written as k1 · · · kn where ki = g and kj = 1 for j = i. By uniqueness of representation in
the internal direct product, hi = ki = 1 for all i, so g = 1.
(2) implies (3): If g belongs to Hi and, in addition, g = h1 · · · hi’1 with hj ∈ Hj , then
g = h1 · · · hi’1 1Hi+1 · · · 1Hn , hence g = 1 by (2).
(3) implies (1): If g ∈ G then since G = H1 · · · Hn we have g = h1 · · · hn with hi ∈ Hi .
Suppose that we have another representation g = k1 · · · kn with ki ∈ Hi . Let i be the
largest integer such that hi = ki . If i < n we can cancel the ht (= kt ), t > i, to get
h1 · · · hi = k1 · · · ki . If i = n then h1 · · · hi = k1 · · · ki by assumption. Now any product
of the Hi is a subgroup of G (as in (1.5.2), hi hj = hj hi for i = j, and the result follows
from (1.3.6)). Therefore
i’1
’1
∈
hi ki Hj ,
j=1
’1 ’1
and since hi ki ∈ Hi , we have hi ki = 1 by (3). Therefore hi = ki , which is a contra
diction. ™
Problems For Section 1.5
In Problems 1“5, Cn is a cyclic group of order n, for example, Cn = {1, a, . . . , an’1 } with
an = 1.
1. Let C2 be a cyclic group of order 2. Describe the multiplication table of the direct
product C2 — C2 . Is C2 — C2 cyclic?
22 CHAPTER 1. GROUP FUNDAMENTALS
2. Show that C2 — C2 is isomorphic to the four group (Section 1.2, Problem 6).
3. Show that the direct product C2 — C3 is cyclic, in particular, it is isomorphic to C6 .
4. If n and m are relatively prime, show that Cn — Cm is isomorphic to Cnm , and is
therefore cyclic.
5. If n and m are not relatively prime, show that Cn — Cm is not cyclic.
6. If p and q are distinct primes and G = p, H = q, show that the direct product G — H
is cyclic.
7. If H and K are arbitrary groups, show that H — K ∼ K — H.
=
8. If G, H and K are arbitrary groups, show that G — (H — K) ∼ (G — H) — K. In fact,
=
both sides are isomorphic to G — H — K.
Chapter 2
Ring Fundamentals
2.1 Basic De¬nitions and Properties
2.1.1 De¬nitions and Comments
A ring R is an abelian group with a multiplication operation (a, b) ’ ab that is associative
and satis¬es the distributive laws: a(b+c) = ab+ac and (a+b)c = ab+ac for all a, b, c ∈ R.
We will always assume that R has at least two elements, including a multiplicative identity
1R satisfying a1R = 1R a = a for all a in R. The multiplicative identity is often written
simply as 1, and the additive identity as 0. If a, b, and c are arbitrary elements of R,
the following properties are derived quickly from the de¬nition of a ring; we sketch the
technique in each case.
(1) a0 = 0a = 0 [a0 + a0 = a(0 + 0) = a0; 0a + 0a = (0 + 0)a = 0a]
(2) (’a)b = a(’b) = ’(ab) [0 = 0b = (a + (’a))b = ab + (’a)b, so (’a)b = ’(ab)]
[0 = a0 = a(b + (’b)) = ab + a(’b), so a(’b) = ’(ab)]
(3) (’1)(’1) = 1 [take a = 1, b = ’1 in (2)]
(4) (’a)(’b) = ab [replace b by ’b in (2)]
(5) a(b ’ c) = ab ’ ac [a(b + (’c)) = ab + a(’c) = ab + (’(ac)) = ab ’ ac]
(6) (a ’ b)c = ac ’ bc [(a + (’b))c = ac + (’b)c) = ac ’ (bc) = ac ’ bc]
(7) 1 = 0 [If 1 = 0 then for all a we have a = a1 = a0 = 0, so R = {0}, contradicting the
assumption that R has at least two elements]
(8) The multiplicative identity is unique [If 1 is another multiplicative identity then
1 = 11 = 1 ]
2.1.2 De¬nitions and Comments
If a and b are nonzero but ab = 0, we say that a and b are zero divisors; if a ∈ R and for
some b ∈ R we have ab = ba = 1, we say that a is a unit or that a is invertible.
Note that ab need not equal ba; if this holds for all a, b ∈ R, we say that R is a
commutative ring.
1
2 CHAPTER 2. RING FUNDAMENTALS
An integral domain is a commutative ring with no zero divisors.
A division ring or skew ¬eld is a ring in which every nonzero element a has a multi
plicative inverse a’1 (i.e., aa’1 = a’1 a = 1). Thus the nonzero elements form a group
under multiplication.
A ¬eld is a commutative division ring. Intuitively, in a ring we can do addition,
subtraction and multiplication without leaving the set, while in a ¬eld (or skew ¬eld) we
can do division as well.
Any ¬nite integral domain is a ¬eld. To see this, observe that if a = 0, the map
x ’ ax, x ∈ R, is injective because R is an integral domain. If R is ¬nite, the map is
surjective as well, so that ax = 1 for some x.
The characteristic of a ring R (written Char R) is the smallest positive integer such
that n1 = 0, where n1 is an abbreviation for 1 + 1 + . . . 1 (n times). If n1 is never 0,
we say that R has characteristic 0. Note that the characteristic can never be 1, since
1R = 0. If R is an integral domain and Char R = 0, then Char R must be a prime
number. For if CharR = n = rs where r and s are positive integers greater than 1, then
(r1)(s1) = n1 = 0, so either r1 or s1 is 0, contradicting the minimality of n.
A subring of a ring R is a subset S of R that forms a ring under the operations of
addition and multiplication de¬ned on R. In other words, S is an additive subgroup of
R that contains 1R and is closed under multiplication. Note that 1R is automatically the
multiplicative identity of S, since the multiplicative identity is unique (see (8) of 2.1.1).
2.1.3 Examples
1. The integers Z form an integral domain that is not a ¬eld.
2. Let Zn be the integers modulo n, that is, Zn = {0, 1, . . . , n ’ 1} with addition
and multiplication mod n. (If a ∈ Zn then a is identi¬ed with all integers a + kn,
k = 0, ±1, ±2, . . . }. Thus, for example, in Z9 the multiplication of 3 by 4 results in 3
since 12 ≡ 3 mod 9, and therefore 12 is identi¬ed with 3.
Zn is a ring, which is an integral domain (and therefore a ¬eld, since Zn is ¬nite) if
and only if n is prime. For if n = rs then rs = 0 in Zn ; if n is prime then every nonzero
element in Zn has a multiplicative inverse, by Fermat™s little theorem 1.3.4.
Note that by de¬nition of characteristic, any ¬eld of prime characteristic p contains
an isomorphic copy of Zp . Any ¬eld of characteristic 0 contains a copy of Z , hence a
copy of the rationals Q.
3. If n ≥ 2, then the set Mn (R) of all n by n matrices with coe¬cients in a ring R
forms a noncommutative ring, with the identity matrix In as multiplicative identity. If we
identify the element c ∈ R with the diagonal matrix cIn , we may regard R as a subring of
Mn (R). It is possible for the product of two nonzero matrices to be zero, so that Mn (R)
is not an integral domain. (To generate a large class of examples, let Eij be the matrix
with 1 in row i, column j, and 0™s elsewhere. Then Eij Ekl = δjk Eil , where δjk is 1 when
j = k, and 0 otherwise.)
4. Let 1, i, j and k be basis vectors in 4dimensional Euclidean space, and de¬ne
multiplication of these vectors by
i2 = j 2 = k 2 = ’1, ji = ’ij, kj = ’jk, ik = ’ki (1)
ij = k, jk = i, ki = j,
2.1. BASIC DEFINITIONS AND PROPERTIES 3
Let H be the set of all linear combinations a + bi + cj + dk where a, b, c and d are real
numbers. Elements of H are added componentwise and multiplied according to the above
rules, i.e.,
(a + bi + cj + dk)(x + yi + zj + wk) = (ax ’ by ’ cz ’ dw) + (ay + bx + cw ’ dz)i
+ (az + cx + dy ’ bw)j + (aw + dx + bz ’ cy)k.
H (after Hamilton) is called the ring of quaternions. In fact H is a division ring; the
inverse of a + bi + cj + dk is (a2 + b2 + c2 + d2 )’1 (a ’ bi ’ cj ’ dk).
H can also be represented by 2 by 2 matrices with complex entries, with multiplication
of quaternions corresponding to ordinary matrix multiplication. To see this, let
1 0 i 0 0 1 0 i
1= , i= , j= , k= ;
’i ’1
0 1 0 0 i 0
a direct computation shows that 1, i, j and k obey the multiplication rules (1) given above.
Thus we may identify the quaternion a + bi + cj + dk with the matrix
a + bi c + di
a1 + bi + cj + dk =
’c + di a ’ bi
√
(where in the matrix, i is ’1, not the quaternion i).
The set of 8 elements ±1, ±i, ±j, ±k forms a group under multiplication; it is called
the quaternion group.
5. If R is a ring, then R[X], the set of all polynomials in X with coe¬cients in R, is
also a ring under ordinary polynomial addition and multiplication, as is R[X1 , . . . , Xn ],
the set of polynomials in n variables Xi , 1 ¤ i ¤ n, with coe¬cients in R. Formally,
the polynomial A(X) = a0 + a1 X + · · · + an X n is simply the sequence (a0 , . . . , an );
the symbol X is a placeholder. The product of two polynomials A(X) and B(X) is a
polynomial whose X k coe¬cient is a0 bk + a1 bk’1 + · · · + ak b0 . If we wish to evaluate a
polynomial on R, we use the evaluation map
a0 + a1 X + · · · + an X n ’ a0 + a1 x + · · · + an xn
where x is a particular element of R. A nonzero polynomial can evaluate to 0 at all points
of R. For example, X 2 + X evaluates to 0 on Z2 , the ¬eld of integers modulo 2, since
1 + 1 = 0 mod 2. We will say more about evaluation maps in Section 2.5, when we study
polynomial rings.
