∈ M0 ⊆ Mxϕ. Since ϕn+1(x) = a0ϕ(x) + · · · + an−1ϕn(x) and ϕn(x) ∈ M0, ϕn+1(x) ∈ M0 and inductively, ϕp(x) ∈ M0 for all p ∈ N. Thus Mxϕ = M0.
Let us denote by Fin(ϕ) the set of all ϕ-integral elements of M . As a consequence of Proposition 1.1 we get
Proposition 1.3. The set Fin(ϕ) is a ϕ-invariant submodule of M .
2 Local finiteness and local nilpotence.
Let, as in the previous section, k be a ring, M a k-module , and let ϕ be a k- endomorphism of M .
We say that ϕ is locally finite if Fin(ϕ) = M . Equivalently (by Proposition 1.1), ϕ is locally finite if and only if M =S
i∈IVi, where each Vi is a finite ϕ-invariant submodule of M .
If ϕ is locally finite and a ∈ k then the endomorphism aϕ is also locally finite. The following example shows that the sum and the composition of two locally finite endomor- phism are not locally finite in general.
Example 2.1. Let M be a vector space over a field k such that a set B of the form B = {x1, x2, . . . , y1, y2, . . . } is a basis of M . Let ϕ and ψ be two k-endomorphisms of M defined as follows:
ϕ(xi) = yi+1, ψ(xi) = 0, ϕ(yi) = 0, ψ(yi) = xi+1.
Then ϕ and ψ are locally finite, but ϕ + ψ and ϕψ are not locally finite.
Proposition 2.2. Let M be a module over a ring k and let ϕ and ψ be locally finite k-endomorphisms of M . If ϕψ = ψϕ then the enodomorphisms ϕ + ψ and ϕψ are locally finite.
Proof. Let x ∈ M . Since ϕ is locally finite, there exists a finite ϕ-invariant submodule W of M containing x. Let {y1, . . . , yn} be a finite set of generators of W and let U = Myψ
1 + · · · + Myψn. Then U is a finite ψ-invariant submodule of M containing W . Since ϕ and ψ commute, we see (by Proposition 1.1) that U is also ϕ-invariant. Therefore, for every x ∈ M , there exists a finite (ϕ + ψ)-invariant and (ϕψ)-invariant submodule U of M containing x. This means that ϕ + ψ and ϕψ are locally finite.
Let us denote by Nil(ϕ) the set of all elements x ∈ M such that ϕn(x) = 0 for some n ∈ N0. It is clear that Nil(ϕ) is a ϕ-invariant submodule of M . Moreover, Ker(ϕ) ⊆ Nil(ϕ) ⊆ Fin(ϕ).
We say that ϕ is locally nilpotent if Nil(ϕ) = M . Thus, every locally nilpotent k- endomorphism is locally finite. The endomorphisms ϕ and ψ from Example 2.1 are locally nilpotent. So, we see by this example, that the sum and the composition of locally nilpotent endomorphisms are not locally nilpotent (even locally finite) in general.
However, it is easy to see that if ϕ and ψ are such locally nilpotent endomorphisms that ϕψ = ψϕ then it the endomorphisms ϕ + ψ and ϕψ are also locally nilpotent.
3 Semisimple modules and endomorphisms.
In the present section we recall some well known notion and facts from the linear algebra and the theory of modules.
Let k be a ring and let M be a k-module. The module M is said to be simple if 0 and M are the only submodules of M . A module which is a direct sum of simple modules
is called a semisimple module. The following three conditions are equivalent (see, for example, [6] or [4]): (1) M is semisimple; (2) M is a sum (not necessarily direct) of simple modules; (3) every submodule of M is a direct summand of M . Submodules and factor modules of a semisimple module are semisimple.
Let ϕ : M −→ M be a k-endomorphism of M . Let us recall (see Section 1) that then M is a k[t]-module and every ϕ-invariant submodule of M is a k[t]-submodule of M .
