When is an irreducible scheme quasi-compact? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T13:46:42Z http://mathoverflow.net/feeds/question/34620 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34620/when-is-an-irreducible-scheme-quasi-compact When is an irreducible scheme quasi-compact? David Rydh 2010-08-05T12:55:59Z 2010-08-06T23:42:35Z <p>The standard examples of schemes that are not quasi-compact are either non-noetherian or have an infinite number of irreducible components. It is also easy to find non-separated irreducible examples. But are there other examples?</p> <p><strong>Question</strong>: Let $X$ be a locally noetherian scheme and assume that $X$ is irreducible (or has a finite number of irreducible components) and separated. Is $X$ quasi-compact (i.e., noetherian)?</p> <p>If the answer is no in general, what conditions on $X$ are sufficient? Locally of finite type over a noetherian base scheme $S$? Fraction field finitely generated over a base? What if $X$ is regular? In general, the question is easily reduced to the case where $X$ is normal and integral.</p> <p>It certainly feels like the answer is yes when $X$ is locally of finite type over $S$. Idea of proof: Choose an open dense affine $U\subseteq X$, choose a compactification $\overline{U}$ and modify $X$ and $\overline{U}$ such that the gluing $Y=X\cup_U \overline{U}$ is separated. Then, $Y=\overline{U}$ (by density and separatedness) is proper and hence quasi-compact.</p> <p><strong>Remark 1</strong>: If $X\to S$ is a proper morphism, then the irreducible components of the Hilbert scheme Hilb(X/S) are proper. The subtle point (in the non-projective case) is the quasi-compactness of the components (which can be proven by a similar trick as outlined above).</p> <p><strong>Remark 2</strong>: If $X\to S$ is universally closed, then $X\to S$ is quasi-compact. This is question <a href="http://mathoverflow.net/questions/23337/is-a-universally-closed-morphism-of-schemes-quasi-compact" rel="nofollow">23337</a>.</p> http://mathoverflow.net/questions/34620/when-is-an-irreducible-scheme-quasi-compact/34656#34656 Answer by BCnrd for When is an irreducible scheme quasi-compact? BCnrd 2010-08-05T17:01:37Z 2010-08-05T17:20:23Z <p>There are smooth counterexamples. Let $S_0$ be a smooth separated irreducible scheme over a field $k$ with dimension $d > 1$, and $s_0 \in S_0(k)$. Blow up $s_0$ to get another such scheme $S_1$ with a $\mathbf{P}^{d-1}_k$ over $s_0$. Blow up a $k$-point $s_1$ over $s_0$ to get $S_2$, and keep going. Get pairs $(S_n, s_n)$ so that the open complement $U_n$ of $s_n$ in $S_n$ is open in $U_ {n+1}$ and is strictly contained in it. Glue them together in the evident manner, to get a smooth irreducible $k$-scheme. It is locally of finite type, but is not quasi-compact (since the $U_n$ are an open cover with no finite subcover). This is separated (either by direct consideration of affine open overlaps, or by using the valuative criterion). </p> http://mathoverflow.net/questions/34620/when-is-an-irreducible-scheme-quasi-compact/34665#34665 Answer by Qing Liu for When is an irreducible scheme quasi-compact? Qing Liu 2010-08-05T17:39:23Z 2010-08-05T17:39:23Z <p>I had in mind the example of a separated, locally noetherian, irreducible regular scheme of dimension 1 which is not quasi-compact, I hope it is correct: let $X$ be a separated integral (locally) noetherian scheme of dimension $\ge 1$ such that the set $F$ of the points of codimension 1 in $X$ is infinite ($X$ could be the affine line over a field). Let $\xi$ be the generic point of $X$ and $K(X)=O_{X,\xi}$. We construct a new scheme $X'$ by gluing all the $U_x:={\mathrm Spec} (O_{X,x})$, $x\in F$, along $\xi$. Then $X'$ is locally noetherian regular of dimension $1$ (because $U_x$ is the the spectrum of a DVR), irreducible because $\xi$ is the unique generic point, and separated because the canonical map $O_{X,x}\otimes O_{X, y}\to K(X)$ is surjective if $x\ne y \in F$. But $X'$ is not quasi-compact because the covering { $U_x$ }$_{x\in F}$ can not be refined by a finite covering.