Does reduced+Noetherian space imply Noetherian scheme - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:48:03Z http://mathoverflow.net/feeds/question/31865 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31865/does-reducednoetherian-space-imply-noetherian-scheme Does reduced+Noetherian space imply Noetherian scheme Ying Zhang 2010-07-14T15:16:16Z 2011-01-30T15:35:57Z <p>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?</p> <p>My guess would be no, consider the direct limit of the series of localizations, </p> <p>$$\underline{lim}\ k[x^{1\over 2^n}]_{(x^{1\over 2^n})}$$</p> <p>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.</p> http://mathoverflow.net/questions/31865/does-reducednoetherian-space-imply-noetherian-scheme/31876#31876 Answer by Karl Schwede for Does reduced+Noetherian space imply Noetherian scheme Karl Schwede 2010-07-14T17:05:01Z 2010-07-14T20:55:50Z <p>The answer is no, consider $k[x,xy, xy^2, xy^3, \dots]$. </p> <p>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). </p> <p>EDIT: Your example may be right too, I'm not quite sure I see what's going on there.</p> <p>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 $(x, xy, xy^2, xy^3, \dots)$. </p> <p>Let me know if you have further questions.</p> http://mathoverflow.net/questions/31865/does-reducednoetherian-space-imply-noetherian-scheme/32132#32132 Answer by Laurent Moret-Bailly for Does reduced+Noetherian space imply Noetherian scheme Laurent Moret-Bailly 2010-07-16T08:32:01Z 2010-07-16T08:32:01Z <p>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. </p> <p>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. </p> <p>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.</p> <p>Geometrically, $X$ is obtained from the affine line over $K$ by "crushing" the origin down from Spec($K$) to Spec($k$).</p> <p>Generalization: take for $k$ any noetherian subring of $K$ such that $K$ is not a finitely generated $k$-module, e.g. $K=\mathbb{Q}$, $k=\mathbb{Z}$.</p> http://mathoverflow.net/questions/31865/does-reducednoetherian-space-imply-noetherian-scheme/53791#53791 Answer by ding8191 for Does reduced+Noetherian space imply Noetherian scheme ding8191 2011-01-30T15:35:57Z 2011-01-30T15:35:57Z <p>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 ! </p>