Some rings have the property that they are a union of (proper, nontrivial) ideals.
- $\mathbb{Z}$ is a union of $2\mathbb{Z},3\mathbb{Z}, 5\mathbb{Z}, \dots$
- $R[x]$ is a union of $xR,x^2R, \dots$
But some rings are not, for instance:
- $\mathbb{Z}/n\mathbb{Z}$
- $\mathbb{Q}$
(Those are the ones I can think of right now -- I would love to hear some others!)
Is there a name for rings with this property?
EDIT: As G. Sassatelli points out, no ring has this property because no ideal can contain $1$. In particular, the correct decompositions are:
- $\mathbb{Z}$ is a union of $\{1\}$ and $2\mathbb{Z}, 3\mathbb{Z}, 5\mathbb{Z},\dots$
- $R[x]$ is a union of $R$ and $xR,x^2R, \dots$
So I suppose my question is: what is the name for rings that are a union of ideals and finitely many subrings? (I guess $\mathbb{Z}/n\mathbb{Z}$ has this property under the new definition, but I don't believe $\mathbb{Q}$ does.)
(Not really an answer, but too long for a comment)
As @G.Sassatelli points out, in every ring $A$, the union of all proper ideals is exactly $A\setminus A^*$. It's easy to see : no element of $A^*$ can be in a proper ideal, and any non-invertible elements is in the proper ideal it generates.
In particular, what you say about $\mathbb{Z}$ and $R[x]$ is false : $\mathbb{Z}$ is the union of its prime ideals and $\{\pm 1\}$ (you forgot $-1$) ; and $R[x]$ is not the union of $R$ and the ideal $(x)$ : for instance $1+x$ is in neither of these sets.
So maybe your question can become : "when is $A^*$ contained in the union of finitely many proper subrings ?".