Zero module of differentials implies finite extension? I am in the middle of a proof where the author asserts that for a field extension $K/E$, $\Omega_{K/E} = 0$ implies that $K/E$ is a finite extension. I am aware of results in the other direction with additional strong hypotheses (finite simple algebraic extension that is also separable). The hypotheses we do have are as follows:
Let $K/k$ be an extension of fields with finite transcendence degree $n$. Suppose $f_1, \dots, f_n$ are elements of $K$ such that $df_1, \dots, df_n$ is a $K$-basis of $\Omega _{K/k}$. Let $E = k(f_1, \dots, f_n)$. 
However, here if this were true, then it would moreover imply that it is a separable extension. Therefore, by the primitive element theorem, this would imply it is a finite, separable, algebraic, simple extension. This seems rather strong from just knowing that $\Omega_{K/E} = 0$. 
 A: Let $K/k$ be an extension of fields with finite transcendence degree.  We say that $K/k$ is of finite type if $K = k(f_1, ... , f_m)$ for some $f_i \in K$.
Fact: $K/k$ is of finite type if and only if for any transcendence basis $x_1, ... , x_n$ of $K/k$, the field extension $k(x_1, ... , x_n) \subseteq K$ is finite.  
Proof: This follows from the fact that an algebraic extension of fields $M \subseteq N$ is finite if and only if there exist finitely many elements $y_1, ... , y_t \in N$ such that $N = M(y_1, ... , y_t)$.
Lemma 1: Assume that $K = k(f)$.  Then $\textrm{Dim}_K(\Omega_{K/k}) \leq 1$, and equal to $0$ if and only if $K/k$ is algebraic and separable (that is, finite and separable).
Lemma 2: Let $F$ be a field with $k \subseteq F \subseteq K$.  Assume that $F$ is also of finite type over $k$.  There is an exact sequence of $K$-vector spaces $$K \otimes_F \Omega_{F/k} \xrightarrow{\alpha} \Omega_{K/k} \rightarrow \Omega_{K/F} \rightarrow 0$$
If $K/F$ is separable algebraic (now $K$ is of finite type over $k$, hence over $F$, so this is equivalent to $K/F$ being finite separable), then $\alpha$ is injective.  
Reference for the lemmas: Springer, Linear Algebraic Groups.
Theorem: Let $K/k$ be an extension of fields of finite type, and assume $\Omega_{K/k} = 0$.  Then $K/k$ is finite separable.
Proof: Writing $K = k(f_1, ... , f_m)$, by induction on $m$.  If $m = 1$, this is the first lemma.  If $m \geq 1$, let $F = k(f_1)$.  If we look at the exact sequence in Lemma 2, $\Omega_{K/k}$ being zero implies that $\Omega_{K/F}$ is also zero.  But $K = F(f_2, ... , f_m)$, so by induction we conclude that $K/F$ is finite separable.  
Since $K/F$ is finite separable, Lemma 2 implies that $\alpha$ is injective, so $K \otimes_F \Omega_{F/k}$ injects into $\Omega_{K/k} = 0$, so $K \otimes_F \Omega_{F/k} = 0$, hence $\Omega_{F/k} = 0$.  Hence $F = k(f_1)$ is finite separable over $k$.  We have shown that $f_2, ... , f_n$ are separable over $k(f_1)$, and that $f_1$ is separable over $k$.  Hence $f_1, ... , f_n$ are separable over $k$. 
A: This is not true. For any algebraic and separable extension, $\Omega_{K/E} = 0.$
This is because every element $\alpha \in K$ is the root of a separable polynomial $p(X) \in E[X]$, so $$0 = \mathrm{d}(p(\alpha)) = p'(\alpha) \mathrm{d}\alpha,$$ where $p'(\alpha) \ne 0$, so $\mathrm{d}\alpha = 0.$
