An $R$-module $M$ that is not finitely generated but $M/xM$ is finitely generated.

328 Views Asked by At

Let $R = k[x]_{(x)}$ be the localization of a polynomial ring over a field $k$ at the prime ideal $(x).$ I want to find a module $M$ that is not finitely generated but $M/xM$ is finitely generated. I think that $Q(R)$ the quotient field is such a module. $Q(R)$ is an $R$-algebra. Hence, if $Q(R)$ is finitely generated as a module then $Q(R)$ is finitely generated as an algebra by finitely many integral elements over $R.$ So then $Q(R)$ is in fact integral over $R$ but an element $\frac{a}{b}$ in the field of fractions is only integral if $b \mid a.$ Clearly this is not the case for all elements of $Q(R)$ so $Q(R)$ is not finitely generated. However, since it is a field $xQ(R) = Q(R)$ and thus $Q(R)/xQ(R) = 0$ which is finitely generated.

I guess I am afraid that my argument that $Q(R)$ is not finitely generated is incorrect.

Any help is appreciated. Thank you!