Modules and vector spaces

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13

Modules and vector spaces

In this chapter, we introduce the basic definitions and results concerning modules over a ringRand vector spaces over a field F. The reader may have seen some of these notions before, but perhaps only in the context of vector spaces over a specific field, such as the real or complex numbers, and not in the context of, say, finite fields likeZp.

13.1 Definitions, basic properties, and examples

Throughout this section,Rdenotes a ring (i.e., a commutative ring with unity).

Definition 13.1. AnR-moduleis a set M together with an addition operation on M and a function µ:R×M →M, such that the set M under addition forms an abelian group, and moreover, for all c,d∈Rand α,β ∈M, we have:

(i) µ(c,µ(d,α))=µ(cd,α);

(ii) µ(c+d,α) =µ(c,α)+µ(d,α);

(iii) µ(c,α+β) =µ(c,α)+µ(c,β);

(iv) µ(1R,α)=α.

One may also call anR-moduleM a module overR, and elements ofR are sometimes calledscalars. The functionµin the definition is called ascalar multiplication map, and the value µ(c,α) is called the scalar productof c and α.

Usually, we shall simply writecα (orc·α) instead ofµ(c,α). When we do this, properties (i)–(iv) of the definition may be written as follows:

c(dα) =(cd)α, (c+d)α=cα+dα, c(α+β) =cα+cβ, 1Rα=α.

Note that there are two addition operations at play here: addition in R (such as c+d) and addition inM (such asα+β). Likewise, there are two multiplication operations at play: multiplication inR(such ascd) and scalar multiplication (such

358

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13.1 Definitions, basic properties, and examples 359 ascα). Note that by property (i), we may writecdαwithout any ambiguity, as both possible interpretations,c(dα) and (cd)α, yield the same value.

For fixed c ∈ R, the map that sendsα ∈ M tocα ∈ M is a group homomorphism with respect to the additive group operation ofM (by property (iii) of the definition); likewise, for fixed α ∈ M, the map that sends c ∈ R tocα ∈ M is a group homomorphism from the additive group of R into the additive group of M (by property (ii)). Combining these observations with basic facts about group homomorphisms (see Theorem 6.19), we may easily derive the following basic facts aboutR-modules:

Theorem 13.2. If M is a module overR, then for all c∈R,α ∈M, and k∈Z, we have:

(i) 0R·α=0M; (ii) c·0M =0M;

(iii) (−c)α=−(cα)=c(−α);

(iv) (kc)α=k(cα)=c(kα).

Proof. Exercise. 2

AnR-moduleMmay betrivial, consisting of just the zero element 0M. IfRis the trivial ring, then anyR-moduleM is trivial, since for everyα ∈ M, we have α=1Rα=0Rα=0M.

Example 13.1. The ring R itself can be viewed as an R-module in the obvious way, with addition and scalar multiplication defined in terms of the addition and multiplication operations ofR. 2

Example 13.2. The set R^×n, which consists of all of n-tuples of elements ofR, forms an R-module, with addition and scalar multiplication defined component- wise: forα=(a1, . . . ,a_n)∈R^×n,β=(b1, . . . ,b_n)∈R^×n, andc∈R, we define

α+β :=(a₁+b₁, . . . ,an+bn) and cα:=(ca₁, . . . ,can). 2

Example 13.3. The ring of polynomialsR[X] over Rforms anR-module in the natural way, with addition and scalar multiplication defined in terms of the addition and multiplication operations of the polynomial ring. 2

Example 13.4. As in Example 7.39, letf be a non-zero polynomial overRwith lc(f) ∈R^∗, and consider the quotient ring E := R[X]/(f). ThenEis a module overR, with addition defined in terms of the addition operation ofE, and scalar multiplication defined byc[g]f :=[c]f·[g]f =[cg]f, forc∈Randg∈R[X]. 2 Example 13.5. Generalizing Example 13.3, if E is any ring containing R as a

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subring (i.e.,Eis an extension ring ofR), thenEis a module overR, with addition and scalar multiplication defined in terms of the addition and multiplication operations ofE. 2

Example 13.6. Any abelian groupG, written additively, can be viewed as a Z- module, with scalar multiplication defined in terms of the usual integer multiplication map (see Theorem 6.4). 2

Example 13.7. LetGbe any group, written additively, whose exponent dividesn.

Then we may define a scalar multiplication that maps [k]n ∈Znandα∈Gtokα.

That this map is unambiguously defined follows from the fact thatGhas exponent dividingn, so that ifk ≡ k⁰ (modn), we havekα−k⁰α = (k−k⁰)α = 0G, since n|(k−k⁰). It is easy to check that this scalar multiplication map indeed makesG into aZn-module. 2

Example 13.8. Of course, viewing a group as a module does not depend on whether or not we happen to use additive notation for the group operation. If we specialize the previous example to the groupG = Z^∗p, where pis prime, then we may viewG as a Zp−1-module. However, since the group operation itself is written multiplicatively, the “scalar product” of [k]_p−1 ∈ Zp−1 andα ∈ Z^∗p is the powerα^k. 2

Example 13.9. If M₁, . . . ,M_k are R-modules, then so is their direct product M₁× · · · ×Mk, where addition and scalar product are defined component-wise. If M =M₁=· · ·=Mk, we write this asM^×k. 2

Example 13.10. IfI is an arbitrary set, andM is anR-module, then Map(I,M), which is the set of all functionsf : I → M, may be naturally viewed as anR- module, with point-wise addition and scalar multiplication: forf,g ∈Map(I,M) andc∈R, we define

(f+g)(i) :=f(i)+g(i) and (cf)(i) :=cf(i) for alli∈I. 2

13.2 Submodules and quotient modules

Again, throughout this section,R denotes a ring. The notions of subgroups and quotient groups extend in the obvious way toR-modules.

