The Stacks project

Lemma 53.15.4. Let $k$ be an algebraically closed field. Let $f : X' \to X$ be a finite morphism algebraic $k$-schemes such that $\mathcal{O}_ X \subset f_*\mathcal{O}_{X'}$ and such that $f$ is an isomorphism away from a finite set of points. Then there is a factorization

\[ X' = X_ n \to X_{n - 1} \to \ldots \to X_1 \to X_0 = X \]

such that each $X_ i \to X_{i - 1}$ is either the glueing of two points or the squishing of a tangent vector (see Examples 53.15.2 and 53.15.3).

Proof. Let $U \subset X$ be the maximal open set over which $f$ is an isomorphism. Then $X \setminus U = \{ x_1, \ldots , x_ n\} $ with $x_ i \in X(k)$. We will consider factorizations $X' \to Y \to X$ of $f$ such that both morphisms are finite and

\[ \mathcal{O}_ X \subset g_*\mathcal{O}_ Y \subset f_*\mathcal{O}_{X'} \]

where $g : Y \to X$ is the given morphism. By assumption $\mathcal{O}_{X, x} \to (f_*\mathcal{O}_{X'})_ x$ is an isomorphism onless $x = x_ i$ for some $i$. Hence the cokernel

\[ f_*\mathcal{O}_{X'}/\mathcal{O}_ X = \bigoplus \mathcal{Q}_ i \]

is a direct sum of skyscraper sheaves $\mathcal{Q}_ i$ supported at $x_1, \ldots , x_ n$. Because the displayed quotient is a coherent $\mathcal{O}_ X$-module, we conclude that $\mathcal{Q}_ i$ has finite length over $\mathcal{O}_{X, x_ i}$. Hence we can argue by induction on the sum of these lengths, i.e., the length of the whole cokernel.

If $n > 1$, then we can define an $\mathcal{O}_ X$-subalgebra $\mathcal{A} \subset f_*\mathcal{O}_{X'}$ by taking the inverse image of $\mathcal{Q}_1$. This will give a nontrivial factorization and we win by induction.

Assume $n = 1$. We abbreviate $x = x_1$. Consider the finite $k$-algebra extension

\[ A = \mathcal{O}_{X, x} \subset (f_*\mathcal{O}_{X'})_ x = B \]

Note that $\mathcal{Q} = \mathcal{Q}_1$ is the skyscraper sheaf with value $B/A$. We have a $k$-subalgebra $A \subset A + \mathfrak m_ A B \subset B$. If both inclusions are strict, then we obtain a nontrivial factorization and we win by induction as above. If $A + \mathfrak m_ A B = B$, then $A = B$ by Nakayama, then $f$ is an isomorphism and there is nothing to prove. We conclude that we may assume $B = A + \mathfrak m_ A B$. Set $C = B/\mathfrak m_ A B$. If $C$ has more than $2$ $k$-subalgebras, then we obtain a subalgebra between $A$ and $B$ by taking the inverse image in $B$. Thus we may assume $C$ has exactly $2$ $k$-subalgebras. Thus $C = k \times k$ or $C = k[\epsilon ]$ by Lemma 53.15.1. In this case $f$ is correspondingly the glueing two points or the squishing of a tangent vector. $\square$


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 0C1L. Beware of the difference between the letter 'O' and the digit '0'.