Relative differentials under base change Let $X \to Y$ be a morphism of varieties over a field $K$ of characteristic zero. Let $K \subseteq L$ be a field extension and consider the induced morphism $X_L \to Y_L$ (where $X_L=X \times_K \textrm{Spec} L$ and the same for $Y_L$) and let $\pi: X_L \to X$ be the projection. Let $x \in X_L$. Is it true that $(\Omega_{X_L/Y_L})_x =0$ if and only if $(\Omega_{X/Y})_{\pi(x)} =0$?
 A: Lemma. For $z \in X$, we have $(\Omega_{X/Y})_z = 0$ if and only if $(\Omega_{X_L/Y_L})_w = 0$ for all $w \in \pi^{-1}(z)$.
Proof. By Hartshorne, Proposition II.8.10 (this is essentially Eisenbud, Proposition 16.4), we have
$$\pi^* \Omega_{X/Y} = \Omega_{X_L/Y_L}.$$
We have a commutative diagram
$$\begin{array}{ccccccccc} \pi^{-1}(z) & \xrightarrow{z_L} & X_L & \to & Y_L & \to & \operatorname{Spec} L\ \\ \downarrow & & \downarrow & & \downarrow & & \downarrow \\ z & \xrightarrow{z} & X & \to & Y & \to & \operatorname{Spec} K, \end{array}$$
all of whose squares are pullbacks. Since $L \to K$ is faithfully flat, so are all other vertical arrows. By Nakayama's lemma (since everything is finitely generated), we have
$$(\Omega_{X/Y})_{z} = 0 \Longleftrightarrow (\Omega_{X/Y})_{z} \otimes_{\mathcal O_{X,z}} \kappa(z) = 0,$$
where $\kappa(z)$ denotes the residue field. The right hand side can also be written as
$$z^* \Omega_{X/Y} = 0.$$
Since $\pi^{-1}(z) \to z$ is faithfully flat, we conclude that
$$z^*\Omega_{X/Y} = 0 \Longleftrightarrow z_L^* \Omega_{X_L/Y_L} = 0.$$
This is equivalent to $(\Omega_{X_L/Y_L})_w = 0$ for all $w \in \pi^{-1}(z)$, by an argument similar to the above. $\square$
Remark. The scheme $\pi^{-1}(z)$ is the spectrum of $\kappa(z) \otimes_K L$. Thus, if $L$ or $\kappa(z)$ is finite over $K$, then $\pi^{-1}(z)$ is finite over $\kappa(z)$ or $L$ (respectively). In particular, the set $\pi^{-1}(z)$ is a finite discrete space.
On the other hand, if both $K \to L$ and $K \to \kappa(z)$ are infinite, then in general the set $\pi^{-1}(z)$ will not be finite (for example, try computing $\operatorname{Spec} K(t) \otimes_K K(t)$, where $t$ is a transcendental element).
Remark. We know (Hartshorne Exercise II.2.15(a)) that $z$ is closed if and only if $K \to \kappa(z)$ is a finite field extension. Then $\pi^{-1}(z) \to \operatorname{Spec} L$ is finite as well. Thus, there are finitely many $w$ above $z$, and for any such $w$ the field extension $L \to \kappa(w)$ is finite. That is, $\pi^{-1}(z)$ is a finite set of closed points. Conversely, if $\pi^{-1}(z)$ is a finite set of closed points, then it is finite over $L$. By fpqc descent, you can prove that this implies that $\kappa(z)$ is finite over $K$, i.e. $z$ is closed.
Example. If $L/K$ is finite, then so is $\kappa(z) \subseteq \kappa(w)$ for any $w \in \pi^{-1}(z)$. The tower $K \subseteq \kappa(z) \subseteq \kappa(w)$ shows that $\kappa(z)$ is finite over $K$ if and only if $\kappa(w)$ is, and the tower $K \subseteq L \subseteq \kappa(w)$ shows that $\kappa(w)$ is finite over $K$ if and only if it is finite over $L$. Thus, $z$ is a closed point if and only $w$ is.
This already fails for $L = K(x)$: indeed, the is $X = \mathbb A^1_K = \operatorname{Spec} K[y]$, then the point $w \in X_L$ corresponding to the prime ideal $(y - x) \subseteq K(x)[y]$ maps to the prime ideal $(0) \subseteq K[y]$. Thus, $w$ is closed, but $z$ is not.
To answer you question, I will stick to the case where $K \to L$ is a finite extension.