Definition 13.3. LetMbe anR-module. A subsetNofM is asubmodule (over R) of Mif

(i) Nis a subgroup of the additive groupM, and

(ii) cα∈Nfor all c∈Rand α∈N(i.e.,N is closed under scalar multiplication).

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13.2 Submodules and quotient modules 361 It is easy to see that a submoduleNof anR-moduleM is also anR-module in its own right, with addition and scalar multiplication operations inherited fromM. Expanding the above definition, we see that a non-empty subset N ofM is a submodule if and only if for allc∈Rand allα,β ∈N, we have

α+β ∈N, −α∈N, and cα∈N.

Observe that the condition−α ∈Nis redundant, as it is implied by the condition cα∈Nwithc=−1R.

Clearly, {0M}andM are submodules ofM. Fork ∈ Z, it is easy to see that not only arekMandM{k}subgroups ofM (see Theorems 6.7 and 6.8), they are also submodules ofM. Moreover, forc∈R,

cM :={cα:α∈M} and M{c}:={α∈M :cα=0M} are also submodules ofM. Further, forα∈M,

Rα:={cα:c∈R}

is a submodule ofM. Finally, ifN₁ andN₂are submodules ofM, thenN₁+N₂ andN₁∩N₂ are not only subgroups ofM, they are also submodules ofM. We leave it to the reader to verify all these facts: they are quite straightforward.

Letα₁, . . . ,αk ∈M. The submodule

Rα₁+· · ·+Rαk

is called the submodule (overR) generated byα₁, . . . ,α_k. It consists of allR- linear combinations

c₁α₁+· · ·+ckαk,

where thec_i’s are elements ofR, and is the smallest submodule ofMthat contains the elementsα₁, . . . ,α_k. We shall also write this submodule ashα₁, . . . ,α_ki_R. As a matter of definition, we allow k = 0, in which case this submodule is {0M}.

We say that M isfinitely generated (over R) if M = hα₁, . . . ,α_ki_R for some α₁, . . . ,αk ∈M.

Example 13.11. For a given integer`≥0, defineR[X]<_`to be the set of polynomials of degree less than`. The reader may verify thatR[X]<_`is a submodule of theR-moduleR[X], and indeed, is the submodule generated by 1,X, . . . ,X^`⁻¹. If

`=0, then this submodule is the trivial submodule{0R}. 2

Example 13.12. LetG be an abelian group. As in Example 13.6, we can view G as a Z-module in a natural way. Subgroups of G are just the same thing as submodules ofG, and fora₁, . . . ,a_k ∈G, the subgroupha₁, . . . ,a_kiis the same as the submoduleha₁, . . . ,a_ki_Z. 2

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Example 13.13. As in Example 13.1, we may view the ring R itself as an R- module. With respect to this module structure, ideals ofRare just the same thing as submodules ofR, and fora₁, . . . ,a_k ∈R, the ideal (a1, . . . ,a_k) is the same as the submoduleha₁, . . . ,a_ki_R. Note that fora ∈ R, the ideal generated by amay be written either asaR, using the notation introduced in §7.3, or asRa, using the notation introduced in this section. 2

Example 13.14. If E is an extension ring of R, then we may view E as anR- module, as in Example 13.5. It is easy to see that every ideal ofEis a submodule;

however, the converse is not true in general. Indeed, the submodule R[X]<_` of R[X] discussed in Example 13.11 is not an ideal of the ringR[X]. 2

IfN is a submodule ofM, then in particular, it is also a subgroup ofM, and we can form the quotient groupM/N in the usual way (see §6.3), which consists of all cosets [α]N, where α ∈ M. Moreover, because N is closed under scalar multiplication, we can also define a scalar multiplication on M/N in a natural way. Namely, forc∈Randα∈M, we define

c·[α]N :=[cα]N.

As usual, one must check that this definition is unambiguous, which means that cα ≡ cα⁰ (modN) wheneverα ≡ α⁰ (mod N). But this follows (as the reader may verify) from the fact that N is closed under scalar multiplication. One can also easily check that with scalar multiplication defined in this way, M/N is an R-module; it is called thequotient module (overR) ofM moduloN.

Example 13.15. Suppose E is an extension ring of R, and I is an ideal of E.