6. If R is a ring, then R[[X]], the set of formal power series
a0 + a1 X + a2 X 2 + . . .
with coe¬cients in R, is also a ring under ordinary addition and multiplication of power
series. The de¬nition of multiplication is purely formal and convergence is never men
tioned; we simply de¬ne the coe¬cient of X n in the product of a0 + a1 X + a2 X 2 + . . .
and b0 + b1 X + b2 X 2 + . . . to be a0 bn + a1 bn’1 + · · · + an’1 b1 + an b0 .
In Examples 5 and 6, if R is an integral domain, so are R[X] and R[[X]]. In Example 5,
look at leading coe¬cients to show that if f (X) = 0 and g(X) = 0, then f (X)g(X) = 0.
In Example 6, if f (X)g(X) = 0 with f (X) = 0, let ai be the ¬rst nonzero coe¬cient of
f (X). Then ai bj = 0 for all j, and therefore g(X) = 0.
4 CHAPTER 2. RING FUNDAMENTALS
2.1.4 Lemma
The generalized associative law holds for multiplication in a ring. There is also a gener
alized distributive law :
m n
(a1 + · · · + am )(b1 + · · · + bn ) = ai bj .
i=1 j=1
Proof. The argument for the generalized associative law is exactly the same as for groups;
see the beginning of Section 1.1. The generalized distributive law is proved in two stages.
First set m = 1 and work by induction on n, using the left distributive law a(b + c) =
ab + ac. Then use induction on m and the right distributive law (a + b)c = ac + bc on
(a1 + · · · + am + am+1 )(b1 + · · · + bn ). ™
2.1.5 Proposition
n
n
(a + b)n = ak bn’k , the binomial theorem, is valid in any ring, if ab = ba.
k=0 k
Proof. The standard proof via elementary combinatorial analysis works. Speci¬cally,
(a + b)n = (a + b) . . . (a + b), and we can expand this product by multiplying an element
(a or b) from object 1 (the ¬rst (a + b)) times an element from object 2 times . . . times an
element from object n, in all possible ways. Since ab = ba, these terms are of the form
ak bn’k , 0 ¤ k ¤ n. The number of terms corresponding to a given k is the number of
n
ways of selecting k objects from a collection of n, namely k . ™
Problems For Section 2.1
1. If R is a ¬eld, is R[X] a ¬eld always? sometimes? never?
2. If R is a ¬eld, what are the units of R[X]?
3. Consider the ring of formal power series with rational coe¬cients.
(a) Give an example of a nonzero element that does not have a multiplicative inverse,
and thus is not a unit.
(b) Give an example of a nonconstant element (one that is not simply a rational
number) that does have a multiplicative inverse, and therefore is a unit.
√
4. Let Z[i] be the ring of Gaussian integers a + bi, where i = ’1 and a and b are
integers. Show that Z[i] is an integral domain that is not a ¬eld.
5. Continuing Problem 4, what are the units of Z[i]?
6. Establish the following quaternion identities:
(a) (x1 + y1 i + z1 j + w1 k)(x2 ’ y2 i ’ z2 j ’ w2 k)
= (x1 x2 + y1 y2 + z1 z2 + w1 w2 ) + (’x1 y2 + y1 x2 ’ z1 w2 + w1 z2 )i
+ (’x1 z2 + z1 x2 ’ w1 y2 + y1 w2 )j + (’x1 w2 + w1 x2 ’ y1 z2 + z1 y2 )k
(b) (x2 + y2 i + z2 j + w2 k)(x1 ’ y1 i ’ z1 j ’ w1 k)
= (x1 x2 + y1 y2 + z1 z2 + w1 w2 ) + (x1 y2 ’ y1 x2 + z1 w2 ’ w1 z2 )i
+ (x1 z2 ’ z1 x2 + w1 y2 ’ y1 w2 )j + (x1 w2 ’ w1 x2 + y1 z2 ’ z1 y2 )k
2.2. IDEALS, HOMOMORPHISMS, AND QUOTIENT RINGS 5
(c) The product of a quaternion h = a + bi + cj + dk
and its conjugate h— = a ’ bi ’ cj ’ dk is a2 + b2 + c2 + d2 .
If q and t are quaternions, then (qt)— = t— q — .
7. Use Problem 6 to establish Euler™s Identity for real numbers xr , yr , zr , wr , r = 1, 2:
(x2 + y1 + z1 + w1 )(x2 + y2 + z2 + w2 ) = (x1 x2 + y1 y2 + z1 z2 + w1 w2 )2
2 2 2 2 2 2
1 2
+ (x1 y2 ’ y1 x2 + z1 w2 ’ w1 z2 )2
+ (x1 z2 ’ z1 x2 + w1 y2 ’ y1 w2 )2
+ (x1 w2 ’ w1 x2 + y1 z2 ’ z1 y2 )2
8. Recall that an endomorphism of a group G is a homomorphism of G to itself. Thus
if G is abelian, an endomorphism is a function f : G ’ G such that f (a + b) =
f (a) + f (b) for all a, b ∈ G. De¬ne addition of endomorphisms in the natural way,
(f +g)(a) = f (a)+g(a), and de¬ne multiplication as functional composition, (f g)(a) =
f (g(a)). Show that the set End G of endomorphisms of G becomes a ring under these
operations.
9. Continuing Problem 8, what are the units of End G?
10. It can be shown that every positive integer is the sum of 4 squares. A key step is to
prove that if n and m can be expressed as sums of 4 squares, so can nm. Do this
using Euler™s identity, and illustrate for the case n = 34, m = 54.
11. Which of the following collections of n by n matrices form a ring under matrix addition
and multiplication?
(a) symmetric matrices
(b) matrices whose entries are 0 except possibly in column 1
(c) lower triangular matrices (aij = 0 for i < j)
(d) upper triangular matrices (aij = 0 for i > j)
2.2 Ideals, Homomorphisms, and Quotient Rings
Let f : R ’ S, where R and S are rings. Rings are, in particular, abelian groups under
addition, so we know what it means for f to be a group homomorphism: f (a + b) =
f (a) + f (b) for all a, b in R. It is then automatic that f (0R ) = 0S (see (1.3.11)). It is
natural to consider mappings f that preserve multiplication as well as addition, i.e.,
f (a + b) = f (a) + f (b) and f (ab) = f (a)f (b) for all a, b ∈ R.
But here it does not follow that f maps the multiplicative identity 1R to the multiplicative
identity 1S . We have f (a) = f (a1R ) = f (a)f (1R ), but we cannot multiply on the left by
f (a)’1 , which might not exist. We avoid this di¬culty by only considering functions f
that have the desired behavior.
2.2.1 De¬nition
If f : R ’ S, where R and S are rings, we say that f is a ring homomorphism if f (a+b) =
f (a) + f (b) and f (ab) = f (a)f (b) for all a, b ∈ R, and f (1R ) = 1S .
6 CHAPTER 2. RING FUNDAMENTALS
2.2.2 Example
Let f : Z ’ Mn (R), n ≥ 2, be de¬ned by f (n) = nE11 (see 2.1.3, Example 3). Then
we have f (a + b) = f (a) + f (b), f (ab) = f (a)f (b), but f (1) = In . Thus f is not a ring
homomorphism.
In Chapter 1, we proved the basic isomorphism theorems for groups, and a key ob
servation was the connection between group homomorphisms and normal subgroups. We
can prove similar theorems for rings, but ¬rst we must replace the normal subgroup by
an object that depends on multiplication as well as addition.
2.2.3 De¬nitions and Comments
Let I be a subset of the ring R, and consider the following three properties:
(1) I is an additive subgroup of R.
(2) If a ∈ I and r ∈ R then ra ∈ I; in other words, rI ⊆ I for every r ∈ R.
(3) If a ∈ I and r ∈ R then ar ∈ I; in other words, Ir ⊆ I for every r ∈ R.
If (1) and (2) hold, I is said to be a left ideal of R. If (1) and (3) hold, I is said to be a
right ideal of R. If all three properties are satis¬ed, I is said to be an ideal (or twosided
ideal ) of R, a proper ideal if I = R, a nontrivial ideal if I is neither R nor {0}.
If f : R ’ S is a ring homomorphism, its kernel is
ker f = {r ∈ R : f (r) = 0};
exactly as in (1.3.13), f is injective if and only if ker f = {0}.
Now it follows from the de¬nition of ring homomorphism that ker f is an ideal of R.
The kernel must be a proper ideal because if ker f = R then f is identically 0, in particular,
f (1R ) = 1S = 0S , a contradiction (see (7) of 2.1.1). Conversely, every proper ideal is the
kernel of a ring homomorphism, as we will see in the discussion to follow.
2.2.4 Construction of Quotient Rings
Let I be a proper ideal of the ring R. Since I is a subgroup of the additive group of
R, we can form the quotient group R/I, consisting of cosets r + I, r ∈ R. We de¬ne
multiplication of cosets in the natural way:
(r + I)(s + I) = rs + I.
