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.
 A: 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. $K=\mathbb{Q}$, $k=\mathbb{Z}$.
A: 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.
A: 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 !     
