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 $\Omega_{K/E} = 0$.

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 $k$.

• If you assume a priori that $K/k$ is of finite type, then isn't that assuming that $K/E$ is finite already? I believe this is an OK assumption to make since $K/k$ is a function field and thus is of finite type. However, my remaining concern is that $\Omega_{K/E} = 0$ does not play a role in showing that $K/E$ is finite. It does come in to say that $K/k$ is separable, as you noted, but from the prose of the author, that is not what one is led to believe. Since it is logically correct now though, I am willing to consider the matter closed. Thank you again! Commented Apr 12, 2016 at 15:30
• Yes, if $K/k$ is of finite type, then if $E$ is a maximal purely transcendental extension of $k$, then $K/E$ will not only be algebraic, but also of finite. It has nothing to do with $\Omega_{K/k}$ or $\Omega_{K/E}$. That is the "Fact" I mentioned.
– D_S
Commented Apr 12, 2016 at 17:32

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 $\mathrm{d}\alpha = 0.$

• You seem to be referring to the converse of my question. I am wondering what can we conclude assuming that $\Omega_{K/E} = 0$. Commented Apr 12, 2016 at 3:02
• @ Prospect: ??? Commented Apr 12, 2016 at 17:55
• @Prospect I think I answered the original question. I don't think the new assumptions change anything. Commented Apr 13, 2016 at 1:29