A ϕ-invariant submodule M0 of M is said to be simple if M0 is simple as the k[t]- module. The endomorphism ϕ is called simple (resp. semisimple) if the k[t]-module M is simple (resp. semisimple). Thus, ϕ is simple if and only if 0 and M are the only ϕ-invariant submodule of M . Moreover, from the above facts we get
Proposition 3.1. Let M be a module over a ring k and let ϕ be a k-endomorphism of M . The following conditions are equivalent:
(1) ϕ is semisimple;
(2) M is a direct sum of simple ϕ-invariant submodules;
(3) M is a sum (not necessarily direct) of simple ϕ-invariant submodules;
(4) for every ϕ-invariant submodule M0 of M there exists a ϕ-invariant submodule M00 of M such that M = M0 ⊕ M00.
Now it is easy to check that if ϕ is semisimple and M0 is a ϕ-invariant submodule of M then the k-endomorphisms ϕ|M0 : M0 −→ M0 and ϕ : M/M0 −→ M/M0 are semisimple.
Moreover, if M = P
i∈IMi is a sum (not necessarily direct) of ϕ-invariant submodules, then ϕ is semisimple if and only if every k-endomorphism ϕ|Mi is semisimple.
We shall say that an element x ∈ M is semisimple with respect to ϕ if there exists a finite ϕ-invariant k-submodule W of M such that x ∈ W and the endomorphism ϕ|W is semisimple. Let us denote by Sem(ϕ) the set of all semisimple (with respect to ϕ) elements of M .
Proposition 3.2. The set Sem(ϕ) is a ϕ-invariant k-submodule of M .
Proof. It is clear that if x ∈ Sem(ϕ) then ϕ(x) ∈ Sem(ϕ) and, for a ∈ k, ax ∈ Sem(ϕ). Assume that x, y ∈ Sem(ϕ). We shall show that x + y ∈ Sem(ϕ). We know, by Proposition 1.1 and the above remarks, that the submodules A = Mxϕ and B = Myϕ are finite and the endomorphisms ϕ|A, ϕ|B are semisimple. Put C = A + B. Then C is a finite ϕ-invariant submodule of M and, by Proposition 3.1, C is a sum of simple ϕ- invariant submodules. Thus, again by Proposition 3.1, ϕ|C is semisimple and x + y ∈ C.
This means that x + y ∈ Sem(ϕ).
Assume now that k is a field, M = V is a finite vector space over k, and ϕ : V −→ V is a k-endomorphism. Note the following well known fact from the linear algebra.
Proposition 3.3. If ϕ is an endomorphism of a finite vector space over a field k then the following two conditions are equivalent:
(1) ϕ is semisimple,
(2) The minimal polynomial of ϕ is a product of pairwise different irreducible monic factors.
The endomorphism ϕ is called nilpotent if ϕs = 0 for some s ∈ N. The minimal polyno- mial of a nilpotent endomorphism is of the form tr, for some r ∈ N so, by Proposition 3.3, we get
Corollary 3.4. If ϕ is nilpotent then ϕ is semisimple if and only if ϕ = 0.
Let us recall that ϕ is said to be diagonalizable if there exists a basis B of V such that ϕ(b) = αbb for any b ∈ B with αb ∈ k, that is, if the matrix of ϕ with respect to B is diagonal. From the linear algebra we have:
Proposition 3.5. If ϕ is an endomorphism of a finite vector space over a field k then the following two conditions are equivalent:
(1) ϕ is diagonalizable,
(2) The minimal polynomial of ϕ is a product of pairwise different linear factors. From this proposition one can easily deduce that if ϕ : V −→ V is diagonalizable and V0 is a ϕ-invariant k-subspace of V then the k-endomorphisms ϕ|V0 : V0 −→ V0 and ϕ : V /V0 −→ V /V0 are also diagonalizable. Moreover, from Propositions 3.3 and 3.5 we get
Corollary 3.6. Let ϕ be an endomorphism of a finite vector space over a field k. If the characteristic polynomial of ϕ is a product of linear factors then ϕ is semisimple if and only if ϕ is diagonalizable.