ViewingE as an R-module,I is a submodule ofE, and hence the quotient ring E/I may naturally be viewed as anR-module, with scalar multiplication defined byc·[α]I :=[cα]I forc∈Randα ∈E. Example 13.4 is a special case of this, applied to the extension ringR[X] and the ideal (f). 2

EXERCISE 13.1. Show that ifN is a submodule of anR-moduleM, then a set P ⊆N is a submodule ofM if and only ifP is a submodule ofN.

EXERCISE 13.2. LetM₁ andM₂ be R-modules, and letN₁ be a submodule of M₁andN₂a submodule ofM₂. Show thatN₁×N₂is a submodule ofM₁×M₂. EXERCISE 13.3. Show that if R is non-trivial, then the R-module R[X] is not finitely generated.

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13.3 Module homomorphisms and isomorphisms 363

13.3 Module homomorphisms and isomorphisms

Again, throughout this section,Ris a ring. The notion of a group homomorphism extends in the obvious way toR-modules.

Definition 13.4. Let M and M⁰ be modules over R. An R-module homomor- phismfromM toM⁰is a functionρ:M →M⁰, such that

(i) ρis a group homomorphism from M toM⁰, and (ii) ρ(cα)=cρ(α)for allc∈Rand α∈M.

An R-module homomorphism is also called an R-linear map. We shall use this terminology from now on. Expanding the definition, we see that a map ρ : M → M⁰ is an R-linear map if and only ifρ(α+β) = ρ(α) + ρ(β) and ρ(cα)=cρ(α) for allα,β ∈M and allc∈R.

Example 13.16. IfN is a submodule of anR-moduleM, then the inclusion map i:N →M is obviously anR-linear map. 2

Example 13.17. SupposeNis a submodule of anR-moduleM. Then thenatural map(see Example 6.36)

ρ: M →M/N α7→[α]N

is not just a group homomorphism, it is also easily seen to be anR-linear map. 2 Example 13.18. LetM be an R-module, and let k be an integer. Then the k- multiplication map onM (see Example 6.38) is not only a group homomorphism, but it is also easily seen to be anR-linear map. Its image is the submodulekM, and its kernel the submoduleM{k}. 2

Example 13.19. LetM be anR-module, and letcbe an element ofR. The map ρ: M →M

α7→cα

is calledc-multiplication map on M, and is easily seen to be anR-linear map whose image is the submodule cM, and whose kernel is the submodule M{c}.

The set of allc∈Rfor whichcM ={0M}is called theR-exponent ofM, and is easily seen to be an ideal ofR. 2

Example 13.20. LetM be anR-module, and letαbe an element ofM. The map ρ: R→M

c7→cα

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is easily seen to be anR-linear map whose image is the submoduleRα (i.e., the submodule generated byα). The kernel of this map is called theR-order ofα, and is easily seen to be an ideal ofR. 2

Example 13.21. Generalizing the previous example, letM be anR-module, and letα₁, . . . ,αk be elements ofM. The map

ρ: R^×k →M

(c1, . . . ,c_k)7→c₁α₁+· · ·+c_kα_k

is easily seen to be anR-linear map whose image is the submoduleRα₁+· · ·+Rαk

(i.e., the submodule generated byα₁, . . . ,α_k). 2

Example 13.22. Suppose thatM₁, . . . ,M_k are submodules of an R-moduleM. Then the map

ρ: M₁× · · · ×Mk →M

(α₁, . . . ,α_k) 7→α₁+· · ·+α_k

is easily seen to be anR-linear map whose image is the submoduleM₁+· · ·+Mk.2 Example 13.23. LetE be an extension ring ofR. As we saw in Example 13.5, Emay be viewed as anR-module in a natural way. Letα ∈ E, and consider the α-multiplication map onE, which sendsβ ∈Etoαβ ∈E. Then it is easy to see that this is anR-linear map. 2

Example 13.24. LetE andE⁰ be extension rings ofR, which may be viewed as R-modules as in Example 13.5. Suppose thatρ:E→E⁰is a ring homomorphism whose restriction toRis the identity map (i.e.,ρ(c) =cfor allc∈R). Thenρis an R-linear map. Indeed, for everyc∈Randα,β∈E, we haveρ(α+β)=ρ(α)+ρ(β) andρ(cα)=ρ(c)ρ(α)=cρ(α). 2

Example 13.25. LetG andG⁰ be abelian groups. As we saw in Example 13.6, GandG⁰may be viewed asZ-modules. In addition, every group homomorphism ρ:G→G⁰is also aZ-linear map. 2

Since anR-module homomorphism is also a group homomorphism on the under- lying additive groups, all of the statements in Theorem 6.19 apply. In particular, an R-linear map is injective if and only if the kernel is trivial (i.e., contains only the zero element). However, in the case ofR-module homomorphisms, we can extend Theorem 6.19, as follows:

Theorem 13.5. Letρ:M →M⁰be anR-linear map. Then:

(i) for every submodule N of M,ρ(N) is a submodule of M⁰; in particular (setting N:=M),Imρis a submodule of M⁰;

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13.3 Module homomorphisms and isomorphisms 365 (ii) for every submoduleN⁰ of M⁰,ρ⁻¹(N⁰) is a submodule of M⁰; in partic-

ular (settingN⁰ :={0M⁰}),Kerρis a submodule of M. Proof. Exercise. 2

Theorems 6.20 and 6.21 have naturalR-module analogs, which the reader may easily verify:

Theorem 13.6. If ρ:M →M⁰andρ⁰:M⁰→M⁰⁰are R-linear maps, then so is their compositionρ⁰◦ρ:M →M⁰⁰.

