Section 10.2.
Homework #2
Masaya Sato
2. Show that the relation ”is R-module isomorphic isomorphic to” is an equivalenc equivalencee relation relation on any set
of R-modules. Proof. (Reflexive) For every R-module M consider the identity R-module map idM : M →
M . Then idM is an isomorphism and M is isomorphic to itself. (Symmetric) Suppose R-modules M and N are isomorphic via an R-module isomorphism isomorphism 1 ϕ : M → N . Then ϕ : N → M is also an isomorphism and thus N is isomorphic to M . (Transitive) (Transitive) Suppose M is isomorphic to N , N , and N is isomorphic to L via R-module isomorphisms ϕ : M → N and ψ : N → L, respectiv respectively ely.. Then their composite composite map ψ ◦ ϕ is an isomorphism between M and L. Thus M is isomorphic to L. −
4. Let A be any
Z-module,
let a be any element of A and let n be a positive integer. Prove that the map ϕa : Z/nZ → A given by ϕa (k) = ka is a well defined Z-module homomorphism ∼ An , where An = {a ∈ A|na = 0} (so An if and only if na if na = 0. Prove that Hom Z (Z/nZ, A) = is the annihilator in A of the ideal (n ( n) of Z). Proof. (⇒) Suppose first that ϕa is a well-defined
Z-module
homomorphism. Then
0a = 0 na = ϕa (n) = ϕa (0) = 0a as desired. (⇐) Conversely suppose that na = 0. Then let l ∈ k be an element in the equivalence class k. Then for some m ∈ Z k = l + nm and (nm))a ϕa (k) = ϕa(l + nm) nm) = (l + nm) nm)a = la + (nm = la + (mn (mn))a = la + m(na) na) = la = ϕa (l), so ϕa is a well-defined map. Moreover for all k, l ∈ Z/nZ ϕa(k + l) = ϕa (k + l) = (k + l)a = ka + la = ϕa (k) + ϕa (l) and for all r ∈ Z ϕa (rk) rk ) = ϕa (rk) rk ) = (rk) rk )a = r (ka) ka) = rϕa(k ) and thus ϕa is a Z-module homomorphism. ∼ An , where An = {a ∈ A|na = 0}. Sinc Now we claim that Hom Z(Z/nZ, A) = Sincee ev every ery homomorphism ϕ ∈ HomZ (Z/nZ, A) is of the form ϕa(k ) = ka, ka, where a ∈ A, consider a map Ψ : HomZ (Z/nZ, A) → An defined by Ψ(ϕ Ψ(ϕa) = ϕa (n) = na. na. Abstract Abstract Algebra Algebra by Dummit and Foote 1
Section 10.2.
Homework #2
Masaya Sato
Then for all ϕa and ϕa in HomZ (Z/nZ, A)
Ψ(ϕ Ψ(ϕa + ϕa ) = (ϕa + ϕa )(n )(n) = ϕa (n) + ϕa (n) = Ψ(ϕ Ψ(ϕa ) + Ψ(ϕ Ψ(ϕa )
and for all r ∈ Z Ψ(rϕ Ψ(rϕa ) = (rn) Ψ(ϕa ) rn)a = r(na) na) = rΨ(ϕ and thus Ψ is a Z-module -module homomo homomorph rphism ism.. Moreo Moreove verr the argument argument above above implies implies that that na = 0 if and only if each ϕa is a well-defined homomorphism. Therefore Ψ is bijective and HomZ (Z/nZ, A) and An are isomorphic. 5. Exhibit all
Z-module
Solution: For ϕ :
homomorphisms from Z/30Z to Z/21Z.
Z/30Z
→ Z/21Z observe that a Z-module homomorphism ϕ is a homomorphism from a finite cyclic group Z/30Z to another cyclic group Z/21Z. The order k of ϕ(1) divides 21, the order of Z/21Z. Moreover 0 = kϕ(1) kϕ(1) = ϕ(k1) = ϕ(k ),
so k ∈ ker ϕ and k divides 30, which is the order of Z/30Z. Therefore ϕ(1) is 0, 7, or 14 and then they all determine distinct homomorphisms.
