# Good argument that $\Omega_{X/k} = 0$ implies that $X$ is finite.

Let $$X$$ be a scheme of finite type over a field $$k$$. I'm pretty sure that $$\Omega_{X/k} = 0$$ implies that $$\dim X = 0$$, i.e. $$X$$ is finite. In other words, if $$\dim X > 0$$, then $$\Omega_{X/k} \neq 0$$. Here is my argument:

Since $$\Omega$$ commutes with base change, we may assume that $$k$$ is algebraically closed. We may also assume that $$X$$ is reduced, because $$X_{\operatorname{red}}$$ is a closed subscheme, and hence $$\Omega_{X_{\operatorname{red}}}$$ is a quotient of $$\Omega_X|_{X_{\operatorname{red}}}$$ (by the 2nd fundamental sequence of Kähler differentials). And now $$X$$ contains a dense open subset $$U$$ which is regular, and hence smooth over $$k$$. Now $$\Omega_{U/k} = \Omega_{X/k}|_U$$ is a free $$\mathcal O_U$$-module of dimension $$\dim U = \dim X > 0$$, hence $$\Omega_{X/k} \neq 0$$.

1. Did I make a mistake anywhere?
2. Is there a "more algebraic" proof which shows for a finitely generated $$k$$-algebra $$A$$ that if $$\Omega_{A/k} = 0$$, then $$A$$ is a finite $$k$$-algebra?

I know that if we choose a quotient $$0 \to (f_1, \dotsc, f_m) \to k[x_1, \dotsc, x_n] \to A \to 0,$$ then there is an exact sequence $$\sum_i A \cdot \left(\sum_j \frac{\partial f_i}{\partial x_j} dx_j\right) \to \bigoplus_i A \cdot dx_i \to \Omega_{A/k} \to 0,$$

but I didn't know how to use the fact that the $$\sum_j \frac{\partial f_i}{\partial x_j} dx_j$$ generate $$\bigoplus_i A \cdot dx_i$$.

• For question 2: it is well known that a locally finite-type k-morphism $X \to Y$ is unramified if and only if $\Omega_{Y/X} = 0$ (see, e.g., Milne's "Etale Cohomology", Proposition I.3.5). Your statement is then equivalent to: a $k$-algebra $A$ which is unramified and of finite type over $k$ is finite over $k$. And indeed, there is a nice purely algebraic proof of this. I can post details if you are interested. Jun 28 '21 at 9:23

I think this is correct. Here’s a more algebraic proof: we can assume that $$k$$ is algebraically closed.

Assume that $$A$$ has a maximal ideal $$\mathfrak{m}$$ so that $$A/\mathfrak{m}=k$$. Let $$e_1,\ldots,e_n$$ be a $$k$$-basis of $$\mathfrak{m}/\mathfrak{m}^2$$, so that we have a decomposition $$A=k\oplus \bigoplus_{i=1}^n{ke_i}\oplus \mathfrak{m}^2$$. Then $$V:=\bigoplus_{i=1}^n{ke_i}\cong \mathfrak{m}/\mathfrak{m}^2$$ is a $$A$$-module and the second projection $$A \rightarrow V$$ is a surjective $$k$$-derivation. But $$\Omega^1_{A/k}=0$$ so that $$V=0$$ and $$\mathfrak{m}=\mathfrak{m}^2$$.

As $$A$$ is Noetherian, ring theory shows that $$\mathfrak{m}=fA$$ for some idempotent $$f \in A$$, and thus $$\{\mathfrak{m}\}$$ is an open subset of the spectrum of $$A$$. As said spectrum is Noetherian (as a topological space), this means that $$A$$ has finitely many maximal ideals. By the Nullstellensatz, this means that zero is a product of maximal ideals of $$A$$, and thus that $$A$$ injects in a finite product of $$A/\mathfrak{r}^k$$ with $$\mathfrak{r}$$ being a maximal ideal and $$k \geq 1$$, which proves that $$A$$ has a finite dimension as a $$k$$-vector space.

Question: "Is there a "more algebraic" proof which shows for a finitely generated $$k$$-algebra $$A$$ that if $$Ω_{A/k}=0$$, then $$A$$ is a finite k-algebra?"

Answer: The result may also be proved using a result on commutative algebra found in Hartshorne and a basic property of the module of differentials:

Let $$k \rightarrow A$$ be a finitely generated $$k$$-algebra with $$k$$ algebraically closed. The following holds for any maximal ideal $$\mathfrak{m}$$ in $$A$$: $$dim(A_{\mathfrak{m}}) \leq dim_k(\mathfrak{m}_{\mathfrak{m}}/\mathfrak{m}_{\mathfrak{m}}^2)$$. This is may be found in Hartshorne Prop. I.5.2A since $$A_{\mathfrak{m}}$$ is a Noetherian local ring with residue field $$k$$.

Hence if you are able to find a maximal ideal $$\mathfrak{m} \subseteq A$$ with $$\mathfrak{m}^2=\mathfrak{m}$$, it follows

$$dim(A)=dim(A_{\mathfrak{m}})\leq dim_k(\mathfrak{m}_{\mathfrak{m}}/\mathfrak{m}_{\mathfrak{m}}^2)=0.$$

Hence you claim also follows from HH.I.5.2A.

Question: "How do you conclude that such m exists, from $$Ω_{A/k}=0$$? – red_trumpet 4 hours ago"

Note: Since you may assume $$k$$ to be algebraically closed, it follows any maximal ideal $$\mathfrak{m}$$ satisfies

$$0:=\Omega^1_{A/k}\otimes_A \kappa(\mathfrak{m})\cong \mathfrak{m}/\mathfrak{m}^2$$

hence for any maximal ideal $$\mathfrak{m}$$ it follows $$\mathfrak{m}^2=\mathfrak{m}$$. The fiber of the "cotangent module/bundle" at a $$k$$-rational point is the cotangent space.

• How do you conclude that such $\mathfrak m$ exists, from $\Omega_{A/k} = 0$? Jun 29 '21 at 13:33