unique non-negative integer p 0 such that ker · = (p); this p is called the characteristic of R

and denoted char R.

Lemma 1.9. If R is an integral domain, its characteristic char R is a prime.

Proof. Consider p = char R. If p = 0 we are done. So suppose that p > 0. The quotient

monomorphism · : Z/ ker · ’’ R identi¬es Z/ ker · with the subring im · = im · of the integral

domain R. But every subring of an integral domain is itself an integral domain, hence Z/ ker · is

an integral domain. Now by Proposition 1.7(i), ker · = (p) is prime ideal and so by Example 1.8,

p is a prime.

Remark 1.10. When discussing a ring with unit R, we can consider it as containing as a

subring of the form Z/(char R) since the quotient homomorphism · : Z/(char R) ’’ R gives

an isomorphism Z/(char R) ’’ im ·, allowing us to identify these rings. In particular, every

integral domain contains as a subring either Z = Z/(0) (if char R = 0) or Z/(p) if p = char R > 0

is a non-zero prime. This subring is sometimes called the characteristic subring of R. The rings

Z and Z/n = Z/(n) for n > 0 are often called core rings. When considering integral domains,

the rings Z and Fp = Z/p = Z/(p) for p > 0 a prime are called prime rings.

Here is a useful and important fact about rings containing a ¬nite prime ring.

Theorem 1.11 (Idiot™s Binomial Theorem). Let R be a commutative ring with unit con-

taining Fp for some prime p > 0. If u, v ∈ R, then

(u + v)p = up + v p .

Proof. We have p1 = 0 in R. Now the Binomial Theorem gives

p p’1 p p’2 2 p

(u + v)p = up + uv p’1 + v p .

u v+ u v + ··· +

1 2 p’1

Suppose that 1 j p ’ 1. Then we have

p p (p ’ 1)!

= .

j j! (p ’ j)!

There are no factors of p appearing in (p ’ 1)!, j! or (p ’ j)!, so since this number is an integer

it must be divisible by p, i.e.,

p

(1.1a) p| ,

j

1.1. RECOLLECTIONS ON INTEGRAL DOMAINS AND FIELDS 3

or equivalently

p

(1.1b) ≡ 0 (mod p).

j

Hence in R we have

p

1 = 0.

j

Combining this fact with the above expansion, we obtain the required equation in R,

(u + v)p = up + v p .

A commutative ring with unit k is a ¬eld if every non-zero element u ∈ k is a unit in k i.e.,

there is an element v ∈ k such that uv = vu = 1. We usually write u’1 for this element v which

is necessarily unique and called the multiplicative inverse of u in k.

Example 1.12. If n 0, the quotient ring Z/n is a ¬eld if and only if n is a positive prime.

Proposition 1.13. If k is a ¬eld then it is an integral domain.

Proof. Suppose that u, v ∈ k and uv = 0. Then if u = 0 we can multiply this equation by

u’1 to obtain

v = u’1 uv = 0,

hence v = 0. So at least one of u, v must be 0.

A non-zero element p ∈ R is irreducible if whenever p = uv with u, v ∈ R, either u or v is a

unit.

Lemma 1.14. Let R be an integral domain. If p ∈ R is a non-zero prime then it is an

irreducible.

Proof. Suppose that p = uv for some u, v ∈ R. Then p | u or p | v, and we might as well

assume that u = tp for some t ∈ R. Then (1 ’ tv)p = 0 and so tv = 1, showing that v is a unit

with inverse t.

Now let D be an integral domain. A natural question to ask is whether D can be found as

a subring of a ¬eld. This is certainly true for the integers Z which are contained in the ¬eld of

rational numbers Q, and for a prime p > 0, the prime ring Fp is itself a ¬eld.

Definition 1.15. The ¬elds Q and Fp where p > 0 is prime are the prime ¬elds.

Of course, we can view Z as a subring of any sub¬eld of the complex numbers so an answer

to this question may not be unique! However, there is always a ˜smallest™ such ¬eld which is

unique up to an isomorphism.

Theorem 1.16. Let D be an integral domain.

(i) There is a ¬eld of fractions of D, Fr(D), which contains D as a subring.

(ii) If • : D ’’ F is a ring monomorphism into a ¬eld F , there is a unique homomorphism

• : Fr(D) ’’ F such that •(t) = •(t) for all t ∈ D ⊆ Fr(D).

• /

D <F

inc

∃! •

Fr(D)

™

Proof. (i) Consider the set

™™¦

™

P(D) = {(a, b) : a, b ∈ D, b = 0}.

Now introduce an equivalence relation ∼ on P(D), namely

(a , b ) ∼ (a, b) ⇐’ ab = a b.

4 1. INTEGRAL DOMAINS, FIELDS AND POLYNOMIAL RINGS

Of course, it is necessary to check that this relation is an equivalence relation; this is left as an

exercise. We denote the equivalence class of (a, b) by [a, b] and the set of equivalence classes by

Fr(D).

We will de¬ne addition and multiplication on Fr(D) by

[a, b] + [c, d] = [ad + bc, bd], [a, b][c, d] = [ac, bd].