In particular, if k is algebraically closed then the notions ”semisimple” and ”diagonal- izable” coincide.
Now let us assume that k is a field of characteristic zero. In such a case every field extension of k is separable so, if a polynomial f ∈ k[t] is a product of pairwise different irreducible factors from k[t] and L is a field containing k then f , as a polynomial from L[t], is a product of pairwise different irreducible factors from L[t]. Therefore, from the above facts we obtain
Proposition 3.7. Let ϕ be an endomorphism of a finite vector space V over a field k of characteristic zero. Then the following two conditions are equivalent:
(1) ϕ is semisimple,
(2) There exists a field L containing k such that ϕ is diagonalizable over L (that is, the L-endomorphism ϕ ⊗ 1 of the space V ⊗kL is diagonalizable).
Note also the following
Proposition 3.8. Let V be a finite vector space over a field k of characteristic zero, and let ϕ, ψ be two k-endomorphisms of V which commute. If ϕ, ψ are semisimple then ϕ + ψ and ϕψ are also semisimple.
Proof. We may assume, by Proposition 3.7, that k is algebraically closed. Therefore, by Corollary 3.6, ϕ and ψ are diagonalizable and we must show that ϕ + ψ and ϕψ are also diagonalizable.
Let α1, . . . , αn be pairwise different eigenvalues of ϕ. Then, V = V1⊕ · · · ⊕ Vn, where Vi = {x ∈ V ; ϕ(x) = αix} for i = 1, . . . , n. Each Vi is a ϕ-invariant subspace of V and, since ϕ and ψ commute, it is also a ψ-invariant subspace. Thus, for every i ∈ {1, . . . , n}, we have the diagonalizable mapping ψ|Vi :−→ Vi so, we have a set {e(i)1 , . . . , e(i)ri} of eigenvectors of ψ|Vi which is a basis of Vi. Each e(i)j is, of course, also an eigenvector of ψ. Therefore, the set {e(1)1 , . . . , e(1)r1 , . . . , e(n)1 , . . . , e(n)rn } is a basis of V and every element of this set is an eigenvector of ϕ and simultaneously of ψ. This implies, in particular, that ϕ + ψ and ϕψ are diagonalizable.
4 Jordan-Chevalley decomposition.
Let k be a field, V a finite dimensional linear k-space, and let ϕ : V −→ V be a k-endomorphism of V .
It is well known that if k is algebraically closed then there exists a basis B of V such that the matrix of ϕ with respect to B has the Jordan canonical form, that is, the matrix is a sum of blocks of the form J = A + B, where
A =
a 0 0 · · · 0 0 0 a 0 · · · 0 0 0 0 a · · · 0 0
· · · · 0 0 0 · · · a 0 0 0 0 · · · 0 a
, B =
0 1 0 · · · 0 0 0 0 1 · · · 0 0 0 0 0 · · · 0 0
· · · · 0 0 0 · · · 0 1 0 0 0 · · · 0 0
.
Since AB = BA, ϕ is a sum of a diagonalizable and a nilpotent endomorphism which commute.
Such a situation is not only for algebraically closed fields. It is true also if the charac- teristic polynomial of ϕ is a product of linear factors. We are interested only in the case when k is a field of characteristic zero. In such a case the following theorem gives a more precise information
Theorem 4.1. Let V be a finite dimensional linear space over a field k of characteristic zero, and let ϕ : V −→ V be a k-endomorphism. Then:
(1) There exist unique k-endomorphisms ϕs and ϕn of V such that (a) ϕ = ϕs+ ϕn,
(b) ϕs is semisimple, (c) ϕn is nilpotent, (d) ϕsϕn= ϕnϕs.
Moreover, the endomorphisms ϕs and ϕn have the following property:
(2) ϕs = u(ϕ) and ϕn= v(ϕ), for some polynomials u, v ∈ k[t] without constant terms.
