# Modules of finite type over Noetherian rings

Let M be a unitary module of finite type over a commutative Noetherian ring R with a unit. Can M then always be represented as a quotient of a pair of free R-modules of finite-type?

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## 1 Answer

Suppose $M$ is generated by $m_1, \ldots, m_n$. Then, there is an evident surjective homomorphism $p : R^{\oplus n} \twoheadrightarrow M$ mapping the canonical generators $e_1, \ldots, e_n$ of $R^{\oplus n}$ to the chosen generators of $M$. Consider the kernel of $p$: this is a submodule of $R^{\oplus n}$ and is therefore also finitely generated, since $R$ is noetherian. We therefore obtain an exact sequence of the form $$R^{\oplus k} \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0$$ as required. However, note that $\ker p$ need not be a free $R$-module. (Take, for example, $R = \mathbb{Z} [\sqrt{-5}]$, $M = R / (2, 1 + \sqrt{-5})$, and note that the ideal $(2, 1 + \sqrt{-5})$ is projective but not free.)

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