Unramified morphism

DEFINITION 10.2   A separated morphism $\phi:X\to Y$ of finite type is said to be unramified if $\Omega^1 _{X/Y}=0$ .

LEMMA 10.3   Let $A$ be a $B$ -algebra. Assume $A$ is generated by $\{x_\lambda\}_{\lambda\in \Lambda}$ as an $A$ -algebra. Then $I_{A/B}=\operatorname{Ker}(A\otimes_B A\to A)$ is generated by

$$S=\{x_\lambda\otimes 1-1\otimes x_\lambda;\quad \lambda\in \Lambda\}.$$

as an ideal of $A\otimes_B A$ .

PROOF.. Let us denote by $J$ the ideal of $A\times_B A$ generated by $S$ . Then we define a subset $T$ of $A$ as follows.

$$T=\{x \in A;\quad x\otimes 1 - 1 \otimes x\in J\}$$

Now we claim the following facts.

1. $T$ is closed under addition.
2. $T$ is stable under multiplication by any element of $B$ .
3. $1\in T$ .
4. $x_\lambda \in T\qquad (\forall \lambda \in \Lambda)$ .
5. $T$ is closed under multiplication.

The only (5) may require proof. For any elements $t_1,t_2 \in T$ , we have

$$t_1 t_2 \otimes 1 - 1\otimes t_1 t_2$$

$$= (t_1\otimes 1)( t_2 \otimes 1 - 1\otimes t_2) -(t_1 \otimes 1 -1\otimes t_1) (1\otimes t_2) \in J.$$

So the subset $T$ is a $B$ -subalgebra of $A$ containing the generators $\{x_\lambda\}$ of $A$ . Thus we have $T=A$ .

$\qedsymbol$

PROPOSITION 10.4   A separated morphism $\phi:X\to Y$ of finite type is unramified if and only if the diagonal map $X \cong \Delta_{X/Y}\hookrightarrow X\otimes_Y X$ is an open immersion.

PROOF.. Let us first prove the ``if'' part. Assume $\Delta_{X/Y}$ is open. then $\Delta_{X/Y}$ is a clopen (``closed and open'') subset of $X\otimes_Y X$ . Namely,

$$X\otimes_Y X= (\Delta_{X/Y}) \cup (\complement \Delta_{X/Y})$$

is a decomposition of the scheme $X\otimes_Y X$ into two Zariski open set. Thus we have

$$\mathcal{O}_{X\otimes_Y X} =\mathcal I_{\Delta_{X/Y}} \oplus I_{\complement \Delta_{X/Y}}.$$

We then note in particular that $\mathcal I_{\Delta_{X/Y}}$ has a distinguished global section (``the identity'') $u$ defined by

$$u=\begin{cases} 1 & \text{on } \Delta_{X/Y}\\ 0 & \text{on } \complement \Delta_{X/Y}. \end{cases}$$

Then we see that

$$\mathcal I_{\Delta_{X/Y}} =u\mathcal I_{\Delta_{X/Y}} \subset \mathcal I_{\Delta_{X/Y}}^2.$$

So we have

$$\Omega^1_{X/Y}=\mathcal I_{\Delta_{X/Y}} /\mathcal I_{\Delta_{X/Y}}^2=0$$

as required.

Let us now prove the ``only if'' part. The question is local on $X$ and on $Y$ . So we may assume that $f$ is of the form

$$f:X=\operatorname{Spec}(A)\to Y=\operatorname{Spec}(B)$$

where $A$ is a finitely generated algebra over $B$ . Let $I=I_{\Delta_{X/Y}}$ be the ideal of definition of the diagonal. The previous Lemma tells us that $I$ is finitely generated over $A\otimes_B A$ . By the assumption we have

$$I/I^2=0.$$

Now we use the Nakayama's lemma (theorem below) to find an element $c\in I$ such that

$$c x=x \qquad (\forall x \in I).$$

Then it is easy to see that $c$ is an idempotent and that $I=c (A\otimes_B A)$ is its range. $\qedsymbol$