In particular,
(i) ϕs and ϕn commute with any k-endomorphism of V commuting with ϕ.
(ii) If V0 ⊆ V1 ⊆ V are k-subspaces and ϕ maps V1 into V0 then ϕs and ϕn also map V1 into V0.
The decomposition ϕ = ϕs+ ϕn is called the Jordan-Chevalley decomposition of ϕ.
If the characteristic polynomial of ϕ is a product of linear factors (in particular, if k is algebraically closed) then this theorem is well known and its proof we may find for example in [2] (p. 17) or [6] (XV exercise 14).
However if k is not algebraically closed (but char(k) = 0) then the author never has seen its precise formulation and its proof. A deduction from the algebraically closed case seems to be difficult. We sketch here a proof based on some facts, given as exercises, in [5].
An essential role it this proof play the following
Lemma 4.2. Let k be a field of characteristic zero and let p(t) ∈ k[x] rk be an irreducible polynomial. Then, for every m ∈ N, there exists a polynomial f (t) ∈ k[x] such that
p(t + f (t)p(t)) = p(t)mg(t) for some g(t) ∈ k[t].
Proof. If m = 1 then it is obvious. Let m > 2 and let d denote the partial derivative
∂
∂t in k[t]. Since char(k) = 0, d(p) 6= 0 and deg d(p) < deg p so, the polynomials d(p) and pm−1 are relatively prime. Therefore,
1 = upm−1 − gd(p), (4.1)
for some u, g ∈ k[t].
Let p =Pn
i=0aiti. Then d(p) =Pn
i=1iaiti−1 and we have p(t + gp) = Pn
i=0ai(t + gp)i
= Pn
i=0ai(ti + igpti−1+ p2vi) (where v0, . . . , vn ∈ k[t])
= p + d(p)gp + vp2 (for some v ∈ k[t])
= (1 + d(p)g)p + vp2
= upm+ vp2 (by (4.1)).
Thus, we proved that there exist polynomials g(t), u(t), v(t) ∈ k[t] such that
p(t + g(t)p(t)) = p(t)2v(t) + p(t)mu(t). (4.2) Now, substituting t 7→ t + g(t)p(t), in (4.2) , and using (4.2), we get
p(t + gp + g(t + gp)p(t + gp)) = p(t + gp)2v(t + gp) + p(t + gp)mu(t + gp),
that is,
p(t + gp + (p2v + pmu)g(t + gp)) = (p2v + pmu)2v(t + gp) + (p2v + pmu)mu(t + gp).
So, we see that there exist polynomials g1, u1, v1 ∈ k[t] such that
p(t + g1(t)p(t)) = p(t)4v1(t) + p(t)mu1(t). (4.3) Substituting t 7→ t + g(t)p(t), in (4.3) , and using again (4.2), we get
p(t + g2(t)p(t)) = p(t)8v2(t) + p(t)mu2(t),
for some g2, u2, v2 ∈ k[t]. Repeating this procedure sufficiently many times we obtain our lemma.
Proof of Theorem 4.1. Let h be the minimal polynomial of ϕ. Assume that h = pm1 1· · · pmrr, where p1, . . . , pr are pairwise relatively prime monic irreducible polynomials in k[t] and m1, . . . , mr > 1.
Put Wi = Ker pmi i(ϕ), for i = 1, . . . , r. Then W1, . . . , Wr are ϕ-invariant subspaces of V , V = W1⊕ · · · ⊕ Wr, and the minimal polynomial of ϕi = ϕ|Wi is equal to pmi i, for all i = 1, . . . , r.
