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Let $R$ be a ring, $I$, $J$ ideals then how does an element of $I^n(R/J)$ looks like. I know that an element of $I^n$ is of the form $(x_1x_2...x_n)$ for $x_i\in I$ but then do I multiply it with $r+J$ for some $r\in R$?

Thanks for your help

user1294729
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  • Note that $R/J$ is an $R$-module, and the structure of an ideal times an $R$-module is described here: https://math.stackexchange.com/questions/90283/what-exactly-is-the-structure-im-for-i-an-ideal-and-m-a-module in short, it's all finite sums of elements of $I^n$ times elements of $R/J$. – wormram Jun 15 '22 at 18:07
  • You simply add some element of $J$ to it, namely $I^n(R/J)$ is the coset ${x_1x_2\cdots x_n + J: x_i \in R}$ of $R/J$. – WhatsUp Jun 15 '22 at 18:07
  • @WhatsUp is it also true if I say $(x_1\cdot \cdot \cdot x_n)+J$? or is this something different – user1294729 Jun 15 '22 at 18:10
  • Yes it's the same. I omitted $()$ because $\cdot$ has higher priority than $+$. – WhatsUp Jun 15 '22 at 18:11
  • @WhatsUp sorry but I mean the () because of ideal generating by $x_1\cdot \cdot \cdot x_n$ not because of "normal" brackets – user1294729 Jun 15 '22 at 18:12
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    An element of $I^n$ is a sum of such products. – Randall Jun 15 '22 at 18:13
  • @Randall sorry I'm somehow a bit lost. So $I^n={r_1x_1\cdot \cdot \cdot x_n+r_2x_1\cdot \cdot \cdot x_n...:r_i\in R, x_i\in I}$? – user1294729 Jun 15 '22 at 18:15
  • $I^n$ is the product of $I$ with itself $n$ times (the ideal product, not the cartesian product). See https://math.stackexchange.com/questions/290229/explaining-the-product-of-two-ideals – wormram Jun 15 '22 at 18:16
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    The reason it's the set of sums of products of elements is so that the resulting object $I^n$ is actually an ideal: it needs to be closed under addition. – wormram Jun 15 '22 at 18:18
  • @MichaelMorrow so $I^n=\left{\sum_{j=1} x_{i_1}y_{i_2}...z_{i_n}\right}$ where the sum needs to be finite. Is this true? – user1294729 Jun 15 '22 at 18:20
  • @MichaelMorrow sorry I'm really lost right now. Could you please explain why WhatsUp only wrote $x_1\cdot \cdot \cdot x_n$ and not as a sum like you are saying – user1294729 Jun 15 '22 at 18:30
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    I'm not sure. $I^n$ is supposed to be the ideal generated by all elements of the form $x_1\cdots x_n$ where each $x_i\in I$. Or in otherwords, the set of all sums of elements of that form just like you wrote in your previous comment. See https://commalg.subwiki.org/wiki/Product_of_ideals for some equivalent definitions. – wormram Jun 15 '22 at 18:32
  • @MichaelMorrow okey thanks I see this now but what I still do not see is why an element in $I^n(R/J)$ is $(x_1\cdot \cdot \cdot x_n) +J$? Could you also explain me this? – user1294729 Jun 15 '22 at 18:34
  • If you click the link I posted in my first comment up at the top, it will describe what the elements look like. In summary, $I^n(R/J)$ is the set of all finite sums of the form $a_1m_1+\ldots+a_km_k$ where $k\in\mathbb{N}$, $a_i\in I^n$ and $m_i\in R/J$. This is a special case of "ideal times module". Do you know about modules? You don't need to know about modules to understand this construction, but it may help. – wormram Jun 15 '22 at 18:39
  • @MichaelMorrow yes I know about modules I'm only a bit confused since formally these $a_i$ would also be sums so we would have like the sum of sums right? That's what confuses me – user1294729 Jun 15 '22 at 18:40
  • Yep, in this case it's like you have sums of sums. Each $a_i$ can be written as a sum, and the elements $a_1m_1+\ldots+a_km_k\in I^n(R/J)$ are also sums. So for example, an element of $I^n(R/J)$ might look like $(x_1\cdots x_n+y_1\cdots y_n)(r_1+J)+z_1\cdots z_n(r_2+J)$ where the $x_i,y_i,z_i\in I$ and the $r_1,r_2\in R$. – wormram Jun 15 '22 at 18:47
  • @MichaelMorrow Ah okey but this is completely useless if I want to who's that $I^n(R/J)=(I/J)^n$ right? because there would be too many indexes – user1294729 Jun 15 '22 at 18:49
  • @MichaelMorrow so would it be easier to calculate with $(x_1\cdot \cdot \cdot x_n)+J$? – user1294729 Jun 15 '22 at 18:57
  • @MichaelMorrow so I think I should get something like $(x_1...x_n)+J=(x_1+J)...(x_n+J)$ but I don't see what property I need to use – user1294729 Jun 15 '22 at 19:05

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