Download Ideals Definition (Ideal). A subring A of a ring R is called a (two

CHAPTER 14
Ideals and Factor Rings
Ideals
Definition (Ideal). A subring A of a ring R is called a (two-sided) ideal
of R if for every r 2 R and every a 2 A, ra 2 A and ar 2 A.
Note.
(1) A “absorbs” elements of R by multiplication.
(2) Ideals are to rings as normal subgroups are to groups.
Definition. An ideal A of R is a proper ideal if A is a proper subset of R.
Theorem (Ideal Test). A nonempty subset A of a ring R is an ideal of
R if
(1) a, b 2 A =) a
b 2 A.
(2) a 2 A and r 2 R =) ar 2 A and ra 2 A.
Proof.
Follows directly from the definition of ideal and Theorem 12.3 (Subring Test).
⇤
Example.
(1) {0} (the trivial ideal) and R itself are ideals of R.
(2) For any positive integer n, nZ = {0, ±n, ±2n, ±3n, . . . } is an ideal of Z.
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14. IDEALS AND FACTOR RINGS
(3) Let R be a commutative ring with identity and let a 2 R. The set
hai = {ra|r 2 R}
is an ideal of R called the principal ideal generated by a.
Note that the commutative assumption is necessary here. Also, context will
distinguish between this use of hai and its use in cyclic groups.
(4) Let R[x] denote the set of all poynomials with real coefficients and let A be
the subset of all polynomials with constant term 0. Then A is an ideal of R[x]
and A = hxi.
(5) Let R be a commutative ring with unity and let a1, a2, . . . , an 2 R. Then
I = ha1, a2, . . . , ani = {r1a1 + r2a2 + · · · + rnan|ri 2 R}
is an ideal of R called the ideal generated by a1, a2, . . . , an.
Proof.
If r1a1 + r2a2 + · · · + rnan, r10 a1 + r20 a2 + · · · + rn0 an 2 I,
(r1a1 + r2a2 + · · · + rnan)
(r1a1
(r10 a1 + r20 a2 + · · · + rn0 an) =
r10 a1) + (r2a2
(r1
r20 a2) + · · · + (rnan
r10 )a1 + (r2
If r1a1 + r2a2 + · · · + rnan 2 I and r 2 R,
rn0 an) =
r20 )a2 + · · · + (rn
rn0 )an 2 I.
r(r1a1 + r2a2 + · · · + rnan) = (r1a1 + r2a2 + · · · + rnan)r =
(rr1)a1 + (rr2)a2 + ı · · · + (rrn)an 2 I.
Therefore I is an ideal by Theorem 14.1.
⇤
14. IDEALS AND FACTOR RINGS
157
(6) Let Z[x] be the ring of all polynomials with integer coefficients and let I be
the subset of Z[x] of all polynomials with even constant term. Then
I = hx, 2i
is an ideal of Z[x].
Proof.
[To show I = hx, 2i.]
If f (x) 2 hx, 2i, f (x) = xg(x) + 2h(x) where g(x), h(x) 2 R. Then
f (0) = 0 · g(0) + 2 · h(0) = 2h(0),
so f (x) 2 I. Also, if f (x) 2 I,
f (x) = anxn + an 1xn
1
+ · · · a1x + 2k =
x(anxn
Thus, by mutual inclusion, I = hx, 2i.
1
+ an 1xn
2
+ · · · a1) + 2k 2 hx, 2i.
