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When $X$ is a nonsingular variety over a field $k$ and Z is a closed nonsingular subvariety, it is known that the conormal sequence $$ 0\to\mathscr{I}/\mathscr{I}^2\to \mathscr{O}_Z \otimes_{\mathscr{O}_X}\Omega_X\to\Omega_Z \to 0 $$ is short exact as $\mathscr{O}_Z$-module. When the condition that $X$ and $Z$ are nonsingular is dropped, it is only assured to be right exact. I wanted to know what is happening when $X$ is singular. For simplicity, I only tried when $X$ is affine.

  1. $X=\mathrm{Spec}\ k[x,y,z,w]/(xy-zw)$, $I=(x,z)$, $Z=\mathrm{Spec}\ k[y,w]$. I wrote $R=k[x,y,z,w]/(xy-zw)$, $S=R/I$.

The computation is as followed.

  • $I/I^2= \frac{Sx\oplus Sw}{xy=zw}$
  • $S\otimes \Omega_R = \frac{Sdx\oplus Sdy\oplus Sdz \oplus Sdw}{ydx=wdz}$
  • $\Omega_S = Sdy \oplus Sdw$

It seems that the sequence is still left exact.

  1. $X=\mathrm{Spec}\ k[x,y,z]/(xz-y^2)$, $I=(x,y)$, $Z=\mathrm{Spec}\ k[z]$. I wrote $R=k[x,y,z]/(xz-y^2)$, $S=R/I$.
  • $I/I^2 = kx \oplus ky \oplus kzy \oplus kz^2 y \oplus \cdots = kx \oplus Sy $
  • $S\otimes \Omega_R = kdx \oplus Sdy \oplus Sdz$
  • $\Omega_S = Sdz$

It again seems the sequence is still left exact.

I first thought those sequences (or at least one of them) should not be left exact, but in my calculation, it was not the case. I debt my calculation is wrong, but I could not find anything wrong. Could you tell me where is wrong in my calculation? Or is this really right?

Thanks in advance.

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  • $\begingroup$ Your examples are correct, and point to a more general pattern: both are hypersurface singularities, and thus $I/I^2$ is locally free (of rank 1). Since the map $I/I^2\to\Omega_X |_Z$ is generically injective when $Z$ is reduced (since $Z$ is generically nonsingular), the kernel must be torsion; if $I/I^2$ is torsionfree (e.g., because it's locally free), the kernel is then 0. The same thing will happen for complete intersection rings, again because $I/I^2$ is locally free of rank the codimension of $Z$. $\endgroup$
    – Devlin Mallory
    Commented Jan 8, 2021 at 3:48
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    $\begingroup$ @DevlinMallory But isn't $x\in I/I^2$ in the second example a torsion since $zx=y^2=0$, viewed as $k[z]$-module? What am I missing? $\endgroup$
    – J1U
    Commented Jan 8, 2021 at 3:55
  • $\begingroup$ $x$ does not lie in $I/I^2$; $I/I^2$ is (freely) generated by the generator of $I$, i.e., $y^2-xz$. $\endgroup$
    – Devlin Mallory
    Commented Jan 8, 2021 at 4:40
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    $\begingroup$ @DevlinMallory But here $I$ is $(x,y)$. I am considering the case that the ambient space itself is singular, $X=\mathrm{Spec} k[x,y,z]/(xz-y^2)$, and $Z$ is the zero set of $(x,y)$. $\endgroup$
    – J1U
    Commented Jan 8, 2021 at 5:07
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    $\begingroup$ Oh my apologies, I totally misread! I don’t know what one can say in this case, as there will often be torsion in the differentials... hopefully someone else knows more. $\endgroup$
    – Devlin Mallory
    Commented Jan 8, 2021 at 5:53

1 Answer 1

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It is easier to think in terms of smoothness instead of nonsingularity (by Grothendieck's EGA 0$_{IV}$ 22.5.8, both comcepts are the same if the base field $k$ is perfect). Then EGA 0$_{IV}$ 22.6.1, 22.6.2, etc. will let you see what is happening. A less complete reference but including the cited results is Matsumura, Commutative Ring Theory, section 28.

If you want to know what is at the left of this sequence in the non-smooth case, the reference is EGA 0$_{IV}$ 20.6.22, or for a more exhaustive study (a whole long homology exact sequence whose last three terms are the ones of your exact sequence) the keyword is "cotangent complex" or "André-Quillen homology". You will can find this long exact sequence for example in the book M. André, Homologie des Algèbres Commutatives, Théorème 5.1.

These references are very general. They do not require any finiteness hypothesis and that is the reason for taking into account the topologies. If you are interested only in schemes of finite type over a field, there are simpler sources, for instance Bourbaki, Algèbre Commutative, chapitre X.

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