The Stacks project

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

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

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

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. $\square$