Product of two ideals

Question

I am trying to understand what the meaning of product of ideals is. From this site: http://commalg.subwiki.org/wiki/Product_of_ideals I have figured out that it should be:

$$ IJ= \sum_{i = 1}^n (a_i b_i) {\rm\ with\ } a_i\in I {\rm\ and \ } b_i\in J, $$

where I and J is ideals of ring A. And I don't understand where the n comes from.

If we set $A = \mathbf{Z}/12$, $I = 2\mathbf{Z}$ and $J = 3\mathbf{Z}$, we will get:

$$IJ= 2\times 3+4\times 6+6\times 9+... \quad \text{(how big is $n$?)}$$

My confusion, is that it seems like the expression will give a number instead of a set of elements. I assume it should be a set, because in my book, it stands that if $I+J= A$, then $IJ= I\cap J$.

Answer 1

Your confusion comes from mistaking a set of elements with a single element. The product $IJ$ is the set of all products of the given form for any possible $n$ (including 0 for the zero element), making the definition $IJ=\left\{\sum_{i=0}^n a_i*b_i\mid a_i\in I, b_i\in J, n\ge 0\right\}$

So in your example the set contains $2*3$, $4*6$, $2*3+4*6$, $2*3+4*6+6*9$, and so on.

Answer 2

Let $A$ be a commutative ring. Given any subset $S\subseteq A$, there is the ideal generated by $S$: $$\langle S\rangle=\bigcap_{\substack{\text{ideals }I\\ \text{with }I\,\supseteq \,S\strut}}\!\!\!I$$ It is comprised of

1. elements of $S$

2. any finite sum of elements of $S$

3. any product of an element of $A$ with an element of $S$

4. any finite sum of elements from the previous steps

5. any product of an element of $A$ with an element from the previous steps

6. any finite sum of elements from the previous steps

7. any product of an element of $A$ with an element from the previous steps

8. $\vdots$

As it turns out, this process is redundant (doesn't produce any new elements) after step 4. Thus, $$\langle S\rangle=\{\text{finite sums each of whose terms is something from $A$ times something from $S$}\}$$ which is usually written in symbols as $$\langle S\rangle=\left\{\sum_{r=1}^na_rs_r:\;a_r\in A,\; s_r\in S,\;n\in\mathbb{N}\right\}$$

Then, given two ideals $I, J$ of a ring $A$, their product is the ideal generated by the set of products: $$P=\{ij:i\in I, j\in J\}$$ That is, $$IJ=\langle P\rangle=\left\{\sum_{r=1}^na_ri_rj_r:\;a_r\in A,\; i_r\in I,\;j_r\in J,\;n\in\mathbb{N}\right\}$$ Since $I$ and $J$ are ideals, they are closed under multiplication by elements of $A$, so any particular $$\begin{align*} \sum_{r=1}^na_ri_rj_r&=\underbrace{(a_1i_1)}_{{}\in I}\underbrace{j_1}_{{}\in J}+\cdots + \underbrace{(a_ni_n)}_{{}\in I}\underbrace{j_n}_{{}\in J}\\\\ \text{or just as well,}&\quad= \underbrace{i_1}_{{}\in I}\underbrace{(a_1j_1)}_{{}\in J}+\cdots + \underbrace{i_n}_{{}\in I}\underbrace{(a_nj_n)}_{{}\in J} \end{align*}$$ Therefore for a product of two ideals, we don't have to write the sums with coefficients from $A$: $$IJ=\langle P\rangle=\left\{\sum_{r=1}^ni_rj_r:\; i_r\in I,\;j_r\in J,\;n\in\mathbb{N}\right\}$$

Answer 3

The purpose of the $n$ is more or less to indicate that the elements in the product $IJ$ come from FINITE linear combinations of elements of $I$ and $J$, i.e. we say that the product of the ideals is the set of all formal linear combinations of elements from $I$ and $J$. If the sum were infinite, then the product of the ideals would not be so well defined, and indeed we may run into a lot of trouble deducing properties of the product.

You are correct that the product is a set of elements - those elements all come from expressions like the one you have above.