To show that multiplication is wellde¬ned, suppose that r + I = r + I and s + I = s + I,
so that r ’ r is an element of I, call it a, and s ’ s is an element of I, call it b. Thus
r s = (r + a)(s + b) = rs + as + rb + ab,
and since I is an ideal, we have as ∈ I, rb ∈ I, and ab ∈ I. Consequently, r s +I = rs+I,
so the multiplication of two cosets is independent of the particular representatives r and
s that we choose. From our previous discussion of quotient groups, we know that the
cosets of the ideal I form a group under addition, and the group is abelian because R
2.2. IDEALS, HOMOMORPHISMS, AND QUOTIENT RINGS 7
itself is an abelian group under addition. Since multiplication of cosets r + I and s + I
is accomplished simply by multiplying the coset representatives r and s in R and then
forming the coset rs + I, we can use the ring properties of R to show that the cosets of I
form a ring, called the quotient ring of R by I. The identity element of the quotient ring
is 1R + I, and the zero element is 0R + I. Furthermore, if R is a commutative ring, so is
R/I. The fact that I is proper is used in verifying that R/I has at least two elements.
For if 1R + I = 0R + I, then 1R = 1R ’ 0R ∈ I; thus for any r ∈ R we have r = r1R ∈ I,
so that R = I, a contradiction.
2.2.5 Proposition
Every proper ideal I is the kernel of a ring homomorphism.
Proof. De¬ne the natural or canonical map π : R ’ R/I by π(r) = r + I. We already
know that π is a homomorphism of abelian groups and its kernel is I (see (1.3.12)). To
verify that π preserves multiplication, note that
π(rs) = rs + I = (r + I)(s + I) = π(r)π(s);
since
π(1R ) = 1R + I = 1R/I ,
π is a ring homomorphism. ™
2.2.6 Proposition
Suppose f : R ’ S is a ring homomorphism and the only ideals of R are {0} and R. (In
particular, if R is a division ring, then R satis¬es this hypothesis.) Then f is injective.
Proof. Let I = ker f , an ideal of R (see 2.2.3). If I = R then f is identically zero, and is
therefore not a legal ring homomorphism since f (1R ) = 1S = 0S . Thus I = {0}, so that
f is injective.
If R is a division ring, then in fact R has no nontrivial left or right ideals. For suppose
that I is a left ideal of R and a ∈ I, a = 0. Since R is a division ring, there is an element
b ∈ R such that ba = 1, and since I is a left ideal, we have 1 ∈ I, which implies that
I = R. If I is a right ideal, we choose the element b such that ab = 1. ™
2.2.7 De¬nitions and Comments
If X is a nonempty subset of the ring R, then X will denote the ideal generated by X,
that is, the smallest ideal of R that contains X. Explicitly,
X = RXR = the collection of all ¬nite sums of the form ri xi si
i
with ri , si ∈ R and xi ∈ X. To show that this is correct, verify that the ¬nite sums of the
given type form an ideal containing X. On the other hand, if J is any ideal containing X,
then all ¬nite sums i ri xi si must belong to J.
8 CHAPTER 2. RING FUNDAMENTALS
If R is commutative, then rxs = rsx, and we may as well drop the s. In other words:
ri xi , ri ∈ R, xi ∈ X.
In a commutative ring, X = RX = all ¬nite sums
i
An ideal generated by a single element a is called a principal ideal and is denoted by a
or (a). In this case, X = {a}, and therefore:
In a commutative ring, the principal ideal generated by a is a = {ra : r ∈ R},
the set of all multiples of a, sometimes denoted by Ra.
2.2.8 De¬nitions and Comments
In an arbitrary ring, we will sometimes need to consider the sum of two ideals I and J,
de¬ned as {x + y : x ∈ I, y ∈ J}. It follows from the distributive laws that I + J is also
an ideal. Similarly, the sum of two left [resp. right] ideals is a left [resp. right] ideal.
Problems For Section 2.2
1. What are the ideals in the ring of integers?
2. Let Mn (R) be the ring of n by n matrices with coe¬cients in the ring R. If Ck is the
subset of Mn (R) consisting of matrices that are 0 except perhaps in column k, show
that Ck is a left ideal of Mn (R). Similarly, if Rk consists of matrices that are 0 except
perhaps in row k, then Rk is a right ideal of Mn (R).
3. In Problem 2, assume that R is a division ring, and let Eij be the matrix with 1 in
row i, column j, and 0™s elsewhere.
(a) If A ∈ Mn (R), show that Eij A has row j of A as its ith row, with 0™s elsewhere.
(b) Now suppose that A ∈ Ck . Show that Eij A has ajk in the ik position, with 0™s
elsewhere, so that if ajk is not zero, then a’1 Eij A = Eik .
jk
(c) If A is a nonzero matrix in Ck with ajk = 0, and C is any matrix in Ck , show that
n
cik a’1 Eij A = C.
jk
i=1
4. Continuing Problem 3, if a nonzero matrix A in Ck belongs to the left ideal I of Mn (R),
show that every matrix in Ck belongs to I. Similarly, if a nonzero matrix A in Rk
belongs to the right ideal I of Mn (R), every matrix in Rk belongs to I.
5. Show that if R is a division ring, then Mn (R) has no nontrivial twosided ideals.
6. In R[X], express the set I of polynomials with no constant term as f for an appro
priate f , and thus show that I is a principal ideal.
7. Let R be a commutative ring whose only proper ideals are {0} and R. Show that R is
a ¬eld.
8. Let R be the ring Zn of integers modulo n, where n may be prime or composite. Show
that every ideal of R is principal.
2.3. THE ISOMORPHISM THEOREMS FOR RINGS 9
2.3 The Isomorphism Theorems For Rings
The basic ring isomorphism theorems may be proved by adapting the arguments used in
Section 1.4 to prove the analogous theorems for groups. Suppose that I is an ideal of the
ring R, f is a ring homomorphism from R to S with kernel K, and π is the natural map,
as indicated in Figure 2.3.1. To avoid awkward analysis of special cases, let us make a
blanket assumption that any time a quotient ring R0 /I0 appears in the statement of a
theorem, the ideal I0 is proper.
GS
f
R
b


π
 f
R/I
Figure 2.3.1
2.3.1 Factor Theorem For Rings
Any ring homomorphism whose kernel contains I can be factored through R/I. In other
words, in Figure 2.3.1 there is a unique ring homomorphism f : R ’ S that makes the
diagram commutative. Furthermore,
(i) f is an epimorphism if and only if f is an epimorphism;
(ii) f is a monomorphism if and only if ker f = I;
(iii) f is an isomorphism if and only if f is an epimorphism and ker f = I.
Proof. The only possible way to de¬ne f is f (a + I) = f (a). To verify that f is well
de¬ned, note that if a + I = b + I, then a ’ b ∈ I ⊆ K, so f (a ’ b) = 0, i.e.,f (a) = f (b).
Since f is a ring homomorphism, so is f . To prove (i), (ii) and (iii), the discussion in
(1.4.1) may be translated into additive notation and copied. ™
2.3.2 First Isomorphism Theorem For Rings
If f : R ’ S is a ring homomorphism with kernel K, then the image of f is isomorphic
to R/K.
Proof. Apply the factor theorem with I = K, and note that f is an epimorphism onto its
image. ™
2.3.3 Second Isomorphism Theorem For Rings
Let I be an ideal of the ring R, and let S be a subring of R. Then:
(a) S + I(= {x + y : x ∈ S, y ∈ I}) is a subring of R;
(b) I is an ideal of S + I;
10 CHAPTER 2. RING FUNDAMENTALS
(c) S © I is an ideal of S;
(d) (S +I)/I is isomorphic to S/(S ©I), as suggested by the “parallelogram” or“diamond”
diagram in Figure 2.3.2.
S+I
hh
yy hh
yy hh
yy hh
yy h
S ii I
yy
ii y
ii yy
ii yy
i yy
S©I
Figure 2.3.2
Proof. (a) Verify directly that S + I is an additive subgroup of R that contains 1R (since
1R ∈ S and 0R ∈ I) and is closed under multiplication. For example, if a ∈ S, x ∈ I,
b ∈ S, y ∈ I, then (a + x)(b + y) = ab + (ay + xb + xy) ∈ S + I.
(b) Since I is an ideal of R, it must be an ideal of the subring S + I.
(c) This follows from the de¬nitions of subring and ideal.
(d) Let π : R ’ R/I be the natural map, and let π0 be the restriction of π to S.
Then π0 is a ring homomorphism whose kernel is S © I and whose image is {a + I :
a ∈ S} = (S + I)/I. (To justify the last equality, note that if s ∈ S and x ∈ I we have
(s + x) + I = s + I.) By the ¬rst isomorphism theorem for rings, S/ ker π0 is isomorphic
to the image of π0 , and (d) follows. ™
2.3.4 Third Isomorphism Theorem For Rings
Let I and J be ideals of the ring R, with I ⊆ J. Then J/I is an ideal of R/I, and
R/J ∼ (R/I)/(J/I).
=
Proof. De¬ne f : R/I ’ R/J by f (a+I) = a+J. To check that f is wellde¬ned, suppose
that a + I = b + I. Then a ’ b ∈ I ⊆ J, so a + J = b + J. By de¬nition of addition and
multiplication of cosets in a quotient ring, f is a ring homomorphism. Now
ker f = {a + I : a + J = J} = {a + I : a ∈ J} = J/I
and
im f = {a + J : a ∈ R} = R/J
(where im denotes image). The result now follows from the ¬rst isomorphism theorem for
rings. ™
2.3. THE ISOMORPHISM THEOREMS FOR RINGS 11
2.3.5 Correspondence Theorem For Rings
If I is an ideal of the ring R, then the map S ’ S/I sets up a onetoone correspondence
between the set of all subrings of R containing I and the set of all subrings of R/I, as well
as a onetoone correspondence between the set of all ideals of R containing I and the
set of all ideals of R/I. The inverse of the map is Q ’ π ’1 (Q), where π is the canonical
map: R ’ R/I.
