Download 5.3 Ideals and Factor Rings

5.3
5.3
J.A.Beachy
1
Ideals and Factor Rings
from A Study Guide for Beginner’s by J.A.Beachy,
a supplement to Abstract Algebra by Beachy / Blair
27. Give an example to show that the set of all zero divisors of a commutative ring need
not be an ideal of the ring.
Solution: The elements (1, 0) and (0, 1) of Z×Z are zero divisors, but if the set of zero
divisors were closed under addition it would include (1, 1), an obvious contradiction.
28. Show that in R[x] the set of polynomials whose graph passes through the origin and
is tangent to the x-axis at the origin is an ideal of R[x].
Solution: We can characterize the given set as I = {f (x)R[x] | f (0) = 0 and f ′ (x) =
0}. Using this characterization of I, it is easy to check that if f (x), g(x) ∈ I, then
f (x) ± g(x) ∈ I. If g(x) ∈ R[x] and f (x) ∈ I, then g(0)f (0) = g(0) · 0 = 0, and
Dx (g(x)f (x)) = g ′ (x)f (x) + g(x)f ′ (x), and so evaluating at x = 0 gives g′ (0)f (0) +
g(0)f ′ (0) = g′ (0) · 0 + g(0) · 0 = 0. This shows that I is an ideal of R[x].
29. To illustrate Proposition 5.3.7 (b), give the lattice diagram of ideals of Z100 = Z/ h100i,
and the lattice diagram of ideals of Z that contain h100i.
Solution: Since multiplication in Z100 corresponds to repeated addition, each subgroup of Z100 is an ideal, and the lattice diagram is the same as that of the subgroups
of Z100 , given in the following diagram.
Z100
@
5Z100
2Z100
@
4Z100
@
@
10Z100
@
25Z100
50Z100
20Z100
@
h0i
The ideals of Z that contain h100i correspond to the positive divisors of 100.
Z
@
h2i
h5i
@
@
h4i
h10i
@
@
h20i
@
h50i
h100i
30. Let R be the ring Z2 [x]/ x3 + 1 .
h25i
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Comment: Table 5.1 gives the multiplication table, though it is not necessary to
compute it in order to solve the problem.
Table 5.1: Multiplication in Z2 [x]/ x3 + 1
x2 +x x+1 x2 +1 x2 +x+1
1
x
x2
2
2
2
2
x +x x+1 x +1
0
x +x
x +1
x+1
2
2
2
x+1 x +1 x +x
0
x+1
x +x
x2 +1
x+1
2
2
2
2
x +1
x +1 x +x x+1
0
x +1
x+1
x2 +x
0
0
0
x2 +x+1 x2 +x+1 x2 +x+1 x2 +x+1
x2 +x+1
2
2
1
x +x x+1 x +1 x2 +x+1
1
x
x2
2
2
2
2
x +1 x +x x+1 x +x+1
x
x
1
x
2
2
2
2
2
x
x+1 x +1 x +x x +x+1
x
1
x
×
x2 +x
(a) Find all ideals of R.
Solution: By
Proposition
5.3.7 (b), the ideals of R correspond to the ideals of Z2 [x]
3
that contain x + 1 . We have the factorization x3 + 1 = x3 − 1 = (x − 1)(x2 + x + 1),
so the only proper, nonzero ideals are the principal ideals given below.
A = h[x + 1]i = {[0], [x2 + x], [x + 1], [x2 + 1]}
B = [x2 + x + 1] = {[0], [x2 + x + 1]}
Comment: We note that R = A ⊕ B, accounting for its 8 elements.
(b) Find the units of R.
Solution: We have [x]3 = [1], so [x] and [x2 ] are units. To show that the only units are
1, [x], and [x2 ], note that [x + 1][x2 + x + 1] = [x3 + 1] = [0], so [x + 1] and [x2 + x + 1]
cannot be units. This also excludes [x2 + x] = [x][x + 1] and [x2 + 1] = [x2 ][1 + x].
(c) Find the idempotent elements of R.
Solution: Using the general fact that (a + b)2 = a2 + 2ab + b2 = a2 + b2 (since Z2 [x]
has characteristic 2) and the identities [x3 ] = [1] and [x4 ] = [x], it is easy to see that
the idempotent elements of R are [0], [1], [x2 + x + 1], and [x2 + x].
