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flatness

We define here notion of ``flatness". For non commutative algebras, we need to distinguish ``left-flat" and ``right-flat''. Ironically, Bourbaki's book ``commutative algebra''[1] is one of the best source available on this subject. For the time being, we define flatness for commutative algebras only.

DEFINITION 7.1   Let $ A$ be a commutative rings. An $ A$ -module $ M$ is flat if for any exact sequence

$\displaystyle 0\to N_1\to N_2\to N_3\to 0,
$

a sequence

$\displaystyle 0\to M\otimes_A N_1\to M\otimes_A N_2\to M\otimes_A N_3\to 0
$

is also exact.

Note that we have the following

LEMMA 7.2   Let $ A$ be a commutative rings. Let $ M$ be an $ A$ -module. Then for any exact sequence

$\displaystyle 0\to N_1\overset{f_1}{\to} N_2\overset{f_2}{\to} N_3\to 0
$

of $ A$ -modules, a sequence

$\displaystyle M\otimes_A N_1\overset{\operatorname{id}_M\otimes f_1}{\to}
M\otimes_A N_2\overset{\operatorname{id}_M\otimes f_2}{\to}
M\otimes_A N_3\to 0
$

is also exact. Thus $ M$ is flat over $ A$ if and only if for any module $ N_2$ and for any submodule $ N_1$ of $ N_2$ , the map

$\displaystyle M\otimes_A N_1
\overset{\operatorname{id}_M\otimes \operatorname{inclusion}}{\longrightarrow}
M\otimes_A N_2
$

is injective.

PROOF.. Given any element $ \sum_i m_i \otimes n_i$ of $ M\otimes N_3$ , we take a lift $ \hat {n_i}$ of $ n_i$ to $ N_2$ . Then we have

$\displaystyle (\operatorname{id}_M\otimes f_2) (\sum_i m_i \otimes \hat{n_i})=\sum_i m_i \otimes n_i.
$

Thus the map $ \operatorname{id}_M\otimes f_2$ is surjective. To see the exactness in the middle, we first notice that

$\displaystyle (\operatorname{id}_M\otimes f_2)
\circ (\operatorname{id}_M\otimes f_2)=\operatorname{id}_M \otimes(f_2\circ f_1)=0.
$

Thus $ \operatorname{id}\otimes f_2$ yields an $ A$ -module homomorphism

$\displaystyle \phi: (M\otimes_A N_2)
/(M\otimes_A N_1)\to M\otimes_A N_3.
$

On the other hand, for any element $ (m,n)\in M\times N_3$ , we take a lift $ \hat {n}$ of $ n$ to $ N_2$ and define

$\displaystyle \alpha: M\times N_3 \ni (m,n) \mapsto [m\otimes\hat n] \in
(M\otimes_A N_2)
/(M\otimes_A N_1).
$

We may easily check that $ \alpha$ is well-defined (independent of the choice of the lift $ \hat n$ of $ n$ ) and is $ A$ -bilinear. So $ \alpha$ defines an $ A$ -module homomorphism

$\displaystyle \psi: M\otimes N_3 \ni (m\otimes n) \mapsto [m\otimes \hat n] \in
(M\otimes_A N_2)
/(M\otimes_A N_1).
$

Then it is easy to show that the homomorphisms $ \phi$ and $ \psi$ are inverse to each other.

(See for example [14, Appendix A] or [1].)

$ \qedsymbol$

DEFINITION 7.3   An $ A$ -algebra $ B$ is flat if it is flat as an $ A$ -module.

A morphism of affine schemes is flat if the corresponding ring homomorphism is flat.

EXAMPLE 7.4   $ \mathbb{Z}/3\mathbb{Z}$ is not flat over $ \mathbb{Z}$ . Indeed, we consider an exact sequence

$\displaystyle 0\to \mathbb{Z}\overset{\times 3}{\to} \mathbb{Z}\to \mathbb{Z}/3\mathbb{Z}\to 0.
$

Then by tensoring with $ \mathbb{Z}/3\mathbb{Z}$ we obtain a sequence

$\displaystyle 0\to \mathbb{Z}/3\mathbb{Z}\overset{\times 3}{\to} \mathbb{Z}/3\mathbb{Z}\to \mathbb{Z}/3\mathbb{Z}\to 0
$

which is not exact.

Let us view it as a homomorphism $ \phi: \tilde{\mathbb{Z}} \overset{3}{\to} \tilde{\mathbb{Z}}$ of quasi coherent sheaf on $ \operatorname{Spec}(\mathbb{Z})$ with the keyword ``section wise'' and ``fiber wise'' in mind.

  1. $ \phi$ is injective if and only if it is injective section wise. Thus our $ \phi$ is injective.
  2. $ \phi$ is ``not injective'' at the fiber of the point $ 3$ .

EXAMPLE 7.5   Let $ k$ be a field. Then a ring homomorphism $ k[X]\to k[X,Y]/(XY,Y^2)$ is not flat. The reason is almost the same with the one above. (We consider an exact sequence

$\displaystyle 0\to k[X] \overset{\cdot X}{\to} k[X] \to k[X]/X k[X] \to 0.
$

In this example, an embedded prime $ (X,Y)$ of $ I=(XY,Y^2)$ in $ k[X,Y]$ falls into the zero locus of $ X$ .)


next up previous
Next: Various kind of morphisms. Up: Topics in Non commutative Previous: fiber products of schemes
2007-12-11