It follows from Lemma 4.2 that there exist polynomials f1, . . . , fr ∈ k[t] such that, for every i = 1, . . . , r,
pi(t + fi(t)pi(t)) ≡ 0 (mod pmi i). (4.4) Since the polynomial pm11, . . . , pmrr are pairwise relatively prime, there exists (by the Chinese Remainder Theorem for the ring k[t]) a polynomial f ∈ k[t] such that
pifi ≡ f (mod pmi i) (4.5)
for all i = 1, . . . , r, and
0 ≡ f (mod t). (4.6)
Notice that the last congruence is superfluous if t ∈ {p1, . . . , pr}, while otherwise t is relatively prime to the other moduli. Set
u(t) = t + f (t), v(t) = −f (t) and let
ϕs= u(ϕ) = ϕ + f (ϕ), ϕn= v(ϕ) = −f (ϕ).
Then ϕs+ ϕn = ϕ, ϕsϕn = ϕnϕs, and the polynomials u, v have zero constant terms (since t|f ).
Now we shall show that ϕn is nilpotent. Let m = max{m1, . . . , mr} and let x ∈ V . Then x = x1+ · · · + xr, where x1 ∈ W1, . . . , xr ∈ Wr, and we have
ϕmn(x) = ϕmn(x1) + · · · + ϕmn(xr).
Fix an i ∈ {1, . . . , r}. Since pmi i is the minimal polynomial of ϕ|Wi, pmi (ϕ)(xi) = 0 and so, by (4.5), we have
ϕmn(xi) = (−1)mfm(ϕ)(xi)
= (−1)mfim(ϕ)pmi (ϕ)(xi)
= (−1)mfim(ϕ)(0) = 0.
Thus, ϕmn(x) = Pr
i=1ϕmn(xi) = 0, that is, ϕn is nilpotent.
Let g = p1· · · pr. We shall show that g(ϕs) = 0. For this aim let us observe that if xi ∈ Wi then ϕ(xi) = ϕ|Wi(xi) and, by (4.5) and (4.4),
pi(ϕ + f (ϕ))(xi) = pi(ϕ + fi(ϕ)pi(ϕ))(xi) = 0.
Therefore, if x ∈ V then x = x1 + · · · + xr, where x1 ∈ W1, . . . , xr ∈ Wr, and we have
g(ϕs)(x) =
n
X
i=1
g(ϕs)(xi) =
n
X
i=1
p1(ϕs) · · · \pi(ϕs) · · · pr(ϕs)pi(ϕ + f (ϕ))(xi) = 0.
Thus, g(ϕs) = 0. This means that the minimal polynomial of ϕs is a product of pairwise different irreducible factors. This implies, by Proposition 3.3, that ϕs is semisimple.
So we see that the pair (ϕs, ϕn) satisfies (a), (b), (c), (d) and also (2).
It remains only to prove the uniqueness assertion in (1). Let (ϕs, ϕn) be the pair constructed as above an suppose that (ψs, ψn) is another such a pair satisfying (a), (b), (c) and (d). Then ϕψs= ψsϕ. In fact:
ϕψs = (ψs+ ψn)ψs = ψsψs+ ψnψs = ψsψs+ ψsψn= ψsϕ.
Analogously ϕψn = ψnϕ. So, by (2), ϕsψs = ψsϕs and ϕnψn= ψnϕn. Hence, ϕn− ψn = ψs− ϕs is a nilpotent endomorphism which is, by Proposition 3.8, semisimple and hence, by Corollary 3.4, ϕs = ψs and ϕn= ψn. This completes the proof of Theorem 4.1.
It is well known (and easy to be proved) that if char(k) = 0 and ϕ is an endomorphism of a finite vector space over k, then the characteristic polynomial of ϕ is a divisor of a power of the minimal polynomial of ϕ. Thus, from the above proof we get
Proposition 4.3. Let ϕ be a k-endomorphism of a finite vector space over k, where k is a field of characteristic zero. Assume that the characteristic polynomial of ϕ is of the form pn11· · · pnrr, where p1, . . . , pr are pairwise relatively prime irreducible monic polynomials from k[t]. Then the minimal polynomial of ϕs is equal to p1· · · pr.
Note the following easy consequence of Theorem 4.1.
