Lemma 53.15.1. Let $k$ be an algebraically closed field. Let $k \subset A$ be a ring extension such that $A$ has exactly two $k$-sub algebras, then either $A = k \times k$ or $A = k[\epsilon ]$.

## 53.15 Glueing and squishing

Below we will indicate $k[\epsilon ]$ the algebra of dual numbers over $k$ as defined in Varieties, Definition 33.16.1.

**Proof.**
The assumption means $k \not= A$ and any subring $k \subset C \subset A$ is equal to either $k$ or $A$. Let $t \in A$, $t \not\in k$. Then $A$ is generated by $t$ over $k$. Hence $A = k[x]/I$ for some ideal $I$. If $I = (0)$, then we have the subalgebra $k[x^2]$ which is not allowed. Otherwise $I$ is generated by a monic polynomial $P$. Write $P = \prod _{i = 1}^ d (t - a_ i)$. If $d > 2$, then the subalgebra generated by $(t - a_1)(t - a_2)$ gives a contradiction. Thus $d = 2$. If $a_1 \not= a_2$, then $A = k \times k$, if $a_1 = a_2$, then $A = k[\epsilon ]$.
$\square$

Example 53.15.2 (Glueing points). Let $k$ be an algebraically closed field. Let $f : X' \to X$ be a morphism of algebraic $k$-schemes. We say $X$ is obtained by glueing $a$ and $b$ in $X'$ if the following are true:

$a, b \in X'(k)$ are distinct points which map to the same point $x \in X(k)$,

$f$ is finite and $f^{-1}(X \setminus \{ x\} ) \to X \setminus \{ x\} $ is an isomorphism,

there is a short exact sequence

\[ 0 \to \mathcal{O}_ X \to f_*\mathcal{O}_{X'} \xrightarrow {a - b} x_*k \to 0 \]where arrow on the right sends a local section $h$ of $f_*\mathcal{O}_{X'}$ to the difference $h(a) - h(b) \in k$.

If this is the case, then there also is a short exact sequence

where arrow on the right sends a local section $h$ of $f_*\mathcal{O}_{X'}^*$ to the multiplicative difference $h(a)h(b)^{-1} \in k^*$.

Example 53.15.3 (Squishing a tangent vector). Let $k$ be an algebraically closed field. Let $f : X' \to X$ be a morphism of algebraic $k$-schemes. We say $X$ is obtained by squishing the tangent vector $\vartheta $ in $X'$ if the following are true:

$\vartheta : \mathop{\mathrm{Spec}}(k[\epsilon ]) \to X'$ is a closed immersion over $k$ such that $f \circ \vartheta $ factors through a point $x \in X(k)$,

$f$ is finite and $f^{-1}(X \setminus \{ x\} ) \to X \setminus \{ x\} $ is an isomorphism,

there is a short exact sequence

\[ 0 \to \mathcal{O}_ X \to f_*\mathcal{O}_{X'} \xrightarrow {\vartheta } x_*k \to 0 \]where arrow on the right sends a local section $h$ of $f_*\mathcal{O}_{X'}$ to the coefficient of $\epsilon $ in $\vartheta ^\sharp (h) \in k[\epsilon ]$.

If this is the case, then there also is a short exact sequence

where arrow on the right sends a local section $h$ of $f_*\mathcal{O}_{X'}^*$ to $\text{d}\log (\vartheta ^\sharp (h))$ where $\text{d}\log : k[\epsilon ]^* \to k$ is the homomorphism of abelian groups sending $a + b\epsilon $ to $b/a \in k$.

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

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

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. $\square$

Lemma 53.15.5. Let $k$ be an algebraically closed field. If $f : X' \to X$ is the glueing of two points $a, b$ as in Example 53.15.2, then there is an exact sequence

The first map is zero if $a$ and $b$ are on different connected components of $X'$ and injective if $X'$ is proper and $a$ and $b$ are on the same connected component of $X'$.

**Proof.**
The map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X')$ is surjective by Varieties, Lemma 33.38.7. Using the short exact sequence

we obtain

We have $H^1(X, f_*\mathcal{O}_{X'}^*) \subset H^1(X', \mathcal{O}_{X'}^*)$ (for example by the Leray spectral sequence, see Cohomology, Lemma 20.13.4). Hence the kernel of $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X')$ is the cokernel of $ab^{-1} : H^0(X', \mathcal{O}_{X'}^*) \to k^*$. If $a$ and $b$ are on different connected components of $X'$, then $ab^{-1}$ is surjective. Because $k$ is algebraically closed any regular function on a reduced connected proper scheme over $k$ comes from an element of $k$, see Varieties, Lemma 33.9.3. Thus $ab^{-1}$ is zero if $X'$ is proper and $a$ and $b$ are on the same connected component. $\square$

Lemma 53.15.6. Let $k$ be an algebraically closed field. If $f : X' \to X$ is the squishing of a tangent vector $\vartheta $ as in Example 53.15.3, then there is an exact sequence

and the first map is injective if $X'$ is proper and reduced.

**Proof.**
The map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X')$ is surjective by Varieties, Lemma 33.38.7. Using the short exact sequence

of Example 53.15.3 we obtain

We have $H^1(X, f_*\mathcal{O}_{X'}^*) \subset H^1(X', \mathcal{O}_{X'}^*)$ (for example by the Leray spectral sequence, see Cohomology, Lemma 20.13.4). Hence the kernel of $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X')$ is the cokernel of the map $\vartheta : H^0(X', \mathcal{O}_{X'}^*) \to k$. Because $k$ is algebraically closed any regular function on a reduced connected proper scheme over $k$ comes from an element of $k$, see Varieties, Lemma 33.9.3. Thus the final statement of the lemma. $\square$

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