Proof. The correspondence theorem for groups yields a onetoone correspondence be
tween additive subgroups of R containing I and additive subgroups of R/I. We must
check that subrings correspond to subrings and ideals to ideals. If S is a subring of R
then S/I is closed under addition, subtraction and multiplication. For example, if s and
s belong to S, we have (s+I)(s +I) = ss +I ∈ S/I. Since 1R ∈ S we have 1R +I ∈ S/I,
proving that S/I is a subring of R/I. Conversely, if S/I is a subring of R/I, then S is
closed under addition, subtraction and multiplication, and contains the identity, hence is
a subring or R. For example, if s, s ∈ S then (s + I)(s + I) ∈ S/I, so that ss + I = t + I
for some t ∈ S, and therefore ss ’ t ∈ I. But I ⊆ S, so ss ∈ S.
Now if J is an ideal of R containing I, then J/I is an ideal of R/I by the third
isomorphism theorem for rings. Conversely, let J/I be an ideal of R/I. If r ∈ R and
x ∈ J then (r + I)(x + I) ∈ J/I, that is, rx + I ∈ J/I. Thus for some j ∈ J we have
rx ’ j ∈ I ⊆ J, so rx ∈ J. A similar argument shows that xr ∈ J, and that J is an
additive subgroup of R. It follows that J is an ideal of R. ™
We now consider the Chinese remainder theorem, which is an abstract version of a
result in elementary number theory. Along the way, we will see a typical application of
the ¬rst isomorphism theorem for rings; in fact the development of any major theorem of
algebra is likely to include an appeal to one or more of the isomorphism theorems. The
following observations may make the ideas easier to visualize.
2.3.6 De¬nitions and Comments
(i) If a and b are integers that are congruent modulo n, then a ’ b is a multiple of n.
Thus a ’ b belongs to the ideal In consisting of all multiples of n in the ring Z of integers.
Thus we may say that a is congruent to b modulo In . In general, if a, b ∈ R and I is an
ideal of R, we say that a ≡ b mod I if a ’ b ∈ I.
(ii) The integers a and b are relatively prime if and only if the integer 1 can be expressed
as a linear combination of a and b. Equivalently, the sum of the ideals Ia and Ib is the
entire ring Z. In general, we say that the ideals I and J in the ring R are relatively prime
if I + J = R.
(iii) If Ini consists of all multiples of ni in the ring of integers (i = 1, . . . k), then the
intersection ©k Ini is Ir where r is the least common multiple of the ni . If the ni are
i=1
relatively prime in pairs, then r is the product of the ni .
(iv) If R1 , . . . , Rn are rings, the direct product of the Ri is de¬ned as the ring of n
tuples (a1 , . . . , an ), ai ∈ Ri , with componentwise addition and multiplication, that is,
12 CHAPTER 2. RING FUNDAMENTALS
with
(a1 , . . . , an ) + (b1 , . . . , bn ) = (a1 + b1 , . . . , an + bn ),
(a1 , . . . , an )(b1 , . . . , bn ) = (a1 b1 , . . . , an bn ).
The zero element is (0, . . . , 0) and the multiplicative identity is (1, . . . , 1).
2.3.7 Chinese Remainder Theorem
Let R be an arbitrary ring, and let I1 , . . . , In be ideals in R that are relatively prime in
pairs, that is, Ii + Ij = R for all i = j.
(1) If a1 = 1 (the multiplicative identity of R) and aj = 0 (the zero element of R)
for j = 2, . . . , n, then there is an element a ∈ R such that a ≡ ai mod Ii for all
i = 1, . . . , n.
(2) More generally, if a1 , . . . , an are arbitrary elements of R, there is an element a ∈ R
such that a ≡ ai mod Ii for all i = 1, . . . , n.
(3) If b is another element of R such that b ≡ ai mod Ii for all i = 1, . . . , n, then b ≡ a
mod I1 © I2 © · · · © In . Conversely, if b ≡ a mod ©n Ii , then b ≡ ai mod Ii for all i.
i=1
n n
(4) R/ i=1 Ii is isomorphic to the direct product R/Ii .
i=1
Proof. (1) If j > 1 we have I1 + Ij = R, so there exist elements bj ∈ I1 and cj ∈ Ij such
that bj + cj = 1; thus
n
(bj + cj ) = 1.
j=2
Expand the left side and observe that any product containing at least one bj belongs to
n
I1 , while c2 . . . cn belongs to j=2 Ij , the collection of all ¬nite sums of products x2 . . . xn
n
with xj ∈ Ij . Thus we have elements b ∈ I1 and a ∈ j=2 Ij (a subset of each Ij ) with
b + a = 1. Consequently, a ≡ 1 mod I1 and a ≡ 0 mod Ij for j > 1, as desired.
(2) By the argument of part (1), for each i we can ¬nd ci with ci ≡ 1 mod Ii and
ci ≡ 0 mod Ij , j = i. If a = a1 c1 + · · · + an cn , then a has the desired properties. To see
this, write a ’ ai = a ’ ai ci + ai (ci ’ 1), and note that a ’ ai ci is the sum of the aj cj , j = i,
and is therefore congruent to 0 mod Ii .
(3) We have b ≡ ai mod Ii for all i i¬ b’a ≡ 0 mod Ii for all i, that is, i¬ b’a ∈ ©n Ii , i=1
and the result follows.
n
(4) De¬ne f : R ’ i=1 R/Ii by f (a) = (a + I1 , . . . , a + In ). If a1 , . . . , an ∈ R, then
by part (2) there is an element a ∈ R such that a ≡ ai mod Ii for all i. But then
f (a) = (a1 + I1 , . . . , an + In ), proving that f is surjective. Since the kernel of f is the
intersection of the ideals Ij , the result follows from the ¬rst isomorphism theorem for
rings. ™
The concrete version of the Chinese remainder theorem can be recovered from the abstract
result; see Problems 3 and 4.
2.4. MAXIMAL AND PRIME IDEALS 13
Problems For Section 2.3
1. Show that the group isomorphisms of Section 1.4, Problems 1 and 2, are ring isomor
phisms as well.
2. Give an example of an ideal that is not a subring, and a subring that is not an ideal.
3. If the integers mi , i = 1, . . . , n, are relatively prime in pairs, and a1 , . . . , an are arbitrary
integers, show that there is an integer a such that a ≡ ai mod mi for all i, and that
any two such integers are congruent modulo m1 . . . mn .
4. If the integers mi , i = 1, . . . , n, are relatively prime in pairs and m = m1 . . . mn ,
n
show that there is a ring isomorphism between Zm and the direct product i=1 Zmi .
Speci¬cally, a mod m corresponds to (a mod m1 , . . . , a mod mn ).
5. Suppose that R = R1 — R2 is a direct product of rings. Let R1 be the ideal R1 — {0} =
{(r1 , 0) : r1 ∈ R1 }, and let R2 be the ideal {(0, r2 ) : r2 ∈ R2 }. Show that R/R1 ∼ R2
=
∼ R1 .
and R/R2 =
6. If I1 , . . . , In are ideals, the product I1 . . . In is de¬ned as the set of all ¬nite sums
i a1i a2i . . . ani , where aki ∈ Ik , k = 1, . . . , n. [See the proof of part (1) of 2.3.7; a
brief check shows that the product of ideals is an ideal.]
Assume that R is a commutative ring. Under the hypothesis of the Chinese remainder
theorem, show that the intersection of the ideals Ii coincides with their product.
7. Let I1 , . . . , In be ideals in the ring R. Suppose that R/ ©i Ii is isomorphic to i R/Ii
via a + ©i Ii ’ (a + I1 , . . . , a + In ). Show that the ideals Ii are relatively prime in pairs.
2.4 Maximal and Prime Ideals
If I is an ideal of the ring R, we might ask“What is the smallest ideal containing I” and
“What is the largest ideal containing I”. Neither of these questions is challenging; the
smallest ideal is I itself, and the largest ideal is R. But if I is a proper ideal and we ask
for a maximal proper ideal containing I, the question is much more interesting.
2.4.1 De¬nition
A maximal ideal in the ring R is a proper ideal that is not contained in any strictly larger
proper ideal.
2.4.2 Theorem
Every proper ideal I of the ring R is contained in a maximal ideal. Consequently, every
ring has at least one maximal ideal.
Proof. The argument is a prototypical application of Zorn™s lemma. Consider the col
lection of all proper ideals containing I, partially ordered by inclusion. Every chain
{Jt , t ∈ T } of proper ideals containing I has an upper bound, namely the union of the
chain. (Note that the union is still a proper ideal, because the identity 1R belongs to
none of the ideals Jt .) By Zorn, there is a maximal element in the collection, that is, a
14 CHAPTER 2. RING FUNDAMENTALS
maximal ideal containing I. Now take I = {0} to conclude that every ring has at least
one maximal ideal. ™
We have the following characterization of maximal ideals.
2.4.3 Theorem
Let M be an ideal in the commutative ring R. Then M is a maximal ideal if and only if
R/M is a ¬eld.
Proof. Suppose M is maximal. We know that R/M is a ring (see 2.2.4); we need to ¬nd
the multiplicative inverse of the element a + M of R/M , where a + M is not the zero
element, i.e., a ∈ M . Since M is maximal, the ideal Ra + M , which contains a and is
/
therefore strictly larger than M , must be the ring R itself. Consequently, the identity
element 1 belongs to Ra + M . If 1 = ra + m where r ∈ R and m ∈ M , then
(r + M )(a + M ) = ra + M = (1 ’ m) + M = 1 + M (since m ∈ M ),
proving that r + M is the multiplicative inverse of a + M .