∼ Z/(n, m)Z. 6. Prove that Hom Z (Z/nZ, Z/mZ) = Proof. Let ϕ :
(1). Then Then the → Z/mZ be a Z-module homomorphism, and let k = ϕ(1). order of k is m/( ( m, k) is the g.c.d of m and k , and the order divides both n m/(m, k), where (m, and m. So if d if d is the g.c.d of n and m, then Z/nZ
n = n d and
m = m d,
where n and m are relatively prime. Then
m/( m/(m, k)|n, i.e. there exists some p ∈ Z such that p(m/( m/(m, k)) = n. So
pm = (m, k)n ⇒ pm d = (m, k)n d ⇒ pm = (m, k)n .
Since n and m are relatively prime, m divides (m, (m, k) and hence m divides k. In oth other er words, m/d = m/( Now consider consider a homomorph homomorphism ism ϕ : Z/nZ → Z/mZ m/(m, n) divides k . Now explicitly defined by ϕ(x) = (km/d) km/d)x,
where k = 0, · · · , d − 1. Therefore there are d homomorphisms and they form a cyclic group of order d = (n, m). Hence HomZ (Z/nZ, Z/mZ) ∼ = Zd = Z(n,m) ∼ = Z/(n, m)Z.
Abstract Abstract Algebra Algebra by Dummit and Foote 2
Section 10.2.
Homework #2
Masaya Sato
7. Let z be a fixed element of the center of R. Prove that the map m → zm is an R-module
homomorphism from M to itself. Sho Show w that for a commutativ commutativee ring R the map from R to EndR (M ) given by r → rI is a ring homomorphism (where I is the identity endomorphism). Proof. We first claim that ϕ : M → M defined by ϕ(m) = zm, zm , where z is some fixed element
of the center Z (R) of R, is an R-module homomorp homomorphism. hism. Then for all all m, n ∈ M and all r∈R ϕ(m + rn) rn) = z (m + rn) rn) = zm + z (rn) rn) = zm + (zr (zr))n = zm + (rz (rz))m = zm + r (zm) zm ) = ϕ(m) + rϕ( rϕ(m) since z ∈ Z (R), i.e. z commutes with every element in R under multiplic multiplication. ation. So ϕ is an R-module homomorphism. Next suppose that R is a commutative ring and let Φ : R → EndR (M ) be a map defined by Φ(r Φ(r) = rI , where I : M → M is the identity endomorphism. Then for all r, s ∈ R Φ(r Φ(r + s) = (r + s)I = rI + sI = Φ(r Φ(r) + Ψ(s Ψ(s) and Φ(rs Φ(rs)) = (rs) )(sI ) = Ψ(r Ψ(r)Ψ(s )Ψ(s) rs)I = (sr) sr)I = (rI )(sI since R is commutative, I 2 = I , and (rI )(sI )(sI )(m )(m) = (rI )(s )(s(I (m))) = (rI (rI )( )(I (rI )(sm )(sm)) = (rsI )(m )(m) I (sm)) sm)) = (rI for every m ∈ M . 9. Let R be a commutative ring. Prove that Hom R (R, M ) and M are isomorphic as left R-
modules. [Show that each element of Hom R (R, M ) is determined by its value on the identity of R. Proof. Observe first that Hom R (R, M ) is an R-module, and define an R-module map Ψ :→
HomR (R, M ) → M by Ψ(f Ψ(f )) = f (1), (1), where 1 ∈ R is the multiplicative identity. Also denote o ∈ HomR (R, M ) is the zero function, i.e. o(r) = 0 for all r ∈ R. If f If f ∈ HomR (R, M ) is not the zero function, then Ψ(f Ψ( f )) = f (1) 0 f (1) = since f (1) f (1) = 0 implies that (1) = 0 ∀r ∈ R. f ( f (r) = f ( f (r.1) r.1) = r.f (1) Abstract Abstract Algebra Algebra by Dummit and Foote 3
Section 10.2.
