Question

Let R be a ring. For n ≥ 0, let I_{n} = {a ∈ R |
5^{n}a = 0}. Show that I = ⋃ I_{n} is an ideal of
R.

Please use the strategies from Chapter 14 in Joseph Gallian's "Contemporary Abstract Algebra."

Answer #1

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 > 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, 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.

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)...

Let R be a commutative ring and let a ε R be a non-zero element.
Show that Ia ={x ε R such that ax=0} is an ideal of R. Show that if
R is a domain then Ia is a prime ideal

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.

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 commutative ring with unity. Let A consist of all
elements in A[x] whose constant term is equal to 0. Show that A is
a prime ideal of A[x]

Let R be a ring and let M and N be
right R-modules. Assume that the only
R-homorphisms M → N and N →
M are 0 maps. Prove that
EndR(M⊕N) ∼=
EndR(M) ⊕ EndR(N) (direct
sum of rings). Remember the convention used for composition of
R-homomorphismps.

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