Understanding the ideal $IJ$ in $R$

Question

I'm a little confused why our book defines an ideal $IJ$ in $R$ (where $I$ and $J$ are ideals) in such a complicated way:

$$IJ=\left\{ \displaystyle\sum\limits_{i=1}^n a_i b_i\mid n\geq 1 ,a_i\in I, b_i\in J \right\}$$

How is this different that defining more simply as:

$$\{ab\mid a\in I,b\in J\}$$ Since $\{ab\}$ is closed on addition then elements of the form $a_1b_1+\cdots +a_mb_m\in\{ab\}$, which sure does seem like it is equal to $IJ$. Am I missing something here?

Thanks.

Why do you think ${ab}$ is closed under addition? It doesn't need to be. — anon, Mar 27 '15 at 13:29
Oh... I first assumed it was an ideal and then assumed it was a subgroup of $(R,+)$ after that. I see my mistake — Patrick Shambayati, Mar 27 '15 at 13:34

score 3 · Accepted Answer · answered Mar 27 '15 at 13:33

Consider the four-variable polynomial ring $\Bbb C[a,b,c,d]$ and ideals $I=(a,b)$, $J=(c,d)$.

Then $IJ$ contains the elements $ac$ and $bd$ hence contains $ac+bd$, but the set

$$\{xy:x\in I,y\in J\}$$

does not contain $ac+bd$, even though it contains $ac$ and $bd$. So no, this set is not closed under addition, and it is not $IJ$. The problem is that summing products of elements in $I$ and $J$ does not generally yield something that can be simplified to a single such product.

By the way, try proving $\{xy:x\in I,y\in J\}$ doesn't contain $ac+bd$ as an exercise.

Thank you! I see the flaw in my logic now – Patrick Shambayati Mar 27 '15 at 13:36 — Patrick Shambayati, Mar 27 '15 at 13:36

Understanding the ideal $IJ$ in $R$

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