Majed Albaity
09/12/2011
The Radical of an Ideal
Abstract: The radical of an ideal plays an important role in commutative
algebra, when we are concerned with the geometry aspects. This is due to the
bijection existing between varieties and radical ideals. In particular, radicals
are an important part of the statement of Hilberts Nullstellensatz. Therefore,
in this presentation, we are going to shed light on some properties of the radical
of an ideal.
1.1.1 Denition:
p Let I be an ideal in a commutative ring R; the radical of
I, denoted by I; is the collection of elements in R some power of which lie in
I; i.e. p
I = r 2 R j rn 2 I for some n 2 Z +
1.1.2 Remarks: p
a) In some
p books, the radical of an ideal
p denoted by rad(I), I, r(I):
b) I I:(For r 2 I ) r1 2 I ) r 2 I)
c) The radical of the zero ideal is called the nilradical of R.
Recall: a 2 R is called nilpotent if there exists n 2 Z + , an = 0:
c) a 2 R is in the nilradical if and only if some power of a is 0, so the
nilradical of R is the set of all nilpotent elements of R:
1.1.3 Notice that, we can rewrite the denition as the following:
p
I = r 2 R j rn 2 I for some n 2 Z +
= r 2 R j (rn + I) = I for some n 2 Z +
= r 2 R j (r + I)n = I for some n 2 Z +
= r 2 R j rn = 0 for some n 2 Z +
= fr 2 R j r is a nilpotent element of R=I g
p
1.1.4 Theorem: Let R be a ring and I be an ideal of R, then I is an
ideal of R. p p p
Proof. rst of all, 0 2 I since 0 = 01 2 I: Suppose x; y 2 I, then xn 2 I
for some n 1, and y m 2 I for some m 1: Let N = m + n, then
P
N
N
(x y)N = ( 1)k k xN k k
y :
k=0
For each k, 0 k N; either k m or N k = n + (m k) n: Thus y k 2 I
or xN k 2 I for every k. Since I is n ideal, it follows that (x y)N 2 I: Thus
1
p p n
x y 2 I: Suppose that x 2 I and r 2 R, then p x 2 I for some n1 1; and
n n n
then (rx) = r x 2 I; therefore; rx 2 I: Hence I is an ideal of R:
Some properties of the radical of an ideal:
1.1.5 Proposition: Let R be a commutative ring and I; J be two ideals of
R. Then p p
(i) if I Jpfor n 2 Z + ; then I J:
p p
(ii) I = I:p
p p p
(iii) I + J = I + J:
p
Proof. (i) Let a 2p I, then there p exists
p n 1 such that an 2 I pJ: Thus
n
a 2 J, n 1; so a 2 p J: Hence I J: (ii) Since we know that I I;pthen
p p p
from (i), we get I I: Now to prove the secondpinclusion, let a 2 I;
this implies that there exists n 1 such that an 2 I;and then there qp exists
nm n m
p p
m 1 such that a = (a ) 2 I: Therefore, a 2 I: Thus I I, so
p pp p p p p
I= I: (iii) SincepI I and J J; then I +J I + J; and then by
p p p
(i), we get I + J I + J:pNow letp us prove the
p second
p inclusion, since
I I + J and J I + J,p then I I
pp + J and J I + J: Therefore,
p p p p p
pIp+ Jp I + J: Thus
p
I+ J I + J: However, from (ii), we get
I+ J I + J:
1.1.6 Proposition: Let R be a commutative ring and I; J be ideals of R,
then p p p
I \ J = I \ J:
p n
Proof. Let a 2 I \ J, then there exists n 1 such that p a 2 I \p J: This
n n
means that
p a p 2 I and a 2 J: p np 1; a 2 I andp a 2p J;and
p So for some
then a 2 I \ p J: Therefore,p I \ J I \ J: Now let a 2 I \ J; this
means that a 2 I and a 2 J: Then there exist n; m 1 such that an 2 I
and am 2 J: Thus an+m = anp am 2 IJ p I \ J:pHence,
pa
n+m
2 I \ J; and for
some n + m 1; we have a 2 I \ J: So I \ J I \ J:
1.2.1
p Denition: An ideal I of a commutative ring R is called radical if
I = I:
p
Note: (ii) of the proposition 1.1.5 shows that I is radical, and so is the
smallest radical ideal containing I.
1.2.2 Theorem: An ideal of a commutative ring R is radical if and only if
the only nilpotent in R=I is I = 0.
1 In some books [2], some authors report that "The easiest way to prove the radical of I
of a ring R is an ideal is to note that it is the preimage of the ideal of nilpotent elements in
R=I."
