9.18. Let Ij C I2 C I3 C …… be ideals of R. Let I = U, In. 1. Show that I is an ideal of R. 2. Suppose that R/I is commutative. Show that for every a, b e R, there exists an n e N such that ab – ba e In.
9.18. Let Ij C I2 C I3 C …… be ideals of R. Let I = U, In. 1. Show that I is an ideal of R. 2. Suppose that R/I is commutative. Show that for every a, b e R, there exists an n e N such that ab – ba e In.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Could you explain how to show 9.18 in detail? I included list of theorems and definitions from the textbook.
![Definition 9.3. Let R be a ring and I an ideal of R. Then the factor ring (or quotient
ring), R/I, is the set of all left cosets {a + I : a e R} together with the operations
(a + I) + (b +I) = a +b+ I and (a + I)(b+ I) = ab + I, for all a, b e R.
Theorem 9.6. For any ring R and ideal I, the factor ring R/I is a ring.
Example 9.9. Let R = Z and I = (5) = 5Z. Then R/I = {0+ I, 1 + I, 2 +
I,3 + 1,4 + I} and, for instance, (2 + I) + (4 + I)
(3 + I)(4+ I) = 12 + I = 2 +I.
= 6+ I = 1 + I and
%3D
Example 9.10. Let R = M2(Z) and let I be the ideal consisting of all matrices
whose entries are even. Then notice that for any a;; e Z, we have
aji a12
b11 b12
+1 =
+1,
a21 a22
b21 b22,
where bij is 0 if a¡; is even and 1 if a;j is odd. Thus, R/I consists of the sixteen
different elements (bii b12
b21 b22
+1,b¡¡ € {0, 1}. We perform arithmetic in the following
fashion:
((49) +*) * (6 ) + +) – (; 2) +- - (")-
(6:) + ') = (; )
+ I
1 0
+ 1) +
+1:
and
(::) -·) (6) +-) - (; ) + -(;) --
)(:) --) - (:)
+ I.
1 2
Example 9.11. Let R = R[x] and I = (x² + 3). Readers familiar with polynomial
long division will know that if f (x) e R, then f (x) = (x² + 3)q(x) +r(x), where
q and r are polynomials, with r(x)
unfamiliar with polynomial long division can peek ahead to Section 10.1, where it
will be discussed in more generality.) Since (x² + 3)q(x) e I by absorption, we
know that elements of R/I are of the form a + bx + I, with a, b e R. Addition
behaves as expected; for instance, (2+3x + I)+(7 – 4x + I) = 9 – x+ I. To deal
with multiplication, observe that x2 – (-3) e I; thus, x² + I = -3+I. Therefore,
we have calculations such as
= a + bx, for some a, b E R. (Those who are
(5 + 4x + I)(-7+2x + I) = -35 – 18x + 8x² + I
= -35 – 18x + 8(-3) + I
= -59 – 18x + I.
Let us also record a few basic facts about factor rings.
Theorem 9.7. Let R be a ring and I an ideal. Then
1. if R is commutative, then so is R/I;
2. if R has an identity, then so does R/I; and
3. if u is a unit of R, then u + I is a unit of R/I.
Theorem 9.8. Let R be a ring with ideals I and J, such that I C J. Then J/I is
an ideal of R/I.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2F4d145006-9585-48cf-8c8d-46294ac0aed3%2Fe28bstnm_processed.png&w=3840&q=75)
Transcribed Image Text:Definition 9.3. Let R be a ring and I an ideal of R. Then the factor ring (or quotient
ring), R/I, is the set of all left cosets {a + I : a e R} together with the operations
(a + I) + (b +I) = a +b+ I and (a + I)(b+ I) = ab + I, for all a, b e R.
Theorem 9.6. For any ring R and ideal I, the factor ring R/I is a ring.
Example 9.9. Let R = Z and I = (5) = 5Z. Then R/I = {0+ I, 1 + I, 2 +
I,3 + 1,4 + I} and, for instance, (2 + I) + (4 + I)
(3 + I)(4+ I) = 12 + I = 2 +I.
= 6+ I = 1 + I and
%3D
Example 9.10. Let R = M2(Z) and let I be the ideal consisting of all matrices
whose entries are even. Then notice that for any a;; e Z, we have
aji a12
b11 b12
+1 =
+1,
a21 a22
b21 b22,
where bij is 0 if a¡; is even and 1 if a;j is odd. Thus, R/I consists of the sixteen
different elements (bii b12
b21 b22
+1,b¡¡ € {0, 1}. We perform arithmetic in the following
fashion:
((49) +*) * (6 ) + +) – (; 2) +- - (")-
(6:) + ') = (; )
+ I
1 0
+ 1) +
+1:
and
(::) -·) (6) +-) - (; ) + -(;) --
)(:) --) - (:)
+ I.
1 2
Example 9.11. Let R = R[x] and I = (x² + 3). Readers familiar with polynomial
long division will know that if f (x) e R, then f (x) = (x² + 3)q(x) +r(x), where
q and r are polynomials, with r(x)
unfamiliar with polynomial long division can peek ahead to Section 10.1, where it
will be discussed in more generality.) Since (x² + 3)q(x) e I by absorption, we
know that elements of R/I are of the form a + bx + I, with a, b e R. Addition
behaves as expected; for instance, (2+3x + I)+(7 – 4x + I) = 9 – x+ I. To deal
with multiplication, observe that x2 – (-3) e I; thus, x² + I = -3+I. Therefore,
we have calculations such as
= a + bx, for some a, b E R. (Those who are
(5 + 4x + I)(-7+2x + I) = -35 – 18x + 8x² + I
= -35 – 18x + 8(-3) + I
= -59 – 18x + I.
Let us also record a few basic facts about factor rings.
Theorem 9.7. Let R be a ring and I an ideal. Then
1. if R is commutative, then so is R/I;
2. if R has an identity, then so does R/I; and
3. if u is a unit of R, then u + I is a unit of R/I.
Theorem 9.8. Let R be a ring with ideals I and J, such that I C J. Then J/I is
an ideal of R/I.
![9.18. Let I1 C I2 C 13 C . .. be ideals of R. Let I =U, In.
n=1
1. Show that I is an ideal of R.
2. Suppose that R/I is commutative. Show that for every a, b e R, there exists an
n e N such that ab – ba E In.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2F4d145006-9585-48cf-8c8d-46294ac0aed3%2Fi5axntg_processed.png&w=3840&q=75)
Transcribed Image Text:9.18. Let I1 C I2 C 13 C . .. be ideals of R. Let I =U, In.
n=1
1. Show that I is an ideal of R.
2. Suppose that R/I is commutative. Show that for every a, b e R, there exists an
n e N such that ab – ba E In.
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