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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?

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  • $\begingroup$ Write out very carefully your definition of "finitely generated" as a ring. $\endgroup$
    – Randall
    Feb 23 at 13:19
  • $\begingroup$ An ideal in R is finitely generated if and only if there exists a finite set of elements in R that generates it. $\endgroup$ Feb 23 at 13:40
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    $\begingroup$ 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? $\endgroup$
    – Randall
    Feb 23 at 13:48
  • $\begingroup$ 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? $\endgroup$ Feb 23 at 13:54

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Your problem lies in the fact that you are confusing being finitely generated as an ideal and being finitely generated as a module.

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).

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  • $\begingroup$ Okay thank you, so Q is finitely generated as a ring? That's right? $\endgroup$ Feb 23 at 13:57
  • $\begingroup$ 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). $\endgroup$
    – Temoi
    Feb 23 at 14:01
  • $\begingroup$ 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? $\endgroup$ Feb 23 at 14:08
  • $\begingroup$ Yep, you are right (btw $\mathbb{Q}$ is a noetherian ring since it is a field) $\endgroup$
    – Temoi
    Feb 23 at 14:09
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    $\begingroup$ Thank you very much, well understand. $\endgroup$ Feb 23 at 14:56

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