Theorem 13.7. Let ρi :M → M_i⁰, for i = 1, . . . ,k, be R-linear maps. Then the map

ρ: M →M₁⁰ × · · · ×M_k⁰ α7→(ρ1(α), . . . ,ρ_k(α)) is an R-linear map.

If an R-linear mapρ : M → M⁰ is bijective, then it is called anR-module isomorphismofM withM⁰. If such anR-module isomorphismρexists, we say thatMis isomorphic toM⁰, and writeM ∼=M⁰. Moreover, ifM =M⁰, thenρis called anR-module automorphismonM.

Theorems 6.22–6.26 also have naturalR-module analogs, which the reader may easily verify:

Theorem 13.8. If ρis anR-module isomorphism of Mwith M⁰, then the inverse functionρ⁻¹is an R-module isomorphism of M⁰with M.

Theorem 13.9 (First isomorphism theorem). Let ρ :M → M⁰be an R-linear map with kernel K and imageN⁰. Then we have anR-module isomorphism

M/K∼=N⁰. Specifically, the map

ρ: M/K →M⁰ [α]K 7→ρ(α) is an injective R-linear map whose image isN⁰.

Theorem 13.10. Letρ:M →M⁰be anR-linear map. Then for every submodule Nof M with N ⊆Kerρ, we may define an R-linear map

ρ: M/N →M⁰ [α]N 7→ρ(α).

Moreover,Imρ=Imρ, and ρis injective if and only if N =Kerρ.

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Theorem 13.11 (Internal direct product). LetMbe an R-module with submodules N₁,N₂, whereN₁∩N₂={0M}. Then we have anR-module isomorphism

N₁×N₂∼=N₁+N₂

given by the map

ρ: N₁×N₂→N₁+N₂ (α₁,α₂)7→α₁+α₂.

Theorem 13.12. Let M and M⁰ be R-modules, and consider the R-module of functionsMap(M,M⁰)(see Example 13.10). Then

HomR(M,M⁰) :={σ∈Map(M,M⁰) :σis an R-linear map}

is a submodule of Map(M,M⁰).

Example 13.26. Consider again the R-module R[X]/(f) discussed in Exam- ple 13.4, wheref ∈ R[X] is of degree` ≥ 0 and lc(f) ∈ R^∗. As an R-module, R[X]/(f) is isomorphic to R[X]<` (see Example 13.11). Indeed, based on the observations in Example 7.39, the map ρ : R[X]<` → R[X]/(f) that sends a polynomialg∈R[X] of degree less than`to [g]f ∈R[X]/(f) is an isomorphism ofR[X]<_`withR[X]/(f). Furthermore,R[X]<_`is isomorphic as anR-module to R^×^`. Indeed, the map ρ⁰ : R[X]<_` → R^×^`that sendsg = P_`−1

i=0aiXⁱ ∈R[X]<_`to (a₀, . . . ,a_`₋₁)∈R^×^`is an isomorphism ofR[X]<_`withR^×^`. 2

EXERCISE 13.4. Verify that the “is isomorphic to” relation onR-modules is an equivalence relation; that is, for allR-modulesM₁,M₂,M₃, we have:

(a) M₁∼=M₁;

(b) M₁∼=M₂impliesM₂∼=M₁;

(c) M₁∼=M₂andM₂∼=M₃impliesM₁∼=M₃.

EXERCISE 13.5. Letρi : Mi → M_i⁰, fori = 1, . . . ,k, beR-linear maps. Show that the map

ρ: M₁× · · · ×M_k →M₁⁰ × · · · ×M_k⁰ (α₁, . . . ,αk)7→(ρ₁(α₁), . . . ,ρk(αk)) is anR-linear map.

EXERCISE 13.6. Letρ:M →M⁰be anR-linear map, and letc∈R. Show that ρ(cM) =cρ(M).

EXERCISE 13.7. Letρ:M →M⁰be anR-linear map. LetNbe a submodule of

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13.4 Linear independence and bases 367 M, and letτ :N → M⁰be the restriction ofρtoN. Show thatτ is anR-linear map and that Kerτ=Kerρ∩N.

EXERCISE 13.8. SupposeM₁, . . . ,Mk are R-modules. Show that for eachi = 1, . . . ,k, the projection mapπi:M₁× · · · ×Mk → Mithat sends (α₁, . . . ,αk) to αiis a surjectiveR-linear map.

EXERCISE 13.9. Show that ifM = M₁ ×M₂ forR-modulesM₁ andM₂, and N₁is a subgroup ofM₁ andN₂is a subgroup ofM₂, then we have anR-module isomorphismM/(N1×N₂)∼=M₁/N₁×M₂/N₂.

EXERCISE 13.10. LetM be an R-module with submodulesN₁ andN₂. Show that we have anR-module isomorphism (N1+N₂)/N2∼=N₁/(N1∩N₂).