We need to verify that these operations are well de¬ned. For example, if [a , b ] = [a, b] and

[c , d ] = [c, d], then

(a d + b c )bd = a d bd + b c bd = ab d d + b bcd = (ad + bc)b d ,

and so (a d + b c , b d ) ∼ (ad + bc, bd), hence addition is well de¬ned. A similar calculation

shows that (a c , b d ) ∼ (ac, bd), so multiplication is also well de¬ned. It is now straightforward

to show that Fr(D) is a commutative ring with zero 0 = [0, 1] and unit 1 = [1, 1]. In fact, as we

will soon see, Fr(D) is a ¬eld.

Let [a, b] ∈ Fr(D). Then [a, b] = [0, 1] if and only if (0, 1) ∼ (a, b) which is equivalent to

requiring that a = 0; notice that for any b = 0, [0, b] = [0, 1]. We also have [a, b] = [1, 1] if and

only if a = b.

Now let [a, b] ∈ Fr(D) be non-zero, i.e., a = 0. Then b = 0 and [a, b], [b, a] ∈ Fr(D) satisfy

[a, b][b, a] = [ab, ba] = [1, 1] = 1,

so [a, b] has [b, a] as an inverse. This shows that Fr(D) is a ¬eld.

We can view D as a subring of Fr(D) using the map

j : D ’’ Fr(D); j(t) = [t, 1]

which is a ring homomorphism; it is easy to check that it is a monomorphism. Therefore we

may identify t ∈ D with j(t) = [t, 1] ∈ Fr(D) and D with the subring im j ⊆ Fr(D).

(ii) Consider the function

¦(a, b) = •(a)•(b)’1 .

¦ : P(D) ’’ F ;

If (a , b ) ∼ (a, b) then

¦(a , b ) = •(a )•(b )’1 = •(a )•(b)•(b)’1 •(b )’1

= •(a b)•(b)’1 •(b )’1

= •(ab )•(b )’1 •(b)’1

= •(a)•(b )•(b )’1 •(b)’1

= •(a)•(b)’1 = ¦(a, b),

so ¦ is constant on each equivalence class of ∼. Hence we can de¬ne the function

• : Fr(D) ’’ F ; •([a, b]) = ¦(a, b).

It is straightforward to verify that • is a ring homomorphism which agrees with • on the subring

D ⊆ Fr(D).

The next three corollaries are left as an exercise. Corollary 1.19 is sometimes said to imply

that the construction of Fr(D) is functorial in the integral domain D.

Corollary 1.17. If F is a ¬eld then F = Fr(F ).

Corollary 1.18. If D is a subring of a ¬eld F , then Fr(D) ⊆ Fr(F ) = F is the smallest

sub¬eld of F containing D.

Corollary 1.19. If D1 and D2 are integral domains and • : D1 ’’ D2 a ring monomor-

phism, there is a unique induced ring homomorphism •— : Fr(D1 ) ’’ Fr(D2 ) for which •— (t) =

1.2. POLYNOMIAL RINGS 5

•(t) if t ∈ D1 ⊆ Fr(D1 ).

• / D2

D1

inc inc

Fr(D1 ) • / Fr(D2 )

—

Moreover, this construction has the properties

• if • : D1 ’’ D2 and θ : D2 ’’ D3 are monomorphisms between integral domains then

θ— —¦ •— = (θ —¦ •)— as homomorphisms Fr(D1 ) ’’ Fr(D3 );

• for any integral domain D, the identity homomorphism id : D ’’ D induces the iden-

tity homomorphism (id)— = id : Fr(D) ’’ Fr(D).

• θ id

/ D2 / D3 /D

D1 D

inc inc

inc inc inc

/ Fr(D)

Fr(D1 ) • / Fr(D2 ) / Fr(D3 ) Fr(D)

id— = id

θ—

—

Remark 1.20. When working with a ¬eld of fractions it is usual to adopt the notation

a

= a/b = [a, b]

b

for the equivalence class of (a, b). The rules for algebraic manipulation of such symbols are the

usual ones for working with fractions, i.e.,

a1 a2 a1 b2 + a2 b1 a1 a2 a1 a2 a1 a2

+ = , — = = .

b1 b2 b1 b2 b1 b2 b1 b2 b1 b2

The ¬eld of fractions of an integral domain is sometimes called its ¬eld of quotients, however as

the word quotient is also associated with quotient rings we prefer to avoid using that terminology.

1.2. Polynomial rings

Let R be a commutative ring. We will make frequent use of the ring R[X] of polynomials

over R in an indeterminate X. This consists of elements of form

p(X) = p0 + p1 X + · · · + pm X m

where m 0 and p0 , p1 , . . . , pm ∈ R; such p(X) are called polynomials. Addition and multipli-

cation in R[X] are de¬ned by

(p0 + p1 X + · · · + pm X m ) + (q0 + q1 X + · · · + qm X m ) =

(p0 + q0 ) + (p1 + q1 )X + · · · + (pm + qm )X m ),

and

(p0 + p1 X + · · · + pm X m )(q0 + q1 X + · · · + qm X m ) =

(p0 q0 ) + (p0 q1 + p1 q0 )X + · · · + (p0 qm + p1 qm’1 + · · · + pm’1 q1 + pm q0 )X 2m .

