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Let I and J be ideals of a ring R. Show by example that the set of products {xy | x ∈ I, y ∈ J} need not be an ideal, but that the set of finite sums of products of elements of I and J is an ideal. This ideal is called the product ideal, and is denoted by IJ.

I want to check if my work on the first part makes any sense. I am not sure if I am picking the right example.

If I let R = Z and I = (2) and J = (3) then I has multiples of 2, and J has multiples of 3

6 will divide the product IJ if we add 2 and 3 though we get 5 in IJ, but I am not sure if 5 Divides IJ.

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    You chose a bad situation: the integers are a principal ideal domain, so in this case the set of products is an ideal. But it's not in general rings. – 2017-01-24
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    ohhhh okay, thanks for pointing that out – 2017-01-25

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Let $I=J=(x,y)\subseteq \mathbb R[x,y]$ notice that $x^2$ and $y^2$ are both in $IJ$ while $x^2+y^2$ is not, because this polynomial is irreducible (so it is no the product of two elements in $I$, because no unit is in $I$).

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    you mean $\mathbb{R}[x,y]$ probably, right? – 2017-01-24
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    oops, my bad. Fixed, thanks! – 2017-01-24
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    Sorry, what is (x,y) again? The definition I know is all polynomials of x,y in R such that the constant term is 0, is that right? – 2017-01-25
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    so you're saying it's not in IJ because we cannot express x^2 + y^2 as a product of two polynomials? – 2017-01-25
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    As a product of two polynomials of degree larger than 0 – 2017-01-25
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    Okay, so that makes sense then. So literally two polynomials with variables in them by the definition of the ideal, since the constant term is 0. – 2017-01-25
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    Ok, one final final question. To add them, don't I have to show that the addition of something like this: i1j1 + i2j2 + i3j3 + i4j4 can be also expressed as that kind of product? Where i1j1 + i2j2 is one element, and i3j3 + i4j4 is another element in the new ideal. I'm not sure how to factor those algebraically. Or is that the wrong place to start to begin with? – 2017-01-25
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    What do you mean? Is your question why $x^2+y^2$ is the sum of two elements in $IJ$? – 2017-01-25
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    No, we said that sum is not in IJ. I'm asking why the set of finite sums of products of elements of I and J is an ideal. So I was trying to show that it's closed under addition – 2017-01-25
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    oh ok, it is clear that it is closed under addition, it just becomes a longer sum of products of of elements of $I$ and $J$. – 2017-01-25
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    Yea, ok, I figured I might need a stronger statement, or to say it in a more formal way but i guess this will be good enough – 2017-01-25
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Jorge's post gives a nice counterexample, but perhaps it is more pedagogically correct to point out that the above set fails to be a new ideal because we have problem with it being a subgroup of the ring $R$. Indeed, being ideal first of all requires being a subgroup of the underlying group structure (abelian) of $R$. By taking the above set only, the sum of two elements will give you for instance an element of the form $xy+zw$, where $x,z \in I$, $y,w \in J$. Though the last might not be written as $xy+zw= \lambda \tau$, for $\lambda \in I$, $\tau \in J$, hence need not be a subgroup necessarily, so neither an ideal!

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    Thank you so much! so for showing the sum, do I add the elements that are generated by 6? it seems that if I have (3) and (2) and I want to add their product, then wouldn't that be (6), which is basically an ideal? – 2017-01-24
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    You're welcome! The above set $\{xy | x \in I, y \in J\}$ isn't an ideal, but is a subset of the ideal $IJ$ (which as you mentioned consists of finite sums of elements). My suggestion is, that there is no reason to think about adding, or removing elements. It is propably better to understand the definition (and why your case doesn't work) and work out through it, its good that you try to give some intuition but in your example you just have integers, if you work in future with more nasty things it will definetly go away. So try to cultivate a more general intuition for those things! – 2017-01-24
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    Okay, but when it says "show by example" doesn't that mean I need to give a specific example? or is that just another way of saying "prove"? Also when you say my case doesn't work, do you mean my example is just wrong? – 2017-01-24
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    Lana, firstly I told you that Jorge's example works fine (in fact provides you an infinite number of counterexamples), also I didn't say that your example does or doesn't work. I said that the whole idea of removing or adding elements doesn't help. But just answering your question, for ideals in $\mathbb{Z}$ of the form $I=(m),J=(n)$, the product $IJ$ is the ideal generated by the element $(mn)$, fill out the details. (Check Atiyah-Macdonald if I recall correct there is an example or something in it). But again this is very special situation! – 2017-01-25