Theorem 14.12. Let ρ be an R-linear map from M into M . Then for any

submodule N contained in ker(ρ), the map ρ : M/N ’ img(ρ) that sends the

¯

coset ± + N for ± ∈ M to ρ(±) is unambiguously de¬ned and is an R-linear

map from M/N onto img(ρ) with kernel ker(ρ)/N .

Proof. Exercise. 2

Theorem 14.13. Let M be an R-module with submodules N1 , N2 . Then

the map ρ : N1 — N2 ’ N1 + N2 that sends (±1 , ±2 ) to ±1 + ±2 is a surjec-

tive R-linear map. Moreover, if N1 © N2 = {0M }, then ρ is an R-module

isomorphism of N1 — N2 with N1 + N2 .

Proof. Exercise. 2

Example 14.13. Let M be an R-module, and let m be an integer. Then

the m-multiplication on M is not only a group homomorphism, but it is an

R-linear map. 2

Example 14.14. Let M be an R-module, and let a be an element of R.

The a-multiplication map on M is the map that sends ± ∈ M to a± ∈ M .

This is an R-linear map whose image is aM , and whose kernel is M {a}. The

set of all a ∈ R for which aM = {0M } is called the R-exponent of M , and

is easily seen to be an ideal of R (verify). 2

Example 14.15. Let M be an R-module, and let ± be an element of M .

Then the map ρ : R ’ M given by ρ(a) = a± is an R-linear map. The

image of this map is ± R . The kernel of this map is called the R-order of

±, and is easily seen to be an ideal of R (verify). 2

Example 14.16. Consider again the R-module R[X]/(f ) discussed in Ex-

ample 14.3, where f is monic of degree . As an R-module, R[X]/(f ) is

isomorphic to R[X]< (see Example 14.9). Indeed, based on the observations

in Example 9.34, the map ρ : R[X]< ’ R[X]/(f ) that sends a polynomial

g ∈ R[X] of degree less than to [g]f ∈ R[X]/(f ) is an isomorphism of R[X]<

with R[X]/(f ). Furthermore, R[X]< is isomorphic as an R-module to R— .

306 Modules and vector spaces

’1

Indeed, the map ρ : R[X]< ’ R— that sends g = i=0 gi Xi ∈ R[X]< to

(g0 , . . . , g ’1 ) ∈ R— is an isomorphism of R[X]< with R— . 2

Example 14.17. Let E and E be ring extensions of the ring R. As we

saw in Example 14.4, E and E may be viewed as R-modules in a natural

way. Suppose that ρ : E ’ E is a ring homomorphism whose restriction to

R is the identity map (i.e., ρ(a) = a for all a ∈ R). Then ρ is an R-linear

map. Indeed, for any a ∈ R and ±, β ∈ E, we have ρ(± + β) = ρ(±) + ρ(β)

and ρ(a±) = ρ(a)ρ(±) = aρ(±). 2

14.4 Linear independence and bases

Throughout this section, R denotes a ring.

De¬nition 14.14. We say that an R-module M is ¬nitely generated

(over R) if it is spanned by a ¬nite number of elements, which is to say

that M = ±1 , . . . , ±n R for some ±1 , . . . , ±n ∈ M .

We say that a collection of elements ±1 , . . . , ±n in M is linearly de-

pendent (over R) if there exist a1 , . . . , an ∈ R, not all zero, such that

a1 ±1 + · · · an ±n = 0M ; otherwise, we say that ±1 , . . . , ±n are linearly in-

dependent (over R).

We say that a collection ±1 , . . . , ±n of elements in M is a basis for M

(over R) if it is linearly independent and spans M .

Note that in the above de¬nition, the collection of elements ±1 , . . . , ±n

may contain duplicates; the collection may also be empty (i.e., n = 0),

in which case, by de¬nition, it is a basis for the trivial submodule {0M }.

Note that the ordering of the elements ±1 , . . . , ±n makes no di¬erence in

any aspect of the de¬nition.

Example 14.18. Consider the R-module R—n . De¬ne ±1 , . . . , ±n ∈ R—n

as follows:

±1 := (1, 0, . . . , 0), ±2 := (0, 1, 0, . . . , 0), . . . , ±n := (0, . . . , 0, 1);

that is, ±i has a 1 in position i and is zero everywhere else. It is easy to

see that ±1 , . . . , ±n form a basis for R—n . Indeed, for any a1 , . . . , an ∈ R,

we have a1 ±1 + · · · + an ±n = (a1 , . . . , an ), from which it is clear that the ±i

span R—n and are linearly independent. The vectors ±1 , . . . , ±n form what

is called the standard basis for R—n . 2

Example 14.19. Consider the Z-module Z—3 . In addition to the standard

14.4 Linear independence and bases 307

basis

(1, 0, 0), (0, 1, 0), (0, 0, 1),

the vectors

±1 := (1, 1, 1), ±2 := (0, 1, 0), ±3 := (2, 0, 1)

also form a basis. To see this, ¬rst observe that for a1 , a2 , a3 , b1 , b2 , b3 ∈ Z,

we have

(b1 , b2 , b3 ) = a1 ±1 + a2 ±2 + a3 ±3

if and only if

b1 = a1 + 2a3 , b2 = a1 + a2 , and b3 = a1 + a3 . (14.1)

