3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Can you please check the language in the first paragraph ? Can you please write the first paragraph more perfectly ? $\endgroup$
    – Why
    Nov 3 '20 at 14:02
  • $\begingroup$ @MabudAli Sorry for my imprudence. I've edited. $\endgroup$
    – XT Chen
    Nov 3 '20 at 14:08
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.