Thread: Sums and Products of Ideals

Let I and J be ideals of a ring R

Prove that if R is commutative and if I + J = R then IJ = I J

what does it mean for I + J = R?

it means EVERY element of R is of the form x + y, for some x in I, and y in J. in particular, 1 = x + y, for some x in I, and y in J (note: what you are trying to prove is NOT TRUE for rings without unity).

now it is clear that IJ is always contained in I∩J (see other thread for proof). what would it mean to say I∩J is contained in IJ?

well, it turns out we can do "algebra with ideals" (which simplifies having to consider just elements, as the expressions can get long and ugly).

the key result we will need is this:

(I+J)(I∩J) = I(I∩J) + J(I∩J)

suppose r is in (I+J)(I∩J).

then r = Σ_k (a_kb_k), where a_k is in I+J, for each k, and b_k is in I∩J for each k.

thus we can write a_k = x_k + y_k, where for each k, x_k is in I, and y_k is in J.

so r = Σ_k (a_kb_k) = Σ_k ((x_k+y_k)b_k) = Σ_k (x_kb_k) + Σ_k (y_kb_k),

which is clearly in I(I∩J) + J(I∩J).

on the other hand, suppose that r is instead in I(I∩J) + J(I∩J).

then r = Σ_k x_ka_k + Σ_m y_mb_m, for x_k in I, y_m in J, and a_k,b_m in I∩J for each value of k and m.

suppose we let n = max(k,m), and add 0 terms for whichever sum comes up short. we can then re-write out sum as:

r = Σ_n (x_na_n + y_nb_n).

note that the term x_ja_j + y_jb_j is in (I+J)(I∩J), since x_ja_j = (x_j+0)(a_j), and y_jb_j = (0+y_j)(b_j),

and that this is true for EVERY j = 1,2...,n. since ideals are closed under addition, this means that r is in (I+J)(I∩J), so we have equality of (I+J)(I∩J) and I(I∩J) + J(I∩J).

ok, back to the main proof now:

note that for any ideal A, RA = A (clearly A is contained in RA, since for any a in A, a = 1a is in RA. on the other hand, an element of RA is r₁a₁+...+r_na_n, and each term r_ja_j is in A, since A is an ideal, so their sum is likewise in A...note how we needed a unity for R to assert this).

therefore, when R = I+J:

I∩J = R(I∩J) = (I+J)(I∩J) = I(I∩J) + J(I∩J) ⊆ IJ + JI = IJ + IJ (since R is commutative)

= IJ (since IJ is closed under addition).