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Chapter 3: Problem 3

A morphism \(f: X \rightarrow Y\) of schemes of finite type over \(k\) is 茅tale if it is smooth of relative dimension 0. It is unramified if for every \(x \in X\), letting \(y=f(x)\), we have \(\mathrm{m}_{\mathrm{y}} \cdot \mathscr{C}_{x}=\mathrm{m}_{\mathrm{x}},\) and \(k(x)\) is a separable algebraic extension of \(k\left(y^{\prime}\right) .\) Show that the following conditions are equivalent: (i) \(f\) is 茅tale (ii) \(f\) is flat, and \(\Omega_{X, Y}=0\) (iii) \(f\) is flat and unramified.

Short Answer

Expert verified
The equivalence of these three conditions comes down to understanding and applying the different properties of each condition. \(f\) being 茅tale, flat with \(\Omega_{X, Y} = 0\), and flat and unramified are equivalent under the rules of scheme theory in algebraic geometry.

Step by step solution

01

Proving (i) implies (ii)

Assume \(f\) is 茅tale, which by definition, means it is smooth of relative dimension 0. This implies \(f\) is flat and its sheaf of relative differentials \(\Omega_{X/Y}\) is the zero sheaf since the definion of a smooth morphism tells us that \(\Omega_{X/Y}\) is free of rank equal to the relative dimension, which in this case is 0.
02

Proving (ii) implies (iii)

Assume \(f\) is flat and \(\Omega_{X, Y} = 0\). Flatness is given by assumption. To prove \(f\) is unramified, we need to check the two properties in the definition. In the stalk at each point \(x\), we have \(m_yf'^*(\mathscr{C}_x) = m_x\) since \(m_{y}f'^{+}\) can be thought of as \(\Omega_{X, Y}\). However, we know \(\Omega_{X, Y} = 0\). On the other hand, the fiber \(k(x)\) is a geometrically regular extension of \(k\left(y'\right)\), which by definition is separable. Therefore, \(f\) is unramified.
03

Proving (iii) implies (i)

Assume \(f\) is flat and unramified. To show that \(f\) is 茅tale, we need to prove it is smooth of relative dimension 0. We know that \(f\) is flat. Given the condition in the definition of being unramified, we can conclude the sheaf of differentials \(\Omega_{X/Y}\) is zero again. Hence, \(f\) is smooth of relative dimension 0. Consequently, \(f\) is 茅tale.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Flat Morphism
A flat morphism is a type of morphism between schemes that maintains the property of flatness. In algebraic geometry, the notion of flatness is crucial. It signifies that the structure maps between the schemes preserve certain 'noetherian' properties, ensuring that the fibers vary nicely in family as we move along the base scheme.
Flatness is particularly fundamental when considering families of schemes, as it guarantees that the dimension of fibers is consistent across all points. Here's a simple way to think about flatness:
  • Flat morphisms preserve the number of dimension across fibers. This means as you change the location or point in the base scheme, the complexity or shape of the fiber remains stable.
  • They're valuable when studying parametrized sets, such as families of algebraic curves or surfaces.
  • Flat morphisms serve as an underlying property for reasons ranging from simple continuity to deeper geometric properties.
This property has implications in not just geometry but also in understanding the cohomology of schemes and various algebraic structures over commutative rings.
Smooth Morphism
A smooth morphism can be intuitively understood as a morphism behaving like a smooth function. In the realm of algebraic geometry, smooth morphisms share qualities with differentiable functions in calculus, being the geometric analog to a function comprised of polynomials.
When we describe a morphism as smooth, we generally mean:
  • It preserves the dimensionality while allowing the scheme to be locally modeled by smooth, even functions or modules.
  • In the context of schemes, this smoothness involves the vanishing of certain ideals: the sheaf of differentials.
  • The relative dimension is small, often zero, for 茅tale morphisms, which indeed makes them part of smooth morphisms.
This property means morphisms are free of singularities in the local neighborhood, providing great utility in proving various approximation theorems as well as offering insight when resolving singularities.
Unramified Morphism
An unramified morphism is another specific kind of morphism between schemes. These morphisms resemble the concept of an unbranched cover found in topology. In algebraic geometry, being unramified means that at a local level, morphisms exhibit behavior analogous to separable extensions of algebraically closed fields.
Important features of unramified morphisms include:
  • They have the property that the differentials vanish, indicating no 'branch over' or 'ramify' more than expected over the base.
  • These morphisms ensure that the fibers at all points are neatly intertwined without overlap or ramification.
  • A connection exists between unramified, flat, and smooth morphisms, particularly when the morphism also happens to be flat and gives a vanishing sheaf of differentials, establishing that the morphism is etale.
This significance is fundamental in the classification of morphisms as it helps in understanding how geometrically clean and uniform the map behaves across diverse sections of the base.

