If local rings are Noetherian, is scheme locally-Noetherian?

Question

If all the local rings of a scheme are Noetherian, is the scheme locally-Noetherian?

Answer 1

Take an algebraically closed field $k$, so that $k$ is infinite, and take a sequence $(\alpha_i)_{i\geq 1}$ of distinct elements in $k$. Consider $A = k[S,T_1,\ldots,T_n,\ldots]$ and the ideal $I$ of $A$ generated by all $(S-\alpha_i)T_{i+1} - T_i$ and $T_i^2$ for $i\in\mathbf{N}^*$. Finally, take $B = A / I$.

Then $B$ is not a noetherian ring (exercise !) and $B_{\mathfrak{p}}$ is noetherian for all prime ideals $\mathfrak{p}$ of $B$. As noetherianess is preserved by localization, it suffice to check that $B_{\mathfrak{p}}$ is noetherian for all maximal ideals $\mathfrak{p}$ of $B$. Now, the nilradical $N$ of $B$ is clearly the ideal generated by the images of the $T_i$'s (as those are clearly in $N$ and as the ideal $J$ they generated verifies that $B/J$ is a domain) and $B/N \simeq k[S]$, so the maximal ideals are (because $k$ is algebraically closed) of the form $\mathfrak{m} = (S-\lambda, T_1,\ldots,T_n,\ldots)$ for $\lambda\in k$. Take $\mathfrak{m}$ such an ideal. If $\lambda$ is an $\alpha_i$ then for $j\not=i$ you have $T_{j+1} = T_{j} / (S - \alpha_i)$ in $B_{\mathfrak{m}}$ so that $B_{\mathfrak{m}}$ appears as the localization of a ring generated by two elements $S$ and $T_i$, and is therefore noetherian. If $\lambda$ is not an $\alpha_i$, an argument analogue to the one of the previous case shows that $B_{\mathfrak{m}}$ appears as a localization of a ring generated by $S$ and $T_1$ and is also noetherian.

Now, as an affine scheme is noetherian if and only if its ring of global sections is a noetherian ring, we have shown that $X=\textrm{Spec}(B)$ is a non-noetherian affine scheme (and if fact non locally noetherian : would it be locally noetherian it would be noetherian also, as it is quasi-compact being affine) whose local rings are all noetherian rings.

Answer 2

Take any infinite boolean ring, such as $\mathbb{F}_2 \times \mathbb{F}_2 \times \dotsc$. All localizations at prime ideals are local boolean rings, hence $\mathbb{F}_2$, but infinite boolean rings are not noetherian.

More generally, one can show that a commutative ring $R$ is von Neumann regular if and only if $R_\mathfrak{p}$ is a field for all prime ideals $\mathfrak{p}$. There are lots of von Neumann regular rings which are not noetherian.

This is one of many reasons why localizations at prime ideals are not as good as localizations at elements. In fact, we have the following result (which is useful for the definition of locally noetherian schemes):

If $A$ is a commutative ring and $f_1,\dotsc,f_n$ are elements of $A$ generated the unit ideal such that each $A_{f_i}$ is noetherian, then $A$ is noetherian, too.

The geometric explanation is that $A_{f_i}$ captures the information of the whole basic-open subset $D(f_i)$ of $\mathrm{Spec}(A)$, whereas $A_{\mathfrak{p}}$ only captures the information of sufficiently small neighborhoods of the point $\mathfrak{p}$.

The notion of a noetherian scheme is not so good as one might think. There are noetherian schemes, even varieties, whose ring of global sections is not noetherian (math.SE/818578).