[Show I is an ideal.] Suppose k(x) 2 Z[x] and f (x) = xg(x) + 2h(x) 2 I.
Then
p(x) = f (x)k(x) = k(x)f (x) = xk(x)g(x) + 2k(x)h(x) 2 I.
Also, if f (x) = xf1(x) + 2f2(x) and g(x) = f (x) = xg1(x) + 2g2(x),
f (x)
g(x) = x (f1(x)
so I is an ideal by Theorem 14.1.
g1(x) + 2 (f2(x)
g2(x) 2 I,
⇤
(7) Let R be the ring of all real-valued functions of a real variable. Let S be
the subset of all di↵erentiable functions (this means for f 2 S, f 0(x) is defined
for all real x. S is not an ideal of R.
8
>
x>0
<1,
Let f (x) = 1 2 S and let g(x) = sgn x = 0,
x = 0.
>
: 1, x < 0
h(x) = g(x)f (x) = sgn x 62 S. Thus S is not an ideal, but is a subring of R.
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14. IDEALS AND FACTOR RINGS
Factor Rings
Theorem (14.2 — Existence of Factor Rings). Let R be a ring and A a
subring of R. The set of cosets {r+A|r 2 R} is a ring under the operations
(s + A) + (t + A) = s + t + A and (s + A)(t + A) = st + A ()
A is an ideal of R.
[In this case , we say R/A is a factor ring of R.]
Proof.
We know the set of cosets form a group under addition. If our multiplication
is well-defined, i.e., multiplication is a binary operation, it is clear that the
multiplication is associative and distributive over addition.
[To show multiplication is well-defined () A is an ideal of R.]
((=) Suppose A is an ideal of R and let s + A = s0 + A and t + A = t0 + A.
Now s = s0 + a and t = t” + b where a, b 2 A. Then
st = (s0 + a)(t0 + b) = s0t0 + s0b + at0 + ab =)
st + A = s0t0 + s0b + at0 + ab + A = s0t0 + A
since s0b + at0 + ab 2 A. Thus multiplication is well-defined.
(=)) (using contrapositive) Suppose A is a subring of R that is not an ideal.
then 9 a 2 A and r 2 R 3 ar 62 A or ra 62 A. WLOG, assume ar 2
6 A.
Consider a + A = 0 + A and r + A.
(a + A)(r + A) = ar + A, but (0 + A)(r + A) = 0 · r + A = A 6= ar + A,
so multiplication is not well-defined and R/A is not a ring.
⇤
14. IDEALS AND FACTOR RINGS
159
Example.
(1) Z/5Z = {0 + 5Z, 1 + 5Z, 2 + 5Z, 3 + 5Z, 4 + 5Z} is a factor ring since 5Z
is an ideal of Z.
(3 + 5Z) + (4 + 5Z) = 7 + 5Z = 2 + 5 + 5Z = 2 + 5Z
and
(3 + 5Z)(4 + 5Z) = 12 + 5Z = 2 + 10 + 5Z = 2 + 5Z.
We have essentially modular 5 arithmetic.
(2) 3Z/9Z = {0 + 9Z, 3 + 9Z, 6 + 9Z}is a factor ring since 9Z is an ideal of
3Z, the arithmetic essentially modulo 9.
(6 + 9Z) + (6 + 9Z) = 12 + 9Z = 3 + 9 + 9Z = 3 + 9Z
and
(6 + 9Z)(6 + 9Z) = 36 + 9Z = 9Z.
⇢
a1 a2
(3) Let R =
ai 2 Z and I be the subring of R consisting of
a3 a4
matrices with even entries. I is an ideal of R.
Proof.