EXERCISE13.11. LetMbe anR-module with submodulesN₁,N₂, andA, where N₂⊆N₁. Show that (N₁∩A)/(N₂∩A) is isomorphic to a submodule ofN₁/N₂. EXERCISE 13.12. Letρ:M →M⁰be anR-linear map with kernelK. LetN be a submodule ofM. Show that we have anR-module isomorphismM/(N+K) ∼= ρ(M)/ρ(N).

EXERCISE13.13. Letρ:M →M⁰be a surjectiveR-linear map. LetS be the set of all submodules ofM that contain Kerρ, and letS⁰ be the set of all submodules ofM⁰. Show that the setsS andS⁰are in one-to-one correspondence, via the map that sendsN ∈ S toρ(N) ∈ S⁰.

13.4 Linear independence and bases Throughout this section,Rdenotes a ring.

Definition 13.13. LetM be anR-module, and let {α_i}ⁿ_i₌₁be a family of elements of M. We say that {α_i}ⁿ_i₌₁

(i) islinearly dependent (over R) if there exist c₁, . . . ,c_n ∈R, not all zero, such thatc₁α₁+· · ·+c_nα_n=0M;

(ii) islinearly independent (over R)if it is not linearly dependent;

(iii) spansM (over R)if for everyα∈M, there existc₁, . . . ,c_n ∈Rsuch that c₁α₁+· · ·+c_nα_n =α;

(iv) is abasis for M (overR)if it is linearly independent and spansM.

The family{α_i}ⁿ_i₌₁always spans some submodule ofM, namely, the submodule Ngenerated byα₁, . . . ,α_n. In this case, we may also callN thesubmodule (over R) spanned by{α_i}ⁿ_i₌₁.

The family{α_i}ⁿ_i₌₁may contain duplicates, in which case it is linearly dependent

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(unlessRis trivial). Indeed, if, say,α₁ = α₂, then settingc₁ := 1,c₂ := −1, and c₃:=· · ·:=c_n:=0, we have the linear relationPn

i=1c_iα_i=0M.

If the family {α_i}ⁿ_i₌₁ contains 0M, then it is also linear dependent (unless Ris trivial). Indeed, if, say,α₁ =0M, then settingc₁:=1 andc₂:=· · ·:=c_n :=0, we have the linear relationPn

i=1ciαi=0M.

The family{αi}ⁿ_i₌₁may also be empty (i.e.,n= 0), in which case it is linearly independent, and spans the submodule{0M}.

In the above definition, the ordering of the elementsα₁, . . . ,αnmakes no differ- ence. As such, when convenient, we may apply the terminology in the definition to any family{α_i}_i∈I, whereIis an arbitrary, finite index set.

Example 13.27. Consider theR-moduleR^×n. Defineα₁, . . . ,αn∈R^×nas follows:

α₁:= (1, 0, . . . , 0), α₂ :=(0, 1, 0, . . . , 0), . . . , αn:=(0, . . . , 0, 1);

that is,α_ihas a 1 in position iand is zero everywhere else. It is easy to see that {α_i}ⁿ_i₌₁is a basis forR^×n. Indeed, for allc₁, . . . ,c_n∈R, we have

c₁α₁+· · ·+cnαn=(c₁, . . . ,cn),

from which it is clear that {αi}ⁿ_i₌₁ spans R^×n and is linearly independent. The family{αi}ⁿ_i₌₁is called thestandard basisforR^×n. 2

Example 13.28. Consider theZ-module Z^×3. In addition to the standard basis, which consists of the tuples

(1, 0, 0), (0, 1, 0), (0, 0, 1), the tuples

α₁:=(1, 1, 1), α₂ :=(0, 1, 0), α₃:=(2, 0, 1)

also form a basis. To see this, first observe that for allc₁,c₂,c₃,d₁,d₂,d₃ ∈Z, we have

(d1,d₂,d₃) =c₁α₁+c₂α₂+c₃α₃ if and only if

d₁=c₁+2c3, d₂ =c₁+c₂, andd₃=c₁+c₃. (13.1) If (13.1) holds withd₁ = d₂ =d₃ = 0, then subtracting the equationc₁+c₃ = 0 fromc₁+2c3=0, we see thatc₃=0, from which it easily follows thatc₁=c₂=0.

This shows that the family{αi}³_i₌₁ is linearly independent. To show that it spans Z^×3, the reader may verify that for any givend₁,d₂,d₃∈Z, the values

c₁:=−d₁+2d₃, c₂:=d₁+d₂−2d₃, c₃ :=d₁−d₃

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13.4 Linear independence and bases 369 satisfy (13.1).

The family of tuples (1, 1, 1), (0, 1, 0), (1, 0, 1) is not a basis, as it is linearly dependent: the third tuple is equal to the first minus the second.

