(1) Let I be a proper ideal of a ring R; then there is a maximal ideal m ...

Solutions to exercise 2
(1) Let I be a proper ideal of a ring R; then there is a maximal ideal m of
R such that I ⊆ m.
Proof. Let S be the set of all proper ideals which contain I; then S is
a partially ordered set with respect to ⊆. It is easy to check that S
satisfies the assumptions of Zorn’s Lemma. Therefore S has a maximal
element, say m. Of course m is a maximal ideal of R and I ⊆ m.
(2) If R is a U.F.D. then every nonzero prime ideal of R contains a nonzero
principal prime ideal.
Proof. Let P be a nonzero prime ideal of R; then there is an element
a ∈ P such that a �= 0 and a is not a unit. Since R is a U.F.D.,
a = b1 · · · bn for some irreducible elements bi. As P is prime and a ∈ P ,
bi ∈ P for some i. Therefore (bi) is a nonzero principal prime ideal
contained in P .
(3) Find a ring R and a multiplicative closed subset S of R such that there
are ideals I �= J of R such that IS = JS.
Solution: Let R = Z and S = Z − {0}. Let I = 2Z and J = 3Z; then
IS = JS = RS = Q.
(4) Let p be a prime number. Prove that any (p)-primary ideal of Z is of
the form (p n ).
Proof. Let Q be a (p)-primary ideal; then Q = (k) for some positive
integer k as Z is a P.I.D.. Since p m ∈ (k) for some positive integer m,
k is a factor of p m . Therefore k = p n for some positive integer n.
(5) Let R be a ring and S be a multiplicative subset of R. Let I be an ideal
of R; then ( √ I)S = √ IS.
Proof. Let a ∈ ( √ I)S; then there are b ∈ √ I, s ∈ S such that a = b/s,
so that b n ∈ I, it follows that a n = b n /s n ∈ IS. Therefore a ∈ √ IS.
Conversely, if a/s ∈ √ IS, then a n /s n = b/t for some n and for some
b ∈ I, t ∈ S, so that there is an element u ∈ S such that u(a n t − bs n ) =
0, it follows that a n tu ∈ I. Therefore atu ∈ √ I and then a/s ∈ ( √ I)S.
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(6) Let R be a ring. Then R is local if and only if for all r, s ∈ R, r + s = 1
implies r or s is a unit.
Proof. Let T be the set of non-units of R.
⇒ Let R be a local ring; then T is an ideal of R. Let r, s ∈ R such
that r + s = 1 and s ∈ T . If t ∈ T , then 1 ∈ T . Therefore t /∈ T and t
is a unit.
⇐ It is enough to show that T is an ideal of R, or equivalently, T is
an additive subgroup of R. For this, let r, s ∈ T . If r + s is a unit
then ar + as = 1, where a = (r + s) −1 , so that ar or as is a unit by
assumption, it follows that r or s is a unit. Therefore r + s ∈ T .
(7) Find the inverse of 1 + x in the power series ring F [[x]].
Solution: It is easy to check that (1 + x) −1 ∞�
=
i=0
(−1) n x n .
(8) Let M be an R-module; prove that annRM = {r ∈ R | rx = 0 ∀x ∈ M}
is an ideal of R.
Proof. Let a, b ∈ annRM; then aM = bM = 0, so that x(a − b) =
xa−xb = 0 for every x ∈ M, it follows that a−b ∈ annRM. Moreover,
let a ∈ annRM and r ∈ R; then raM ⊆ aM = 0, so that ra ∈ annRM.
We conclude that annRM is an ideal.
(9) Let R be a ring.
(a) Prove that if a, b ∈ R are NZDs, then so is ab.
(b) Let Σ be the set of all ideals in which every element is a zero-divisor.
Prove that the set Σ has maximal elements and that every maximal
element of Σ is a prime ideal.
Proof. (a) If (ab)c = 0 for some element c ∈ R, then bc = 0 as a is a
NZD, so that c = 0 as b is a NZD. Thus ab is a NZD.
(b) Let ⊆ be the order of Σ; then Σ is a partially ordered set. It is easy
to check that Σ satisfies the assumptions of Zorn’s Lemma. Therefore
Σ has maximal elements. Let P be a maximal element of Σ and there
are a, b ∈ R such that ab ∈ P , a /∈ P and b /∈ P ; then P is properly
contained in the ideals (a) + P and (b) + P , so that by the choice of P ,
there are r, s ∈ R and c, d ∈ P such that ar + c and bs + d are NZDs, it
follows that (ar + c)(bs + d) ∈ P , which is a NZD by (a). Contradicts
to the fact that P ∈ Σ. Therefore P is prime.
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(10) Let M be an R-module and I be an ideal of R. If IM = 0, then M
can be viewed as R/I module.
Proof. Define x · (r + I) = xr for x ∈ M and r ∈ R. Then it is easy to
see that · is well-defined and M is an R/I-module.
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