31. Let S be the ring Z2 [x]/ x3 + x .
Comment: It isn’t necessary to construct Table 5.2 in order to answer (a) and (b).
(a) Find all ideals of S.
Solution: Over Z2 we have the factorization x3 + x = x(x2 + 1) = x(x + 1)2 , so by
Proposition 5.3.7 (b) the proper nonzero ideals of S are the principal ideals generated
by [x], [x + 1], [x2 + 1] = [x + 1]2 , and [x2 + x] = [x][x + 1].
A = h[x]i = {[0], [x2 ], [x], [x2 + x]} ⊇ C = [x2 + x] = {[0], [x2 + x]}
B = [x2 + 1] = {[0], [x2 + 1]}
B + C = h[x + 1]i = {[0], [x + 1], [x2 + 1], [x2 + x]}
Comment: We note that S = A ⊕ B.
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Table 5.2: Multiplication in Z2 [x]/ x3 + x
x2
×
x2
x2
x
x
2
2
x +x
x +x
0
x2 +1
x+1
x2 +x
x2
1
2
x +x+1 x2
x
x
x2
x2 +x
0
x2 +x
x
x
x2 +x
x2 +x
x2 +x
0
0
0
x2 +x
x2 +x
x2 +1
0
0
0
x2 +1
x2 +1
x2 +1
x2 +1
x+1
x2 +x
x2 +x
0
x2 +1
x2 +1
x+1
x+1
1
x2 +x+1
x2
x2
x
x
2
2
x +x
x +x
x2 +1
x2 +1
x+1
x+1
2
1
x +x+1
x2 +x+1
1
(b) Find the units of S.
Solution: Since no unit can belong to a proper ideal, it follows from part (a) that we
only need to check [x2 + x + 1]. This is a unit since [x2 + x + 1]2 = [1].
(c) Find the idempotent elements of S.
Solution: Since [x3 ] = [x], we have [x2 ]2 = [x2 ], and then [x2 + 1]2 = [x2 + 1]. These,
together with [0] and [1], are the only idempotents.
32. Show that the rings R and S in Problems 30 and 31 are isomorphic as abelian groups,
but not as rings.
Solution: Both R and S are isomorphic to Z2 × Z2 × Z2 , as abelian groups. They
cannot be isomorphic as rings since S has a nonzero nilpotent element, [x2 + x], but R
does not. We can also prove this by noting that the given multiplication tables show
that R has 3 units, while S has only 2.
33. Let I, J be ideals of the commutative ring R, and for r ∈ R, define the function
φ : R → R/I ⊕ R/J by φ(r) = (r + I, r + J).
(a) Show that φ is a ring homomorphism, with ker(φ) = I ∩ J.
Solution: The fact that φ is a ring homomorphism follows immediately from the
definitions of the operations in a direct sum and in a factor ring. Since the zero
element of R/I ⊕ R/J is (0 + I, 0 + J), we have r ∈ ker(φ) if and only if r ∈ I and
r ∈ J, so ker(φ) = I ∩ J.
(b) Show that if I + J = R, then φ is onto, and thus R/(I ∩ J) ∼
= R/I ⊕ R/J.
Solution: If I + J = R, then we can write 1 = x + y, for some x ∈ I and y ∈ J.
Given any element (a + I, b + J) ∈ R/I ⊕ R/J, consider r = bx + ay, noting that
a = ax + ay and b = bx + by. We have a − r = a − bx − ay = ax − bx ∈ I, and
b − r = b − bx − ay = by − ay ∈ J. Thus φ(r) = (a + I, b + J), and φ is onto. The
isomorphism follows from the fundamental homomorphism theorem.
Comment: This result is sometimes called the Chinese remainder theorem for commutative rings. It is interesting to compare this proof with the one given for Theorem 1.3.6.
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34. Let I, J be ideals of the commutative ring R. Show that if I +J = R, then I 2 +J 2 = R.