The family of tuples (1, 0, 12), (0, 1, 30), (0, 0, 18) is linearly independent, but does not spanZ^×3: the last component of anyZ-linear combination of these tuples must be divisible by gcd(12, 30, 18) =6. However, this family of tuples is a basis for theQ-moduleQ^×3. 2

Example 13.29. Consider again the submodule R[X]<_` of R[X], where _` ≥ 0, consisting of all polynomials of degree less than _` (see Example 13.11). Then {Xⁱ⁻¹^}^`_i₌₁is a basis forR[X]<_`overR. 2

Example 13.30. Consider again the ringE = R[X]/(f), wheref ∈ R[X] with deg(f) =`≥0 and lc(f)∈R^∗. As in Example 13.4, we may naturally viewEas a module overR. From the observations in Example 7.39, it is clear that{ξⁱ⁻¹}^`_i₌₁ is a basis forEoverR, whereξ :=[X]f ∈E. 2

The next theorem highlights a critical property of bases:

Theorem 13.14. If {αi}ⁿ_i₌₁is a basis for anR-moduleM, then the map ε: R^×n→M

(c₁, . . . ,cn)7→c₁α₁+· · ·+cnαn

is an R-module isomorphism. In particular, every element of Mcan be expressed in a unique way asc₁α₁+· · ·+cnαn, for c₁, . . . ,cn∈R.

Proof. We already saw thatεis anR-linear map in Example 13.21. Since{αi}ⁿ_i₌₁ is linearly independent, it follows that the kernel ofεis trivial, so thatεis injective.

Thatεis surjective follows immediately from the fact that{αi}ⁿ_i₌₁spansM. 2 The following is an immediate corollary of this theorem:

Theorem 13.15. Any twoR-modules with bases of the same size are isomorphic.

The following theorem develops an important connection between bases and linear maps.

Theorem 13.16. Let {α_i}ⁿ_i₌₁be a basis for anR-moduleM, and let ρ:M →M⁰ be anR-linear map. Then:

(i) ρis surjective if and only if {ρ(αi)}ⁿ_i₌₁spansM⁰;

(ii) ρis injective if and only if {ρ(αi)}ⁿ_i₌₁is linearly independent;

(iii) ρis an isomorphism if and only if {ρ(αi)}ⁿ_i₌₁is a basis forM⁰.

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Proof. By the previous theorem, we know that every element ofM can be written uniquely as P

ic_iα_i, where the c_i’s are in R. Therefore, every element in Imρ can be expressed as ρ(P

ic_iα_i) = P

ic_iρ(αi). It follows that Imρ is equal to the subspace ofM⁰spanned by{ρ(αi)}ⁿ_i₌₁. From this, (i) is clear.

For (ii), consider a non-zero elementP

iciαiofM, so that not allci’s are zero.

Now,P

iciαi ∈Kerρif and only ifP

iciρ(αi)=0M⁰, and thus, Kerρis non-trivial if and only if{ρ(αi)}ⁿ_i₌₁is linearly dependent. That proves (ii).

(iii) follows from (i) and (ii). 2

EXERCISE 13.14. LetM be an R-module. Suppose {α_i}ⁿ_i₌₁ is a linearly independent family of elements ofM. Show that for everyJ ⊆ {1, . . . ,n}, the subfamily{α_j}_j∈J is also linearly independent.

EXERCISE 13.15. Supposeρ:M →M⁰is anR-linear map. Show that if{α_i}ⁿ_i₌₁ is a linearly dependent family of elements of M, then {ρ(αi)}ⁿ_i₌₁ is also linearly dependent.

EXERCISE 13.16. Supposeρ : M → M⁰ is an injectiveR-linear map and that {αi}ⁿ_i₌₁is a linearly independent family of elements ofM. Show that{ρ(αi)}ⁿ_i₌₁is linearly independent.

EXERCISE 13.17. Suppose that{αi}ⁿ_i₌₁spans anR-moduleM and thatρ:M → M⁰is anR-linear map. Show that:

(a) ρis surjective if and only if{ρ(αi)}ⁿ_i₌₁spansM⁰; (b) if{ρ(αi)}ⁿ_i₌₁is linearly independent, thenρis injective.

13.5 Vector spaces and dimension Throughout this section,F denotes a field.

A module over a field is also called avector space. In particular, anF-module is called anF-vector space, or avector space overF.

For vector spaces overF, one typically uses the termssubspaceandquotient space, instead of (respectively) submodule and quotient module; likewise, one usually uses the termsF-vector space homomorphism,isomorphismandauto- morphism, as appropriate.

We now develop the basic theory of dimension for finitely generated vector spaces. Recall that a vector space V over F is finitely generated if we have V =hα₁, . . . ,αni_F for someα₁, . . . ,αnofV. The main results here are that

• every finitely generated vector space has a basis, and

• all such bases have the same number of elements.

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13.5 Vector spaces and dimension 371 Throughout the rest of this section,V denotes a vector space overF. We begin with a technical fact that will be used several times throughout this section:

Theorem 13.17. Suppose that{α_i}ⁿ_i₌₁is a linearly independent family of elements that spans a subspace W ( V, and that α_n₊₁ ∈ V \W. Then {α_i}ⁿ_i₌⁺¹₁ is also linearly independent.

