First some definitions:

Let $k$ be algebraically closed. Let $X$ be a topological space. We define an algebraic variety $X$ to be a $k$-space $(X,\mathcal O_X)$, such that for each $x\in X$, we have an open $U\subset X$ with $x\in U$, such that $(U,\mathcal O_X\vert_U)\cong (Z,\mathcal O_Z)$, where $Z\subset\mathbb A^n$ is an (embedded) affine variety, and $\mathcal O_Z$ is its sheaf of regular functions. In short: $X$ is locally affine.

An (embedded) affine variety $Z\subset\mathbb A^n$ is a closed subset of $\mathbb A^n$ (with the Zariski topology), and for open $U\subset Z$, $\mathcal O_Z(U)$ is the $k$-algebra consisting of functions $U\to k$ that locally can be written as $f/g$ with $f,g\in k[x_1,\dots,x_n]$.

Given an algebraic variety $X$ we have the so-called sheaf of 1-forms $\Omega^1_X$, which is defined to be the sheafification of the presheaf $U\mapsto \Omega^1_{\mathcal O_X(U)}$, where $\Omega^1_{\mathcal O_X(U)}$ is the module of Kähler differentials of $\mathcal O_X(U)$ (the regular functions on $U$).

Now, my lecture notes state the following two results without proof:

For an affine open $U\subset X$, we have $\Omega^1_X(U)\cong\Omega^1_{\mathcal O_X(U)}$.

For affine opens $V\subset U\subset X$, we have $\Omega^1_{\mathcal O_X(V)}\cong\Omega^1_{\mathcal O_X(U)}\otimes_{\mathcal O_X(U)}\mathcal O_X(V)$, and furthermore, the restriction map $\Omega^1_X(U)\to\Omega^1_X(V)$ is given by $\omega\mapsto\omega\otimes 1$.

When they say the restriction map is given this way, do they mean that we have the following commutative diagram:

enter image description here

Here, the restriction mapping $\Omega^1_{\mathcal O_X(U)}\to\Omega^1_{\mathcal O_X(V)}$ is the one induced from $\iota^*$ (exactly such that this diagram commutes).

  • $\begingroup$ Giving a good explanation for your first item requires knowing the definition you're using. Please add this to the post. $\endgroup$
    – KReiser
    Feb 3, 2021 at 10:21
  • $\begingroup$ @KReiser If I need to add more (e.g., how we're defining an algebraic variety), let me know. $\endgroup$
    – Sha Vuklia
    Feb 3, 2021 at 10:28
  • 1
    $\begingroup$ The short answer is yes, you get the commutative diagram. $\endgroup$ Feb 3, 2021 at 10:29

1 Answer 1


Your question: "For an affine open $U \subseteq X$, we have

$$\text{O1}. \text{ }\Omega^1_X(U) \cong \Omega^1_{\mathcal{O}_X(U)}.$$

For affine open sets $V\subseteq U \subseteq X$ we have

$$\text{O2}. \text{ } \Omega^1_{\mathcal{O}_X(V)} \cong \Omega^1_{\mathcal{O}_X(U)}\otimes_{\mathcal{O}_X(U)} \mathcal{O}_X(V),$$

and furthermore, the restriction map $\Omega^1_X(U) \rightarrow \Omega^1_X(V)$ is given by $ω \rightarrow ω\otimes 1$.

It seems to me the following is true:

Question O1. Since your sheaf $\Omega^1_X$ is defined as the sheafification $F^{+}$ of the pre-sheaf $F(U):= \Omega^1_{\mathcal{O}_X(U)}$, it seems to me $F^{+}$ and $F$ should have she same sections on affine open subsets $U:=Spec(A) \subseteq X$. Hence

$$F(U)=F^{+}(U) \cong \Omega^1_{A/k}:=\Omega^1_{\mathcal{O}_X(U)}.$$

Question O2. The construction of the module of differentials in functorial in the sense that for any open subscheme $U \subseteq X$ it follows $i^*(\Omega^1_{X/k}) \cong \Omega^1_{U/k}$ where $i: U \rightarrow X$ is the inclusion map.

Moreover for any pair of maps of commutative rings $A \rightarrow B$ and $A \rightarrow S$ let $B_S :=S\otimes_A B$. There is (Matsumura's book "Commutative ring theory", Exercise 25.4 page 198) a canonical isomorphism

$$ B_S \otimes_B \Omega^1_{B/A} \cong \Omega^1_{B_S/S}.$$

Hence for any inclusion of basic open sets $D(f) \subseteq D(g) \subseteq U:=Spec(A)$ you should get the following:

$$(\Omega^1_{D(g)/k})_{D(f)}:= \Omega^1_{A_g/k} \otimes_{A_g} A_f \cong \Omega^1_{A_{fg}/k}\cong \Omega^1_{A_f/k}.$$

By $(\Omega^1_{D(g)/k})_{D(f)}$ I mean the restriction of $\Omega^1_{D(g)/k}$ to the open set $D(f)$. Here we have used that $D(f)\cap D(g):=D(fg) =D(f)$ since $D(f) \subseteq D(g)$. The restriction map

$$\rho: \Omega^1_{D(g)/k}\rightarrow (\Omega^1_{D(g)/k})_{D(f)} := \Omega^1_{D(g)/k}\otimes_{A_g} A_f\cong \Omega^1_{D(f)/k}$$

is the canonical map $\rho(\omega):=\omega \otimes 1$ where $1\in A_f$ is the multiplicative identity. Formula O2 should follow since $U$ and $V$ have open covers on the form $D(f)$.

  • $\begingroup$ Thanks for the elaborate answer! I'm not able to follow every detail (which is entirely on me), but I think I got the idea. Am I correct in saying that you only explained that the diagram commutes as it does for basic opens (instead of for general affine opens)? $\endgroup$
    – Sha Vuklia
    Feb 3, 2021 at 11:39

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