a a
Clearly, subtraction is closed in I. So let 1 2 2 R and
a3 a4


2b1 2b2
b b
= 2 1 2 2 I. Then
2b3 a4
b3 b4


✓

◆
a1 a2
b b
a1 a2 b1 b2
(2) 1 2 = 2
2I
a3 a4
b3 b4
a3 a4 b3 b4
and
✓ 
◆
✓

◆
b1 b2
a1 a2
b1 b2 a1 a2
2
=2
2 I,
b3 b4
a3 a4
b3 b4 a3 a4
so I is an ideal.
⇤
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14. IDEALS AND FACTOR RINGS
What is |R/I|?
Solution.

2a1 + r1 2a2 + r2
Every member of R can be written in the form
where
2a3 + r3 2a4 + r4
ai 2 Z and ri 2 {0, 1}. But




2a1 + r1 2a2 + r2
2a1 2a2
r r
r r
+I =
+ 1 2 + I = 1 2 + I.
2a3 + r3 2a4 + r4
2a3 2a4
r3 r4
r3 r4

r r
Thus there are 24 choices for 1 2 or 16 choices for the lements of R/I. ⇤
r3 r4
Example. Consider R = Z[i]/h2 ii, the factor ring of the Gaussian
integers over h2 ii. What are its elements?
For r 2 R, r = a + bi + h2 ii as a start. Now 2 i + h2 ii = 0 + h2
so we can consider 2 i = 0 mod h2 ii or i = 2. Then, as an example,
3 + 4i + h2
ii = 3 + 8 + h2
Similarly, for all r 2 R, r = a + h2
ii = 11 + h2
ii,
ii.
ii where a 2 Z.
Next, we also have, for i = 2, i2 = 4 or
1 = 4 or 0 = 5. Thus
7+6i+h2 ii = 7+12+h2 ii = 19+h2 ii = 4+5·3+h2 ii = 4+h2 ii.
It follows that
{0 + h2
ii, 1 + h2
ii, 2 + h2
5(1 + h2
ii) = 5 + h2
Are any of these the same?
so |1 + h2
ii| is 1 or 5. If |1 + h2
1 + h2
ii = 0 + h2
ii, 3 + h2
ii, 4 + h2
ii = 0 + h2
ii}.
ii
ii| = 1, then
ii =) 1 2 h2
ii or 1 = (2
i)(a + bi)
where a + bi 2 Z[i]. Then 2a + b + ( a + 2b)i = 1 or 2a + b = 1 or a + 2b = 0.
1
Then 5b = 1 =) b = , a contradiction. Thus |1 + h2 ii| = 5 and |R| = 5.
5
14. IDEALS AND FACTOR RINGS
161
Example. Let R[x] be the ring of polynomials with real coefficients and
hx2 + 1i the principal ideal generated by x2 + 1. Then
Now
hx2 + 1i = {f (x)(x2 + 1)|f (x) 2 R[x]}.
R[x]/hx2 + 1i = {g(x) + hx2 + 1i|g(x) 2 R[x]},
and (using the division algorithm for real polynomials),
g(x) = q(x)(x2 + 1) + r(x)
where r(x) = 0 or the degree of r(x) is less than 2, the degree of x2 + 1.
Thus r(x) = ax + b where a, b 2 R. Thus g(x) = q(x)(x2 + 1) + ax + b and
so
g(x) + hx2 + 1i = ax + b + q(x)(x2 + 1) + hx2 + 1i = ax + b + hx2 + 1i,
R[x]/hx2 + 1i = {ax + b + hx2 + 1i|a, b 2 R}.
How to multiply in R[x]/hx2 + 1i?
Since x2 + 1 + hx2 + 1i = 0 + hx2 + 1i, x2 + 1 = 0 mod x2 + 1 or x2 =
Thus
(2x + 3 + hx2 + 1i)(5x
2 + hx2 + 1i) =
10x2 + 11x
6 + hx2 + 1i = 11x
1.
16 + hx2 + 1i.
Note that, with x playing the role of i, this ring is isomorphic to the complex
numbers.
Prime Ideals and Maximal Ideals
Definition (Prime Ideal, Maximal Ideal). A prime ideal A of a commutative ring R is a proper ideal of R such that a, b 2 R and ab 2 A implies a 2 A
or b 2 A. A maximal ideal A of R is a proper ideal of R if, whenever B is an
ideal of R and A ✓ B ✓ R, then B = A or B = R.
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14. IDEALS AND FACTOR RINGS
Example. Consider the ring Z.
{0} is a prime ideal of Z. If ab 2 A, then ab = 0 =) a = 0 or b = 0 since Z
is an integral domain =) a 2 {0} or b 2 {0}.
For n > 1, nZ is a prime ideal () n is prime.
Proof.
((=) Suppose n is prime. Recalling hni = nZ, suppose a, b 2 Z with ab 2 n.
Then n|ab =) (Euclid’s Lemma) n|a or n|b =) a 2 hni or b 2 hni. Thus
hni is prime.
(=)) (by contrapositive) Suppose n is not prime. Then n = st where s < n
