Proving I1*I2 is an Ideal in Commutative Ring R

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Given a commutative ring R with a unit, how do you prove that the product of two ideals, I1 and I2, is also an ideal?
The product of course is defined to be {x*y | x in I1, y in I2}, where * is the multiplication in the ring R.
I'm having trouble proving that I1*I2 is a group under addition.

Thanks,
Chen
 
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The product of course is defined to be {x*y | x in I1, y in I2}, where * is the multiplication in the ring R.

Are you sure?
 
http://planetmath.org/encyclopedia/ProductOfIdeals.html
 
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Well I can see this definition doesn't match mine, but at any rate - I was asked to prove that this group:
{x*y | x in I1, y in I2}
is an ideal.
 
I'm working on the similar problem right now, and I was going through many threads in this forum and couldn't find the answer whether the product as Chen defined it is an ideal. I wanted to find the counterexample that {x*y | x in I1, y in I2} is an ideal, but was not able to come up with anything. So, does anybody have any ideas of how to do it, or can you at list give me a hint whether or not it is an ideal? Thanks!
 
Chen said:
Well I can see this definition doesn't match mine, but at any rate - I was asked to prove that [EDIT: I changed Chen's notation--- CH]
<br /> I \, J = \{ fg | f \in I, g \in J \}<br />
is an ideal.

Is it possible that you are confusing an abelian subgroup with a subring with an ideal? A good short textbook which should help is Herstein, Abstract Algebra. See the excellent and very readable textbook by Cox, Little, and OShea, Ideals, Varieties, and Algorithms, for much more about such constructions as IJ, \, I \cap J, \, I+J, \, I:J, \, \sqrt{I}.
 
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I suspect that the question had meant that IJ is generated by the set {fg : f in I, g in J}, and not equal to it. Occasionally this is expressed by writing IJ = <fg : f in I, g in J>.
 
Ditto morphism (this point and others are well explained in IVA).
 
your definition is wrong. the product of ideals is tautologically an ideal, as it is defined as the ideal generated by those products, or equivalently as all sums of them.
 
  • #10
Hi,

I saw many people have given a wrong definition for the product of ideals. The following is the correct definition as appeared in mathematical text:

I*J = {i[1]*j[1]+i[2]*j[2]+...+i[n]*j[n]:i[n] in I and j[n] in J where n is finite}
 
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