# Product of ideals

• December 25th 2012, 12:49 AM
Bernhard
Product of ideals
Prove that IJ is an ideal contained in I $\cap$J and, further, give an example where IJ $\ne$ I $\cap$ J

• December 25th 2012, 12:26 PM
Deveno
Re: Product of ideals
the elements of IJ are sums of elements of the form xy, where x is in I, y is in J (just elements of R of the form xy aren't "enough" to form an ideal, as such elements may not be closed under addition).

to prove IJ ⊆ I∩J we need to show that every element of IJ is both in I, AND in J.

so suppose r is in IJ.

then r = x1y1+...+xnyn, where xk is in I, and yk is in J, for each k = 1,2,..,n.

since I is an ideal, xk in I, means xkyk is in I, for each k. since ideals are closed under addition, the sum over k of xkyk is likewise in I.

thus r is in in I.

a similar line of reasoning (based on the fact that J is an ideal) shows r is likewise in J, hence r is in I∩J. that's all there is to it (it's really quite simple).

for examples, it's often good to look at rings you are familiar with.

in the ring of integers, an ideal is of the form (m), for some integer m (since Z is a principal ideal domain). the ideal (m)(n) is thus all integers of the form Σk(akm)(bkn).

such a sum can be written as (mn)Σk (akbk), so clearly every element of (m)(n) is in (mn).

also we have mn = (1*m)(1*n), so (mn) is contained in (m)(n). so in Z, (m)(n) = (mn).

however, in Z, (m)∩(n) = (lcm(m,n)), so if gcd(m,n) > 1, this is strictly larger than (mn).

for example, take m = 6, n= 15. then (6)(15) = (90), whereas (6)∩(15) = (30). it is clear that 30 is in (6) (as 5*6) and in (15) (as 2*15), so 30 is clearly in (6)∩(15).

suppose we could write 30 = (6a1)(15b1) +....+(6ak)(15bk) = 90(a1b1+...+akbk)

since 90 (= 6*15) divides every term of the RHS, 90 must divide the sum, and thus 90 divides 30, which is clearly a contradiction.