or t < n. We have st 2 hni, but s 62 hni and t 62 hni, so hni is not prime. ⇤
Example. Consider the lattice of ideals of Z100.
The diagram shows that h2i and h5i are the maximal ideals.
14. IDEALS AND FACTOR RINGS
163
Problem (Page 275 # 15). If A is an ideal of a ring R and 1 2 A, then
A = R.
Proof. Let r 2 R. Then r = r · 1 2 A.
⇤
Example. hx2 + 1i is a maximal ideal of R[x].
Proof.
Suppose A is an ideal of R[x] and hx2 + 1i & A,
[To show 9 c 2 R, c 6= 0, with c 2 A.] Let f (x) 2 A, f (x) 62 hx2 + 1i. Then
f (x) = q(x)(x2 + 1) + r(x)
where r(x) 6= 0 and deg r(x) < 2. Then r(x) = ax + b where a and b are not
both 0, and
ax + b = r(x) = f (x) q(x)(x2 + 1) 2 A.
Thus
a2x2 b2 = (ax + b)(ax b) 2 A, and a2(x2 + 1) 2 A
since hx2 + 1i ✓ A. Then
0 6= a2 + b2 = (a2x2 + a2)
(a2x2 b2) 2 A.
1
1
Let c = a2 + b2. Since c 2 A, c 2 R[x] =) 2 R[x], so 1 = · c 2 A. By
c
c
2
Page 275 # 15, A = R[x], and so hx + 1i is a maximal ideal of R[x].
⇤
Example. hx2 + 1i is not prime in Z2[x], since it contains
(x + 1)2 = x2 + 2x + 1 = x2 + 1,
but does not contain x + 1.
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14. IDEALS AND FACTOR RINGS
Problem (Page 275 # 26). If R is a commutative ring with unity and A
is a proper ideal of R, then R/A is a commutative ring with unity.
Proof.
Note that
(b + A)(c + A) = bc + A = cb + A = (c + A)(b + A).
Thus R/A is commutative. Also, if 1 is the unit of R, 1 + A is the unit of
R/A.
⇤
Theorem (14.3 — R/A is an Integral Domain () A is Prime). Let R
be a commutative ring with identity and let A be an ideal of R. Then R/A
is an integral domain () A is prime.
Proof.
(=)) Suppose R/A is an integral domain and ab 2 A. Then
(a+A)(b+A) = ab+A = A =) a+A = A or b+A = A =) a 2 A or b 2 A.
Thus A is prime.
((=) Note that R/A is a commutative ring with unity for any proper ideal A
from Pge 275 # 26. Suppose A is prime and
(a + A)(b + A) = ab + A = 0 + A = A.
Then ab 2 A =) a 2 A or b 2 A since A is prime. Thus
a + A = A or b + A = A =) R/A has no zero-divisors and is thus an integral
domain.
⇤
14. IDEALS AND FACTOR RINGS
165
Problem (Page 275 # 25). Let R be a commutative ring with unity and
let A be an ideal of R. If b 2 R andB = {br + a|r 2 R, a 2 A}, then B is an
ideal of R.
Proof.
For br1 + a1, br2 + a2 2 B and and r, r0 2 R,
(br1 + a1)
(br2 + a2) = b(r1
r2) + (a1
and
a2) 2 B
r0(br + a) = b(r0r) + r0a 2 B.
Thus B is an ideal by the ideal test.
⇤
Theorem (14.4 — R/A is a Field () A is maximal). Let R be a
commutative ring with unity and let A be an ideal of R. Then R/A is a
field () A is maximal.
Proof.
(=)) Suppose R/A is a field and B is an ideal of R with A & B. Let b 2 B,
b 62 A. Then b + A 6= A, so 9 c 2 A 3 (b + A)(c + A) = 1 + A, the
multiplicative identity of R/A. Since b 2 B, bc 2 B. Because
1 + A = (b + A)(c + A) = bc + A,
1 bc 2 A & B. Thus 1 = (1
A is maximal.
bc) + bc) 2 B. By Page 275 # 15, B = R, so
((=) Suppose A is maximal and let b 2 R, b 62 A.
[To show b + A has a multiplicative inverse.]
Consider B = {br + a|r 2 R, a 2 A}. B is an ideal of R by Page 275 #25,
and A & B Since A is maximal, B = R. Thus 1 2 B, say 1 = bc + a0 where
a0 2 A. Then
1 + A = bc + a0 + A = bc + A = (b + A)(c + A).
Thus R/A is a field.
Corollary. A maximal ideal is a prime ideal.
⇤