7.4. A DIRECT SUM DECOMPOSITION 157

vector space over F. As usual, for α ∈ F [x]≡ D,αm≡ α (L)m. For p a monic irreduciblepolynomial,

Vp ≡{

m ∈V : pkm = 0 for some k ∈ N}

That is, eventually pkm = 0. It might be possible that k could change for different m ∈ V .Note that if p is invertible, then Vp = 0 because xpm = 0 if and only if m =

(x−1)p 0 = 0, so

nothing is lost from considering only irreducible non constant polynomials. It is obviousthat Vp is a subgroup of the module V and is itself a module. Indeed, if m∈Vp so that pkm=0 for some k, then 0 = α pkm = pkαm = 0 also and so αm ∈ V . If m, m̂ ∈ Vp, then lettingkm,km̂ be the exponents for m, m̂, let k ≥max(km,km̂) and pk (m+ m̂) = pkm+ pkm̂ = 0 sothe sum m+ m̂ is in Vp if m, m̂ are.

Proposition 7.4.1 Let p1, · · · , pn be monic irreducible nonconstant polynomials. Let V bea finite dimensional vector space. Then(

Vp1 + · · ·+Vp j−1 +Vp j+1 + · · ·+Vpn

)∩Vp j = 0

and so ∑i Vpi =⊕

i Vpi .

Proof: This follows from the observation that ∏i̸= j pkii and p

k jj are relatively prime. If

q is monic and divides the second, then it is of the form pm jj ,m j ≤ k j. If q divides the

first, then q is ∏i̸= j pmii ,mi ≤ ki. Thus ∏i ̸= j pmi

i = pm jj contradicting Theorem 1.13.9 about

uniqueness of factorization. Since the irreducible polynomials are distinct, we must have allm j,mi equal to 0 and q = 1 so these two, ∏i̸= j pki

i and pk jj are relatively prime as claimed. If

m ∈(

Vp1 + · · ·+Vp j−1 +Vp j+1 + · · ·+Vpn

)∩Vp j , then m = ∑i ̸= j mi and so there exist ki,k j

such that pkii mi = 0 and p

k jj m = 0. Since ∏i̸= j pki

i and pk jj are relatively prime, there exist

σ ,τ such that1 = σ ∏

i ̸= jpki

i + τ pk jj (*)

Then do both sides of ∗ to m.

m =

(σ ∏

i ̸= jpki

i

)(=m

∑i ̸= j

mi

)+ τ p

k jj m = 0

This yields m = 0 and verifies the conclusion of the proposition.It follows from Lemma 6.0.2 that if mi ∈ Vpi , and if ∑i mi = 0, then each mi = 0 so

∑i Vpi =⊕

i Vpi . ■

Lemma 7.4.2 Let V be a vector space over a field F and let L ∈L (V,V ) have minimumpolynomial α (x) = ∏

ni=1 pki

i (x) in which the pi are distinct irreducible monic polynomials.Let D≡ F [x] and for α ∈ D,α (m)≡ α (L)(m). Let

V̂pi ≡{

m ∈V : pkii m = 0

}⊆Vpi , (7.4)

(Note that here ki is fixed. ) Then

Dm⊆ V̂p1 ⊕·· ·⊕V̂pn ⊆Vp1 ⊕·· ·⊕Vpn (7.5)

7.4. A DIRECT SUM DECOMPOSITION 157vector space over F. As usual, for & € F [x] = D, am = a (L) m. For p a monic irreduciblepolynomial,Y= {meV : pkm=0 for some k e N}That is, eventually p‘m = 0. It might be possible that k could change for different m € V.Note that if p is invertible, then V, = 0 because x?m = 0 if and only if m= (x7! )’0 = 0, sonothing is lost from considering only irreducible non constant polynomials. It is obviousthat V,, is a subgroup of the module V and is itself a module. Indeed, if m € V, so that pkm=0 for some k, then 0 = ap*m = pam = 0 also and so am € V. If m,n € V,, then lettingkn, ki be the exponents for m, 71, let k > max (kn,kj) and p* (m+ m) = p‘m+ ph =0 sothe sum m+ mis in V, if m, 7m are.Proposition 7.4.1 Let p,,--- , Pp, be monic irreducible nonconstant polynomials. Let V bea finite dimensional vector space. Then(Von +o V pp a Vpn ++ Vpn) OV) =0and so Yi Vp; = BV p;-Proof: This follows from the observation that []jz; p; and pil are relatively prime. Ifq is monic and divides the second, then it is of the form pi! ,mj; <k;. If q divides thefirst, then q is [];z;p;",mji < kj. Thus [];; p)" = pi! contradicting Theorem 1.13.9 aboutuniqueness of factorization. Since the irreducible polynomials are distinct, we must have allmj,mj; equal to 0 and gq = | so these two, |]; p; and p,! are relatively prime as claimed. Ifim © (Voy +2 Voi 1 FV pj +2°* + Vpy) AVpjs then m = ¥iz;; and so there exist kik;i+such that pi mj; = 0 and pi! m = 0. Since |]; pi and pi! are relatively prime, there existo, 7 such thatk; kj1 =o[]p; + tp; (*)iAjThen do both sides of * to m.kj =" kjm=|o[[p;'] | Ym | +tp’m=0iAj iAjThis yields m = 0 and verifies the conclusion of the proposition.It follows from Lemma 6.0.2 that if m; € V,,, and if )};m; = 0, then each m; = 0 soLiVp; = Di Voie aLemma 7.4.2 Let V be a vector space over a field F and let LE &(V,V) have minimumpolynomial a (x) =TTL, pi (x) in which the p; are distinct irreducible monic polynomials.Let D = F [x] and for a& € D,a(m) = a (L) (m). Let%p, = {meV: piim=0} CVp, (7.4)(Note that here k; is fixed. ) ThenDm CVp, +++ BVp, Vp, B+ OVp, (7.5)