User charley - MathOverflow

Answer by Charley for are deformations of torsion modules always torsion?

2009-10-06T16:24:06Z

[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.]

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.

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.

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'.

I haven't checked that M is flat over k[h], but the definition of M certainly suggests that this should be the case.

Answer by Charley for What is a topos?

2009-10-05T04:58:08Z

There are two concepts which both get called a topos, so it depends on who you ask. The more basic notion is that of an elementary topos, which can be characterized in several ways. The simple definition:

An elementary topos is a category C which has finite limits and power objects.

(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").

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 Sheaves in Geometry and Logic. The main reference work is Johnstone's as-yet-unfinished Sketches of an Elephant, but I certainly wouldn't start there.

The other major notion of topos is that of a Grothendieck topos, which is the category of sheaves of sets on some site (a site is a (decently nice) category with a structure called a Grothendieck topology 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 Sheaves in Geometry and Logic for more info, or you might check out Grothendieck's SGA4 for the algebraic geometry take on things.

Answer by Charley for What is the universal property of normalization?

2009-10-01T20:03:54Z

I've realized that my answer is wrong. Here's the right answer: if Z is a normal scheme and f: Z -- > X is a morphism such that each associated point of Z maps to an associated point of X, then f factors through n.

A counterexample that shows why what I said previously doesn't work: let f be the inclusion of the node into the nodal curve. There is no unique lift of f to the normalization of the nodal curve.

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.

Answer by Charley for What is the universal property of normalization?

2009-10-01T15:53:52Z

Normalization is right adjoint to the inclusion functor from the category of normal schemes into the category of reduced schemes. In other words, if n: Y --> X is the normalization of X and f: Z --> X is any morphism where Z is a normal scheme, then f factors uniquely though n.

Answer by Charley for If \Omega_X/Y is locally free of rank dim(X)-dim(Y), is X->Y smooth?

2009-10-01T15:38:37Z

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.

The differential-wise condition you want is this: for the morphism morphism f: X --> Y to be smooth, you need that the sequence

0 --> f^* Omega_Y --> Omega_X --> Omega_X/Y --> 0

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 dim X = dim Y, Omega_X/Y is 0 if and only if f is unramified. But in this case f^* Omega_Y --> Omega_X can still fail to be injective.

Comment by Charley

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 $d_1 \cdots d_r \cdot \text{deg}(X) \neq 0$.

Comment by Charley

2009-10-12T15:13:57Z

I see what's going on now. I misinterpreted "torsion" as "not torsion-free."

Comment by Charley

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].