Are schemes that "have enough locally frees" necessarily separated - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:07:58Z http://mathoverflow.net/feeds/question/25122 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25122/are-schemes-that-have-enough-locally-frees-necessarily-separated Are schemes that "have enough locally frees" necessarily separated Ariyan Javanpeykar 2010-05-18T13:35:13Z 2010-05-19T18:05:46Z <p>Let me motivate my question a bit. </p> <p><strong>Thm</strong>. Let $X$ be a locally noetherian finite-dimensional regular scheme. If $X$ has enough locally frees, then the natural homomorphism $K^0(X)\longrightarrow K_0(X)$ is an isomorphism.</p> <p>A locally noetherian scheme has enough locally frees if every coherent sheaf is the quotient of a locally free coherent sheaf, $K^0(X)$ denotes the Grothendieck group of vector bundles on $X$ and $K_0(X)$ denotes the Grothendieck group of coherent sheaves on $X$.</p> <p>The above theorem is shown as follows.</p> <p>By the regularity (and finite-dimensionality!) of $X$, we can construct a finite resolution by a standard procedure. (Surject onto the kernel at each stage with a vector bundle.) Then the "Euler characteristic" associated to this resolution is inverse to the natural morphism.</p> <p>Now, I was looking through the literature (Weibel's book basically) and I saw that this theorem appears with the additional condition of separability. (Edit: This is not necessary. The point is that noetherian schemes that have enough locally frees are semi-separated.)</p> <p><strong>Example</strong>. Take the projective plane $X$ with a double origin. Then $K^0(X) \cong \mathbf{Z}^3$ whereas $K_0(X) \cong \mathbf{Z}^4$. </p> <p><strong>Example</strong>. Take the affine plane $X$ with a double origin. Then <code>$K^0(X) \cong \mathbf{Z}$</code>, whereas <code>$K_0(X) \cong \mathbf{Z}\oplus \mathbf{Z}$</code>.</p> <p>So I figured I must be missing something...Thus, I ask:</p> <p><strong>Q</strong>. Are locally noetherian schemes that have enough locally frees separated? </p> <p><strong>EDIT</strong>.</p> <p>The answer to the above question is "No" as the example by Antoine Chambert-Loir shows. </p> <p>From Philipp Gross's answer, we conclude that a noetherian scheme which has enough locally frees is semi-separated. This means that, for every pair of affine open subsets $U,V\subset X$, it holds that $U\cap V$ is affine. Note that separated schemes are semi-separated and that Antoine's example is also semi-separated.</p> <p>Taking a look at Totaro's article cited by Philipp Gross, we see that a regular noetherian scheme which is semi-separated has enough locally frees. (Do regular semi-separated and locally noetherian schemes have enough locally frees?)</p> <p>This was (in a way) also remarked by Hailong Dao. He mentions the result of Kleiman and independently Illuzie. Recently, Brenner and Schroer observed that their proof works also with $X$ noetherian semi-separated locally $\mathbf{Q}$-factorial. See page 4 of Totaro's paper. In short, separated is not really needed but semi-separated is.</p> <p>Thus, we can conclude the following.</p> <p>Suppose that $X$ is a regular and finite-dimensional scheme.</p> <p>If $X$ has enough locally frees, then <code>$K^0(X) \longrightarrow K_0(X)$</code> is an isomorphism. For example, $X$ is noetherian and semi-separated.</p> <p>Anyway, thanks to everybody for their answers. They helped me alot!</p> http://mathoverflow.net/questions/25122/are-schemes-that-have-enough-locally-frees-necessarily-separated/25149#25149 Answer by ACL for Are schemes that "have enough locally frees" necessarily separated ACL 2010-05-18T15:49:02Z 2010-05-18T15:49:02Z <p>This doesn't answer your question but it is may be worth noticing that any vector bundle $E$ on the affine plane $X$ with doubled origine is trivial. Indeed, the inverse image of $E$ via the two natural maps $u_i\colon A^2\rightarrow X$ are vector bundles on $A^2$, so are trivial. The glueing condition on $X\setminus{o_1,o_2}$ ($o_1,o_2$ are the two origins) is an automorphism of the trivial line bundle on $A^2\setminus{o}$, hence extends to an automorphism on $A^2$ by Hartogs. This implies that the initial vector bundle is trivial.</p> <p>By the way, the affine <em>line</em> with doubled origin certainly has enough locally frees...</p> http://mathoverflow.net/questions/25122/are-schemes-that-have-enough-locally-frees-necessarily-separated/25184#25184 Answer by Hailong Dao for Are schemes that "have enough locally frees" necessarily separated Hailong Dao 2010-05-18T23:11:26Z 2010-05-19T12:23:34Z <p>Actually the implication should be reversed: a separated regular Noatherian scheme has enough locally frees (this is Exercise 6.8, Chapter III Hartshorne). So the hypothesis is certainly needed for the proof. </p> <p>EDIT: The statement in Hartshorne assumes X is integral, but this is not needed: see SGA 6, 2.2.3, 2.2.4, 2.2.5 and 2.2.7.1 (page 168-172 <a href="http://www.msri.org/publications/books/sga/sga/6/index.html" rel="nofollow">here</a> ). In particular, you need separatedness and locally factorial (which follows from regular) to show that any coherent sheaf is a quotient of a direct sum of line bundles (for precise statement and example see 2.2.6 and 2.2.6.1). </p> <p>In summary, the statement of the theorem is : For $X$ a regular, Noetherian, separated scheme one has $K_0(X) \cong K^0(X)$. </p> <p>The answer to your question is no, as pointed out by Antoine. </p> http://mathoverflow.net/questions/25122/are-schemes-that-have-enough-locally-frees-necessarily-separated/25224#25224 Answer by Philipp Gross for Are schemes that "have enough locally frees" necessarily separated Philipp Gross 2010-05-19T12:34:24Z 2010-05-19T12:34:24Z <p>The property that every coherent sheaf admits a surjection from a coherent locally free sheaf is also known as the <em>resolution property</em>.</p> <p>The theorem can be refined as follows:</p> <p>Every noetherian, locally $\mathbb Q$-factorial scheme with affine diagonal (equiv. semi-semiseparated) has the resolution property (where the resolving vector bundles are made up from line bundles).</p> <p>This is Proposition 1.3 of the following paper:</p> <p><em>Brenner, Holger; Schröer, Stefan Ample families, multihomogeneous spectra, and algebraization of formal schemes. Pacific J. Math. 208 (2003), no. 2, 209--230.</em> </p> <p>You will find a detailed discussion of the resolution property in</p> <p><em>Totaro, Burt. The resolution property for schemes and stacks. J. Reine Angew. Math. 577 (2004), 1--22</em></p> <p>Totaro proves in Proposition 3.1. that every scheme (or algebraic stack with affine stabilizers) has affine diagonal if it satisfies the resolution property.</p> <p>The converse is also true for smooth schemes:</p> <p><strong>Proposition 8.1</strong> : Let $X$ be a smooth scheme of finite type over a field. Then the following are equivalent:</p> <ol> <li>$X$ has affine diagonal.</li> <li>X has the resolution property.</li> <li>The natural map $K_0^{naive} \to K_0$ is surjective.</li> </ol>