Then R[X] is a commutative ring with the constant polynomials 0 and 1 as its zero and unit.

We identify r ∈ R with the obvious constant polynomial; this allows us to view R as a subring

of R[X] and the inclusion function inc : R ’’ R[X] is a monomorphism.

More generally, we inductively can de¬ne the ring of polynomials in n indeterminates

X1 , . . . , Xn over R,

R[X1 , . . . , Xn ] = R[X1 , . . . , Xn’1 ][Xn ]

for n 1. Again there is an inclusion monomorphism inc : R ’’ R[X1 , . . . , Xn ] which sends

each element of R to the corresponding constant polynomial.

These polynomial rings have an important universal property.

6 1. INTEGRAL DOMAINS, FIELDS AND POLYNOMIAL RINGS

Theorem 1.21 (Homomorphism Extension Property). Let • : R ’’ S be a ring homomor-

phism.

(i) For each s ∈ S there is a unique ring homomorphism •s : R[X] ’’ S for which

• •s (r) = •(r) for all r ∈ R,

• •s (X) = s.

• /S

R =

inc

∃! •s

R[X]

(ii) For n 1 and s1 , . . . , sn ∈ S, there is a unique ring homomorphism

•s1 ,...,sn : R[X1 , . . . , Xn ] ’’ S

for which

• •s1 ,...,sn (r) = •(r) for all r ∈ R,

• •s1 ,...,sn (Xi ) = si for i = 1, . . . , n.

•

8/ S

R

inc

∃! •s1 ,...,sn

R[X1 , . . . , Xn ]

™

Proof. (Sketch)

™™¦

™

(i) For a polynomial p(X) = p0 + p1 X + · · · + pm X m ∈ R[X], we de¬ne

•s (p(X)) = p0 + p1 s + · · · + pm sm ∈ S.

(1.2)

It is then straightforward to check that •s is a ring homomorphism with the stated properties

and moreover is the unique such homomorphism.

(ii) is proved by induction on n using (i).

We will refer to •s1 ,...,sn as the extension of • by evaluation at s1 , . . . , sn . It is standard to

write

p(s1 , . . . , sn ) = •s1 ,...,sn (p(X1 , . . . , Xn )).

An extremely important special case occurs when we start with the identity homomorphism

id : R ’’ R and r1 , . . . , rn ∈ R; then we have the homomorphism

µr1 ,...,rn = idr1 ,...,rn : R[X1 , . . . , Xn ] ’’ R.

Slightly more generally we may take the inclusion of a subring inc : R ’’ S and s1 , . . . , sn ∈ S;

then

µs1 ,...,sn = incs1 ,...,sn : R[X1 , . . . , Xn ] ’’ S

is called evaluation at s1 , . . . , sn and we denote its image by

R[s1 , . . . , sn ] = µs1 ,...,sn R[X1 , . . . , Xn ] ⊆ S.

Then R[s1 , . . . , sn ] is a subring of S, called the subring generated by s1 , . . . , sn over R.

Here is an example illustrating how we will use such evaluation homomorphisms.

Example 1.22. Consider the inclusion homomorphism inc : Q ’’ C. We have the evalua-

tion at i homomorphism µi , for which µi (X) = i. We easily ¬nd that µi Q[X] ⊆ C is a subring

Q[i] ⊆ C consisting of the complex numbers of form a + bi with a, b ∈ Q.

Notice that if we had used ’i instead of i, evaluation at ’i, µ’i , would also give µ’i Q[X] =

Q[i]. These evaluation homomorphisms are related by complex conjugation since

µ’i (p(X)) = µi (p(X)),

1.2. POLYNOMIAL RINGS 7

which is equivalent to the functional equation

µ’i = ( ) —¦ µi .

Notice also that in these examples we have

ker µ’i = ker µi = (X 2 + 1) Q[X],

hence we also have

Q[i] ∼ Q[X]/(X 2 + 1).

=

In fact (X 2 + 1) is actually a maximal ideal and so Q[i] ⊆ C is a sub¬eld; later we will write

Q(i) for this sub¬eld.

Proposition 1.23. Let R be an integral domain.

(i) The ring R[X] of polynomials in an indeterminate X over R is an integral domain.

(ii) The ring R[X1 , . . . , Xn ] of polynomials in n indeterminates X1 , . . . , Xn over R is an

integral domain.

Corollary 1.24. If k is a ¬eld and n 1, the polynomial ring k[X1 , . . . , Xn ] in n indeter-

minates X1 , . . . , Xn is an integral domain.

As we will make considerable use of such rings we describe in detail some of their important

properties. First we recall long division in a polynomial ring k[X] over a ¬eld k; full details can

be found in a basic course on commutative rings or any introductory book on this subject.

Theorem 1.25 (Long Division). Let k be a ¬eld and X an indeterminate. Let f (X), d(X) ∈

k[X] and assume that d(X) = 0 so that deg d(X) > 0. Then there are unique polynomials