Lemma 53.16.1. Let $k$ be a separably closed field. Let $A$ be a $1$-dimensional reduced Nagata local $k$-algebra with residue field $k$. Then

If equality holds, then $A^\wedge $ is as in (53.16.0.1).

In this section we discuss the simplest possible curve singularities.

Let $k$ be a field. Consider the complete local $k$-algebra

53.16.0.1

\begin{equation} \label{curves-equation-multicross} A = \{ (f_1, \ldots , f_ n) \in k[[t]] \times \ldots \times k[[t]] \mid f_1(0) = \ldots = f_ n(0)\} \end{equation}

In the language introduced in Varieties, Definition 33.40.4 we see that $A$ is a wedge of $n$ copies of the power series ring in $1$ variable over $k$. Observe that $k[[t]] \times \ldots \times k[[t]]$ is the integral closure of $A$ in its total ring of fractions. Hence the $\delta $-invariant of $A$ is $n - 1$. There is an isomorphism

\[ k[[x_1, \ldots , x_ n]]/(\{ x_ ix_ j\} _{i \not= j}) \longrightarrow A \]

obtained by sending $x_ i$ to $(0, \ldots , 0, t, 0, \ldots , 0)$ in $A$. It follows that $\dim (A) = 1$ and $\dim _ k \mathfrak m/\mathfrak m^2 = n$. In particular, $A$ is regular if and only if $n = 1$.

Lemma 53.16.1. Let $k$ be a separably closed field. Let $A$ be a $1$-dimensional reduced Nagata local $k$-algebra with residue field $k$. Then

\[ \delta \text{-invariant }A \geq \text{number of branches of }A - 1 \]

If equality holds, then $A^\wedge $ is as in (53.16.0.1).

**Proof.**
Since the residue field of $A$ is separably closed, the number of branches of $A$ is equal to the number of geometric branches of $A$, see More on Algebra, Definition 15.106.6. The inequality holds by Varieties, Lemma 33.40.6. Assume equality holds. We may replace $A$ by the completion of $A$; this does not change the number of branches or the $\delta $-invariant, see More on Algebra, Lemma 15.108.7 and Varieties, Lemma 33.39.6. Then $A$ is strictly henselian, see Algebra, Lemma 10.153.9. By Varieties, Lemma 33.40.5 we see that $A$ is a wedge of complete discrete valuation rings. Each of these is isomorphic to $k[[t]]$ by Algebra, Lemma 10.160.10. Hence $A$ is as in (53.16.0.1).
$\square$

Definition 53.16.2. Let $k$ be an algebraically closed field. Let $X$ be an algebraic $1$-dimensional $k$-scheme. Let $x \in X$ be a closed point. We say $x$ defines a *multicross singularity* if the completion $\mathcal{O}_{X, x}^\wedge $ is isomorphic to (53.16.0.1) for some $n \geq 2$. We say $x$ is a *node*, or an *ordinary double point*, or *defines a nodal singularity* if $n = 2$.

These singularities are in some sense the simplest kind of singularities one can have on a curve over an algebraically closed field.

Lemma 53.16.3. Let $k$ be an algebraically closed field. Let $X$ be a reduced algebraic $1$-dimensional $k$-scheme. Let $x \in X$. The following are equivalent

$x$ defines a multicross singularity,

the $\delta $-invariant of $X$ at $x$ is the number of branches of $X$ at $x$ minus $1$,

there is a sequence of morphisms $U_ n \to U_{n - 1} \to \ldots \to U_0 = U \subset X$ where $U$ is an open neighbourhood of $x$, where $U_ n$ is nonsingular, and where each $U_ i \to U_{i - 1}$ is the glueing of two points as in Example 53.15.2.

**Proof.**
The equivalence of (1) and (2) is Lemma 53.16.1.

Assume (3). We will argue by descending induction on $i$ that all singularities of $U_ i$ are multicross. This is true for $U_ n$ as $U_ n$ has no singular points. If $U_ i$ is gotten from $U_{i + 1}$ by glueing $a, b \in U_{i + 1}$ to a point $c \in U_ i$, then we see that

\[ \mathcal{O}_{U_ i, c}^\wedge \subset \mathcal{O}_{U_{i + 1}, a}^\wedge \times \mathcal{O}_{U_{i + 1}, b}^\wedge \]

is the set of elements having the same residue classes in $k$. Thus the number of branches at $c$ is the sum of the number of branches at $a$ and $b$, and the $\delta $-invariant at $c$ is the sum of the $\delta $-invariants at $a$ and $b$ plus $1$ (because the displayed inclusion has codimension $1$). This proves that (2) holds as desired.

