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 $\Omega_{X/k} \neq 0$.

*

*Did I make a mistake anywhere?

*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 $\bigoplus_i A \cdot dx_i$.
 A: 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.
A: 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.
