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.
Best Answer
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 $A$.
I think that more generally this proves that a filtered colimit of a diagram of flat ring extensions in which every ring is coherent is coherent itself.