# Definition: rng ideal versus algebra ideal in a non-unital $C^*$-algebra

When I first moved into operator algebras, equipped with the basics of abstract algebra, I encountered the following definition of ideals:

Definition (algebra ideals). A left (respectively, right) ideal in an algebra $A$ is a vector subspace $I$ of $A$ such that $$a\in A\quad\text{and}\quad b\in I \quad \implies\quad ab\in I\qquad\text{(respectively, ba\in I)}.$$ An ideal in $A$ is a vector subspace that is simultaneously a left and a right ideal in $A$.

(This definition is taken from Murphy's textbook $C^*$-algebras and operator theory. It should be noted that algebra in this case means associative algebra over $\mathbb{C}$, not necessarily commutative or with unit.)

To a large degree, mathematics is the art of making the right definitions, and it is intuitively clear to me that we want $I$ to be a subspace. Nevertheless, I can't help but wonder whether or not the above is equivalent to the notion of a rng ideal in $A$:

Definition (rng ideals). A left (respectively, right) rng ideal in an algebra $A$ is an additive subgroup $I$ of $A$ such that $$a\in A\quad\text{and}\quad b\in I \quad \implies\quad ab\in I\qquad\text{(respectively, ba\in I)}.$$ A rng ideal in $A$ is an additive subgroup that is simultaneously a left and a right ideal in $A$.

(The only difference is that the subspace assumption is replaced by the weaker subgroup assumption.)

Now, if the algebra is unital, it is easy to show that these two notions coincide. Suppose that $I$ is, say, a left rng ideal, then for all $\lambda \in \mathbb{C}$ and $b\in I$ we have $\lambda b = (\lambda 1)b \in I$. However, it is not immediately clear whether or not these definitions are equivalent if $A$ fails to be unital.

This page in the Encyclopedia of Mathematics gives a pathological example showing that the definitions are not equivalent for general associative algebras (or even Banach algebras): if $ab = 0$ for all $a,b\in A$, then every additive subgroup is a rng ideal, but only the subspaces are algebra ideals. This answers part of my question.

However, since I'm primarily interested in $C^*$-algebras, the following question remains:

Question. Are there examples of rng ideals in $C^*$-algebras that fail to be algebra ideals?

In this setting, the pathological example from before is ruled out, due to the identity $\lVert a^*a\rVert = \lVert a\rVert^2$.

• Partial result: the closure of a (left/right/two-sided) rng ideal in a $C^*$-algebra is automatically a (left/right/two-sided) algebra ideal. (Use approximate units.) – Josse van Dobben de Bruyn Sep 13 '17 at 1:03

Consider $A=c_0$ and let $I$ be the rng ideal generated by the sequence $s$ with $s_n=1/n$. Explicitly, we can describe $I$ as the set of sequences of the form $as+ts$ where $a\in \mathbb{Z}$ and $t\in c_0$. I claim that if $b\in\mathbb{C}\setminus\mathbb{Z}$, then $bs\not\in I$, so $I$ is not an algebra ideal. Indeed, if we had $bs=as+ts$, then $(b-a)s=ts$ where $b-a$ is some nonzero scalar. Since each term of $s$ is nonzero, this can only be true if $t_n=b-a$ for all $n$, which contradicts the requirement that $t\in c_0$.