Assume the equivalent conditions (1) and (2). We may choose an open $U \subset X$ such that $x$ is the only singular point of $U$. Then we apply Lemma 53.15.4 to the normalization morphism

\[ U^\nu = U_ n \to U_{n - 1} \to \ldots \to U_1 \to U_0 = U \]

All we have to do is show that in none of the steps we are squishing a tangent vector. Suppose $U_{i + 1} \to U_ i$ is the smallest $i$ such that this is the squishing of a tangent vector $\theta $ at $u' \in U_{i + 1}$ lying over $u \in U_ i$. Arguing as above, we see that $u_ i$ is a multicross singularity (because the maps $U_ i \to \ldots \to U_0$ are glueing of pairs of points). But now the number of branches at $u'$ and $u$ is the same and the $\delta $-invariant of $U_ i$ at $u$ is $1$ bigger than the $\delta $-invariant of $U_{i + 1}$ at $u'$. By Lemma 53.16.1 this implies that $u$ cannot be a multicross singularity which is a contradiction. $\square$

Lemma 53.16.4. Let $k$ be an algebraically closed field. Let $X$ be a reduced algebraic $1$-dimensional $k$-scheme. Let $x \in X$ be a multicross singularity (Definition 53.16.2). If $X$ is Gorenstein, then $x$ is a node.

**Proof.**
The map $\mathcal{O}_{X, x} \to \mathcal{O}_{X, x}^\wedge $ is flat and unramified in the sense that $\kappa (x) = \mathcal{O}_{X, x}^\wedge /\mathfrak m_ x \mathcal{O}_{X, x}^\wedge $. (See More on Algebra, Section 15.43.) Thus $X$ is Gorenstein implies $\mathcal{O}_{X, x}$ is Gorenstein, implies $\mathcal{O}_{X, x}^\wedge $ is Gorenstein by Dualizing Complexes, Lemma 47.21.8. Thus it suffices to show that the ring $A$ in (53.16.0.1) with $n \geq 2$ is Gorenstein if and only if $n = 2$.

If $n = 2$, then $A = k[[x, y]]/(xy)$ is a complete intersection and hence Gorenstein. For example this follows from Duality for Schemes, Lemma 48.24.5 applied to $k[[x, y]] \to A$ and the fact that the regular local ring $k[[x, y]]$ is Gorenstein by Dualizing Complexes, Lemma 47.21.3.

Assume $n > 2$. If $A$ where Gorenstein, then $A$ would be a dualizing complex over $A$ (Duality for Schemes, Definition 48.24.1). Then $R\mathop{\mathrm{Hom}}\nolimits (k, A)$ would be equal to $k[n]$ for some $n \in \mathbf{Z}$, see Dualizing Complexes, Lemma 47.15.12. It would follow that $\mathop{\mathrm{Ext}}\nolimits ^1_ A(k, A) \cong k$ or $\mathop{\mathrm{Ext}}\nolimits ^1_ A(k, A) = 0$ (depending on the value of $n$; in fact $n$ has to be $-1$ but it doesn't matter to us here). Using the exact sequence

\[ 0 \to \mathfrak m_ A \to A \to k \to 0 \]

we find that

\[ \mathop{\mathrm{Ext}}\nolimits ^1_ A(k, A) = \mathop{\mathrm{Hom}}\nolimits _ A(\mathfrak m_ A, A)/A \]

where $A \to \mathop{\mathrm{Hom}}\nolimits _ A(\mathfrak m_ A, A)$ is given by $a \mapsto (a' \mapsto aa')$. Let $e_ i \in \mathop{\mathrm{Hom}}\nolimits _ A(\mathfrak m_ A, A)$ be the element that sends $(f_1, \ldots , f_ n) \in \mathfrak m_ A$ to $(0, \ldots , 0, f_ i, 0, \ldots , 0)$. The reader verifies easily that $e_1, \ldots , e_{n - 1}$ are $k$-linearly independent in $\mathop{\mathrm{Hom}}\nolimits _ A(\mathfrak m_ A, A)/A$. Thus $\dim _ k \mathop{\mathrm{Ext}}\nolimits ^1_ A(k, A) \geq n - 1 \geq 2$ which finishes the proof. (Observe that $e_1 + \ldots + e_ n$ is the image of $1$ under the map $A \to \mathop{\mathrm{Hom}}\nolimits _ A(\mathfrak m_ A, A)$.) $\square$

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