Proof. Suppose we have a linear relation

0V =c₁α₁+· · ·+c_nα_n+c_n₊₁α_n₊₁,

where thec_i’s are inF. We want to show that all thec_i’s are zero. Ifc_n₊₁6=0, then we have

α_n₊₁=−c⁻¹_n₊₁(c₁α₁+· · ·+c_nα_n)∈W,

contradicting the assumption thatα_n₊₁ ∈/ W. Therefore, we must havec_n₊₁ = 0, and the linear independence of{α_i}ⁿ_i₌₁implies thatc₁=· · ·=c_n =0. 2

The next theorem says that every finitely generated vector space has a basis, and in fact, any family that spans a vector space contains a subfamily that is a basis for the vector space.

Theorem 13.18. Suppose {αi}ⁿ_i₌₁ is a family of elements that spansV. Then for some subsetJ ⊆ {1, . . . ,n}, the subfamily{α_j}_j∈J is a basis for V.

Proof. We prove this by induction onn. Ifn= 0, the theorem is clear, so assume n > 0. Consider the subspace W of V spanned by {αi}ⁿ⁻¹_i₌₁. By the induction hypothesis, for someK ⊆ {1, . . . ,n−1}, the subfamily{αk}k∈Kis a basis forW. There are two cases to consider.

Case 1: αn ∈ W. In this case, W = V, and the theorem clearly holds with J :=K.

Case 2: αn ∈/W. We claim that settingJ := K ∪ {n}, the subfamily {α_j}_j∈J is a basis for V. Indeed, since {α_k}_k∈K is linearly independent, and αn ∈/ W, Theorem 13.17 immediately implies that{α_j}_j∈J is linearly independent. Also, since{α_k}_k∈K spansW, it is clear that{α_j}_j∈J spansW +hα_ni_F =V. 2 Theorem 13.19. If V is spanned by some family of nelements of V, then every family of n+1elements of V is linearly dependent.

Proof. We prove this by induction onn. Ifn= 0, the theorem is clear, so assume thatn > 0. Let{αi}ⁿ_i₌₁ be a family that spansV, and let {βi}ⁿ_i₌⁺₁¹ be an arbitrary family of elements ofV. We wish to show that{β_i}ⁿ_i₌⁺₁¹is linearly dependent.

We know thatβ_n₊₁is a linear combination of theα_i’s, say,

β_n₊₁=c₁α₁+· · ·+c_nα_n. (13.2)

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If all thec_i’s were zero, then we would haveβ_n₊₁=0V, and so trivially,{β_i}ⁿ_i₌₁⁺¹is linearly dependent. So assume that somec_iis non-zero, and for concreteness, say c_n 6= 0. Dividing equation (13.2) through byc_n, it follows thatα_nis an F-linear combination ofα₁, . . . ,α_n−1,β_n₊₁. Therefore,

hα₁, . . . ,α_n−1,βn+1i_F ⊇ hα₁, . . . ,α_n−1i_F +hα_ni_F =V.

Now consider the subspaceW :=hβn+1iF and the quotient spaceV /W. Since the family of elementsα₁, . . . ,α_n−1,βn+1spansV, it is easy to see that{[αi]W}ⁿ⁻¹_i₌₁ spansV /W; therefore, by induction,{[βi]W}ⁿ_i₌₁is linearly dependent. This means that there existd₁, . . . ,dn∈F, not all zero, such thatd₁β₁+· · ·+dnβn≡0 (modW), which means that for somedn+1 ∈F, we haved₁β₁+· · ·+dnβn =dn+1βn+1. That proves that{β_i}ⁿ_i₌⁺₁¹is linearly dependent. 2

An important corollary of Theorem 13.19 is the following:

Theorem 13.20. If V is finitely generated, then any two bases forV have the same size.

Proof. If one basis had more elements than another, then Theorem 13.19 would imply that the first basis was linearly dependent, which contradicts the definition of a basis. 2

Theorem 13.20 allows us to make the following definition:

Definition 13.21. If V is finitely generated, the common size of any basis is called thedimensionof V, and is denoted dimF(V).

Note that from the definitions, we have dimF(V) = 0 if and only ifV is the trivial vector space (i.e., V = {0V}). We also note that one often refers to a finitely generated vector space as afinite dimensionalvector space. We shall give preference to this terminology from now on.

To summarize the main results in this section up to this point: if V is finite dimensional, it has a basis, and any two bases have the same size, which is called the dimension ofV.

Theorem 13.22. Suppose that dimF(V) = n, and that {αi}ⁿ_i₌₁ is a family of n elements of V. The following are equivalent:

(i) {α_i}ⁿ_i₌₁is linearly independent;

(ii) {α_i}ⁿ_i₌₁spans V; (iii) {αi}ⁿ_i₌₁is a basis for V.

Proof. LetW be the subspace ofV spanned by{αi}ⁿ_i₌₁.

First, let us show that (i) implies (ii). Suppose{αi}ⁿ_i₌₁ is linearly independent.

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13.5 Vector spaces and dimension 373 Also, by way of contradiction, suppose thatW (V, and chooseα_n₊₁ ∈V \W. Then Theorem 13.17 implies that {α_i}ⁿ_i₌₁⁺¹ is linearly independent. But then we have a linearly independent family ofn+1 elements ofV, which is impossible by Theorem 13.19.

