Let I and J be ideals of a ring R

Prove that if R is commutative and if I + J = R then IJ = I $\displaystyle \cap$ J

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- Dec 24th 2012, 11:52 PMBernhardSums 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 $\displaystyle \cap$ J - Dec 25th 2012, 12:23 PMDevenoRe: Sums and Products of Ideals
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_{k}b_{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_{k}b_{k}) = Σ_{k}((x_{k}+y_{k})b_{k}) = Σ_{k}(x_{k}b_{k}) + Σ_{k}(y_{k}b_{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_{k}a_{k}+ Σ_{m}y_{m}b_{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_{n}a_{n}+ y_{n}b_{n}).

note that the term x_{j}a_{j}+ y_{j}b_{j}is in (I+J)(I∩J), since x_{j}a_{j}= (x_{j}+0)(a_{j}), and y_{j}b_{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 r_{1}a_{1}+...+r_{n}a_{n}, and each term r_{j}a_{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).