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.



Does reduced+Noetherian space imply Noetherian scheme
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.


