0

Let $R$ be a rng (general ring not requiring existence of identity), and $U\subseteq R$ be a subset. Let $I$ be ideal generated by $U$, which is defined as the intersection of all ideals containing $U$.

Show that if $R$ has an identity, then $I=\{\sum_{i=1}^n r_iu_is_i : n\in\mathbb{N},u_i\in U, r_i,s_i\in R\}$. Here answers that $I=\mathbb{Z}U+RU+UR+RUR$ in the problem, is the two answers equivalent?

This indicates a possibility, that ideals of rng $R$ generated by $U$ may not be represented in such form, is there an example?

spicychicken
  • 43

1 Answers1

0

Intuitively (and formally) the ideal $\langle U\rangle$ generated by $U$ consists of all elements, which are in $U$ or can be built from $U$ using operations of the ring. This at least suggests that $I = \Bbb Z U + RU + UR + RUR$ always is the correct description of the ideal generated by $U$: you can add elements of $U$ to themselves ($\Bbb Z U$ is the abelian subgroup generated by $U$), you can multiply elements of $R$ from the left / right / both sides, and a generic element in the ideal will be a combination (ie. sum) of those options.

Now consider the set $$RUR = \left\{\sum \limits_{i=1}^n r_i u_i s_i \mid r_i,s_i \in R, u_i \in U\right\}$$ By the previous discussion $RUR \subseteq \langle U\rangle$ is always the case. But if the ring does not admit a unit, how can you obtain an element $u\in U$ from an element of the form $ru's$ or a sum of those? This suggests that the inclusion can be strict. And in fact it sometimes is: Consider the sub-rng $R:=2\Bbb Z \subseteq \Bbb Z$ and let $U=\{2\}$. We have $RUR = 8\Bbb Z,$ while $\langle U\rangle = 2\Bbb Z = R$.

Now if the ring does have a unit, we find that $\Bbb Z U \subseteq RUR$, $RU \subseteq RUR$ and $UR \subseteq RUR$ thus $I = RUR$.

Jonas Linssen
  • 11,016