Interpreting the set $IJ = \{\sum_i x_iy_i \mid x_i \in I, y_i \in J\}$ where $I$ and $J$ are ideals Let $I$ and $J$ be ideals in a ring $R$.
Show that $IJ = \left\{\sum_i x_iy_i \mid x_i \in I, y_i \in J\right\}$ is an ideal.
Question
I am not sure how to interpret this question because of the index $i$. Using the same index $i$ for both $x$ and $y$ seems to suggest that there will always be the same number of elements in $I$ and $J$ and that they get multiplied together through some arbitrary ordering, i.e. $x_1y_1 + x_2y_2 + \cdots$
Can anyone make clear up what is meant by the set $IJ = \left\{\sum_i x_iy_i \mid x_i \in I, y_i \in J\right\}$?
 A: You choose arbitrary $k$ and then choose arbitrary $x_1,\ldots,x_k \in I$ and arbitrary $y_1,\ldots,y_k \in J$ and then form the sum $\sum_{i=1}^k x_iy_i$. The particular selection of the $x_i$ and $y_i$ are not fixed, so you can take any choices for them when you define a sum. The ideal $IJ$ is the set of all such sums, for every choice of $k$ and every choice of the particular $x_i,y_i$ for the sum. It is a good exercise to check this $IJ$ is an ideal.
A: Here's a short discussion to motivate that definition.
The notation $IJ$ is meant to suggest the "product of ideals $I$, $J$". We would like the product of two ideals to be again another ideal.
Many beginners at first thing this is just the set of all $ij$ for $i\in I$, $j \in J$. The problem is that this is not necessarily an ideal of $R$: it is often just a subset. Let's call this subset $S$.
It's obvious that the product of any $s\in S$ with another element in $R$ on the right or the left is again in $S$, but the problem comes in when you try to prove the set is closed under addition. The problem is that you can't prove that the sum of two elements of $S$ is in $S$.
This all disappears, though, when you define $IJ$ to be "the set of finite sums of elements of $S$." This set, in contrast, is obviously closed under addition.
A: The sum does not run over all of $I$ nor does it run over all of $J$. Rather, these are arbitrary finite sums. A more explicit notation would be $IJ = \{\sum_i^n x_iy_i | x_i \in I, y_i \in J\}$ but then one would also have to explain that $n\in\mathbb{N}$ is arbitrary.
