The Stacks project

101.26 Flat at a point

We still have to develop the general machinery needed to say what it means for a morphism of algebraic stacks to have a given property at a point. For the moment the following lemma is sufficient.

Lemma 101.26.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$. Consider commutative diagrams

\[ \vcenter { \xymatrix{ U \ar[d]_ a \ar[r]_ h & V \ar[d]^ b \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} } } \quad \text{with points} \vcenter { \xymatrix{ u \in |U| \ar[d] \\ x \in |\mathcal{X}| } } \]

where $U$ and $V$ are algebraic spaces, $b$ is flat, and $(a, h) : U \to \mathcal{X} \times _\mathcal {Y} V$ is flat. The following are equivalent

  1. $h$ is flat at $u$ for one diagram as above,

  2. $h$ is flat at $u$ for every diagram as above.

Proof. Suppose we are given a second diagram $U', V', u', a', b', h'$ as in the lemma. Then we can consider

\[ \xymatrix{ U \ar[d] & U \times _\mathcal {X} U' \ar[l] \ar[d] \ar[r] & U' \ar[d] \\ V & V \times _\mathcal {Y} V' \ar[l] \ar[r] & V' } \]

By Properties of Stacks, Lemma 100.4.3 there is a point $u'' \in |U \times _\mathcal {X} U'|$ mapping to $u$ and $u'$. If $h$ is flat at $u$, then the base change $U \times _ V (V \times _\mathcal {Y} V') \to V \times _\mathcal {Y} V'$ is flat at any point over $u$, see Morphisms of Spaces, Lemma 67.31.3. On the other hand, the morphism

\[ U \times _\mathcal {X} U' \to U \times _\mathcal {X} (\mathcal{X} \times _\mathcal {Y} V') = U \times _\mathcal {Y} V' = U \times _ V (V \times _\mathcal {Y} V') \]

is flat as a base change of $(a', h')$, see Lemma 101.25.3. Composing and using Morphisms of Spaces, Lemma 67.31.4 we conclude that $U \times _\mathcal {X} U' \to V \times _\mathcal {Y} V'$ is flat at $u''$. Then we can use composition by the flat map $V \times _\mathcal {Y} V' \to V'$ to conclude that $U \times _\mathcal {X} U' \to V'$ is flat at $u''$. Finally, since $U \times _\mathcal {X} U' \to U'$ is flat at $u''$ and $u''$ maps to $u'$ we conclude that $U' \to V'$ is flat at $u'$ by Morphisms of Spaces, Lemma 67.31.5. $\square$

Definition 101.26.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$. We say $f$ is flat at $x$ if the equivalent conditions of Lemma 101.26.1 hold.

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0CIB. Beware of the difference between the letter 'O' and the digit '0'.



View Section 101.26 as pdf