# Does reduced+Noetherian space imply Noetherian scheme

In the special case of affine schemes, there is an exercise on Hartshorne saying that when Spec A is a Noetherian topological space A may not be a Noetherian ring. While it is easy to find an example for that when A has nilpotent elements, e.g. $A=k[x_1,...,x_i,...]/(X_1, x_2^2,...,x_i^i,...).$ It is not clear to me that whether we could still find a counter example when A is a domain. In the general case, I am asking for a reduced scheme, if the underlying topological space is Noetherian, is the scheme necessarily a Noetherian scheme?

My guess would be no, consider the direct limit of the series of localizations,

$$\underline{lim}\ k[x^{1\over 2^n}]_{(x^{1\over 2^n})}$$

each process within the limit is a one dimension scheme, and I think the limit is also a one dimension scheme. (In general, will dimension necessarily be held constant, non increasing or non decreasing in a limit process? or none of the above?). However, it is not hard to see the limit ring is not a Noetherian ring.

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sorry, inverse limit should be direct limit, I just edited that. However, though the \underrightarrow command works in the preview, it doesn't work here. Could anyone fix it? Btw, I don't think a general direct inverse of schemes (direct limit on the ring side) will hold dimension for any reason, since we can embed into sth big. –  Ying Zhang Jul 14 '10 at 16:13

The answer is no, consider $k[x,xy, xy^2, xy^3, \dots]$.

Some more details. This is basically a copy of A^2 where all the points of one axis (including the generic point of that axis) are all glued together (into the obvious maximal ideal of that ring).

EDIT: Your example may be right too, I'm not quite sure I see what's going on there.

EDIT2: (More information on the example) Call the ring $R$ and it obviously sits inside $k[x,y]$. The induced map $\mathbb{A}^2 \to Spec R$ contracts the axis $x = 0$ to a ($k$-valued) point, otherwise, it is an isomorphism (to check that, invert $x$). The ideal corresponding to the contracted axis is the maximal ideal $(x, xy, xy^2, xy^3, \dots)$.

Let me know if you have further questions.

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In my example, for each n, if we substitue $y=x^{1\over 2^n}$, then the ring we get is $k[y]_{(y)}$, i.e. k[y] localize at a prime (also maximal ideal) generated by (y). Therefore every step we have a DVR, and the injections of the direct limit are local homomorphisms. btw, thanks for example, but would you please give more information about your example, like which axis and which maximal ideal are you refering to? –  Ying Zhang Jul 14 '10 at 17:49
I see, the point is that the quotient of a Noetherian topological space is still gonna be Noetherian, got it. –  Ying Zhang Jul 14 '10 at 22:21

Another example: Let $K/k$ be an infinite field extension, and let $R$ be the subring of $K[X]$ consisting of polynomials with constant term in k. I claim that $R$ is not noetherian but $X:=$Spec($R$) is a noetherian space.

Consider the obvious evaluation map from $R$ to $k$: it is surjective with kernel $I$ generated by the set $S=K^*.X$. If $J$ is the ideal generated by a finite subset of $S$, then the degree 1 coefficients of all elements of $J$ form a finite-dimensional $k$-subspace of $K$. Thus $I$ is not finitely generated.

The closed subscheme $Y$ defined by $I$ is isomorphic to Spec($k$), hence noetherian. On the other hand, inverting any element of $S$ in $R$ yields the ring $K[X,X^{-1}]$. Hence the open complement of $Y$ is also noetherian.

Geometrically, $X$ is obtained from the affine line over $K$ by "crushing" the origin down from Spec($K$) to Spec($k$).

Generalization: take for $k$ any noetherian subring of $K$ such that $K$ is not a finitely generated $k$-module, e.g. $K=\mathbb{Q}$, $k=\mathbb{Z}$.

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As an afterthought, here is a more familiar example: just take any nondiscrete, rank one valuation ring, such as the integers of $\mathbb{C}_p$ or the ring of Puiseux series over any field k (i.e. $\bigcup_{n\geq1}k[[t^{1/n}]]$. Such a ring is obviously not noetherian and its spectrum has only two points. –  Laurent Moret-Bailly Jul 16 '10 at 13:49

In fact, the ring considered by Ying zhang is just a non-discrete rank one valuation ring. Therefore it is already a desired example by the last remark maded by Laurent Moret-Bailly. But I think the above two examples are very interesting themselves. So many thanks to Karl Schwede and Laurent Moret-Bailly !

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