homomorphism maps subalgebras to subalgebras. Second, Theorems 9.22,

9.23, 9.24, 9.25, 9.26, and 9.27 carry over mutatis mutandis from rings to

R-algebras.

17.1 Algebras 361

Example 17.5. Since C contains R as a subring, we may naturally view C

as an R-algebra. The complex conjugation map on C that sends a + bi to

a ’ bi, for a, b ∈ R, is an R-algebra automorphism on C (see Example 9.5).

2

Example 17.6. Let p be a prime, and let F be the ¬eld Zp . If E is an

F -algebra, with associated map „ : F ’ E, then the map ρ : E ’ E that

sends ± ∈ E to ±p is an F -algebra homomorphism. To see this, note that

E is either trivial, or contains a copy of Zp . In the former case, there is

nothing really to prove. In the latter case, E has characteristic p, and so

the fact that ρ is a ring homomorphism follows from Example 9.42 (the

“freshman™s dream”); moreover, by Fermat™s little theorem, for all a ∈ F ,

we have „ (a)p = „ (ap ) = „ (a). 2

Polynomial evaluation

Let E be an R-algebra with associated map „ : R ’ E. Any polynomial

g ∈ R[X] naturally de¬nes a function on E: if g = i gi Xi , with each gi ∈ R,

and ± ∈ E, then

„ (gi )±i .

g(±) :=

i

For ¬xed ± ∈ E, the polynomial evaluation map ρ : R[X] ’ E sends

g ∈ R[X] to g(±) ∈ E. It is easily veri¬ed that ρ is an R-algebra homomor-

phism (where we naturally view R[X] as an R-algebra via inclusion). The

image of ρ is denoted R[±], and is a subalgebra of E. Indeed, R[±] is the

smallest subalgebra of E containing ±.

Note that if E contains R as a subring, then the notation R[±] has the

same meaning as that introduced in Example 9.39.

We next state a very simple, but extremely useful, fact:

Theorem 17.1. Let ρ : E ’ E be an R-algebra homomorphism. Then for

any g ∈ R[X] and ± ∈ E, we have

ρ(g(±)) = g(ρ(±)).

Proof. Let „ : R ’ E and „ : R ’ E be the associated maps. Let

362 More rings

i ∈ R[X]. Then we have

g= i gi X

„ (gi )±i ) = ρ(„ (gi )±i )

ρ(g(±)) = ρ(

i i

ρ(„ (gi ))ρ(±i ) = „ (gi )ρ(±)i

=

i i

= g(ρ(±)). 2

As a special case of Theorem 17.1, if E = R[·] for some · ∈ E, then every

element of E can be expressed as g(·) for some g ∈ R[X], and ρ(g(·)) =

g(ρ(·)); hence, the action of ρ is completely determined by its action on ·.

Example 17.7. Let E := R[X]/(f ) for some monic polynomial f ∈ R[X], so

that E = R[·], where · := [X]f , and let E be any R-algebra.

Suppose that ρ : E ’ E is an R-algebra homomorphism, and that · :=

ρ(·). The map ρ sends g(·) to g(· ), for g ∈ R[X]. Also, since f (·) = 0E ,

we have 0E = ρ(f (·)) = f (· ). Thus, · must be a root of f .

Conversely, suppose that · ∈ E is a root of f . Then the polynomial

evaluation map from R[X] to E that sends g ∈ R[X] to g(· ) ∈ E is an R-

algebra homomorphism whose kernel contains f , and this gives rise to the

R-algebra homomorphism ρ : E ’ E that sends g(·) to g(· ), for g ∈ R[X].

One sees that complex conjugation is just a special case of this construction

(see Example 9.44). 2

R-algebras as R-modules

If E is an R-algebra, with associated map „ : R ’ E, we may naturally

view E as an R-module, where we de¬ne a scalar multiplication operation

as follows: for a ∈ R and ± ∈ E, de¬ne

a · ± := „ (a)±.

The reader may easily verify that with scalar multiplication so de¬ned, E is

an R-module.

Of course, if E is an algebra over a ¬eld F , then it is also a vector space

over F .

Exercise 17.1. Show that any ring E may be viewed as a Z-algebra.

Exercise 17.2. Show that the only R-algebra homomorphisms from C into

itself are the identity map and the complex conjugation map.