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Most popular questions from this chapter

Let \(\left(X, \mathcal{O}_{X}\right)\) be a ringed space, and let \(\mathscr{F}^{\prime}, \mathscr{F}^{\prime \prime} \in \mathbb{M}_{\mathrm{O}} \mathfrak{d}(X) .\) An extension of \(\mathscr{F}^{\prime \prime} \mathrm{by}\) \(\mathscr{F}^{\prime}\) is a short exact sequence $$0 \rightarrow \mathscr{F}^{\prime} \rightarrow \mathscr{F} \rightarrow \mathscr{F}^{\prime \prime} \rightarrow 0$$ in \(\operatorname{Mod}(X) .\) Two extensions are isomorphic if there is an isomorphism of the short exact sequences, inducing the identity maps on \(\mathscr{F}^{\prime}\) and \(\mathscr{F}^{\prime \prime} .\) Given an extension as above consider the long exact sequence arising from Hom \(\left(\mathscr{F}^{\prime \prime}, \cdot\right),\) in particular the map $$\delta: \operatorname{Hom}\left(\mathscr{F}^{\prime \prime}, \mathscr{F}^{\prime \prime}\right) \rightarrow \operatorname{Ext}^{1}\left(\mathscr{F}^{\prime \prime}, \mathscr{F}^{\prime}\right)$$ and let \(\xi \in \operatorname{Ext}^{1}\left(\mathscr{F}^{\prime \prime}, \mathscr{F}^{\prime}\right)\) be \(\delta\left(1_{\mathscr{F}^{\prime \prime}}\right) .\) Show that this process gives a one-to-one correspondence between isomorphism classes of extensions of \(\mathscr{F}^{\prime \prime}\) by \(\mathscr{F}^{\prime},\) and elements of the group \(\operatorname{Ext}^{1}\left(\mathscr{F}^{\prime \prime}, \mathscr{F}^{\prime}\right) .\) For more details, see, e.g., Hilton and Stammbach [1, Ch. III].

Let \(X=\mathbf{P}_{k}^{1},\) with \(k\) an infinite field. (a) Show that there does not exist a projective object \(\mathscr{P} \in \mathfrak{M}\) od \((X)\), together with a surjective map \(\mathscr{P} \rightarrow \mathscr{O}_{X} \rightarrow 0 .\) [Hint: Consider surjections of the form \(\mathscr{O}_{V} \rightarrow\) \(k(x) \rightarrow 0,\) where \(x \in X\) is a closed point, \(V\) is an open neighborhood of \(x\) and \(\mathscr{O}_{V}=j_{!}\left(\left.\mathcal{O}_{X}\right|_{V}\right),\) where \(j: V \rightarrow X\) is the inclusion. (b) Show that there does not exist a projective object \(\mathscr{P}\) in either \(\mathbb{Z} \operatorname{co}(X)\) or \(\operatorname{Cob}(X)\) together with a surjection \(\mathscr{P} \rightarrow \mathscr{O}_{X} \rightarrow 0 .\) [Hint: Consider surjections of the form \(\mathscr{L} \rightarrow \mathscr{L} \otimes k(x) \rightarrow 0,\) where \(x \in X\) is a closed point, and \(\mathscr{L}\) is an invertible sheaf on \(X .]\)

A Nonprojectire Scheme. We show the result of \((\mathrm{Ex} .5 .8)\) is false in dimension 2 Let \(k\) be an algebraically closed field of characteristic \(0,\) and let \(X=\mathbf{P}_{k}^{2}\). Let \(w\) be the sheaf of differential 2 -forms (II, se). Define an infinitesimal extension \(X\) of \(X\) by \((\cdot)\) by giving the element \(\xi \in H^{1}(X,(\cdot) \otimes . \mathscr{J})\) defined as follows (Ex. 4.10 ). Let \(x_{0}, x_{1}, x_{2}\) be the homogeneous coordinates of \(X\). let \(U_{0} . \ell_{1}, \ell_{2}\), be the standard open covering. and let \(\check{\xi}_{1,1}=\left(x_{1}, x_{1}, d / x_{i}, x_{1}\right) .\) This gives a Cech 1 -cocycle with values in \(\Omega_{x}^{1}\). and since dim \(X=2\), we have \((\cdot) \otimes . \bar{J} \cong \Omega^{1}(\mathrm{II}, \mathrm{E} \times .5 .16 \mathrm{b}) .\) Now use the exact sequence $$\ldots \rightarrow H^{1}\left(X,_{(\cdot))} \rightarrow \operatorname{Pic} X^{\prime} \rightarrow \operatorname{Pic} X \stackrel{\bullet}{\rightarrow} H^{2}(X,(\cdot)) \rightarrow \ldots\right.$$ of \((\mathrm{Ex} .4 .6)\) and show \(\delta\) is injective. We have ()\(\cong\left(x^{\prime}-3\right)\) by \((11,8.20 .1) .\) so \(H^{2}(X, \omega) \cong k .\) since char \(h=0,\) you need only show that \(\delta(C(1)) \neq 0,\) which can be done by calculating in Cech cohomology. since \(H^{1}(X, \omega)=0,\) we see that Pic \(X^{\prime}=0 .\) In particular, \(X^{\prime}\) has no ample invertible sheaves, so it is not projective. Note. In fact, this result can be generalized to show that for any nonsingular projective surface \(X\) over an algebraically closed field \(k\) of characteristic 0 , there is an infinitesimal extension \(X^{\prime}\) of \(X\) by \(\omega,\) such that \(X^{\prime}\) is not projective over \(k\) Indeed, let \(D\) be an ample divisor on \(X\). Then \(D\) determines an element \(c_{1}(D) \in\) \(H^{1}\left(X, \Omega^{1}\right)\) which we use to define \(X^{\prime},\) as above. Then for any divisor \(E\) on \(X\) one can show that \(\delta(\mathscr{P}(E))=(D . E) .\) where \((D . E)\) is the intersection number (Chapter \(\mathrm{V}\) ), considered as an element of \(k .\) Hence if \(E\) is ample, \(\delta\left(\mathscr{L}^{\prime}(E)\right) \neq 0 .\) Therefore \(X^{\prime}\) has no ample divisors. On the other hand, over a field of characteristic \(p>0,\) a proper scheme \(X\) is projective if and only if \(X_{\text {rad }}\) is!

