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I'm trying to understand the sheaf of relative differentials $\Omega_{X/Y}$ with a given morphism of schemes $f:X\to Y$.

To make things concrete, I want to consider the example of a parabola embedded in $\mathbb{A}^2$, so take $Y=\mathbb{A}^2 = \operatorname{spec}k[x,y]$ and $X=\operatorname{spec}k[x,y]/(y-x^2)$.

To simplify notation, let's set $A=k[x,y]$, $I=(y-x^2)$, and $\overline{A} = A/I$. My goal really is to compute the module $\Omega_{\overline{A}/A}$.

My understanding is that $\Omega_{\overline{A}/k} = (\overline{A}dx \oplus \overline{A}dy )/(dy-2xdx)$, which comes from the exact sequence $$ \begin{equation} I/I^2 \to \Omega_{A/k}\otimes_k \overline{A} \to \Omega_{\overline{A}/k}\to 0. \end{equation} $$ To calculate $\Omega_{\overline{A}/A}$, we can use the exact sequence $$ \Omega_{A/k}\otimes_A \overline{A} \to \Omega_{\overline{A}/k} \to \Omega_{\overline{A}/A}\to 0. $$ Since $\Omega_{A/k} = Adx \oplus Ady$, the above sequence becomes $$ \overline{A}dx \oplus \overline{A}dy \to (\overline{A}dx \oplus \overline{A}dy)/(dy-2xdx) \to \Omega_{\overline{A}/A}\to 0, $$ Now it seems that the left arrow is surjective, which would imply that $\Omega_{\overline{A}/A}=0$, and hence so is the corresponding sheaf $\Omega_{X/Y}$.

My questions are:

  1. Are my calculations correct?

  2. What does it mean geometrically that $\Omega_{X/Y}$ is zero? To make this question a bit clearer, is there anything about the geometric picture of a parabola embedded in $\mathbb{A}^2$ that would have lead me to expect $\Omega_{X/Y}=0$?

To clarify a bit more, I'm comfortable with $\Omega_{X/Y}$ when $Y=\operatorname{spec}k$, where $k$ is the ground field. It's when $Y$ is more complicated that I don't know how to think about $\Omega_{X/Y}$.

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  • $\begingroup$ Why don't you work more directly? The ring map $A \to A/I$ puts the $A$ module structure on $A/I$ so the relative differential has to be zero right? $\endgroup$ – Shubhodip Mondal Jul 10 '16 at 18:55
  • $\begingroup$ Sheaf of relative differentials being zero + flatness implies etale. $\endgroup$ – Shubhodip Mondal Jul 10 '16 at 19:02
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    $\begingroup$ If $X\subset Y$ is a closed subscheme, then $\Omega^1_{X/Y}=0$. $\endgroup$ – Mohan Jul 10 '16 at 19:11
  • $\begingroup$ I see. I was overthinking it. You have $d(a+I)=d(a)(1+I) +ad(1+I) =0$ for any $A$-derivation out of $A/I$. $\endgroup$ – Andrew Jul 10 '16 at 19:20
  • $\begingroup$ To answer your second question, morally $\Omega$ is supposed to be the dual of the relative tangent bundle, whose fibers are the "tangent spaces to the fibers". In the case of a closed embedding, you should expect this to be zero. $\endgroup$ – CCC Jul 12 '16 at 3:02

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