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I have the following problem:

Let $S\subset R$, where $R$ is a ring. We write $(S)$ for the smallest two sided ideal of $R$ containing $S$, i.e. $$(S)=\left\{\sum_{i=1}^n a_i s_i b_i: a_i, b_i \in R, s_i \in S, n\geq 0\right\}$$

So I mean that's our definition so it is somehow an ideal by definition but I really wanted to check the axioms of an ideal. Therefore I took $$x=\sum_{i=1}^n a_i s_i b_i, \,\,y=\sum_{i=1}^{m} a_i's_i'b_i'\in (S)$$and I wanted to check that $x+y$ is also in $(S)$. But there I struggle. What do I need to use to get to the end?

Thanks for your help

Wuestenfux
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user1294729
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2 Answers2

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$$(S)=\{\sum_{i=1}^{n}a_{i}s_{i}b_{i}\,,\,a_{i},b_{i}\in R, s_{i}\in S,n\in\mathbb{N}\}$$.

In words, $(S)$ is just the set of finite sum of elements of the form $asb$ where $a$ and $b$ are in $R$ and $s$ is in $S$.

Now it is just a matter of notation and checking.

For example you take $x=\sum_{i=1}^{n}a_{i}s_{i}b_{i}$ and $y=\sum_{i=1}^{n'}a'_{i}s_{i}'b_{i}'$.

Now the confusion you might be having is that how should we add them. Or more precisely how should we deal with $s_{i}$ and $s'_{i}$.

No matter you just relabel $s_{i}'$ as $s_{n+1},s_{n+2},...s_{n+n'}$. Note that nowhere in the definition does it say that these $s_{i}$ have to be distinct. In fact we can take many of them to be just $0$ if we want. So we are just changing the index $i$.

So you end up with something like $x+y=\sum_{j=1}^{n+n'}a_{i}''s_{j}''b_{j}''$. Which does lie in $(S)$.

Now for any $r\in R$. $rx=\sum_{i=1}^{n}(ra_{i})s_{i}b_{i}$ lies in $(S)$ and similarly for $sx$. (You just take $a_{i}'=ra_{i}$.

Hence you have an ideal. We should have shown for $x-y$ but no matter. It's again just a matter of notation.

Dovahkiin
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  • so but then I say that $a_i=a_i'$ or how does it work – user1294729 Jan 20 '22 at 08:11
  • You just relabel the coefficients of $y$ to $n+1,n+2,....$. It is fine even if you don't . You can directly say that the sum is of the form of what's required. In fact this is the case for many such constructions like direct sums . – Dovahkiin Jan 20 '22 at 08:13
  • sorry I don't see it yet, we have some primes there, so all the $a_i'$'s ect are different from $a_i$ how can I add them? or do you mean that the second sum starts at $j=n+1$, so we have like two "distinct" sums? – user1294729 Jan 20 '22 at 08:15
  • Like I said, nowhere in the definition does it say that $a_{i}$ and $b_{i}$ have to be distinct. You can add the same element as many times as you want. In fact that is vital such that $ns$ belongs to the ideal. That is you can take $a_{i}=b_{i}=1$ and $s_{i}=s$ for all $i=1,2,..n$ . So what you are really doing is that you are doing is relabeling the elements $s_{i}'=s_{n+i}$ and $a_{i}'=a_{n+i}$ and then calling them $a_{i}''$ and $s_{i}''$ to get rid of writing it again and again. – Dovahkiin Jan 20 '22 at 08:18
  • a perfect thanks! – user1294729 Jan 20 '22 at 08:19
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After fixing the definition of $(S)$, it is clear that the sum $x+y = \sum_{i=1}^n a_is_ib_i + \sum_{j=1}^m a'_js'_jb'_j$ lies in $(S)$

as it can be written as

$\sum_{k=1}^{n+m} u_kt_kv_k$ with

$u_k =a_k$, $t_k=s_k$, $v_k=b_k$ for $1\leq k\leq n$ and

$u_{n+k} ={a'}_k$, $t_{n+k}=s'_k$, $v_{n+k}=b'_k$ for $1\leq k\leq m$.

Wuestenfux
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