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I had a question in which encounters to ideal multiplication. My approach to this problem force me to find out the $I^n$ structure as described below:

Suppose $R$ is a ring and $I\unlhd R$. Identify the $I^n$ set ($n\in\mathbb{N\cup\{0\}}$) by describing it's elements when:

1- $I$ is finitely generated.

2- $I$ is an arbitrary ideal in $R$.

To prove, take $R$ commutative, with identity or without these presumes.

I doubt if there is a clear answer to part 2 above. would you solve this problem?

Thanks in advance.

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RSh
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1 Answers1

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When you have two ideals $I$ and $J$, the product ideal $IJ$ is defined as all elements of the form $\sum_{i=1}^k r_i s_i$, with $r_i\in I$, $s_i\in J$. Likewise a product of three ideals $I$, $J$, $K$ is all elements of the form $\sum_{i=1}^k r_i s_i t_i$, with $r_i\in I$, $s_i\in J$, $t_i\in K$, and so on.

This means elements of the ideal $I^n$ have the form $\sum_{i=1}^k r_{i_1} r_{i_2} ... r_{i_n}$ with $r_{i_j}\in I$. In general there's no other structure.

If we have generators for $I$ we can do a little better. Suppose $I$ is generated by $s_\alpha$ with $\alpha \in S$, where $S$ is some indexing set. Then an element in $I$ is of the form $\sum_{\alpha \in S} r_\alpha s_\alpha$, where $r_\alpha \in R$ are almost all zero, but otherwise arbitrary ring elements. This representation is typically not unique. Now $I^2$ is generated by elements $s_\alpha s_\beta$, with $\alpha, \beta \in S$, and any element in $I^2$ is a finite linear combination of these. Elements of $I^3$ are finite linear combinations of $s_\alpha s_\beta s_\gamma$, with $\alpha, \beta, \gamma \in S$, and so on.

As an example, take $R=\mathbb{Z}[x,y,z]$, and let $I=(x,y,z)$. Then $I^2=(x^2, y^2, z^2, xy, xz, yz)$, $I^3=(xy^2, yx^2, xz^2, zx^2, yz^2, zy^2, x^3, y^3, z^3,xyz)$.

Hope that helps.

Zavosh
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