Let $A$ be a ring and $X=\mathbb P^1_A$. On $D_+(T_0)\cap D_+(T_1)$, we have

$$T^2_0d(T_1/T_0)= -T^2_1d(T_0/T_1).$$

How does one formally compute this transition? Of course this is obvious intuitively if one thinks of differential forms and $\mathbb P^1$ as a manifold. It is even obvious intuitively when thinking algebraically, if you think of the intersection as "the place where you can invert $T_0$ and $T_1$," and then compute $d\left(\frac{T_1}{T_0}\frac{T_0}{T_1}\right)$.

(Recall that we define $\mathbb P^1_A$ as $\operatorname{Proj} A[T_0,T_1]$. We can cover this with two principal open subsets $D_+(T_i)$, with sections $A[T_0,T_1]_{(T_i)}$. It is a relatively exercise to compute that on these "charts" the sheaf of differentials looks like $A[d(T_0/T_1)]$ and $A[d(T_1/T_2)]$. The elements above are then elements of $\mathcal O_X(2)\otimes \Omega^1_{X/A}$.)

The excerpt above appears when Liu computes the sheaf of differentials on $\mathbb P_A^1$. He does this by writing down two differentials on each chart, noting they glue to a global section, then noting this global section is a generator. So in justifying the above, I would like to avoid doing what Hartshorne does, which is to guess that $\Omega^1_{X/A}$ is $\tilde{M}$ for a certain module $M$ and then checking that this is correct.

Also, Liu constructs projective schemes by describing the sheaf $\mathcal O_X$ on principal open sets and then citing the theorem that this is enough to determine a sheaf. I think one can make the algebraic intuition I described above by using a more explicit description of the sheaf (like the one given in Hartshorne), but I'd like to avoid doing this, too. It is distinctly not in Liu's style, and I'm very curious what he intended.

Update. Put another way, my question is how to derive the gluing maps for these "charts" rigorously and how to show how these gluing maps transform the differentials in the expected manner.

  • $\begingroup$ I do not have Liu at hand right now, but explicity $D_{+}(T_{0}) \cap D_{+}(T_{1})= \mathbb{A}^{1}\ 0$. We identify the co-ordinate function $t$ on $ \mathbb{A}^{1}\ 0$ with the co-ordinate function $T_{1}/T_{0}$ on the intersection. Then the relation you mention is the formal fact $t^{2}d(1/t) = -dt$. I am not sure if this answers what you are asking for.. $\endgroup$
    – DBS
    Jul 10, 2013 at 6:50
  • $\begingroup$ @DBS Yeah, that's essentially what you do in the case when $\mathbb P^1$ is a manifold and you're talking about differential forms. My issue is working out the details rigorously. I don't see any theorem on pullbacks in Liu. $\endgroup$
    – Potato
    Jul 10, 2013 at 6:56
  • $\begingroup$ I will take a look at Liu later. But I guess I do not understand the distinction you are making between differential forms and Kahler differentials. $\endgroup$
    – DBS
    Jul 10, 2013 at 7:01
  • $\begingroup$ @DBS Differential forms live on manifolds, and I know how to pull them back. Differentials live on schemes, and I know how to pull them back but can't prove it. :). Thanks for your help. $\endgroup$
    – Potato
    Jul 10, 2013 at 7:03
  • 1
    $\begingroup$ @littleO Algebraic Geometry and Arithmetic Curves by Qing Liu. It's in the Oxford Graduate Texts in Mathematics series. $\endgroup$
    – Potato
    Jul 11, 2013 at 2:26

1 Answer 1


For a relative scheme $X/S$ it is well-known that $\Omega^1_{X/S}$, the sheaf of modules on $X$ which classifies $\mathcal{O}_S$-derivations on $\mathcal{O}_X$ (i.e. it comes equipped with a universal $\mathcal{O}_S$-derivation $d : \mathcal{O}_X \to \Omega^1_{X/S}$), is quasi-coherent and can be described locally on $S$ as well as on $X$. If $S=\mathrm{Spec}(R)$ and $X=\mathrm{Spec}(A)$ are affine, then $\Omega^1_{X/S}$ is associated to $\Omega^1_{A/R}$, the $A$-module which classifies $R$-derivations on $A$. It is also well-known that $\Omega^1_{R[t]/R}$ is free of rank $1$ with generator $d(t)$.

We also have to know that $\Omega^1$ commutes with localization. In particular, $\Omega^1_{R[t,t^{-1}]/R}$ is free of rank $1$ with generator $d(t)$; but also with the generator $d(t^{-1})$ (since $R[t,t^{-1}]$ has an automorphism mapping $t$ to $t^{-1}$). Hence these differ by a unit. Explicitly, we compute

$$0 = d(1) = d(t \cdot t^{-1}) = t \cdot d(t^{-1}) + t^{-1} \cdot d(t)$$

and hence

$$d(t) = -t^2 \cdot d(t^{-1}).$$

Now, the projective line $\mathbb{P}^1_R$ is covered by the two affine lines $D_+(T_0) \cong \mathbb{A}^1_R$ with variable $T_1/T_0$, and $D_+(T_1) \cong \mathbb{A}^1_R$ with variable $T_0/T_1$. It follows that $\Omega^1_{\mathbb{P}^1_R/R}$ is free of rank $1$ on both of these, with generators $d(T_1/T_0)$ resp. $d(T_0/T_1)$. The intersection of the two affine lines is the spectrum of the localization $R[T_0/T_1,T_1/T_0]$. It follows that $\Omega^1_{\mathbb{P}^1_R/R}$ is also free on this intersection, with generator $d(T_0/T_1)$, as well as the generator $d(T_1/T_0)$. I have chosen the same notation as above, because these are really just the restrictions of the generators on the affine lines, by construction of the involved isomorphisms. On the intersection, i.e. in $\Omega^1_{\mathbb{P}^1_R/R}|_{D_+(T_0) \cap D_+(T_1)}$, or equivalently in $\Omega^1_{R[T_0/T_1,T_1/T_0]/R}$, we have computed above that

$$d(T_1/T_0)=-(T_1/T_0)^2 d(T_0/T_1).$$

This is a complete description of $\Omega^1_{\mathbb{P}^1}$. By the way, this calculation shows nothing else than $\Omega^1_{\mathbb{P}^1}=\mathcal{O}(-2)$. More generally, one can show $\Omega^d_{\mathbb{P}^d} = \mathcal{O}(-d-1)$.

  • $\begingroup$ Fantastic answer. I can't award the bounty for another 20 or so hours, but please know I will eventually. Thank you. $\endgroup$
    – Potato
    Jul 12, 2013 at 6:55

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