# Is ring R itself a finitely generated module over $R$?

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..

• Yes, any ring is finitely generated over itself, $R1=R$. – Pedro Tamaroff Jan 20 '15 at 17:02
• And yes, it is trivial. – Christopher Jan 20 '15 at 17:05

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.