Conversely, if R/M is a ¬eld, then M must be a proper ideal. If not, then M = R,
so that R/M contains only one element, contradicting the requirement that 1 = 0 in
R/M (see (7) of 2.1.1). By (2.2.6), the only ideals of R/M are {0} and R/M , so by the
correspondence theorem 2.3.5, there are no ideals properly between M and R. Therefore
M is a maximal ideal. ™
If in (2.4.3) we relax the requirement that R/M be a ¬eld, we can identify another
class of ideals.
2.4.4 De¬nition
A prime ideal in a commutative ring R is a proper ideal P such that for any two elements
a, b in R,
ab ∈ P implies that a ∈ P or b ∈ P.
We can motivate the de¬nition by looking at the ideal (p) in the ring of integers. In this
case, a ∈ (p) means that p divides a, so that (p) will be a prime ideal if and only if
p divides ab implies that p divides a or p divides b,
which is equivalent to the requirement that p be a prime number.
2.4.5 Theorem
If P is an ideal in the commutative ring R, then P is a prime ideal if and only if R/P is
an integral domain.
2.4. MAXIMAL AND PRIME IDEALS 15
Proof. Suppose P is prime. Since P is a proper ideal, R/P is a ring. We must show that
if (a + P )(b + P ) is the zero element P in R/P , then a + P = P or b + P = P , i.e., a ∈ P
or b ∈ P . This is precisely the de¬nition of a prime ideal.
Conversely, if R/P is an integral domain, then, as in (2.4.3), P is a proper ideal. If
ab ∈ P , then (a + P )(b + P ) is zero in R/P , so that a + P = P or b + P = P , i.e., a ∈ P
or b ∈ P . ™
2.4.6 Corollary
In a commutative ring, a maximal ideal is prime.
Proof. This is immediate from (2.4.3) and (2.4.5). ™
2.4.7 Corollary
Let f : R ’ S be an epimorphism of commutative rings. Then:
(i) If S is a ¬eld then ker f is a maximal ideal of R;
(ii) If S is an integral domain then ker f is a prime ideal of R.
Proof. By the ¬rst isomorphism theorem (2.3.2), S is isomorphic to R/ ker f , and the
result now follows from (2.4.3) and (2.4.5). ™
2.4.8 Example
Let Z[X] be the set of all polynomials f (X) = a0 + a1 X + · · · + an X n , n = 0, 1, . . . in
the indeterminate X, with integer coe¬cients. The ideal generated by X, that is, the
collection of all multiples of X, is
X = {f (X) ∈ Z[X] : a0 = 0}.
The ideal generated by 2 is
2 = {f (X) ∈ Z[X] : all ai are even integers.}
Both X and 2 are proper ideals, since 2 ∈ X and X ∈ 2 . In fact we can say
/ /
much more. Consider the ring homomorphisms • : Z[X] ’ Z and ψ : Z[X] ’ Z2 given by
•(f (X)) = a0 and ψ(f (X)) = a0 , where a0 is a0 reduced modulo 2. We will show that
both X and 2 are prime ideals that are not maximal.
First note that by (2.4.7), X is prime because it is the kernel of •. Then observe
that X is not maximal because it is properly contained in 2, X , the ideal generated
by 2 and X.
To verify that 2 is prime, note that it is the kernel of the homomorphism from Z[X]
onto Z2 [X] that takes f (X) to f (X), where the overbar indicates that the coe¬cients of
f (X) are to be reduced modulo 2. Since Z2 [X] is an integral domain (see the comment
at the end of 2.1.3), 2 is a prime ideal. Since 2 is properly contained in 2, X , 2 is
not maximal.
16 CHAPTER 2. RING FUNDAMENTALS
Finally, 2, X is a maximal ideal, since
ker ψ = {a0 + Xg(X) : a0 is even and g(X) ∈ Z[X]} = 2, X .
Thus 2, X is the kernel of a homomorphism onto a ¬eld, and the result follows
from (2.4.7).
2.4.9 Problems For Section 2.4
1. We know from Problem 1 of Section 2.2 that in the ring of integers, all ideals I are
of the form n for some n ∈ Z, and since n ∈ I implies ’n ∈ I, we may take n to
be nonnegative. Let n be a nontrivial ideal, so that n is a positive integer greater
than 1. Show that n is a prime ideal if and only if n is a prime number.
2. Let I be a nontrivial prime ideal in the ring of integers. Show that in fact I must be
maximal.
3. Let F [[X]] be the ring of formal power series with coe¬cients in the ¬eld F (see (2.1.3),
Example 6). Show that X is a maximal ideal.
4. Perhaps the result of Problem 3 is a bit puzzling. Why can™t we argue that just as in
(2.4.8), X is properly contained in 2, X , and therefore X is not maximal?
5. Let I be a proper ideal of F [[X]]. Show that I ⊆ X , so that X is the unique
maximal ideal of F [[X]]. (A commutative ring with a unique maximal ideal is called a
local ring.)
6. Show that every ideal of F [[X]] is principal, and speci¬cally every nonzero ideal is of
the form (X n ) for some n = 0, 1, . . . .
7. Let f : R ’ S be a ring homomorphism, with R and S commutative. If P is a prime
ideal of S, show that the preimage f ’1 (P ) is a prime ideal of R.
8. Show that the result of Problem 7 does not hold in general when P is a maximal ideal.
9. Show that a prime ideal P cannot be the intersection of two strictly larger ideals I
and J.
2.5 Polynomial Rings
In this section, all rings are assumed commutative. To see a good reason for this restric
tion, consider the evaluation map (also called the substitution map) Ex , where x is a ¬xed
element of the ring R. This map assigns to the polynomial a0 + a1 X + · · · + an X n in
R[X] the value a0 + a1 x + · · · + an xn in R. It is tempting to say that “obviously”, Ex is
a ring homomorphism, but we must be careful. For example,
Ex [(a + bX)(c + dX)] = Ex (ac + (ad + bc)X + bdX 2 ) = ac + (ad + bc)x + bdx2 ,
Ex (a + bX)Ex (c + dX) = (a + bx)(c + dx) = ac + adx + bxc + bxdx,
and these need not be equal if R is not commutative.
2.5. POLYNOMIAL RINGS 17
The degree, abbreviated deg, of a polynomial a0 + a1 X + · · · + an X n (with leading
coe¬cient an = 0) is n; it is convenient to de¬ne the degree of the zero polynomial as ’∞.
If f and g are polynomials in R[X], where R is a ¬eld, ordinary long division allows us to
express f as qg + r, where the degree of r is less than the degree of g. We have a similar
result over an arbitrary commutative ring, if g is monic, i.e., the leading coe¬cient of g
is 1. For example (with R = Z), we can divide 2X 3 + 10X 2 + 16X + 10 by X 2 + 3X + 5:
2X 3 + 10X 2 + 16X + 10 = 2X(X 2 + 3X + 5) + 4X 2 + 6X + 10.
The remainder 4X 2 + 6X + 10 does not have degree less than 2, so we divide it by
X 2 + 3X + 5:
4X 2 + 6X + 10 = 4(X 2 + 3X + 5) ’ 6X ’ 10.
Combining the two calculations, we have
2X 3 + 10X 2 + 16X + 10 = (2X + 4)(X 2 + 3X + 5) + (’6X ’ 10)
which is the desired decomposition.
2.5.1 Division Algorithm
If f and g are polynomials in R[X], with g monic, there are unique polynomials q and r
in R[X] such that f = qg + r and deg r < deg g. If R is a ¬eld, g can be any nonzero
polynomial.
Proof. The above procedure, which works in any ring R, shows that q and r exist.If
f = qg + r = q1 g + r1 where r and r1 are of degree less than deg g, then g(q ’ q1 ) = r1 ’ r.
But if q ’ q1 = 0, then, since g is monic, the degree of the left side is at least deg g, while
the degree of the right side is less than deg g, a contradiction. Therefore q = q1 , and
consequently r = r1 . ™
2.5.2 Remainder Theorem
If f ∈ R[X] and a ∈ R, then for some unique polynomial q(X) in R[X] we have
f (X) = q(X)(X ’ a) + f (a);
hence f (a) = 0 if and only if X ’ a divides f (X).
Proof. By the division algorithm, we may write f (X) = q(X)(X ’ a) + r(X) where the
degree of r is less than 1, i.e., r is a constant. Apply the evaluation homomorphism X ’ a
to show that r = f (a). ™
18 CHAPTER 2. RING FUNDAMENTALS
2.5.3 Theorem
If R is an integral domain, then a nonzero polynomial f in R[X] of degree n has at most
n roots in R, counting multiplicity.
Proof. If f (a1 ) = 0, then by (2.5.2), possibly applied several times, we have f (X) =
q1 (X)(X ’ a1 )n1 , where q1 (a1 ) = 0 and the degree of q1 is n ’ n1 . If a2 is another
root of f , then 0 = f (a2 ) = q1 (a2 )(a2 ’ a1 )n1 . But a1 = a2 and R is an integral
domain, so q1 (a2 ) must be 0, i.e. a2 is a root of q1 (X). Repeating the argument, we have
q1 (X) = q2 (X)(X ’a2 )n2 , where q2 (a2 ) = 0 and deg q2 = n’n1 ’n2 . After n applications
of (2.5.2), the quotient becomes constant, and we have f (X) = c(X ’ a1 )n1 . . . (X ’ ak )nk
where c ∈ R and n1 + · · · + nk = n. Since R is an integral domain, the only possible roots
of f are a1 , . . . , ak . ™
2.5.4 Example
Let R = Z8 , which is not an integral domain. The polynomial f (X) = X 3 has four roots
in R, namely 0, 2, 4 and 6.
