# Equivalence of properties defining unramified morphisms

SGA I.3 claims the following three properties of a finite type morphism $X \rightarrow Y$ are equivalent (Let $x \in X$, $y = f(x)$:

Let $A$ be the stalk of $X$ at $x$, and $B$ be the stalk of $Y$ at $y$. Let $m_a, m_b$ be their respective maximal ideals. The morphism of sheaves then determines a morphism $g$ of local rings in the other direction.

i.) $g(m_b)$ generates $m_a$, and $\frac{A}{m_a}$ is a finite separable extension of $\frac{B}{m_b}.$

ii.)The stalk of $\Omega^1_{X/Y}$ is 0 at $x$.

iii.)The diagonal morphism is an open immersion in a neighborhood of $x$.

I don't understand their proof that i.) implies ii.); It is just written that Nakayama's immediately reduces it to the case where X and Y are spectrums of fields. Can anyone please elaborate?

• The "proof" you mention is anything but: the case of fields is trivial and saying that you "instantly" reduce to that case is not in the least helpful (as your very question attests) . Of course you can also say that Hironaka's proof of desingularization of schemes in characteristic zero instantly reduces to a sequence of blowing-ups... – Georges Elencwajg Aug 11 '12 at 12:38
• Hmm, perhaps my translation of SGA is bad; it is written "on est ramené aussitôt par Nakayama au cas où $Y = Spec(k)$, $X = Spec(k').$ I know aussitôt means immediately (which probably would have been a better word than instantly) and ramené means reduced. – only Aug 11 '12 at 16:06

We may assume that $X=Spec(S), Y=Spec(R)$ are affine.
Let $\phi : R\to S$ be the ring morphism corresponding to $f:X=Spec(S)\to Y= Spec(R)$.
The points $x\in X, y\in Y$ correspond to prime ideals $\mathfrak q\subset S$ and $\mathfrak p=f^{-1}(\mathfrak q)\subset S$.
The hypothesis implies that $\phi$ induces a finite separable extension $\bar \phi:\kappa (\mathfrak p)\to \kappa(\mathfrak q)$ of the residual fields, and separability of this finite extension implies that $\Omega_{\kappa(\mathfrak q)/\kappa (\mathfrak p)}=0$.

Now $\Omega_{\kappa(\mathfrak q)/\kappa (\mathfrak p)}=\Omega_{S_\mathfrak q/R_\mathfrak p}/ \phi(\mathfrak p) \Omega_{S_\mathfrak q/R_\mathfrak p}$ by base change and thus $0=\Omega_{\kappa(\mathfrak q)/\kappa (\mathfrak p)}=\Omega_{S_\mathfrak q/R_\mathfrak p}/ \phi(\mathfrak p) \Omega_{S_\mathfrak q/R_\mathfrak p}=\Omega_{S_\mathfrak q/R_\mathfrak p}/ \mathfrak q \Omega_{S_\mathfrak q/R_\mathfrak p}$, the last equality due to $\phi(\mathfrak p)S_{\mathfrak q}=\mathfrak qS_{\mathfrak q}$.
Hence Nakayama may be applied to the module $\Omega_{S_\mathfrak q/R_\mathfrak p}$ over the local ring $S_\mathfrak q$:
from $\Omega_{S_\mathfrak q/R_\mathfrak p}/ \mathfrak q \Omega_{S_\mathfrak q/R_\mathfrak p}=0$ we conclude that $\Omega_{S_\mathfrak q/R_\mathfrak p}=0$.
Finally we need only remark that $(\Omega_{X/Y})_x=\Omega_{S_\mathfrak q/R_\mathfrak p}$ to obtain the required equality $(\Omega_{X/Y})_x=0$

A reminder in field theory
I used that a separable algebraic extension of fields $k\subset K$ has trivial module of Kähler differentials: $\Omega _{K/k}=0$.
Indeed, if $d:K\to \Omega _{K/k}$ is the universal derivation and if $a\in K$ has minimal polynomial $f(x)$ over $k$, then $f(a)=0$ implies $d(f(a))=f'(a)da=0$, and so $da=0$ since $f'(a)\neq 0$ by separability of $a$.
Since $da=0$ holds for all $a\in K$ we conclude that $\Omega _{K/k}=0$.