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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

$\displaystyle 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.

$\displaystyle 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

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

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,

$\displaystyle 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

$\displaystyle \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

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

Then we see that

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

So we have

$\displaystyle \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

$\displaystyle 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

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

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

$\displaystyle 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$



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Next: NAK Up: Étale morphism Previous: morphism of finite type
2007-12-11