most recent 30 from http://mathoverflow.net 2013-05-19T23:33:20Z http://mathoverflow.net/feeds/question/41 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41/if-omega-x-y-is-locally-free-of-rank-dimx-dimy-is-x-y-smooth Anton Geraschenko 2009-09-30T20:14:16Z 2009-10-01T15:38:37Z

<p>Suppose I have a morphism f:X&rarr;Y such that the relative sheaf of differentials &Omega;_X/Y is locally free. Does it follow that f is smooth?</p> <p>The answer is no, but for a silly reason. You could have some non-reducedness (Spec(k[e]/(e^2)) over Spec(k) has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of &Omega;_X/Y is dim(X)-dim(Y)?</p> <p><strong>Edit:</strong> As Jonathan points out in his answer, I was careless with my counterexample. It only works if char(k)=2.</p>

http://mathoverflow.net/questions/41/if-omega-x-y-is-locally-free-of-rank-dimx-dimy-is-x-y-smooth/43#43 Ishaidc 2009-10-01T01:16:23Z 2009-10-01T01:16:23Z

<p>Let X = spec A be an affine integral scheme of dimension one which is not smooth. Smoothness may be checked smooth-locally on the source (given U --> V if there exists a W --> U which is smooth and surjective such that W --> V is smooth then U --> V was smooth). Thus, the normalization X~ = spec A~ cannot be smooth over X. But if d: A~ --> M is any derivation of A~ over A and a/b is an element of A~ then d(a/b)= (bda - adb)/(b^2)= 0; this shows that the rel. differentials are zero, hence in particular loc free of fin rank.</p>

http://mathoverflow.net/questions/41/if-omega-x-y-is-locally-free-of-rank-dimx-dimy-is-x-y-smooth/49#49 Jonathan Wise 2009-10-01T04:49:31Z 2009-10-01T05:43:54Z

<p>A variation on Ishai's example is a closed embedding: its sheaf of relative differentials is <code>0</code>, hence free of finite rank, even though it needn't be smooth.</p> <p>However, <code>k[e] / e^2</code> over <code>k</code> is not actually a counterexample (except in characteristic <code>2</code>). The module of relative differentials of <code>Spec k[e] / e^2</code> over <code>Spec k</code> is not free if the characteristic of <code>k</code> is not 2. Let <code>A = k[e]</code> and <code>B = k[e] / e^2</code>. Then </p> <pre><code>Omega_B = Omega_A (x) B / d(e^2) = k[e] / (e^2, 2e) </code></pre> <p>via the isomorphism <code>Omega_A --&gt; A : dt --&gt; 1</code>. This is not isomorphic to <code>B</code> unless <code>2 = 0</code>.</p> <p>On the other hand, you can conclude that <code>B</code> is smooth if its cotangent complex is a vector bundle in degree 0. In the case of <code>k[e] / e^2</code>, the cotangent complex is </p> <pre><code>[ I_{B/A} / I_{B/A}^2 ---&gt; Omega_A (x) B ] = [ e^2 A / e^4 A ---&gt; B de ] </code></pre> <p>in degrees <code>[-1,0]</code> and the differential is the universal derivation. (I write <code>I_{B/A}</code> for the ideal of <code>B</code> in <code>A</code>.) Even in characteristic <code>2</code>, the differential has a kernel, so the cotangent complex is not concentrated in degree <code>0</code>.</p>

http://mathoverflow.net/questions/41/if-omega-x-y-is-locally-free-of-rank-dimx-dimy-is-x-y-smooth/52#52 Charley 2009-10-01T15:38:37Z 2009-10-01T15:38:37Z

<p>I think Ishai's example is close, but one must be a little careful; the normalization of the node is a good example, but the normalization of the cusp is ramified, and the sheaf of relative differentials in that case is not even locally free.</p> <p>The differential-wise condition you want is this: for the morphism morphism <code>f: X --&gt; Y</code> to be smooth, you need that the sequence</p> <pre><code>0 --&gt; f^* Omega_Y --&gt; Omega_X --&gt; Omega_X/Y --&gt; 0 </code></pre> <p>be exact and locally split (I can't find a reference that says this is sufficient, so it may not be). In the special case when <code>dim X = dim Y</code>, <code>Omega_X/Y</code> is 0 if and only if <code>f</code> is unramified. But in this case <code>f^* Omega_Y --&gt; Omega_X</code> can still fail to be injective.</p>