MATH 113 - WORKSHEET 6
SOLUTIONS
In the problems below, let φ : R → S be an onto ring homomorphism.
Recall that we proved in class that
• if I ⊆ S is an ideal in S, then φ−1 [I] is an ideal in R, and
• if J ⊆ R is an ideal in R, then φ[J] is an ideal in φ[R].
Let Id(S) be the set of ideals in S, and let Idφ (R) be the set of ideals in R containing
ker(φ) (ideals I ⊆ R such that ker(φ) ⊆ I). The goal of this sequence of problems is to
establish that there is an inclusion preserving bijection between Id(S) and Idφ (R).
(1) Define i : Idφ (R) → Id(S) by i(J) = φ[J] and p : Id(S) → Idφ (R) by p(I) = φ−1 [I].
Verify that i and p are well-defined, that is, that i(J) ∈ Id(S) and p(I) ∈ Idφ (R).
i and p stand for image and preimage.
Solution: Given J ∈ Idφ (R), we know that i(J) = φ[J] is an ideal in φ[R]. Now
φ is onto, so φ[R] = S, thus φ[J] ∈ Id(S).
Given I ∈ Id(S), we know that p(I) = φ−1 [I] is an ideal in R. We need to show
that ker(φ) ⊆ φ−1 [I]. Let a ∈ ker(φ). Then φ(a) = 0 ∈ I, so a ∈ φ−1 [I].
(2) Show that i and p are inclusion preserving, that is, if J ⊆ J 0 , then i(J) ⊆ i(J 0 ), and
if I ⊆ I 0 , then p(I) ⊆ p(I 0 ).
Solution: Suppose J ⊆ J 0 ⊆ R. Let a ∈ i(J) = φ[J]. Then there exists a0 ∈ J
such that φ(a0 ) = a. But then also a0 ∈ J 0 , so a = φ(a) ∈ φ[J 0 ] = i(J 0 ). Hence
i(J) ⊆ i(J 0 ).
Suppose I ⊆ I 0 ⊆ S. Let a ∈ p(I) = φ−1 [I]. Then φ(a) ∈ I. But then also
φ(a) ∈ I 0 , so a ∈ φ−1 [I 0 ] = p(I 0 ). Hence p(I) ⊆ p(I 0 ).
(3) Show that i ◦ p is the identity on Id(S), that is, i(p(I)) = I for all I ∈ Id(S).
Solution: Let I ∈ Id(S). Then i(p(I)) = φ[φ−1 [I]].
φ[φ−1 [I]] ⊆ I: Let a ∈ φ[φ−1 [I]]. Then there exists a0 ∈ φ−1 [I] such that φ(a0 ) = a.
But if a0 ∈ φ−1 [I], then φ(a0 ) ∈ I. Hence a ∈ I.
I ⊆ φ[φ−1 [I]]: Let a ∈ I. Since φ is onto, there exists a0 ∈ R such that φ(a0 ) = a.
Then a0 ∈ φ−1 [I], so a = φ(a0 ) ∈ φ[φ−1 [I]].
(4) Show that p ◦ i is the identity on Idφ (R), that is, p(i(J)) = J for all J ∈ Idφ (R).
Solution: Let J ∈ Idφ (R). Then p(i(J)) = φ−1 [φ[J]].
φ−1 [φ[J]] ⊆ J: Let b ∈ φ−1 [φ[J]]. Then φ(b) ∈ φ[J]. So there exists c ∈ J such
that φ(c) = φ(b). But then φ(b − c) = φ(b) − φ(c) = 0, so b − c ∈ ker(φ) ⊆ J. Since
b = (b − c) + c, and J is closed under addition, b ∈ J.
J ⊆ φ−1 [φ[J]]: Let b ∈ J. Then φ(b) ∈ φ[J], so b ∈ φ−1 [φ[J]].
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(5) Conclude that i and p give a bijection between Id(S) and Idφ (R).
Consider the onto ring homomorphism φ : Z → Z6 , φ(n) = n. Write down the
elements of the sets Id(Z6 ) and Idφ (Z) and the bijection between them.
What happens to ideals in Z which do not contain ker(φ)? For example, what is
φ[h4i]? What about φ[h5i]? φ[h12i]?
Solution: We’ve seen that i and p are inverse functions between Id(S) and Idφ (R),
so these sets are in bijection.
Id(Z6 ) = {h0i, h1i, h2i, h3i}. Note that h4i = h2i and h5i = h1i = Z6 (the multiples
of 4 and 2 are the same in Z6 , as are the multiples of 1 and 5).
We have ker(φ) = h6i ⊂ Z. The ideals in Z containing h6i are Idφ (Z) = {h6i, h2i, h3i, h1i}.
The bijection is:
h6i ↔ h0i
h2i ↔ h2i
h3i ↔ h3i
h1i ↔ h1i
i(h4i) = h2i, i(h5i) = h1i, and i(h12i) = h0i. So other ideals in Z get mapped
forward to ideals in Z6 , but not in a one-to-one way.