2
p
Proof. Let I be a radical, this means I = I: Consider a + I is a nilpotent
n
element in R=I; then there exists n 1 such that (a + I)p = I = 0: Therefore,
n n
a + I = I; and then a 2 I for some n 1: Thus a 2 I = I; and so a 2 I
which means a + I = I: Hence a = 0 is nilpotent
p equal to zero. Nowplet I is the
onlypnilpotent in R=I: We know that I I, so it remains to prove I I: Let
a 2 I; then there exists n 1 such that an 2 I;but this means that an + I = I
which meant (a + I)n = I: So an = 0. p Therefore, a + I is nilpotent, and then
a + I = I which implies a 2 I: Thus I I; and hence I is radical.
1.2.3 Lemma: Let R be a commutative ring then every prime ideal is
radical. p
Proof. Supposep P is a prime ideal. We want p to show that pP = P : One
can see that P P ; so it remains tho prove P P: Let x 2 P , then there
exists n 1 such that xn 2 P . Let us show x 2 P by induction on n. The
n = 1 case being trivial. Suppose n 1, xn 2 P implies x 2 P: Now since
n+1
x = x :x 2 P; and P is a prime, then either x 2 P or xn 2 P: Since the
n
later implies
p x 2 P by the inductive hypothesis, we conclude x 2 P in any case.
Thus P P: Hence P is radical.2
1.2.4 Corollary: Since every maximal ideal is prime, then the maximal
ideal is radical.
k1 k2 km
p 1.2.5 Proposition: In Z, Let 1 6= n 2 Z, and n = p1 p2 :::pm :Then,
(n) = (p1 p2 :::pm ):
1.2.6 Examples: Consider the ring Z of integers,
a. The radical of the ideal 4Z is 2Z.
b. The radical of 5Z is 5Z.
c. The radical of 12Z is 6Z.
1.2.7 Theorem: Let R and S be rings, and let ' : R ! S be a surjective
homomorphism, then p p
1) if I is an ideal of R with Ker'p I; '( p I) = '(I):
1 1
2) if J is an ideal of S,
pthen ' ( J) = ' (J):
Proof. (1) Let r 2 '(I), then by denition of the radical there exists
n 1 such that rn 2 '(I). This means that there exists i 2 I such that
n
r = '(i), p hence ' 1 (rn ) I since p Ker' p I: Thus (' p1
(r))n I; and then
1
' (r) I: This means
p r 2 '( I);
p so '(I) '( I): For the second
inclusion, let r 2 '( I); ' 1 (r) I: Thus there exists n 1 such
p that
(' 1 (r))np I, then p ' 1 n
(r ) I: Therefore,
p r n
2 '(I); then r 2 '(I):
Hence '( I) '(I): (2) Let r 2 ' 1 (J); then there exists n 1 such
2 Another proof, Note that If P is a prime ideal, then R=P is an integral domain, so it
cannot have zero divisors, and in particular it cannot have nonzero nilpotents. Hence, the
nilradical of R=P is f0g, and its preimage, being P , is a radical ideal.
3
that rn 2 ' 1 (J); and n
p then '(r ) J: This pimplies that ('(r))
n
J for some
1
n 1; then '(r) p J: Therefore, rp2 ' ( J): To prove the second inclusion,
consider r 2 ' 1 ( J); then '(r) J: Now by denition of the radical, there
exists n 1 such that ('(r))n pJ; it implies that '(r n
p ) J:
n
pSo r 2 ' (J)
1
1 1 1
and then
p forpsome n 1; r 2 ' (J): Thus ' ( J) ' (J): Hence,
' 1 ( J) = ' 1 (J):
1.2.8 Proposition: p If R is Noetherian ring then for any ideal I some
positive power of I is contained in I. In particular, the nilradical, N , of a
Noetherian ring is a nilpotent ideal: N k = 0 for some k: p
Proof. Since R is a Noetherian ring then for any p ideal I; the ideal I is
nitely generated. Let a1; :::::; am be generators of I; then by the denition
of the radical, for each i, we have aki i p2 I for some ki 1: Let k be the
maximum of all the ki : Then the ideal ( I)km is generated by elements of the
form ad11 ad22 ::::admm where d1 + ::: + dm = km; and each of these elements p
has at
least one factorpadi i with di k: Then adi i 2 I; hence each generator of ( I)km
is in I. Thus ( I)km I:3
References
[1] Dummit, D.S., Foote, R.M., Abstract Algebra, Third Edition, John Wiley
Sons, Inc., 2004.
[2] Malik, D.S., Mordeson J.M., and Sen M.K., Fundamentals of Abstract
Algebra, First Edition, The McGraw-Hill Companies, Inc., 1997.
Acknowledgement
I am indebted to Prof. S. Rankin for his illuminating guidance at every step
while I was preparing to this presentation.
3 This proof is exactely from Abstract Algebra (Dummit and Foote)