Example. Assume $L/K$ is finite Galois, with group $G$. Then $X_L \to X$ is a $G$-torsor (since $L/K$ is), so $G$ acts transitively on the preimage $\pi^{-1}(z)$ of any point $z \in X$. In this case, we know
$$(\Omega_{X_L/Y_L})_w = (\Omega_{X_L/Y_L})_{\sigma(w)},$$
as $\kappa(w) \cong \kappa(\sigma(w))$-vector space. Thus, if one of them is trivial, then all of them are (by transitivity of the Galois action). Thus, in this case the answer to your question is positive!
More generally, if $L/K$ is finite separable (e.g. a finite extension of characteristic $0$ fields), we can take the Galois closure $M/K$ of $L$, and a straightforward argument deduces the result for $L/K$ from that of $M/K$ (which we know by the above).
Remark. To elaborate a little bit on the transitivity argument: write $\underline G$ for the constant group scheme $\operatorname{Spec}(K^G)$. That is, $\underline G$ is the set $G$, where each point is a $\operatorname{Spec} K$, with the obvious group (scheme) structure. We get a right action
$$\mu \colon \operatorname{Spec} L \times_{\operatorname{Spec} K} \underline G \to \operatorname{Spec} L,$$
given by the Hopf algebra action $L \to L \otimes_K K^G$ given by $\ell \mapsto (\sigma(\ell))_{\sigma \in G}$. This is a torsor: one can prove by hand that the induced map
$$\operatorname{Spec} L \times_{\operatorname{Spec} K} \underline G \stackrel{(\operatorname{id},\mu)}\longrightarrow \operatorname{Spec} L \times_{\operatorname{Spec} K} \operatorname{Spec} L$$
is an isomorphism (use the primitive element theorem and the Chinese remainder theorem).
But then the base change $\pi^{-1}(z) \times_{\operatorname{Spec} K} \underline G \to \pi^{-1}(z)$ is again a $G$-torsor: the map
$$\pi^{-1}(z) \times_{\operatorname{Spec} K} \underline G \to \pi^{-1}(z) \times_z \pi^{-1}(z)$$
is still an isomorphism. Show that this implies that the $G$-action on $\pi^{-1}(z)$ is (set-theoretically) transitive by looking at the isomorphism on $\overline {\kappa(z)}$-points.
Remark. It would be interesting to see if we can construct a counterexample when the field extension is not finite separable (e.g. purely inseparable, or transcendental).
A: Let me reorganise and improve my previous answer a bit. I realised that it's much easier, and that a much more general result holds.
Lemma. Let $f \colon Z \to X$ be a morphism of schemes. Let $\mathscr F$ be a coherent sheaf on $X$. Let $z \in Z$, and let $x = f(z)$. Then $\mathscr F_x = 0$ if and only if $(f^*\mathscr F)_z = 0$.
Remark. We will apply it to $Z = X_L$, where $f = \pi$ is the projection, and $\mathscr F = \Omega_{X/Y}$. Note that $f^*\Omega_{X/Y} = \Omega_{X_L/Y_L}$, so that the lemma does indeed give the result you want (with no restrictions on the field extension $K \subseteq L$).
Proof of Lemma. We consider the diagram
$$\begin{array}{rrrccc} z & \to & Z_x :=\!\!\! & Z \times_X x & \to & Z \\ & & & \downarrow & & \downarrow \\ & & & x & \to & X. \end{array}$$
Since $\mathscr F$ is coherent, Nakayama's lemma gives $\mathscr F_x = 0$ if and only if $x^*\mathscr F = 0$, and similarly $(f^*\mathscr F)_z = 0$ if and only if $z^*f^*\mathscr F = 0$. But $x^* \mathscr F$ is a finite dimensional vector space over $\kappa(x)$, so
$$x^* \mathscr F \cong \kappa(x)^r$$
for some $r \in \mathbb N$. It follows from commutativity of the diagram that $z^*f^* \mathscr F \cong \kappa(z)^r$. Thus,
\begin{align*}
&\ &\ &\ &\ & x^*\mathscr F = 0 \Longleftrightarrow r= 0 \Longleftrightarrow z^*f^* \mathscr F = 0. &\ &\ &\ &\ \square
\end{align*}
Remark. If you like, we're using that the field extension $\kappa(x) \subseteq \kappa(z)$ is faithfully flat, whereas previously we only used that $K \subseteq L$ is faithfully flat.
