examples of radicals of ideals in commutative rings
Let R be a commutative ring. Recall, that ideals I,J in R are called coprime iff I+J=R. It can be shown, that if I,J are coprime, then IJ=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=pα11⋯pαnn, |
for some prime elements pi∈R, αi∈ℕ and assume that p1,…,pn are coprime. Denote by
ˉx=p1⋯pn. |
We shall denote by r(I) the radical of an ideal I⊆R.
Proposition. r((x))=(ˉx).
Proof. ,,⊇” Let α=max(α1,…,αn). Then we have
ˉxα=(p1⋯pn)α=pα1⋯pαn=pα-α11⋯pα-αnnpα11⋯pαnn=yx |
and thus ˉxα∈(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 ˉx divides y and therefore y∈(ˉx), which completes 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 integers 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 |
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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 |