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Considering the product ideal $IJ = \{ \sum_{i=1}^n a_ib_i | a_i \in I, b_i \in J \forall i\}$, I've always seen it written that the more naive notion $IJ = \{ ij | i \in I, j \in J\}$ is not an ideal because in general it is not additively closed and is thus not a subring. I am unable to come up with a proof or any examples.

Any ideas?

walkar
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  • If one tries to write $ij + kl$ with $i,k \in I$ and $j,l \in J$ as a product $mn$ with $m \in I$ and $n \in J$ this is not possible in general. – Rogelio Molina Nov 17 '14 at 02:01
  • Yes, but that's just an assertion. Can you provide a proof or example? – walkar Nov 17 '14 at 02:02
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    If one of the ideals is principal, what you wrote is the same thing as the normal $IJ$, so you need to try in a non-principal ring, for example $k[x,y]$. Perhaps the easiest would be to use $k[x,y,z,w]$. – Mike Nov 17 '14 at 02:09
  • Following Mike's suggestion consider the following exemple: $K[x,y]$, take for instance $xy + 1$, this one cannot factor as product of polynomials $p(x) q(y)$. – Rogelio Molina Nov 17 '14 at 02:10
  • I just noticed this was a duplicate. I couldn't find it with the search feature but I found it in the sidebar after I'd posted it... – walkar Nov 17 '14 at 02:53

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I think I figured out an example thanks to Mike's comment and another post found here: Defining the Product of Ideals

Let $\mathbb{R}[x,y]$ be a two-variable polynomial ring over $\mathbb{R}$, and consider the ideal $(x,y)$. Notably $x^2$ is in $(x,y).(x,y) = \{ p(x,y)q(x,y) | p(x,y) \in (x,y) \text{ and } q(x,y) \in (x,y)\}$ as is $y^2$, but $x^2+y^2$ is not in $(x,y).(x,y)$ (as it cannot be written as the product of two polynomials, it is irreducible). So then, in general, $I.J = \{ ij | i \in I, j \in J\}$ is not an ideal.

Community
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walkar
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  • Ah, I see. Could you suggest a modification? – walkar Nov 17 '14 at 02:32
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    Good example, depending on the field $F$. Strictly speaking, you mean $x^2 + y^2$ can't be written as a product of non-constant polynomials. For $F = \mathbb{C}$, the example is incorrect, however, as $x^2 + y^2$ can be factored. I was also going to suggest the ideals $(x,y)$ and $(z,w)$ and the element $xz + yw$. – Mike Nov 17 '14 at 03:06
  • When I learned "irreducible," it meant cannot be factored nontrivially (neither factor can be a unit). Over $\mathbb{C}$, every constant is a unit, so factoring out a constant doesn't help anything. – walkar Nov 17 '14 at 03:08
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    Sorry I wasn't clearer. My second sentence was only nitpicking about terminology. My third sentence was a real objection if $F = \mathbb{C}$, as there is a nontrivial factorization of $x^2 + y^2$. The fact that the only units are the constants is true over any field. Using $x^2 + y^3$ solves this problem. – Mike Nov 17 '14 at 03:13
  • I changed it to $\mathbb{R}[x,y]$ so that works. Thanks for all your help! – walkar Nov 17 '14 at 03:19
  • Nb. to see that $x^2+y^2$ is irreducible over ${\bf R}$, notice that it is not a square (not even up to a constant), so if it was reducible, it would be negative for some $x,y\in {\bf R}$. – tomasz Nov 17 '14 at 03:29