Download Integral Domains While the set of integers is one of our prototypical

13. Integral Domains and Fields
1
Integral Domains
While the set of integers is one of our prototypical
examples of a ring, there are too many important
properties of Z unaccounted for in the definition of
ring; besides being commutative and having a
unity element, the most important arithmetical
property of Z not captured by satisfying the
definition of a ring is that while only 1 and –1 are
units, there is still a cancellation property:
ab = ac ⇒ b = c . This leads to the following
definitions.
The element a in a ring is called a zero divisor if
there exists a nonzero b in the ring so that ab = 0.
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A commutative ring with unity is called an
integral domain if it contains no zero divisors.
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Examples:
• Z is an integral domain (of course!)
• Zn is an integral domain only when n is a prime,
for if n = ab is a nontrivial factorization of n,
then ab = 0 in this ring
• Z[x] is an integral domain
13. Integral Domains and Fields
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Theorem If a, b, and c are elements of an integral
domain D and a ≠ 0, then ab = ac ⇒ b = c .
Proof ab = ac ⇒ a(b − c) = 0 ⇒ b − c = 0 since b – c
cannot be a zero divisor.
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In much the same way that the structure of an
integral domain is more descriptive of the integers
than the basic structure of a ring, the rings Q, R,
and C all share a basic property not identified by
the fact that they are rings: in each example, every
nonzero element is a unit. Any ring in which all
nonzero elements are units is called a field. There
are other rings that have the additional structure
of a field.
Theorem Any finite integral domain is a field.
Proof Let D be a finite integral domain and
suppose that a is any nonzero element. If a = 1,
then a is its own inverse. If not, the list of powers
of a must eventually repeat: there are positive
integers i > j so that a i = a j . By cancellation, we
get a i− j = 1. But a ≠ 1, so i – j > 1 and the inverse of
a is a i− j −1. //
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13. Integral Domains and Fields
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Corollary If p is a prime, then Z p is a field.
Proof All we need to show is that Z p contains no
zero divisors. So suppose
€ ab mod p = 0. Then there
is some integer k so that ab = pk, whence p divdes
the product ab. It follows that either p divides a
(a mod p = 0) or p divides b€(b mod p = 0). So
neither a nor b is a zero divisor. //
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More examples of fields:
• Z3 [i], the set of all polynomial expressions in
powers of i = −1 with coefficients from the field
Z3 (the Gaussian integers mod 3); any element of
this ring has the form a + bi with a,b ∈ Z3 since
i 2€= −1 ∈ Z3 , and every nonzero element is a unit
(see the multiplication table on p. 251)
• Q[ 2], the set of polynomial
€ expressions in
powers of 2 with rational coefficients; again,
any element of this ring has the form a + b 2 for
rational a and b, and all nonzero elements are
units
€ because
1
a+b 2
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=
1
⋅
a −b 2
a +b 2 a −b 2
=
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a
b
−
2
2
2
2
2
a − 2b a − 2b
13. Integral Domains and Fields
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The characteristic of a ring R, denoted char R, is
the smallest positive integer n such that nx = 0 for
all x in the ring; if no such integer exists, we say
that the ring has characteristic 0.
Theorem A ring R with unity 1 has positive
characteristic n if and only if the order of 1 within
the additive group that defines R equals n. The
ring has characteristic 0 if and only if 1 has infinite
order under addition.
Proof 1 has additive order n > 0 ⇔ n ⋅1 = 0 ⇔
nx = x1+4
x2
+ L+
12
+L
4
44
3x = (11+4
4+1
3 )x = (n ⋅1)x = 0 ⇔
n terms
n terms
char R = n. Further, if 1 €
has infinite order then
char R must be 0; the converse is clear. //
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13. Integral Domains and Fields
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Theorem If D is an integral domain, then char D
is either 0 or a prime number.
Proof If 1 has infinite order in D, then char D = 0.
Otherwise, char D equals the finite order n of 1
under addition. But if n is composite, it factors as
n = rs with 0 < r, s < n. So
0 = n ⋅1 = (rs) ⋅1 = r(s ⋅1) = (r ⋅1)(s ⋅1),
and since there are no zero divisors, either r ⋅1 = 0
or s ⋅1 = 0, in violation of the fact that n is the
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smallest
positive integer so that n ⋅1 = 0. Thus n
must be prime. //
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