If (14.1) holds with b1 = b2 = b3 = 0, then subtracting the equation a1 +a3 =

0 from a1 + 2a3 = 0, we see that a3 = 0, from which it easily follows that

a1 = a2 = 0. This shows that the vectors are linearly independent. To show

that they span Z—3 , the reader may verify that for any given b1 , b2 , b3 ∈ Z,

the values

a1 := ’b1 + 2b3 , a2 := b1 + b2 ’ 2b3 , a3 := b1 ’ b3

satisfy (14.1).

The vectors

(1, 1, 1), (0, 1, 0), (1, 0, 1)

do not form a basis, as they are linearly dependent: the third vector is equal

to the ¬rst minus the second.

The vectors (1, 0, 12), (0, 1, 30), (0, 0, 18) are linearly independent, but do

not span Z—3 ” the last component of any Z-linear combination of these

vectors must be divisible by gcd(12, 30, 18) = 6. These vectors do, however,

form a basis for the Q-module Q—3 . 2

Example 14.20. If R is non-trivial, the ring of polynomials R[X] is not

¬nitely generated as an R-module, since any ¬nite set of polynomials spans

only polynomials of some bounded degree. 2

Example 14.21. Consider the submodule R[X]< of R[X], where ≥ 0. If

= 0, then R[X]< is trivial; otherwise, 1, X, . . . , X ’1 form a basis. 2

Example 14.22. Consider again the ring E = R[X]/(f ), where f ∈ R[X] is

monic of degree ≥ 0. If f = 1, then E is trivial; otherwise, 1, ·, · 2 , . . . , · ’1 ,

where · := [X]f ∈ E, form a basis for E over R. 2

The next theorem highlights a critical property of bases:

308 Modules and vector spaces

Theorem 14.15. If ±1 , . . . , ±n form a basis for M , then the map ρ : R—n ’

M that sends (a1 , . . . , an ) ∈ R—n to a1 ±1 + · · · + an ±n ∈ M is an R-module

isomorphism of R—n with M . In particular, every element of M can be

expressed in a unique way as a1 ±1 + · · · + an ±n , for a1 , . . . , an ∈ R.

Proof. To show this, one has to show (1) that ρ is an R-linear map, which

follows immediately from the de¬nitions, (2) that ρ is injective, which follows

immediately from the linear independence of ±1 , . . . , ±n , and (3) that ρ is

surjective, which follows immediately from the fact that ±1 , . . . , ±n span M .

2

The following theorems develop important connections among the notions

of spanning, linear independence, and linear maps.

Theorem 14.16. Suppose that ±1 , . . . , ±n span an R-module M and that

ρ : M ’ M is an R-linear map.

(i) ρ is surjective if and only if ρ(±1 ), . . . , ρ(±n ) span M .

(ii) If ρ(±1 ), . . . , ρ(±n ) are linearly independent, then ρ is injective.

Proof. Since the ±i span M , every element of M can be expressed as i ai ±i ,

where the ai are in R. It follows that the image of ρ consists of all elements

of M of the form ρ( i ai ±i ) = i ai ρ(±i ). That is, the image of ρ is the

submodule of M spanned by ρ(±1 ), . . . , ρ(±n ), which implies (i).

For (ii), suppose that ρ is not injective. Then ρ(±) = 0M for some

± = 0M , and since the ±i span M , we can write ± = i ai ±i , where the ai

are in R. Since ± is non-zero, some of the ai must be non-zero. So we have

0M = ρ( i ai ±i ) = i ai ρ(±i ), and hence ρ(±1 ), . . . , ρ(±n ) are linearly

dependent. 2

Theorem 14.17. Suppose ρ : M ’ M is an injective R-linear map and

that ±1 , . . . , ±n ∈ M are linearly independent. Then ρ(±1 ), . . . , ρ(±n ) are

linearly independent.

Proof. Suppose that 0M = i ai ρ(±i ) = ρ( i ai ±i ). Then, as ker(ρ) =

{0M }, we must have i ai ±i = 0M , and as the ±i are linearly independent,

all the ai must be zero. 2

Theorem 14.18. Let ±1 , . . . , ±n be a basis for an R-module M , and let

ρ : M ’ M be an R-linear map.

(i) ρ is surjective if and only if ρ(±1 ), . . . , ρ(±n ) span M .

(ii) ρ is injective if and only if ρ(±1 ), . . . , ρ(±n ) are linearly independent.