Problems For Section 2.5
In Problems 14, we review the Euclidean algorithm. Let a and b be positive integers,
with a > b. Divide a by b to obtain
a = bq1 + r1 with 0 ¤ r1 < b,
then divide b by r1 to get
b = r1 q2 + r2 with 0 ¤ r2 < r1 ,
and continue in this fashion until the process terminates:
r 1 = r 2 q 3 + r 3 , 0 ¤ r 3 < r2 ,
.
.
.
rj’2 = rj’1 qj + rj , 0 ¤ rj < rj’1 ,
rj’1 = rj qj+1
1. Show that the greatest common divisor of a and b is the last remainder rj .
2. If d is the greatest common divisor of a and b, show that there are integers x and y
such that ax + by = d.
3. De¬ne three sequences by
ri = ri’2 ’ qi ri’1
xi = xi’2 ’ qi xi’1
yi = yi’2 ’ qi yi’1
2.6. UNIQUE FACTORIZATION 19
for i = ’1, 0, 1, . . . with initial conditions r’1 = a, r0 = b, x’1 = 1, x0 = 0, y’1 = 0,
y0 = 1. (The qi are determined by dividing ri’2 by ri’1 .) Show that we can generate
all steps of the algorithm, and at each stage, ri = axi + byi .
4. Use the procedure of Problem 3 (or any other method) to ¬nd the greatest common
divisor d of a = 123 and b = 54, and ¬nd integers x and y such that ax + by = d.
5. Use Problem 2 to show that Zp is a ¬eld if and only if p is prime.
6. If a(X) and b(X) are polynomials with coe¬cients in a ¬eld F , the Euclidean algo
rithm can be used to ¬nd their greatest common divisor. The previous discussion can
be taken over verbatim, except that instead of writing
a = q1 b + r1 with 0 ¤ r1 < b,
we write
a(X) = q1 (X)b(X) + r1 (X) with deg r1 (X) < deg b(X).
The greatest common divisor can be de¬ned as the monic polynomial of highest degree
that divides both a(X) and b(X).
Let f (X) and g(X) be polynomials in F [X], where F is a ¬eld. Show that the ideal I
generated by f (X) and g(X), i.e., the set of all linear combinations a(X)f (X) +
b(X)g(X), with a(X), b(X) ∈ F [X], is the principal ideal J = d(X) generated by
the greatest common divisor d(X) of f (X) and g(X).
7. (Lagrange Interpolation Formula) Let a0 , a1 , . . . , an be distinct points in the ¬eld F ,
and de¬ne
X ’ aj
Pi (X) = , i = 0, 1, . . . , n;
ai ’ aj
j=i
then Pi (ai ) = 1 and Pi (aj ) = 0 for j = i. If b0 , b1 , . . . , bn are arbitrary elements of
F (not necessarily distinct), use the Pi to ¬nd a polynomial f (X) of degree n or less
such that f (ai ) = bi for all i.
8. In Problem 7, show that f (X) is the unique polynomial of degree n or less such that
f (ai ) = bi for all i.
9. Suppose that f is a polynomial in F [X], where F is a ¬eld. If f (a) = 0 for every
a ∈ F , it does not in general follow that f is the zero polynomial. Give an example.
10. Give an example of a ¬eld F for which it does follow that f = 0.
2.6 Unique Factorization
If we are asked to ¬nd the greatest common divisor of two integers, say 72 and 60, one
method is to express each integer as a product of primes; thus 72 = 23 — 32 , 60 =
22 — 3 — 5. The greatest common divisor is the product of terms of the form pe , where
for each prime appearing in the factorization, we use the minimum exponent. Thus
gcd(72, 60) = 22 — 31 — 50 = 12. To ¬nd the least common multiple, we use the maximum
20 CHAPTER 2. RING FUNDAMENTALS
exponent: lcm(72, 60) = 23 — 32 — 51 = 360. The key idea is that every integer (except
0, 1 and ’1) can be uniquely represented as a product of primes. It is natural to ask
whether there are integral domains other than the integers in which unique factorization
is possible. We now begin to study this question; throughout this section, all rings are
assumed to be integral domains.
2.6.1 De¬nitions
Recall from (2.1.2) that a unit in a ring R is an element with a multiplicative inverse.
The elements a and b are associates if a = ub for some unit u.
Let a be a nonzero nonunit; a is said to be irreducible if it cannot be represented as a
product of nonunits. In other words, if a = bc, then either b or c must be a unit.
Again let a be a nonzero nonunit; a is said to be prime if whenever a divides a product
of terms, it must divide one of the factors. In other words, if a divides bc, then a divides b
or a divides c (a divides b means that b = ar for some r ∈ R). It follows from the de¬nition
that if p is any nonzero element of R, then p is prime if and only if p is a prime ideal.
The units of Z are 1 and ’1, and the irreducible and the prime elements coincide.
But these properties are not the same in an arbitrary integral domain.
2.6.2 Proposition
If a is prime, then a is irreducible, but not conversely.
Proof. We use the standard notation rs to indicate that r divides s. Suppose that a is
prime, and that a = bc. Then certainly abc, so by de¬nition of prime, ab or ac, say a  b.
If b = ad then b = bcd, so cd = 1 and therefore c is a unit. (Note that b cannot be 0, for
if so, a = bc = 0, which is not possible since a is prime.) Similarly, if ac with c = ad then
c = bcd, so bd = 1 and b is a unit. Therefore a is irreducible. √
To give an example of an irreducible element that is not prime, consider R = Z[ ’3 ] =
√
{a + ib 3 : a, b ∈ Z}; in R, 2 is irreducible but not prime. To see this, ¬rst suppose that
we have a factorization of the form
√ √
2 = (a + ib 3 )(c + id 3 );
take complex conjugates to get
√ √
2 = (a ’ ib 3 )(c ’ id 3 ).
Now multiply these two equations to obtain
4 = (a2 + 3b2 )(c2 + 3d2 ).
Each factor on the right must be a divisor of 4, and there is no way that a2 + 3b2 can
be 2. Thus one of the factors must be 4 and the other must be 1. If, say, a2 + 3b2 = 1,
then a = ±1 and b = 0. Thus in the original factorization of 2, one of the factors must
√ √
be a unit, so 2 is irreducible. Finally, 2 divides the product (1 + i 3 )(1 ’ i 3 ) ( = 4),
so if 2 were prime, it would divide one of the factors, which means that 2 divides 1, a
contradiction since 1/2 is not an integer. ™
2.6. UNIQUE FACTORIZATION 21
The distinction between irreducible and prime elements disappears in the presence of
unique factorization.
2.6.3 De¬nition
A unique factorization domain (UFD) is an integral domain R satisfying the following
properties:
(UF1) Every nonzero element a in R can be expressed as a = up1 . . . pn , where u is a
unit and the pi are irreducible.
(UF2): If a has another factorization, say a = vq1 . . . qm , where v is a unit and the qi are
irreducible, then n = m and, after reordering if necessary, pi and qi are associates
for each i.
Property UF1 asserts the existence of a factorization into irreducibles, and UF2 asserts
uniqueness.
2.6.4 Proposition
In a unique factorization domain, a is irreducible if and only if a is prime.
Proof. By (2.6.2), prime implies irreducible, so assume a irreducible, and let a divide bc.
Then we have ad = bc for some d ∈ R. We factor d, b and c into irreducibles to obtain
aud1 . . . dr = vb1 . . . bs wc1 . . . ct
where u, v and w are units and the di , bi and ci are irreducible. By uniqueness of
factorization, a, which is irreducible, must be an associate of some bi or ci . Thus a
divides b or a divides c. ™
2.6.5 De¬nitions and Comments
Let A be a nonempty subset of R, with 0 ∈ A. The element d is a greatest common divisor
/
(gcd) of A if d divides each a in A, and whenever e divides each a in A, we have ed.
If d is another gcd of A, we have dd and d d, so that d and d are associates. We will
allow ourselves to speak of “the” greatest common divisor, suppressing but not forgetting
that the gcd is determined up to multiplication by a unit.
The elements of A are said to be relatively prime (or the set A is said to be relatively
prime) if 1 is a greatest common divisor of A.
The nonzero element m is a least common multiple (lcm) of A if each a in A divides m,
and whenever ae for each a in A, we have me.
Greatest common divisors and least common multiples always exist for ¬nite subsets
of a UFD; they may be found by the technique discussed at the beginning of this section.
We will often use the fact that for any a, b ∈ R, we have ab if and only if b ⊆ a .
This follows because ab means that b = ac for some c ∈ R. For short, divides means
contains.
It would be useful to be able to recognize when an integral domain is a UFD. The
following criterion is quite abstract, but it will help us to generate some explicit examples.
22 CHAPTER 2. RING FUNDAMENTALS
2.6.6 Theorem
Let R be an integral domain.
(1) If R is a UFD then R satis¬es the ascending chain condition (acc) on principal ide
als: If a1 , a2 , . . . belong to R and a1 ⊆ a2 ⊆ . . . , then the sequence eventually
stabilizes, that is, for some n we have an = an+1 = an+2 = . . . .
(2) If R satis¬es the ascending chain condition on principal ideals, then R satis¬es UF1,
that is, every nonzero element of R can be factored into irreducibles.
(3) If R satis¬es UF1 and in addition, every irreducible element of R is prime, then R is
a UFD.
Thus R is a UFD if and only if R satis¬es the ascending chain condition on principal
ideals and every irreducible element of R is prime.
Proof. (1) If a1 ⊆ a2 ⊆ . . . then ai+1 ai for all i. Therefore the prime factors of ai+1
consist of some (or all) of the prime factors of ai . Multiplicity is taken into account here;
for example, if p3 is a factor of ai , then pk will be a factor of ai+1 for some k ∈ {0, 1, 2, 3}.
