Example on coherent scheme but not noetherian, $\mathrm{Spec}\underset{n \in \mathbb{N}}{\cup}k[[t^{\frac{1}{n}}]]$.

In class, my teacher gave an example of coherent scheme that is not noetherian, namely $$\mathrm{Spec}\underset{n \in \mathbb{N}}{\cup}k[[t^{\frac{1}{n}}]]$$.

The definition, of a coherent sheaf of module over a scheme $$(X,\mathcal{O}_X)$$, is a sheaf of $$\mathcal{O}_X$$-module locally (on $$\mathrm{Spec}{A} \subset X$$) being $$\tilde{M}$$ with $$M$$ a finitely generated $$A$$-module, and every kernel of arbitrary $$A^{\oplus n} \rightarrow M$$ is finitely generated.

Going back to the example. $$k[[t^{\frac{1}{n}}]]:=A$$ is obviously not noetherian. But I don't know how to show that kernel of arbitrary $$A^{\oplus n} \rightarrow A$$ is finitely generated.

• Can you please check the language in the first paragraph ? Can you please write the first paragraph more perfectly ?
– Why
Nov 3 '20 at 14:02
• @MabudAli Sorry for my imprudence. I've edited. Nov 3 '20 at 14:08

Define $$A_n = k[[t^{\frac{1}{n}}]]$$ for every $$n \in \mathbb{N}$$. Note that if $$n$$ divides $$m$$, then $$A_m$$ is a flat $$A_n$$-module. Indeed, we have $$A_{m} = \bigoplus_{j=1}^{\frac{m}{n}}A_n\cdot t^{\frac{j}{m}}$$
So it is even a free $$A_n$$-module.
Consider a morphism $$\phi:A^{\oplus s} \rightarrow A$$. We denote its kernel by $$K$$. We have $$\phi(e_i) = f_i$$ for $$1\leq i \leq s$$ (here $$\{e_i\}_{1\leq i\leq s}$$ is the standard basis of a free module $$A^{\oplus s}$$). There exists $$n_0$$ such that $$f_1,...,f_s\in k[[t^\frac{1}{n_0}]] = A_{n_0}$$. Define $$\mathcal{F} = \{n\in \mathbb{N}\,|\,n\geq n_0\mbox{ and }n_0\mbox{ divides }n\}$$ For $$n\in \mathcal{F}$$ define $$\phi_n:A_n^{\oplus s}\rightarrow A_n$$ by $$\phi(e_i) = f_i$$ for every $$1\leq i\leq s$$ (this time $$\{e_i\}_{1\leq i\leq s}$$ is the standard basis of a free module $$A_n^{\oplus s}$$). We also denote by $$K_n$$ the kernel of $$\phi_n$$. Note that $$1_{A_n}\otimes_{A_{n_0}}\phi_{n_0} = \phi_n$$ for $$n\in \mathcal{F}$$. Since for $$n\in \mathcal{F}$$ algebra $$A_n$$ is flat over $$A_{n_0}$$, we derive that the tensor product $$A_n\otimes_{A_{n_0}}(-)$$ preserves kernels and hence $$K_n = A_n\otimes_{A_{n_0}}K_{n_0}$$ Thus $$K = \bigcup_{n\in \mathcal{F}}K_n = \mathrm{colim}_{n\in \mathcal{F}}K_n = \mathrm{colim}_{n\in \mathcal{F}}\left(A_n\otimes_{A_{n_0}}K_{n_0}\right) =$$ $$= \left(\mathrm{colim}_{n\in \mathcal{F}}A_n\right)\otimes_{A_{n_0}}K_{n_0} = A\otimes_{A_{n_0}}K_{n_0}$$ Due to the fact that $$K_{n_0}$$ is finitely generated over $$A_{n_0}$$ ($$A_{n_0}$$ is noetherian), we derive that $$K$$ is finitely generated over $$A$$.