User charley - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:37:15Z http://mathoverflow.net/feeds/user/37 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/138/are-deformations-of-torsion-modules-always-torsion/140#140 Answer by Charley for are deformations of torsion modules always torsion? Charley 2009-10-06T16:24:06Z 2009-10-06T16:24:06Z <p>[I'm going to work over k[h] as the base instead; I don't think anything changes, but if I'm wrong you should let me know.]</p> <p>Consider the case M = k[s,t,h]/(st-h^2). Setting h=0 yields M_0 = k[s,t]/(st), which is certainly torsion over k[t]. But inverting h yields M' = k[s,t,h,h^{-1}]/(st-h^2). This is a (Z-)graded k[t]-module, so to show it is torsion-free it suffices to show that there are no homogeneous torsion elements.</p> <p>Let p be any element of k[t] and suppose that pg = f*(st-h^2) for some f in k[s,t,h,h^{-1}] and some homogeneous element g in k[s,t,h,h^{-1}]. Again by homogeneity, we can assume that both f and p are homogeneous, so in particular p = t^k for some k.</p> <p>But now, we have g * t^k = f * (st-h^2). The ring k[s,t,h,h^{-1}] is a UFD, so either t|f or t|(st-h^2). The latter is false; every degree-1 element of this ring looks like (as + bt + ch)(p(s/h, t/h)), and clearing denominators, to have a solution to pt = st-h^2 would be to have p(s,t)(as + bt + ch)t = h^j(st-h^2), which can't happen since the left-hand-side doesn't have any terms of degree > 1 in h. Hence t|f, so repeating this argument, t^k | f. Dividing both sides by t^k shows that g is divisible by (st-h^2), so g=0 as an element of M'.</p> <p>I haven't checked that M is flat over k[h], but the definition of M certainly suggests that this should be the case.</p> http://mathoverflow.net/questions/101/what-is-a-topos/108#108 Answer by Charley for What is a topos? Charley 2009-10-05T04:58:08Z 2009-10-05T05:03:09Z <p>There are two concepts which both get called a <em>topos</em>, so it depends on who you ask. The more basic notion is that of an <em>elementary topos</em>, which can be characterized in several ways. The simple definition:</p> <pre><code>An elementary topos is a category C which has finite limits and power objects. </code></pre> <p>(A power object for A is an object P(A) such that morphisms B --> P(A) are in natural bijection with subobjects of A x B, so we could rephrase the condition "C has power objects" as "the functor Sub(A x -) is representable for every object A in C").</p> <p>The issue with the simple definition is that it doesn't show you why these things are actually interesting. It turns out that a great deal follows from these axioms. For example, C also has finite colimits, exponential objects, has a representable limit-preserving functor P: C^op --> Doct where Doct the category of Heyting algebras such that if f: AxB --> A is the projection map for some objects A and B in C, then P(A) --> P(AxB) has both left and right adjoints considered as a morphism of Heyting algebras, etc etc. What the long-winded definition boils down to is "an elementary topos the the category of types in some world of intuitionistic logic." There's an incredible amount of material here; the best place to start is probably MacLane and Moerdijk's <em>Sheaves in Geometry and Logic.</em> The main reference work is Johnstone's as-yet-unfinished <em>Sketches of an Elephant,</em> but I certainly wouldn't start there.</p> <p>The other major notion of <em>topos</em> is that of a <em>Grothendieck topos</em>, which is the category of sheaves of sets on some site (a site is a (decently nice) category with a structure called a <em>Grothendieck topology</em> which generalizes the notion of "open cover" in the category of open sets in a topological space). Grothendieck topoi are elementary topoi, but the converse is not true; Giraud's Theorem classifies precisely the conditions needed for an elementary topos to be a Grothendieck topos. Depending on your point of view, you might also look at <em>Sheaves in Geometry and Logic</em> for more info, or you might check out Grothendieck's SGA4 for the algebraic geometry take on things.</p> http://mathoverflow.net/questions/46/what-is-the-universal-property-of-normalization/56#56 Answer by Charley for What is the universal property of normalization? Charley 2009-10-01T20:03:54Z 2009-10-01T20:03:54Z <p>I've realized that my answer is wrong. Here's the right answer: if <code>Z</code> is a normal scheme and <code>f: Z -- &gt; X</code> is a morphism such that each associated point of <code>Z</code> maps to an associated point of <code>X</code>, then <code>f</code> factors through <code>n</code>.</p> <p>A counterexample that shows why what I said previously doesn't work: let <code>f</code> be the inclusion of the node into the nodal curve. There is no unique lift of <code>f</code> to the normalization of the nodal curve.</p> <p>What's going on here: taking the total ring of fractions is not a functor for arbitrary morphisms of reduced rings. You need a morphism such that no NZD gets mapped to a ZD, which is equivalent (for Noetherian rings) to saying that the preimage of any associated prime is an associated prime.</p> http://mathoverflow.net/questions/46/what-is-the-universal-property-of-normalization/53#53 Answer by Charley for What is the universal property of normalization? Charley 2009-10-01T15:53:52Z 2009-10-01T15:53:52Z <p>Normalization is right adjoint to the inclusion functor from the category of normal schemes into the category of reduced schemes. In other words, if <code>n: Y --&gt; X</code> is the normalization of <code>X</code> and <code>f: Z --&gt; X</code> is any morphism where <code>Z</code> is a normal scheme, then <code>f</code> factors uniquely though <code>n</code>. </p> http://mathoverflow.net/questions/41/if-omega-x-y-is-locally-free-of-rank-dimx-dimy-is-x-y-smooth/52#52 Answer by Charley for If \Omega_X/Y is locally free of rank dim(X)-dim(Y), is X->Y smooth? 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> http://mathoverflow.net/questions/83075/what-is-the-probability-for-a-random-algebraic-cycle-to-be-homologically-trivial/83128#83128 Comment by Charley Charley 2011-12-10T19:06:41Z 2011-12-10T19:06:41Z Adding to this: in this particular case, say $X$ has dimension $k \leq r$ and your polynomials $f_1,\ldots,f_r$ have degrees $d_1, \ldots, d_r$. Let $H$ be the class of a hyperplane section on $X$. Then the cycle $Z$ that you construct pairs with $H^{k-r}$ to give the value $d_1 \cdots d_r \cdot \text{deg}(X) \neq 0$. http://mathoverflow.net/questions/138/are-deformations-of-torsion-modules-always-torsion/140#140 Comment by Charley Charley 2009-10-12T15:13:57Z 2009-10-12T15:13:57Z I see what's going on now. I misinterpreted &quot;torsion&quot; as &quot;not torsion-free.&quot; http://mathoverflow.net/questions/138/are-deformations-of-torsion-modules-always-torsion/140#140 Comment by Charley Charley 2009-10-07T04:16:49Z 2009-10-07T04:16:49Z A module over a P.I.D. is flat iff it is torsion-free. So the fact that no polynomial in k[h] divides (st-h^2) in k[s,t,h] immediately implies that M is flat over k[h].