Question

9.3.2 Problem. Let R be a ring and I an ideal of R. Let π : R→R/I be the natural projection. Let J be an ideal of R.

- Show that π−1(π(J)) = (I, J).
- Show that if J is a maximal ideal of R with, I not ⊆ J, then π (J) = R/I.
- Suppose that J is an ideal of R with I ⊆ J. Show that J is a maximal ideal of R if and only if π(J) is a maximal ideal of R/I.

Answer #1

Let I be an ideal in a commutative ring R with identity. Prove
that R/I is a field if and only if I ? R and whenever J is an ideal
of R containing I, I = J or J = R.

Let I, M be ideals of the commutative ring R. Show that M is a
maximal ideal of R if and only if M/I is a maximal ideal of
R/I.

Suppose that R is a commutative ring and I is an ideal in R.
Please prove that I is maximal
if and only if R/I is a field.

Let I be an ideal of the ring R. Prove that the reduction map
R[x] → (R/I)[x] is a ring homomorphism.

Let
R be a ring, and let N be an ideal of R.
Let γ : R → R/N be the canonical homomorphism.
(a) Let I be an ideal of R such that I ⊇ N.
Prove that γ−1[γ[I]] = I.
(b) Prove that mapping
{ideals I of R such that I ⊇ N} −→ {ideals of R/N} is a
well-defined bijection between two sets

Let R be a ring, and set I:={(r,0)|r∈R}. Prove that I is an
ideal of R×R, and that (R×R)/I is isomorphism to R.

Let R be a ring. For n > or equal to 0, let In = {a element
of R | 5na = 0}. Show that I = union of In is an ideal of R.

Let R be a ring. For n ≥ 0, let In = {a ∈ R |
5na = 0}. Show that I = ⋃ In is an ideal of
R.
Please use the strategies from Chapter 14 in Joseph Gallian's
"Contemporary Abstract Algebra."

Let R be a ring. For n ≥ 0, let In = {a ∈ R |
5na = 0}. Show that I = ⋃ In is an ideal of
R.
Please use the strategies from Chapter 14 in Joseph Gallian's
"Contemporary Abstract Algebra."

Let P be a commutative PID (principal ideal domain) with
identity. Suppose that there is a surjective ring homomorphism f :
P -> R for some (commutative) ring R. Show that every ideal of R
is principal. Use this to list all the prime and maximal ideals of
Z12.

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