17.2 The ¬eld of fractions of an integral domain 363

Exercise 17.3. Let E be an R-algebra, viewed as an R-module as discussed

above.

(a) Show that for all a ∈ R and ±, β ∈ E, we have a · (±β) = (a · ±)β.

(b) Show that a subring S of E is a subalgebra if and only if it is also

submodule.

(c) Show that if E is another R-algebra, then a ring homomorphism

ρ : E ’ E is an R-algebra homomorphism if and only if it is an

R-linear map.

Exercise 17.4. This exercise develops an alternative characterization of

R-algebras. Let R be a ring, and let E be a ring, together with a scalar

multiplication operation, that makes E into an R-module. Further suppose

that for all a ∈ R and ±, β ∈ E, we have a(±β) = (a±)β. De¬ne the

map „ : R ’ E that sends a ∈ R to a · 1E ∈ E. Show that „ is a ring

homomorphism, so that E is an R-algebra, and also show that „ (a)± = a±

for all a ∈ R and ± ∈ E.

17.2 The ¬eld of fractions of an integral domain

Let D be any integral domain. Just as we can construct the ¬eld of rational

numbers by forming fractions involving integers, we can construct a ¬eld

consisting of fractions whose numerators and denominators are elements of

D. This construction is quite straightforward, though a bit tedious.

To begin with, let S be the set of all pairs of the form (a, b), with a, b ∈ D

and b = 0D . Intuitively, such a pair (a, b) is a “formal fraction,” with

numerator a and denominator b. We de¬ne a binary relation ∼ on S as

follows: for (a1 , b1 ), (a2 , b2 ) ∈ S, we say (a1 , b1 ) ∼ (a2 , b2 ) if and only if

a1 b2 = a2 b1 . Our ¬rst task is to show that this is an equivalence relation:

Lemma 17.2. For all (a1 , b1 ), (a2 , b2 ), (a3 , b3 ) ∈ S, we have

(i) (a1 , b1 ) ∼ (a1 , b1 );

(ii) (a1 , b1 ) ∼ (a2 , b2 ) implies (a2 , b2 ) ∼ (a1 , b1 );

(iii) (a1 , b1 ) ∼ (a2 , b2 ) and (a2 , b2 ) ∼ (a3 , b3 ) implies (a1 , b1 ) ∼ (a3 , b3 ).

Proof. (i) and (ii) are rather trivial, and we do not comment on these any

further. As for (iii), assume that a1 b2 = a2 b1 and a2 b3 = a3 b2 . Multiplying

the ¬rst equation by b3 we obtain a1 b3 b2 = a2 b3 b1 and substituting a3 b2 for

a2 b3 on the right-hand side of this last equation, we obtain a1 b3 b2 = a3 b2 b1 .

Now, using the fact that b2 is non-zero and that D is an integral domain,

we may cancel b2 from both sides, obtaining a1 b3 = a3 b1 . 2

364 More rings

Since ∼ is an equivalence relation, it partitions S into equivalence classes,

and for (a, b) ∈ S, we denote by [a, b] the equivalence class containing (a, b),

and we denote by K the collection of all such equivalence classes. Our

next task is to de¬ne addition and multiplication operations on equivalence

classes, mimicking the usual rules of arithmetic with fractions. We want

to de¬ne the sum of [a1 , b1 ] and [a2 , b2 ] to be [a1 b2 + a2 b1 , b1 b2 ], and the

product of [a1 , b1 ] and [a2 , b2 ] to be [a1 a2 , b1 b2 ]. Note that since D is an

integral domain, if b1 and b2 are non-zero, then so is the product b1 b2 , and

therefore [a1 b2 + a2 b1 , b1 b2 ] and [a1 a2 , b1 b2 ] are indeed equivalence classes.

However, to ensure that this de¬nition is unambiguous, and does not depend

on the particular choice of representatives of the equivalence classes [a1 , b1 ]

and [a2 , b2 ], we need the following lemma.

Lemma 17.3. For (a1 , b1 ), (a1 , b1 ), (a2 , b2 ), (a2 , b2 ) ∈ S with (a1 , b1 ) ∼

(a1 , b1 ) and (a2 , b2 ) ∼ (a2 , b2 ), we have

(a1 b2 + a2 b1 , b1 b2 ) ∼ (a1 b2 + a2 b1 , b1 b2 )

and

(a1 a2 , b1 b2 ) ∼ (a1 a2 , b1 b2 ).