examples of radicals of ideals in commutative rings


Let R be a commutative ring. Recall, that ideals I,J in R are called coprimeMathworldPlanetmathPlanetmath iff I+J=R. It can be shown, that if I,J are coprime, then I⁢J=I∩J. Elements x1,…,xn∈R are called pairwise coprime iff (xi)+(xj)=R for i≠j. It follows by induction, that for pairwise coprime x1,…,xn∈R we have (x1⁢⋯⁢xn)=(x1)∩⋯∩(xn),

Let x∈R be such that

x=p1α1⁢⋯⁢pnαn,

for some prime elementsMathworldPlanetmath pi∈R, αi∈ℕ and assume that p1,…,pn are coprime. Denote by

x¯=p1⁢⋯⁢pn.

We shall denote by r⁢(I) the radicalPlanetmathPlanetmathPlanetmath of an ideal I⊆R.

Proposition. r⁢((x))=(x¯).

Proof. ,,⊇” Let α=max⁢(α1,…,αn). Then we have

x¯α=(p1⁢⋯⁢pn)α=p1α⁢⋯⁢pnα=p1α-α1⁢⋯⁢pnα-αn⁢p1α1⁢⋯⁢pnαn=y⁢x

and thus x¯α∈(x). This shows the first inclusion.

,,⊆” Assume that y∈r⁢((x)) and y≠0. Then there is n∈ℕ such that yn∈(x). Thus x divides yn. Of course for any i∈{1,…,n} we have that pi divides x. Thus pi divides yn and since pi is prime, we obtain that pi divides y. Now for i≠j elements pi and pj are coprime, thus x¯ divides y and therefore y∈(x¯), which completesPlanetmathPlanetmath the proof. □

Remark. If we assume that R is a PID (and thus UFD), then the previous proposition gives us the full characterization of radicals of ideals in R. In particular an ideal in PID is radical if and only if it is generated by an element of the form p1⁢⋯⁢pn, where for i≠j elements pi and pj are not associated primes.

Examples. Consider ring of integersMathworldPlanetmath ℤ. Then we have:

r⁢((12))=(6);
r⁢((9))=(3);
r⁢((7))=(7);
r⁢((1125))=(15).
Title examples of radicals of ideals in commutative rings
Canonical name ExamplesOfRadicalsOfIdealsInCommutativeRings
Date of creation 2013-03-22 19:04:34
Last modified on 2013-03-22 19:04:34
Owner joking (16130)
Last modified by joking (16130)
Numerical id 5
Author joking (16130)
Entry type Example
Classification msc 16N40
Classification msc 14A05
Classification msc 13-00