I can't understand the proof of theorem 8.17 from Hartshorne's "Algebraic Geometry". Namely, he says that we have an exact sequence $$ 0 \to \mathcal J'/\mathcal J'^2 \to \Omega_{X/k} \otimes \mathcal O_{Y'} \to \Omega_{Y'/k} \to 0 $$ and there is $x_1, \dots,x_n$ such that $dx_1, \dots, dx_n$ generate a free subsheaf of rank $r$ in a neighborhood of $y \in Y$. It follows that $\Omega_{Y'/k}$ is locally free. Why?

  • $\begingroup$ Perhaps he means free summand, rather than just a free subsheaf? Looking at the proof of the theorem, the $x_i$ come from a locally split short exact sequence after all. $\endgroup$ – Jake Levinson Jun 25 '14 at 7:54
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    $\begingroup$ A possible source of confusion: he claims there are $r$ sections $x_1,\ldots, x_r$ such that $dx_1,\ldots, dx_r$ generate a rank $r$ free subsheaf. You've written $x_1,\ldots, x_n$, but $n$ is the dimension of $X$. $\endgroup$ – Andrew Jun 25 '14 at 16:44
  • $\begingroup$ Hartshorne starts the construction of $Y'$ with the sheaf $\Omega_{Y/k}$ from $Y$ which is locally free. Hence by construction $\Omega_{Y'/k} = \Omega_{Y/k}$ is locally free. $\endgroup$ – Jo Wehler Jan 30 '15 at 5:41
  • $\begingroup$ Can I offer you an alternative? X and Y are varieties so locally the rings are finitely generated $k$-algebras and you can use Prop 8.7. If you go to the stalks you get a map $${\frak{m}}\big/{\frak{m}}^2\rightarrow(\Omega\otimes{\mathcal{O}}_{Y})_{y}$$ and if you see it $\mod \frak{m}$ is an isomorphism. Thus the composition with the projection $(\Omega\otimes\mathcal{O}_Y)_y\rightarrow \Omega\otimes k$ is an isomorphism. From this follows that the first map is injective. $\endgroup$ – Ignacio Barros Mar 16 '15 at 9:43

It may be that, once I develop a better intuition, this part of the proof will seem natural, but for now, I find it implausible and incomprehensible. I want to propose an alternative for verifying that the relative conormal sequence is exact in the case of a nonsingular subvariety of a nonsingular variety (per the attendant definitions in H) using an argument that feels (to me at least) more substantive. (I owe a lot of this argument to my friend Drew.)

Surely we can agree that the question is local, and so we are in the setting of Proposition 16.12 in Eisenbud.

Proposition 16.12 (Eisenbud): If $\pi:S\rightarrow T$ is an epimorphism of $R$-algebras, with kernel $I$, then in the conormal sequence $$I/I^2\xrightarrow{d}T\otimes_S\Omega_{S/R}\xrightarrow{D\pi}\Omega_{T/R}\rightarrow0$$ the map $d$ is a split injection iff there is a map of $R$-algebras $\tau:T\rightarrow S/I^2$ splitting the projection map $S/I^2\twoheadrightarrow S/I=T$.

In fact, EGA IV Th. 20.5.12 establishes a canonical bijection between left inverses of $d$ and morphisms $\tau$.

Hence to achieve left-exactness of the relative conormal sequence, it is sufficient to establish the existence of such a map $\tau$. Hartshorne discusses the existence of a more general map in ex. II.8.6 under the title The Infinitesimal Lifting Property, but the vocabulary formally smooth is developed in EGA.

EGA IV Déf. 19.3.1 A morphism of algebras $A\rightarrow B$ is formally smooth (for the discrete topology) if, for all $A$-algebra $C$ and all nilpotent ideal $\mathfrak I$ of $C$, every $A$-homomorphism $u:B\rightarrow C/\mathfrak I$ factors as $B\xrightarrow{v}C\xrightarrow{\varphi}C/\mathfrak I$, where $\varphi$ is the canonical epimorphism. ($\Rightarrow$ $C$ satisfies the the infinitesimal lifting property.)

You will notice that in the Hartshorne ex. II.8.6, it is proved that if $A=k=k^{\mathrm{alg}}$ and $C$ is a finitely-generated $A$-algebra such that the associated affine scheme is nonsingular/$k$, then $C$ has the infinitesimal lifting property. Taking $R=k$, $B=S/I$, $C=S/I^2$, and $u$ the identity, we would establish the existence of a desired map $\tau$ and verify the left-exactness of the relative conormal sequence provided we could show that if $I$ corresponds to a nonsingular subvariety, $C=S/I^2$ has the infinitesimal lifting property. The Hartshorne ex. II.8.6 does this for us, but unfortunately it uses the result we are trying to prove. So instead, certainly it would suffice to show that for varieties, nonsingular implies formally smooth.

EGA IV defines (17.3.1) a morphism to be smooth if it is formally smooth and of locally finite presentation; since we are considering a morphism of varieties, a morphism is defined to be smooth if it is formally smooth. The notion of smooth morphism $A\rightarrow B$ is local on both (Rem. 17.3.2), and it is the same to say that a morphism is smooth, or that it is smooth at every point of the source (scheme).

We conclude with the desired remark, namely that a variety/$k$ is formally smooth iff it is a nonsingular variety. This is the content of EGA IV Cor. 17.5.2.


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