Second, let us prove that (ii) implies (i). Let us assume that {αi}ⁿ_i₌₁is linearly dependent, and prove thatW ( V. By Theorem 13.18, we can find a basis for W among the αi’s, and since{αi}ⁿ_i₌₁ is linearly dependent, this basis must contain strictly fewer thannelements. Hence, dimF(W) < dimF(V), and therefore, W (V.

The theorem now follows from the above arguments, and the fact that, by definition, (iii) holds if and only if both (i) and (ii) hold. 2

We next examine the dimension of subspaces of finite dimensional vector spaces.

Theorem 13.23. Suppose that V is finite dimensional and W is a subspace of V. Then W is also finite dimensional, with dimF(W) ≤ dimF(V). Moreover, dimF(W) =dimF(V)if and only if W =V.

Proof. Suppose dimF(V) = n. Consider the set S of all linearly independent families of the form {αi}^m_i₌₁, where m ≥ 0 and each αi is inW. The setS is certainly non-empty, as it contains the empty family. Moreover, by Theorem 13.19, every member ofSmust have at mostnelements. Therefore, we may choose some particular element{αi}^m_i₌₁ofS, wheremis as large as possible. We claim that this family {α_i}^m_i₌₁ is a basis for W. By definition, {α_i}^m_i₌₁ is linearly independent and spans some subspace W⁰ of W. If W⁰ ( W, we can choose an element αm+1 ∈W\W⁰, and by Theorem 13.17, the family{α_i}^m_i₌⁺₁¹is linearly independent, and therefore, this family also belongs toS, contradicting the assumption thatmis as large as possible.

That proves thatW is finite dimensional with dimF(W)≤dimF(V). It remains to show that these dimensions are equal if and only ifW = V. Now, ifW = V, then clearly dimF(W) =dimF(V). Conversely, if dimF(W)=dimF(V), then by Theorem 13.22, any basis forW must already spanV. 2

Theorem 13.24. If V is finite dimensional, and W is a subspace of V, then the quotient spaceV /W is also finite dimensional, and

dimF(V /W) =dimF(V)−dimF(W).

Proof. Suppose that{α_i}ⁿ_i₌₁spansV. Then it is clear that{[αi]W}ⁿ_i₌₁spansV /W. By Theorem 13.18, we know thatV /W has a basis of the form{[αi]W}^`_i₌₁, where

` ≤ n (renumbering theαi’s as necessary). By Theorem 13.23, we know thatW has a basis, say{βj}^m_j₌₁. The theorem will follow immediately from the following:

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Claim.The elements

α₁, . . . ,α_`, β₁, . . . ,βm (13.3) form a basis forV.

To see that this family spans V, consider any element γ of V. Then since {[αi]W}^`_i₌₁spansV /W, we haveγ≡P

iciαi(modW) for somec₁, . . . ,c_` ∈F. If we setβ :=γ−P

iciαi∈W, then since{β_j}^m_j₌₁spansW, we haveβ =P

jdjβjfor somed₁, . . . ,dm ∈F, and henceγ =P

iciαi+P

jdjβj. That proves that the family of elements (13.3) spansV. To prove this family is linearly independent, suppose we have a relation of the form P

ic_iα_i + P

jd_jβ_j = 0V, where c₁, . . . ,c_` ∈ F andd₁, . . . ,d_m ∈ F. If any of thec_i’s were non-zero, this would contradict the assumption that{[αi]W}^`_i₌₁is linearly independent. So assume that all thec_i’s are zero. If any of thed_j’s were non-zero, this would contradict the assumption that {βj}^m_j₌₁ is linearly independent. Thus, all the ci’s and dj’s must be zero, which proves that the family of elements (13.3) is linearly independent. That proves the claim. 2

Theorem 13.25. If V is finite dimensional, then every linearly independent family of elements of V can be extended to form a basis for V.

Proof. One can prove this by generalizing the proof of Theorem 13.18. Alterna- tively, we can adapt the proof of the previous theorem. Let{β_j}^m_j₌₁ be a linearly independent family of elements that spans a subspaceW ofV. As in the proof of the previous theorem, if{[αi]W}^`_i₌₁is a basis for the quotient spaceV /W, then the elements

α₁, . . . ,α_`, β₁, . . . ,β_m form a basis forV. 2

Example 13.31. Suppose thatF is finite, say|F| = q, and thatV is finite dimensional, say dimF(V) = n. Then clearly |V| = qⁿ. If W is a subspace with dimF(W) =m, then|W|=q^m, and by Theorem 13.24, dimF(V /W)=n−m, and hence|V /W|=q^n−m. Just viewingV andW as additive groups, we know that the index ofW inV is [V :W]=|V /W|=|V|/|W|=q^n−m, which agrees with the above calculations. 2

We next consider the relation between the notion of dimension and linear maps.

First, observe that by Theorem 13.15, if two finite dimensional vector spaces have the same dimension, then they are isomorphic. The following theorem is the converse:

Theorem 13.26. If V is of finite dimensionn, andV is isomorphic toV⁰, thenV⁰ is also of finite dimensionn.