8
$\begingroup$

In a commutative ring with unity $1$, call it $R$, the the ideal generated by the set $S=\{a_1,...,a_n\}$ is the smallest ideal of $R$ containing $S$. It can be proven that this ideal is

$$ (a_1,...,a_n)=\left\{\sum_{k=1}^{n} r_ka_k \, : \, r_k \in R \right\} $$

I think to have proved this fact by the standard way: clearly any such ideal containing $S$ must contain this set, and this set is itself an ideal. But if the commutative ring does not have a unit, i cannot see why the same proof does not apply. That is, where we use the fact of a unity's existence to show this?

Thanks.

$\endgroup$
6
$\begingroup$

If $R$ has no multiplicative identity, $(a)$ need not contain $a$ if you take it to be $aR$. You want the ideal generated by a set $S$ to contain $S$.

$\endgroup$
1
  • $\begingroup$ @Martin: Well, yes: that was rather the point of my answer. Apparently I should make the counterfactual more explicit. $\endgroup$ – Brian M. Scott Apr 26 '13 at 12:06
4
$\begingroup$

The elements of the ideal $(S)$ generated by $S$ are finite sums or differences of elements of the form $s$ or $rs$, where $s \in S$ and $r \in R$. So in formulas:

$(S) = \{\sum_{i=1}^{n} r_i s_i : n \in \mathbb{N}, s_i \in S, r_i \in \mathbb{Z} \sqcup R\}$.

Here we use the $\mathbb{Z}$- and the $R$-module structure of $R$. In the non-commutative case, the formula above gives the left ideal generated by $S$. The two-sided ideal is even more complicated, we also have to take elements of the form $rsr'$ where $r,r' \in \mathbb{Z} \sqcup R$.

$\endgroup$
0
$\begingroup$

In the noncommutative case include s(i) t(i) with t(i) in R . here I used s(i) instead of the subscript i.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.