Proving IJ is an Ideal When S is Not: A Case Study

[Ratpigeon]

Homework Statement

Show that if I and J are ideals of the (commutative) ring; R then
S={xy|x in I y in J} is not necessarily an ideal but the set of finite sums
IJ={Ʃ(x_vy_v)|x_vin I y_vin J}
is (and called the product ideal).

Homework Equations

An ideal satisfies the properties
For all x, x' in I, all r in R
(i) x+x' in I
(ii) rx in I

The Attempt at a Solution

This seems wrong to me since if I is generated by i and J is generated by j
xy+x'y'= ai*bj+a'i*b'j=(ab+a'b')(ij)=(ab+a'b')i*j which is in S since (ab+a'b')i is in I and j is in J...
and obviously by the same rule;
r(xy) is in S.
Therefore, it seems to me that S and IJ are equivalent sets (since the elements of S could be split up into a sum of product elements) and I don't see how S could _not_ be an ideal.

I haven't been able to find much information about product ideals; but this is a problem in my textbook - Algebra by Michael Artin (second edition) so I'm disinclined to think the lecturer phrased the question wrong...
Please help?

 

[hedipaldi]

Attached