Solution: If I + J = R, then there exist a ∈ I and b ∈ J with a + b = 1. Cubing both
sides gives us a3 + 3a2 b + 3ab2 + b3 = 1. Then we only need to note that a3 + 3a2 b ∈ I 2
and 3ab2 + b3 ∈ J 2 .
35. Show that x2 + 1 is a maximal ideal of R[x].
2
Solution:
R,
2 Since
x + 1 is irreducible
over
it follows from results in Chapter 4 that
2
R[x]/ x + 1 is a field. Therefore x + 1 is a maximal ideal.
36. Is x2 + 1 a maximal ideal of Z2 [x]?
Solution:
Since x2 + 1 is not irreducible over Z2 , the ideal it generates is not a
maximal ideal. In fact, x + 1 generates a larger ideal since it is a factor of x2 + 1.
37. Let R and S be commutative rings, and let φ : R → S be a ring homomorphism.
(a) Show that if I is an ideal of S, then φ−1 (I) = {a ∈ R | φ(a) ∈ I} is an ideal of R.
Solution: If a, b ∈ φ−1 (I), then φ(a) ∈ I and φ(b) ∈ I, so φ(a ± b) = φ(a) ± φ(b) ∈ I,
and therefore a ± b ∈ φ−1 (I). If r ∈ R, then φ(ra) = φ(r)φ(a) ∈ I, and therefore
ra ∈ φ−1 (I). By Definition 5.3.1, this shows that φ−1 (I) is an ideal of R.
(b) Show that if P is a prime ideal of S, then φ−1 (P ) is a prime ideal of R.
Solution: Suppose that P is a prime ideal of R and ab ∈ φ−1 (P ) for a, b ∈ R. Then
φ(a)φ(b) = φ(ab) ∈ P , so φ(a) ∈ P or φ(b) ∈ P since P is prime. Thus a ∈ φ−1 (P ) or
b ∈ φ−1 (P ), showing that φ−1 (P ) is prime in R.
38. Prove that in a Boolean ring every prime ideal is maximal.
Solution: We will prove a stronger result: if R is a Boolean ring and P is a prime
ideal of R, then R/P ∼
= Z2 . Since Z2 is a field, Proposition 5.3.9 (a) implies that P
is a maximal ideal.
If a ∈ R, then a2 = a, and so a(a − 1) = 0. Since P is an ideal, it contains 0, and
then a(a − 1) ∈ P implies a ∈ P or a − 1 ∈ P , since P is prime. Thus each element
of R is in either P or 1 + P , so there are only two cosets of P in R/P , and therefore
R/P must be the ring Z2 .
39. In R = Z[i], let I = {m + ni | m ≡ n (mod 2)}.
(a) Show that I is an ideal of R.
Solution: Let a + bi and c + di belong to I. Then a ≡ b (mod 2) and c ≡ d (mod 2),
so a + c ≡ b + d (mod 2) and a − c ≡ b − d (mod 2), and therefore (a + bi) ± (c + di) ∈ I.
For m + ni ∈ R, we have (m + ni)(a + bi) = (ma − nb) + (na + mb)i. Then ma − nb ≡
na + mb (mod 2) since n ≡ m (mod 2) and −nb ≡ nb (mod 2). This completes the
proof that I is an ideal of R.
(b) Find a familiar commutative ring isomorphic to R/I.
Solution: Note that m + ni ∈ I if and only if m − n ≡ 0 (mod 2). Furthermore,
m + ni ∈ 1 + I if and only if 1 − (m + ni) ∈ I, and this occurs if and only if
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1 − m ≡ −n (mod 2) if and only if m − n ≡ 1 (mod 2). Since any element of m + ni
R satisfies either m − n ≡ 0 (mod 2) or m − n ≡ 1 (mod 2), the only cosets in R/I
are 0 + I and 1 + I. Therefore R/I ∼
= Z2 .
ANSWERS AND HINTS
45. Let P and Q be maximal ideals of the commutative ring R. Show that
R/(P ∩ Q) ∼
= R/P ⊕ R/Q.
Hint: If you can show that P + Q = R, then you can use Problem 33 (b).
√
√
by 2 is a maximal ideal.
48. Show that in Z[ 2] the principal ideal generated
√
Hint: Define a ring homomorphism from Z[ 2] onto Z2 .