Proposition 4.4. Let k be a field of characteristic zero. Let V be a finite dimensional vector space over k and let ϕ, ψ be k-endomorphisms of V . If ϕψ = ψϕ then (ϕ + ψ)s = ϕs+ ψs and (ϕ + ψ)n = ϕn+ ψn.
5 Locally finite endomorphisms of a vector space.
Let k be a field, V be a vector space over k, and let ϕ : V −→ V be a k-endomorphism.
We do not assume that V is finite.
Lemma 5.1. If W is a ϕ-invariant simple subspace of V then W is finite.
Proof. Let 0 6= x ∈ W and denote by Wx the subspace of W generated by all the elements of the form ϕm(x), where m ∈ N0. Then Wx is ϕ-invariant so, since W is simple, Wx = W for every nonzero x ∈ W . If ϕ(x) = 0 then W = Wx = kx is finite dimensional. Assume that ϕ(x) 6= 0. Then Wx = Wϕ(x) = W and hence, x ∈ Wϕ(x), that is, x = a1ϕ(x) + · · · + anϕn(x), for some natural n and for some a1, . . . , an ∈ k.
Since x 6= 0, we may assume that an 6= 0. Thus, ϕn(x) ∈ kx + kϕ(x) + · · · + kϕn−1 and consequently, W = Wx = kx + kϕ(x) + · · · + kϕn−1 is a finite subspace.
The following two propositions are consequences of Lemma 5.1 and the facts described in the previous sections.
Proposition 5.2. Let ϕ be an endomorphism of a vector space. If ϕ is semisimple then ϕ is locally finite.
Proposition 5.3. Let ϕ be an endomorphism of a vector space V . The following condi- tions are equivalent:
(1) ϕ is semisimple.
(2) Sem(ϕ) = V .
(3) ϕ is locally finite and for every finite ϕ-invariant subspace W of V the endomor- phism ϕ|W is semisimple.
(4) For every x ∈ V there exists a finite ϕ-invariant subspace W of V such that x ∈ W and ϕ|W is semisimple.
Note also the following
Proposition 5.4. Let ϕ be a k-endomorphism of a vector space over k, where k is an algebraically closed field. The following conditions are equivalent:
(1) ϕ is semisimple.
(2) There exists a basis B of V such that ϕ(b) = αbb for any b ∈ B with αb ∈ k.
(3) V =L
α∈kVα, where Vα = {x ∈ V ; ϕ(x) = αx}.
Proof. (1) ⇒ (2). Since ϕ is semisimple, V =L
i∈IWi, where each Wi is a simple ϕ-invariant subspace of V . Then each ϕ|Wi is semisimple and Lemma 5.1 implies that Wi is finite. Thus, each ϕ|Wi is diagonalizable (Corollary 3.6).
(2) ⇒ (3). It is clear that V = P
α∈kVα. This sum is direct because every finite set of eigenvectors belonging to pairwise different eigenvalues is linearly independent.
The implication (3) ⇒ (2) is evident and the implication (2) ⇒ (1) follows from Proposition 3.2.
The next proposition is a generalization of Proposition 3.8.
Proposition 5.5. Let V be a linear space over a field k of characteristic zero, and let ϕ, ψ be two k-endomorphisms of V such that ϕψ = ψϕ. If ϕ, ψ are semisimple then ϕ + ψ and ϕψ are also semisimple.
Proof. Let x ∈ V . We know already, by Proposition 5.2, that ϕ and ψ are locally finite. Therefore, as in the proof of Proposition 2.2, there exists a finite subspace U of V containing x which is ϕ-invariant and ψ-invariant. Hence, we have two semisimple endomorphisms ϕ|U and ψ|U , of the finite vector space U , which commute. Now our proposition follows from Proposition 3.8 and Proposition 5.3.
In the previous sections we introduced three subspaces of V : Fin(ϕ), Nil(ϕ) and Sem(ϕ). Now it is not difficult to prove the following
Theorem 5.6. If ϕ is an endomorphism of a vector space then:
(1) Ker(ϕ) ⊆ Nil(ϕ) ⊆ Fin(ϕ), (2) Ker(ϕ) ⊆ Sem(ϕ) ⊆ Fin(ϕ), (3) Nil(ϕ) ∩ Sem(ϕ) = Ker(ϕ).