Homework #2
Masaya Sato
Now we claim that Ψ is a bijective bijective R-module homomorphism. Then for all f, g ∈ HomR (R, M ) and all r ∈ R Ψ is well-defined and Ψ(f Ψ(f + r.g) r.g) = (f + r.g)(1) r.g)(1) = f (1) f (1) + r.g(1) r.g(1) = f (1) (rg)(1) )(1) f (1) + (rg = f (1) f (1) + r.( r.(g (1)) = Ψ(f Ψ(f )) + r.Ψ( r.Ψ(gg ). So Ψ is an R-module -module homomo homomorph rphism ism.. Next Next suppose that that Ψ(f Ψ(f )) = Ψ (g (g ) for all f, g ∈ HomR (R, M ). ). Then Ψ(f Ψ(f )) = Ψ(g Ψ(g) ⇒ f (1) f (1) = g(1) ⇒ f (1) f (1) − g (1) = 0 ⇒ (f − g)(1) = 0. According to the argument above f − g is the zero function, i.e. f − g = o. Thus f = g and Ψ is injective. Moreover for every m ∈ M construct some f ∈ HomR (R, M ) such that m = f (1). f (1). This map f is well-defined since for all r ∈ R and m1 , m2 ∈ M if (r, m1 ) ∈ f and
(r, m2 ) ∈ f
then m1 = f ( f (r) = f ( f (r1) = rf (1) rf (1) = rm and m2 = f ( f (r) = f ( f (r1) = rf (1) rf (1) = rm, rm, so m1 = m2 . Theref Therefore ore Ψ is surjec surjectiv tivee and hence hence a biject bijectiv ivee R-module homomorphis homomorphism. m. Thus HomR (R, M ) is isomorphic to M as left R-modules. 10. Let R be a commutative ring. Prove that Hom R (R, R) and R are isomorphic as rings.
Proof. Since a commutative ring R is a left module over itself, Hom R (R, R) is isomorphic to
-modules by result from problem problem #9. So all we have have to claim is that an R-module R as left R-modules isomorphism Ψ : HomR (R, R) → R respects multiplication. Then for all f, g ∈ HomR (R, R), where g (1) = r, Ψ(f Ψ(f ◦ g ) = (f ◦ g )(1) = f (r) = rf (1) rf (1) = rΨ(f Ψ(f )) = Ψ(f Ψ(f ))r = Ψ(f Ψ(f )Ψ( )Ψ(gg ) since R is commutativ commutative. e. Therefor Thereforee Ψ is a ring isomorphism isomorphism and hence HomR (R, R) is isomorphic to R. Abstract Abstract Algebra Algebra by Dummit and Foote 4
Section 10.2.
Homework #2
Masaya Sato
11. Let A1 , A1 , · · · , An be R-modules and let Bi be a submodule of A of Ai for each i = 1, 2, · · · , n.
Prove that (A1 × · · · × An )/(B1 × · · · × Bn) ∼ = (A1 /B1 ) × · · · × (A1 /B1 ). Proof. Define an R-module homomorphism π : A1 × · · · × An → (A1 /B1 ) × · · · × (A1 /B1 ) by
π(a1 , . . . , an ) = (a1 + B1 ) × · · · × (an + Bn). Observe that π is well-defined and surjective by its construction. Moreover the kernel ker π is given by ker π = {(a1, . . . , an) ∈ A|π(a1 , . . . , an ) = (0 + B1 ) × . . . (0 + Bn )} = {(a1, . . . , an) ∈ A|(a1 + B1 ) × · · · × (an + Bn ) = (0 + B1) × . . . (0 + Bn )} = {(a1, . . . , an) ∈ A|ai ∈ Bi ∀i = 1, · · · , n} = B1 × · · · × Bn. Therefore by the First Isomorphism Theorem for an R-module homomorphism (A1 × · · · × An )/ ker π ∼ = im π and hence
(A1 × · · · × An )/(B1 × · · · × Bn) ∼ = (A1 /B1 ) × · · · × (A1 /B1 )
as desired. 13. Let I be a nilpotent ideal in a commutative ring R, let M and N be R-modules
and let ϕ : M → N be an R-module -module homomor homomorphi phism. sm. Sho Show w that that if the induced induced map ϕ : M/IM → N/IN is surjective, then ϕ is surjective. Proof. Suppose that I is a nilpotent ideal, i.e. there is an integer r so that I r = {0}.
Since ϕ : M/IM → N/IN is surjective, for every n + I N ∈ N + I N there exists some m + I M ∈ M + I M such that ϕ(m + I M ) = n + I N . N . Observe that ϕ(I M ) ⊂ I N because ϕ is an R-module homomorph homomorphism. ism. So ϕ(m) − n ∈ I N and thus N = ϕ(M ) + I N because ϕ is surjective. Then N = ϕ(M ) + I N = ϕ(M ) + I (ϕ(M ) + I N ) = ϕ(M ) + I 2 N since I ϕ(M ) = ϕ(I M ) = ϕ(M ). ). Therefore at the k-th stage N = ϕ(M ) + I k N and this is true for every non-negative integer k. Hence N = ϕ(M ) + I r N = ϕ(M ) + {0}N . This implies that N = ϕ(M ) and thus ϕ is surjective as desired. Abstract Abstract Algebra Algebra by Dummit and Foote 5
