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$(15)$ If $I,J$ are ideals of $R$, let $IJ$ be the set of all sums of elements of the form $ij$, where $i \in I$ and $j \in J$. Prove that $IJ$ is an ideal of $R$.

This is a question from Abstract Algebra, by Herstein. I don't quite understand the set of all sums of elements of the form $ij$. Is that suppose to mean $i+j$? If not, why the word sums?

From user input, I have refined my understanding of the set $IJ$ as follows. $IJ \doteqdot \big\{ \sum_{i} a_{i}b_{i} \colon a_{i} \in I, b_{i} \in J \big\}$

egreg
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Jack
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2 Answers2

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What Herstein wants to define is the least ideal of $R$ which contains all elements of the form $ij$, for $i\in I$ and $j\in J$.

The answer is, of course, the ideal generated by the set $$ X=\{ij:i\in I,j\in J\} $$ and we want to see how it looks like. In general, the ideal generated by a subset $A$ of $R$ must contain all elements of the form $ras$, for $r,s\in R$ and $a\in A$, and sums thereof. The set of all elements of the form $$ r_1a_1s_1+r_2a_2s_2+\dots+r_na_ns_n $$ for $r_k,s_k\in R$ and $a_k\in A$ is clearly an ideal of $R$ and so it is the ideal generated by $A$.

In the case of the set above, we see that $$ r(ij)s=(ri)(js) $$ and, since both $I$ and $J$ are ideals, we have, for $i\in I$ and $j\in J$, $ri\in I$ and $js\in J$. Thus the ideal generated by the set $X$ consists of all elements of the form $$ i_1j_1+i_2j_2+\dots+i_nj_n $$ for $i_k\in I$ and $j_k\in J$. This is the ideal denoted by $IJ$.

egreg
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$$IJ=\{i_1j_1+\cdots+i_nj_n:n\in\mathbb N, i_1,\ldots,i_n\in I,j_1,\ldots,j_n\in J\}$$

Tim Raczkowski
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  • How do we know that this is a finite sum? Also what about $i_{2}j_{1} + i_{1}j_{2}$ or other combinations? – Jack Jan 22 '16 at 16:27
  • Well it's defined to be a finite sum. As far as the subscripts go they are arbitrary. $i_2j_1+i_1j_2$ would be in $IJ$ but we can rename $j_1$ and $j_2$. The subscripts are not chosen at the outset, but as they are chosen as part of the sum. – Tim Raczkowski Jan 22 '16 at 16:33
  • So is this a correct interpretation? $IJ = \big{ \sum_{i} a_{i}b_{i} \colon a_{i} \in I, b_{i} \in J \big}$. Is it implicitly defined to be a finite sum? It does not say that explicitly. – Jack Jan 22 '16 at 16:34
  • You can't really talk about infinite sums unless there is a concept of convergence. In other words a topology. – Tim Raczkowski Jan 22 '16 at 16:37
  • beyond the scope of the class? haha...so should I write is as $IJ = \big{ \sum_{i = 1}^{n} a_{i}b_{i} \colon a_{i} \in I, b_{i} \in J \big}$ as you did? – Jack Jan 22 '16 at 16:38
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    Yes. The fact that the sums are finite is an assumption which is implicit when deal with an algebraic structure that does not have a corresponding topological structure. – Tim Raczkowski Jan 22 '16 at 16:40
  • Well, this notation is actually a little confusing. Is $n$ fixed or not? I assume $n$ is allowed to range over the non-negative integers, but it's not clear from the notation above. – Viktor Vaughn Jan 22 '16 at 20:01
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    Yes, $n$ is allowed to range over the non-negative integers. I'll edit my answer. – Tim Raczkowski Jan 22 '16 at 20:03