The next theorem is a locally finite version of the Jordan-Chevalley decomposition.
Such a version is mentioned (without proof) in [3] p. 96. A proof gave A. Tyc in [7]. We present here a sketch of the author’s proof.
Theorem 5.7. Let V be a linear space over a field k of characteristic zero and let ϕ : V −→ V be a locally finite k-endomorphism. Then:
(1) There exist unique k-endomorphisms ϕs and ϕn of V such that (a) ϕ = ϕs+ ϕn,
(b) ϕs is semisimple, (c) ϕn is locally nilpotent, (d) ϕsϕn= ϕnϕs.
(2) The endomorphisms ϕsand ϕncommute with any k-endomorphism (not necessarily locally finite) of V commuting with ϕ.
(3) If V0 ⊆ V1 ⊆ V are k-subspaces and ϕ maps V1 into V0 then ϕs and ϕn also map V1 into V0.
Proof. If W is a ϕ-invariant subspace of V then we denote by ϕW the restriction ϕ|W : W −→ W , and we denote by ϕWs , ϕWn (if W is finite) the semisimple and the nilpotent part, respectively, of ϕW. Observe that if W and U are finite ϕ-invariant subspaces of V then
ϕW ∩Us = ϕWs |W ∩ U = ϕUs|W ∩ U and ϕW ∩Un = ϕWn |W ∩ U = ϕUn|W ∩ U. (5.7) Now we define the mappings ϕs and ϕn as follows: if x ∈ V then
ϕs(x) = ϕWs (x) and ϕn(x) = ϕWn (x),
where W is a finite ϕ-subspace of V containing x (such a subspace exists, because ϕ is locally finite). The conditions (5.7) imply that ϕs and ϕn are well defined mappings.
They are k-endomorphisms of V and it is clear that they satisfy (a), (b), (c) and (d).
Now we shall show that the pair (ϕs, ϕn) satisfies (2). Let h : V −→ V be an arbitrary endomorphism such that ϕh = hϕ. Let x ∈ V , and let W be a finite ϕ-subspace containing x. Then h(W ) is a finite ϕ-subspace containing h(x). Set U = W + h(W ). The subspace U is ϕ-invariant and y, h(y) ∈ U for all y ∈ W . Moreover, hϕU(y) = ϕUh(y) for y ∈ W . We know (Theorem 4.1) that there exist polynomials u, v ∈ k[t], without constant term, such that ϕUs = u(ϕU) and ϕUn = v(ϕU). This fact and an easy calculation imply that hϕUs = ϕUsh for y ∈ W . Now we have:
hϕs(x) = hϕUs(x) = ϕUsh(x) = ϕsh(x)
(since x ∈ U and h(x) ∈ U ), that is, hϕs = ϕsh and analogously hϕn = ϕnh. Thus we have (2).
By a similar way (using Theorem 4.1) we show (3). For a proof of the uniqueness assertion in (1) we use Proposition 5.5 and next, we copy the proof of the corresponding part of Theorem 4.1.
References
[1] M. F. Atiyah, I. G. Macdonald, Introduction to Commutative Algebra, Addison–Wesley Publishing Company, 1969.
[2] J. E. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer- Verlag, New York Heidelberg Berlin, 1972.
[3] J. E. Humphreys, Linear Algebraic Groups, Springer-Verlag, New York Heidelberg Berlin, 1981.
[4] F. Kasch, Moduln und Ringe, B. G. Teubner, Stuttgart, 1977.
[5] A. I. Kostrykin, Exercises in Algebra (in Russian), Moskow, Nauka, 1987.
[6] S. Lang, Algebra, Addison–Wesley Publ. Comp. 1965.
[7] A. Tyc, Jordan decomposition and the ring of constants of locally finite derivations, Preprint 1993.