</p> http://mathoverflow.net/questions/34620/when-is-an-irreducible-scheme-quasi-compact/34676#34676 Answer by Torsten Ekedahl for When is an irreducible scheme quasi-compact? Torsten Ekedahl 2010-08-05T19:30:09Z 2010-08-05T19:30:09Z <p>This is just an elaboration on BCnrd's example to illustrate that it is in no way pathological but rather appears very naturally.</p> <p>To begin with we can let the starting variety be $\mathbb P^1\times\mathbb P^1$. This is a toric variety described by the fan in $\mathbb R^2$ given by the four quadrants (and the lattice is $\mathbb Z^2$). The blowing up of $(0,0)$ corresponds to adding the halfline through $(1,1)$ (and the cones on either side). Then blowing $0$ up $(0\colon1)$ on the exceptional curve corresponds to adding the halfline through $(2,1)$. Continuing in this way corresponds to adding the halflines through $(n,1)$ and passing to the limit gives the fan consisting the three quadrant and the fans spanned by $(n,1)$ and $(n+1,1)$ for $n\ge 0$. Concretely the affine toric variety corresponding to the fan spanned by $(n,1)$ and $(n+1,1)$ has affine algebra generated $r_i=xy{-i}$ and $s_i=x^{-1}y^{i+1}$ and (where $x$ and $y$ are generators for the coordinate ring of the torus). Note that we have $r_is_i=y$ a relation in terms of one of the coordinates as well as $r_{i+1}s_{i}=1$. </p> <p>We may make the construction more symmetric by blowing up also the points at $\infty$ of the exceptional divisors. This corresponds to adding the halflines through $(1,n)$ and for the affine rings letting $i$ run over all integers. This construction then works over a discrete valution ring $R$ with generator (say) $\pi$. We then put $r_is_i=\pi$ and embed the ring generated by it into $K[x,x^{-1}]$ by mapping $r_i=x\pi^{-i}$ and $s_ix^{-1}\pi^{i+1}$. This gives a scheme locally of finite type over $\mathrm{Spec}R$ whose generic fibre is $\mathbb P^1_K$ and whose special fibre is an infinite line of $\mathbb P^1$'s (the $(0\colon1)$ identified with $(1\colon0)$ of the next). We have an autormorphism $\varphi$ of this scheme taking $r_i\mapsto r_{i+1}$ and $s_i\mapsto s_{i+1}$ which on the generic fibre takes $x$ to $x\pi^{-1}$. It shifts the $\mathbb P^1$'s. We can now consider the formal completion with respect to the closed point of $\mathrm{Spec}R$. On it $\varphi$ acts properly continuosly so we may construct the quotient. The resulting formal scheme is proper and can be algebraised (when $R$ is complete). The resulting scheme is the Tate curve.</p> <p>A more invariant way of constructing fan is to consider the convex hull of the lattice points of the open first quadrant and then draw halflines through the lattice points at the boundary of this hull. As these points are exactly the $(1,n)$ and $(n,1)$ we get the previous example. We can do the same thing in the following situation: We let $K$ be a real quadratic field, let the cone $C$ be spanned by the totally positive elements of $K$ and let the lattice be the algebraic numbers of $K$. We can then construct the fan by adding halflines through the lattice points on the boundary of the convex hull which gives us a toric scheme locally of finite type. The totally positive units of $K$ acts on this fan and we may take the quotient of the formal completion as before. This gives a resolution of a (particular) cusp of the Hilbert modular surface associated to $K$ (see for instance Oda: Convex bodies and algebraic geometry).</p> http://mathoverflow.net/questions/34620/when-is-an-irreducible-scheme-quasi-compact/34802#34802 Answer by Michael Thaddeus for When is an irreducible scheme quasi-compact? Michael Thaddeus 2010-08-06T22:36:35Z 2010-08-06T23:42:35Z <p>Actually, Nick Proudfoot and I have been talking for years about the irreducible smooth surface constructed from countably many copies of ${\mathbb A}^2$ by gluing $(p,q)$ in the $n$th copy to $(p^2q,p^{-1})$ in the $n+1$th copy. This even has a ${\mathbb G}_m$ action $\lambda \cdot (p,q) = (\lambda p, \lambda^{-1} q)$, a symplectic form $dp \wedge dq$, and a moment map $(p,q) \mapsto pq$ whose zero fiber is an infinite chain of projective lines. This too can be regarded as the toric variety associated to a fan of infinite type in the plane. It appears to be another way of describing Ekedahl's example.</p>