# Is every unitary ring finitely generated?

I'm puzzled by the following:

if $$R$$ is a unitary ring then $$R$$ is generated by $$1_R$$, denoted as $$R = \langle 1_R \rangle.$$ can we conclude that every unitary ring is finitely generated? I know the answer is no, as $$\mathbb{Q}$$ is not finitely generated, although we can express $$\mathbb{Q}$$ as $$\langle 1 \rangle$$. Can somebody help explain what's causing my confusion?

• Write out very carefully your definition of "finitely generated" as a ring. Feb 23 at 13:19
• An ideal in R is finitely generated if and only if there exists a finite set of elements in R that generates it. Feb 23 at 13:40
• The subring generated by a set $X$ and the ideal generated by a set $X$ inside a ring $R$ can be different. Which do you mean? Feb 23 at 13:48
• Are you suggesting that stating that R is finitely generated as a ring is distinct from expressing that the ideal R (as an ideal of it self) is finitely generated? Feb 23 at 13:54

Taking any ring $$R$$, it is true that $$R=(1)$$ as an ideal, which is equivalent to say that $$R=\langle 1\rangle _R$$ as an $$R$$-module.
So $$\mathbb{Q}=\langle 1\rangle _\mathbb{Q}$$ as a $$\mathbb{Q}$$-module (i.e. as an ideal and thus, according to your definition, as a ring).
However $$\mathbb{Q}$$ is also a $$\mathbb{Z}$$-module. If you consider $$\mathbb{Q}$$ with this structure, as you correctly pointed out, it is not finitely generated. But this means that it is not finitely generated as a $$\mathbb{Z}$$-module, not as an ideal ($$=\mathbb{Q}$$-module).
• Usually one does not define the notion of being finitely generated as a ring because if you consider being finitely generated as a ring as being finitely generated as an ideal, then every unitary ring is finitely generated (by $1$). So yes, using this notion $\mathbb{Q}$ is finitely generated as a ring (but as i said one does not usually defines this notion since it is trivially true for any ring). Feb 23 at 14:01
• Thank you. By the way, we can infer that $\mathbb{Q}$ is not a Noetherian $\mathbb{Z}$-module, but we cannot deduce that $\mathbb{Q}$ is not a Noetherian ring. Is this right? Feb 23 at 14:08
• Yep, you are right (btw $\mathbb{Q}$ is a noetherian ring since it is a field) Feb 23 at 14:09