Since a1 has only ¬nitely many prime factors, there will come a time when the prime
factors are the same from that point on, that is, an = an+1 = . . . for some n.
(2) Let a1 be any nonzero element. If a1 is irreducible, we are ¬nished, so let a1 = a2 b2
where neither a2 nor b2 is a unit. If both a2 and b2 are irreducible, we are ¬nished, so
we can assume that one of them, say a2 , is not irreducible. Since a2 divides a1 we have
a1 ⊆ a2 , and in fact the inclusion is proper because a2 ∈ a1 . (If a2 = ca1 then
/
a1 = a2 b2 = ca1 b2 , so b2 is a unit, a contradiction.) Continuing, we have a2 = a3 b3 where
neither a3 nor b3 is a unit, and if, say, a3 is not irreducible, we ¬nd that a2 ‚ a3 . If
a1 cannot be factored into irreducibles, we obtain, by an inductive argument, a strictly
increasing chain a1 ‚ a2 ‚ . . . of principal ideals.
(3) Suppose that a = up1 p2 . . . pn = vq1 q2 . . . qm where the pi and qi are irreducible
and u and v are units. Then p1 is a prime divisor of vq1 . . . qm , so p1 divides one of the
qi , say q1 . But q1 is irreducible, and therefore p1 and q1 are associates. Thus we have, up
to multiplication by units, p2 . . . pn = q2 . . . qm . By an inductive argument, we must have
m = n, and after reordering, pi and qi are associates for each i. ™
We now give a basic su¬cient condition for an integral domain to be a UFD.
2.6.7 De¬nition
A principal ideal domain (PID) is an integral domain in which every ideal is principal,
that is, generated by a single element.
2.6.8 Theorem
Every principal ideal domain is a unique factorization domain. For short, PID implies
UFD.
2.6. UNIQUE FACTORIZATION 23
Proof. If a1 ⊆ a2 ⊆ . . . , let I = ∪i ai . Then I is an ideal, necessarily principal by
hypothesis. If I = b then b belongs to some an , so I ⊆ an . Thus if i ≥ n we have
ai ⊆ I ⊆ an ⊆ ai . Therefore ai = an for all i ≥ n, so that R satis¬es the acc on
principal ideals.
Now suppose that a is irreducible. Then a is a proper ideal, for if a = R then
1 ∈ a , so that a is a unit. By the acc on principal ideals, a is contained in a maximal
ideal I. (Note that we need not appeal to the general result (2.4.2), which uses Zorn™s
lemma.) If I = b then b divides the irreducible element a, and b is not a unit since I is
proper. Thus a and b are associates, so a = b = I. But I, a maximal ideal, is prime
by (2.4.6), hence a is prime. The result follows from (2.6.6). ™
The following result gives a criterion for a UFD to be a PID. (Terminology: the zero
ideal is {0}; a nonzero ideal is one that is not {0}.)
2.6.9 Theorem
R is a PID if and only if R is a UFD and every nonzero prime ideal of R is maximal.
Proof. Assume R is a PID; then R is a UFD by (2.6.8). If p is a nonzero prime ideal
of R, then p is contained in the maximal ideal q , so that q divides the prime p. Since
a maximal ideal must be proper, q cannot be a unit, so that p and q are associates. But
then p = q and p is maximal.
The proof of the converse is given in the exercises. ™
Problems For Section 2.6
Problems 16 form a project designed to prove that if R is a UFD and every nonzero
prime ideal of R is maximal, then R is a PID.
Let I be an ideal of R; since {0} is principal, we can assume that I = {0}. Since R
is a UFD, every nonzero element of I can be written as up1 . . . pt where u is a unit and
the pi are irreducible, hence prime. Let r = r(I) be the minimum such t. We are going
to prove by induction on r that I is principal.
1. If r = 0, show that I = 1 = R.
2. If the result holds for all r < n, let r = n, with up1 . . . pn ∈ I, hence p1 . . . pn ∈ I.
Since p1 is prime, p1 is a prime ideal, necessarily maximal by hypothesis. By (2.4.3),
R/ p1 is a ¬eld. If b belongs to I but not to p1 , show that for some c ∈ R we have
bc ’ 1 ∈ p1 .
3. By Problem 2, bc ’ dp1 = 1 for some d ∈ R. Show that this implies that p2 . . . pn ∈ I,
which contradicts the minimality of n. Thus if b belongs to I, it must also belong
to p1 , that is, I ⊆ p1 .
4. Let J = {x ∈ R : xp1 ∈ I}. Show that J is an ideal.
5. Show that Jp1 = I.
6. Since p1 . . . pn = (p2 . . . pn )p1 ∈ I, we have p2 . . . pn ∈ J. Use the induction hypothesis
to conclude that I is principal.
24 CHAPTER 2. RING FUNDAMENTALS
7. Let p and q be prime elements in the integral domain R, and let P = p and Q = q
be the corresponding prime ideals. Show that it is not possible for P to be a proper
subset of Q.
8. If R is a UFD and P is a nonzero prime ideal of R, show that P contains a nonzero
principal prime ideal.
2.7 Principal Ideal Domains and Euclidean Domains
In Section 2.6, we found that a principal ideal domain is a unique factorization domain,
and this exhibits a class of rings in which unique factorization occurs. We now study some
properties of PID™s, and show that any integral domain in which the Euclidean algorithm
works is a PID. If I is an ideal in Z, in fact if I is simply an additive subgroup of Z, then
I consists of all multiples of some positive integer n; see Section 1.1, Problem 6. Thus Z
is a PID.
Now suppose that A is a nonempty subset of the PID R. The ideal A generated by
ri ai with ri ∈ R and ai ∈ A; see (2.2.7). We show that if
A consists of all ¬nite sums
d is a greatest common divisor of A, then d generates A, and conversely.
2.7.1 Proposition
Let R be a PID, with A a nonempty subset of R. Then d is a greatest common divisor
of A if and only if d is a generator of A .
Proof. Let d be a gcd of A, and assume that A = b . Then d divides every a ∈ A,
ri ai . In particular d divides b, hence b ⊆ d ; that is,
so d divides all ¬nite sums
A ⊆ d . But if a ∈ A then a ∈ b , so b divides a. Since d is a gcd of A, it follows that
b divides d, so d is contained in b = A . We conclude that A = d , proving that d
is a generator of A .
Conversely, assume that d generates A . If a ∈ A then a is a multiple of d, so d  a.
By (2.2.7), d can be expressed as ri ai , so any element that divides everything in A
divides d. Therefore d is a gcd of A. ™
2.7.2 Corollary
If d is a gcd of A, where A is a nonempty subset of the PID R, then d can be expressed
as a ¬nite linear combination ri ai of elements of A with coe¬cients in R.
Proof. By (2.7.1), d ∈ A , and the result follows from (2.2.7). ™
As a special case, we have the familiar result that the greatest common divisor of two
integers a and b can be expressed as ax + by for some integers x and y.
The Euclidean algorithm in Z is based on the division algorithm: if a and b are integers
and b = 0, then a can be divided by b to produce a quotient and remainder. Speci¬cally,
we have a = bq + r for some q, r ∈ Z with r < b. The Euclidean algorithm performs
equally well for polynomials with coe¬cients in a ¬eld; the absolute value of an integer
2.7. PRINCIPAL IDEAL DOMAINS AND EUCLIDEAN DOMAINS 25
is replaced by the degree of a polynomial. It is possible to isolate the key property that
makes the Euclidean algorithm work.
2.7.3 De¬nition
Let R be an integral domain. R is said to be a Euclidean domain (ED) if there is a
function Ψ from R \ {0} to the nonnegative integers satisfying the following property:
If a and b are elements of R, with b = 0, then a can be expressed as bq + r for some
q, r ∈ R, where either r = 0 or Ψ(r) < Ψ(b).
We can replace“r = 0 or Ψ(r) < Ψ(b)” by simply “Ψ(r) < Ψ(b)” if we de¬ne Ψ(0) to
be ’∞.
In any Euclidean domain, we may use the Euclidean algorithm to ¬nd the greatest
common divisor of two elements; see the problems in Section 2.5 for a discussion of the
procedure in Z and in F [X], where F is a ¬eld.
A Euclidean domain is automatically a principal ideal domain, as we now prove.
2.7.4 Theorem
If R is a Euclidean domain, then R is a principal ideal domain. For short, ED implies
PID.
Proof. Let I be an ideal of R. If I = {0} then I is principal, so assume I = {0}. Then
{Ψ(b) : b ∈ I, b = 0} is a nonempty set of nonnegative integers, and therefore has a
smallest element n. Let b be any nonzero element of I such that Ψ(b) = n; we claim that
I = b . For if a belongs to I then we have a = bq + r where r = 0 or Ψ(r) < Ψ(b). Now
r = a ’ bq ∈ I (because a and b belong to I), so if r = 0 then Ψ(r) < Ψ(b) is impossible
by minimality of Ψ(b). Thus b is a generator of I. ™
The most familiar Euclidean domains are Z and F [X], with F a ¬eld. We now examine
some less familiar cases.
2.7.5 Example
√ √
Let Z[ d ] be√ ring of all elements a + b d, where a, b ∈ Z. If d = ’2, ’1, 2 or 3, we
the
claim that Z[ d ] is a Euclidean domain with
√
Ψ(a + b d ) = a2 ’ db2 .
√ √
Since the a + b d are real or complex numbers, there are no zero divisors, and Z[ √d ]
√
is an integral domain. Let ±, β ∈ Z[ d ], β = 0, and divide ± by β to get x + y d.
