If R is a Noetherian ring, then Rⁿ is a Noetherian R-module

Let R be a Noetherian ring. Prove that R^n, as an R-module in the usual way, is also Noetherian.

Recall that a ring is Noetherian if every ideal is finitely generated, and a module is Noetherian if every submodule is finitely generated. (That is, a ring is Noetherian (as a ring) if it is Noetherian (as a module) over itself.)

We will proceed by induction on n.

For the base case, n = 1, R^1 is certainly Noetherian as a module over R.

For the inductive step, suppose R^n is Noetherian. Now let M \subseteq R^{n+1} be a submodule; our goal is to show that M must be finitely generated. To that end, let A = \{r \in R \ |\ (m_i)_0 = r\ \mathrm{for\ some}\ (m_i) \in M \}. That is, A is the collection (in R) of all zeroth coordinates of elements in M.

We claim that A \subseteq R is an ideal. If a,b \in A, then there exist elements (m_i) and (n_i) in M such that m_0 = a and n_0 = b. Now since M \subseteq R^{n+1} is a submodule, we have (m_i) + r(n_i) \in M for all r \in R, so that a+rb \in A. We clearly have 0 \in A, so that by the submodule criterion, A is an ideal of R.

In particular, since R is Noetherian, the ideal A is finitely generated – say by \{a_0, a_1, \ldots, a_k\}. We will let \alpha_i be an element of M whose zeroth coordinate is a_i. Now let m = (m_0,\ldots,m_{n+1}) \in M. Now m_0 = \sum c_i a_i for some c_i \in R, and so m - \sum c_i m_i = (0,t_1,\ldots,t_{n+1}) is an element of M whose first coordinate is 0. In particular, we have M = (m_0,\ldots,m_k) + B, where B = M \cap (0 \times R^n). (We showed the (\subseteq) direction, and the (\supseteq) inclusion is clear.) Now M \cap (0 \times R^n) is an ideal of 0 \times R^n \cong_R R^n, which is Noetherian by the induction hypothesis. So B is finitely generated as an R-module, and thus M is finitely generated over R.

Since M was an arbitrary submodule of R^{n+1}, R^{n+1} is Noetherian as a module.

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