Let \(X\) be a projective scheme over a field \(k,\) and let \(\overline{\mathscr{I}}\) be a coherent sheaf on \(X\) We define the Euler characteristic of \(\mathscr{F}\) by $$\chi(\mathscr{F})=\sum(-1)^{i} \operatorname{dim}_{k} H^{i}(X, \overline{\mathscr{F}})$$ If$$0 \rightarrow \mathscr{F}^{\prime} \rightarrow \mathscr{F} \rightarrow \mathscr{F}^{\prime \prime} \rightarrow 0$$ is a short exact sequence of coherent sheaves on \(X,\) show that \(\chi(\mathscr{F})=Y\left(\mathscr{F}^{\prime}\right)+\) \(x\left(\mathscr{F}^{\prime \prime}\right)\)

On an arbitrary topological space \(X\) with an arbitrary abelian sheaf \(\mathscr{F},\) Cech cohomology may not give the same result as the derived functor cohomology. But here we show that for \(H^{1},\) there is an isomorphism if one takes the limit over all coverings. (a) Let \(\mathfrak{U}=\left(U_{i}\right)_{i e I}\) be an open covering of the topological space \(X\). A refinement of \(\mathbb{U}\) is a covering \(\mathfrak{B}=\left(V_{j}\right)_{\text {je } J}\), together with a \(\operatorname{map} \lambda: J \rightarrow I\) of the index sets, such that for each \(j \in J, V_{j} \subseteq U_{\lambda(j)} .\) If \(\mathfrak{B}\) is a refinement of \(\mathbb{U}\), show that there is a natural induced map on Cech cohomology, for any abelian sheaf \(\mathscr{F},\) and for each \(i\) \\[ \lambda^{i}: \check{H}^{i}(\mathbf{U}, \mathscr{F}) \rightarrow \check{H}^{i}(\mathfrak{Y}, \mathscr{F}) \\] The coverings of \(X\) form a partially ordered set under refinement, so we can consider the Cech cohomology in the limit \\[ \varliminf_{u} \check{H}^{i}(\mathfrak{U}, \mathscr{F}) \\] (b) For any abelian sheaf \(\mathscr{F}\) on \(X,\) show that the natural maps (4.4) for each covering \\[ \check{H}^{i}(\mathbf{u}, \mathscr{F}) \rightarrow H^{i}(X, \mathscr{F}) \\] are compatible with the refinement maps above (c) Now prove the following theorem. Let \(X\) be a topological space, \(\mathscr{F}\) a sheaf of abelian groups. Then the natural map \\[ \varliminf_{u} \check{H}^{1}(\mathfrak{U}, \mathscr{F}) \rightarrow H^{1}(X, \mathscr{F}) \\] is an isomorphism. [Hint: Embed \(\mathscr{F}\) in a flasque sheaf \(\mathscr{G},\) and let \(\mathscr{R}=\mathscr{G} / \mathscr{F}\) so that we have an exact sequence \\[ 0 \rightarrow \mathscr{F} \rightarrow \mathscr{G} \rightarrow \mathscr{R} \rightarrow 0 \\] Define a complex \(D^{\prime}(\mathbb{Z})\) by \\[ 0 \rightarrow C^{\prime}(\mathbf{U}, \mathscr{F}) \rightarrow C^{\prime}(\mathbf{U}, \mathscr{G}) \rightarrow D^{\prime}(\mathbf{U}) \rightarrow 0 \\] Then use the exact cohomology sequence of this sequence of complexes, and the natural map of complexes \\[ D^{\prime}(\mathfrak{U}) \rightarrow C^{\prime}(\mathfrak{U}, \mathscr{R}) \\] and see what happens under refinement.
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