Unfortunately, x and y need not be integers, but at least they are rational numbers. We
can ¬nd integers reasonably close to x and y; let x0 and y0 be integers such that x ’ x0 
and y ’ y0  are at most 1/2. Let
√ √
q = x0 + y0 d, r = β((x ’ x0 ) + (y ’ y0 ) d );
then
√
βq + r = β(x + y d ) = ±.
26 CHAPTER 2. RING FUNDAMENTALS
We must show that Ψ(r) < Ψ(β). Now
√ √ √
Ψ(a + b d ) = (a + b d )(a ’ b d ),
√
and it follows from (Problem 4 that for all γ, δ ∈ Z[ d ] we have
Ψ(γδ) = Ψ(γ)Ψ(δ).
(When d = ’1, this says that the magnitude of the product of two complex numbers is
the product of the magnitudes.) Thus Ψ(r) = Ψ(β)[(x ’ x0 )2 ’ d(y ’ y0 )2 ], and the factor
in brackets is at most 1 + d( 1 ) ¤ 1 + 3 = 1. The only possibility for equality occurs
4 4 4 4
when d = 3 (d = ’3 is excluded by hypothesis) and x ’ x0  = y ’ y0  = 1 . But in this
2
case, the factor in brackets is  1 ’ 3( 1 ) = 1 < 1. We have shown that Ψ(r) < Ψ(β), so
√ 4 4 2
Z[ d ] is a Euclidean domain. √
When d = ’1, we obtain the Gaussian integers a + bi, a, b ∈ Z, i = ’1.
Problems For Section 2.7
1. Let A = {a1 , . . . , an } be a ¬nite subset of the PID R. Show that m is a least common
multiple of A i¬ m is a generator of the ideal ©n ai .
i=1
2. Find the gcd of 11 + 3i and 8 ’ i in the ring of Gaussian integers.
3. Suppose that R is a Euclidean domain in which Ψ(a) ¤ Ψ(ab) for all nonzero elements
a, b ∈ R. Show that Ψ(a) ≥ Ψ(1), with equality if and only if a is a unit in R.
√ √
4. Let R = Z[ d ], where d is any integer, and de¬ne Ψ(a + b d ) = a2 ’ db2 . Show
that for all nonzero ± and β, Ψ(±β) = Ψ(±)Ψ(β), and if d is not a perfect square, then
Ψ(±) ¤ Ψ(±β).
√
5. Let R = Z[ d ] where d is not a perfect square. Show that 2 is not prime in R. (Show
that 2 divides d2 ’ d.)
√
6. If d is a negative integer with d ¤ ’3, show that 2 is irreducible in Z[ d ].
√
7. Let R = Z[ d ] where d is a negative integer. We know (see (2.7.5)) that R is an ED,
hence a PID and a UFD, for d = ’1 and d = ’2. Show that for d ¤ ’3, R is not
a UFD.
8. Find the least common multiple of 11 + 3i and 8 ’ i in the ring of Gaussian integers.
9. If ± = a + bi is a Gaussian integer, let Ψ(±) = a2 + b2 as in (2.7.5). If Ψ(±) is prime
in Z, show that ± is prime in Z[i].
2.8 Rings of Fractions
It was recognized quite early in mathematical history that the integers have a signi¬cant
defect: the quotient of two integers need not be an integer. In such a situation a mathe
matician is likely to say “I want to be able to divide one integer by another, and I will”.
This will be legal if the computation takes place in a ¬eld F containing the integers Z.
Any such ¬eld will do, since if a and b belong to F and b = 0, then a/b ∈ F . How do we
know that a suitable F exists? With hindsight we can take F to be the rationals Q, the
2.8. RINGS OF FRACTIONS 27
reals R, or the complex numbers C. In fact, Q is the smallest ¬eld containing Z, since
any ¬eld F ⊇ Z contains a/b for all a, b ∈ Z, b = 0, and consequently F ⊇ Q.
For an arbitrary integral domain D, the same process that leads from the integers
to the rationals allows us to construc a ¬eld F whose elements are (essentially) fractions
a/b, a, b ∈ D, b = 0. F is called the ¬eld of fractions or quotient ¬eld of D. The
mathematician™s instinct to generalize then leads to the following question: If R is an
arbitrary commutative ring (not necessarily an integral domain), can we still form fractions
with numerator and denominator in R? Di¬culties quickly arise; for example, how do
we make sense of a d when bd = 0? Some restriction must be placed on the allowable
c
b
denominators, and we will describe a successful approach shortly. Our present interest
is in the ¬eld of fractions of an integral domain, but later we will need the more general
development. Since the ideas are very similar, we will give the general construction now.
2.8.1 De¬nitions and Comments
Let S be a subset of the ring R; we say that S is multiplicative if (a) 0 ∈ S, (b) 1 ∈ S,
/
and (c) whenever a and b belong to S, we have ab ∈ S. We can merge (b) and (c) by
stating that S is closed under multiplication, if we regard 1 as the empty product. Here
are some standard examples of multiplicative sets.
(1) The set of all nonzero elements of an integral domain
(2) The set of all elements of a commutative ring R that are not zero divisors,
(3) R \ P , where P is a prime ideal of the commutative ring R
If S is a multiplicative subset of the commutative ring R, we de¬ne the following
equivalence relation on R — S:
(a, b) ∼ (c, d) if and only if for some s ∈ S we have s(ad ’ bc) = 0.
If we are constructing the ¬eld of fractions of an integral domain, then (a, b) is our
¬rst approximation to a/b. Also, since the elements s ∈ S are never 0 and R has no zero
divisors, we have (a, b) ∼ (c, d) i¬ ad = bc, and this should certainly be equivalent to
a/b = c/d.
Let us check that we have a legal equivalence relation. (Commutativity of multiplica
tion will be used many times to slide an element to a more desirable location in a formula.
There is also a theory of rings of fractions in the noncommutative case, but we will not
need the results, and in view of the serious technical di¬culties that arise, we will not
discuss this area.)
Re¬‚exivity and symmetry follow directly from the de¬nition. For transitivity, suppose
that (a, b) ∼ (c, d) and (c, d) ∼ (e, f ). Then for some elements s and t in S we have
s(ad ’ bc) = 0 and t(cf ’ de) = 0.
Multiply the ¬rst equation by tf and the second by sb, and add the results to get
std(af ’ be) = 0,
which implies that (a, b) ∼ (e, f ), proving transitivity.
28 CHAPTER 2. RING FUNDAMENTALS
If a ∈ R and b ∈ S, we de¬ne the fraction a to be the equivalence class of the pair
b
(a, b). The set of all equivalence classes is denoted by S ’1 R, and in view of what we are
about to prove, is called the ring of fractions of R by S. The term localization of R by S
is also used, because it will turn out that in Examples (1) and (3) above, S ’1 R is a local
ring (see Section 2.4, Problem 5).
We now make the set of fractions into a ring in a natural way.
a c ad + bc
addition +=
b d bd
ac ac
multiplication =
bd bd
0 0
( = for any s ∈ S)
additive identity
1 s
’a
a
additive inverse ’( ) =
b b
1 s
( = for any s ∈ S)
multiplicative identity
1 s
2.8.2 Theorem
Let S be a multiplicative subset of the commutative ring R. With the above de¬nitions,
S ’1 R is a commutative ring. If R is an integral domain, so is S ’1 R. If R is an integral
domain and S = R \ {0}, then S ’1 R is a ¬eld (the ¬eld of fractions or quotient ¬eld of R).
Proof. First we show that addition is wellde¬ned. If a1 /b1 = c1 /d1 and a2 /b2 = c2 /d2 ,
then for some s, t ∈ S we have
s(a1 d1 ’ b1 c1 ) = 0 and t(a2 d2 ’ b2 c2 ) = 0 (1)
Multiply the ¬rst equation of (1) by tb2 d2 and the second equation by sb1 d1 , and add the
results to get
st[(a1 b2 + a2 b1 )d1 d2 ’ (c1 d2 + c2 d1 )b1 b2 ] = 0.
Thus
a1 b2 + a2 b1 c1 d2 + c2 d1
= ,
b1 b2 d1 d2
in other words,
a1 a2 c1 c2
+ = +
b1 b2 d1 d2
so that addition of fractions does not depend on the particular representative of an equiv
alence class.
2.8. RINGS OF FRACTIONS 29
Now we show that multiplication is wellde¬ned. We follow the above computation as
far as (1), but now we multiply the ¬rst equation by ta2 d2 , the second by sc1 b1 , and add.
The result is
st[a1 a2 d1 d2 ’ b1 b2 c1 c2 ] = 0
which implies that
a1 a2 c1 c2
= ,
b1 b2 d1 d2
as desired. We now know that the fractions in S ’1 R can be added and multiplied in
exactly the same way as ratios of integers, so checking the de¬ning properties of a commu
tative ring essentially amounts to checking that the rational numbers form a commutative
ring; see Problems 3 and 4 for some examples.
Now assume that R is an integral domain. It follows that if a/b is zero in S ’1 R, i.e.,
a/b = 0/1, then a = 0 in R. (For some s ∈ S we have sa = 0, and since R is an integral
domain and s = 0, we must have a = 0.) Thus if a d = 0, then ac = 0, so either a or c is
c
b
0, and consequently either a/b or c/d is zero. Therefore S ’1 R is an integral domain.
If R is an integral domain and S = R \ {0}, let a/b be a nonzero element of S ’1 R.
Then both a and b are nonzero, so a, b ∈ S. By de¬nition of multiplication we have
’1
R is a ¬eld. ™
1
ab
b a = 1 . Thus a/b has a multiplicative inverse, so S
When we go from the integers to the rational numbers, we don™t lose the integers