Lemma 53.19.4. Let $k$ be a field. Let $(A, \mathfrak m, \kappa )$ be a Nagata local $k$-algebra. The following are equivalent

1. $k \to A$ is as in Lemma 53.19.3,

2. $\kappa /k$ is separable, $A$ is reduced of dimension $1$, the $\delta$-invariant of $A$ is $1$, and $A$ has $2$ geometric branches.

If this holds, then the integral closure $A'$ of $A$ in its total ring of fractions has either $1$ or $2$ maximal ideals $\mathfrak m'$ and the extensions $\kappa (\mathfrak m')/k$ are separable.

Proof. In both cases $A$ and $A^\wedge$ are reduced. In case (2) because the completion of a reduced local Nagata ring is reduced (More on Algebra, Lemma 15.43.6). In both cases $A$ and $A^\wedge$ have dimension $1$ (More on Algebra, Lemma 15.43.1). The $\delta$-invariant and the number of geometric branches of $A$ and $A^\wedge$ agree by Varieties, Lemma 33.39.6 and More on Algebra, Lemma 15.108.7. Let $A'$ be the integral closure of $A$ in its total ring of fractions as in Varieties, Lemma 33.39.2. By Varieties, Lemma 33.39.5 we see that $A' \otimes _ A A^\wedge$ plays the same role for $A^\wedge$. Thus we may replace $A$ by $A^\wedge$ and assume $A$ is complete.

Assume (1) holds. It suffices to show that $A$ has two geometric branches and $\delta$-invariant $1$. We may assume $A = \kappa [[x, y]]/(ax^2 + bxy + cy^2)$ with $q = ax^2 + bxy + cy^2$ nondegenerate. There are two cases.

Case I: $q$ splits over $\kappa$. In this case we may after changing coordinates assume that $q = xy$. Then we see that

$A' = \kappa [[x, y]]/(x) \times \kappa [[x, y]]/(y)$

Case II: $q$ does not split. In this case $c \not= 0$ and nondegenerate means $b^2 - 4ac \not= 0$. Hence $\kappa ' = \kappa [t]/(a + bt + ct^2)$ is a degree $2$ separable extension of $\kappa$. Then $t = y/x$ is integral over $A$ and we conclude that

$A' = \kappa '[[x]]$

with $y$ mapping to $tx$ on the right hand side.

In both cases one verifies by hand that the $\delta$-invariant is $1$ and the number of geometric branches is $2$. In this way we see that (1) implies (2). Moreover we conclude that the final statement of the lemma holds.

Assume (2) holds. More on Algebra, Lemma 15.106.7 implies $A'$ either has two maximal ideals or $A'$ has one maximal ideal and $[\kappa (\mathfrak m') : \kappa ]_ s = 2$.

Case I: $A'$ has two maximal ideals $\mathfrak m'_1$, $\mathfrak m'_2$ with residue fields $\kappa _1$, $\kappa _2$. Since the $\delta$-invariant is the length of $A'/A$ and since there is a surjection $A'/A \to (\kappa _1 \times \kappa _2)/\kappa$ we see that $\kappa = \kappa _1 = \kappa _2$. Since $A$ is complete (and henselian by Algebra, Lemma 10.153.9) and $A'$ is finite over $A$ we see that $A' = A_1 \times A_2$ (by Algebra, Lemma 10.153.4). Since $A'$ is a normal ring it follows that $A_1$ and $A_2$ are discrete valuation rings. Hence $A_1$ and $A_2$ are isomorphic to $\kappa [[t]]$ (as $k$-algebras) by More on Algebra, Lemma 15.38.4. Since the $\delta$-invariant is $1$ we conclude that $A$ is the wedge of $A_1$ and $A_2$ (Varieties, Definition 33.40.4). It follows easily that $A \cong \kappa [[x, y]]/(xy)$.

Case II: $A'$ has a single maximal ideal $\mathfrak m'$ with residue field $\kappa '$ and $[\kappa ' : \kappa ]_ s = 2$. Arguing exactly as in Case I we see that $[\kappa ' : \kappa ] = 2$ and $\kappa '$ is separable over $\kappa$. Since $A'$ is normal we see that $A'$ is isomorphic to $\kappa '[[t]]$ (see reference above). Since $A'/A$ has length $1$ we conclude that

$A = \{ f \in \kappa '[[t]] \mid f(0) \in \kappa \}$

Then a simple computation shows that $A$ as in case (1). $\square$

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