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It seems trivial that ring $R$ itself is a $R$-module. But then can we say R is finitely generated by multiplicative identity? That seems so trivial..

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    $\begingroup$ Yes, any ring is finitely generated over itself, $R1=R$. $\endgroup$ – Pedro Tamaroff Jan 20 '15 at 17:02
  • $\begingroup$ And yes, it is trivial. $\endgroup$ – Christopher Jan 20 '15 at 17:05
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Yes. We can write $R = (1)$, but notice that it's not necessary that every ideal in $R$ is finitely generated:

Choose $R = k[x_1, x_2, \ldots]$ and let $I = (x_1, x_2, \ldots)$. $R$ is finitely generated since $R = (1)$, but $I$ is not finitely generated.

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