The Stacks project

Proof. The spectrum of $A$ has finitely many primes $\mathfrak p_1, \ldots , \mathfrak p_ n$ besides $\mathfrak m$ and these are all minimal, see Algebra, Lemma 10.31.6. Then the equivalence of (1) and (2) follows from Algebra, Lemma 10.17.2. Clearly, (3) implies (2). Conversely, if (2) is true, then the spectrum of $A_ f$ is the subset $\{ \mathfrak p_1, \ldots , \mathfrak p_ n\} $ of $\mathop{\mathrm{Spec}}(A)$ with induced topology, see Algebra, Lemma 10.17.5. This is a finite discrete topological space. Hence $A_ f = \prod _{\mathfrak p\text{ minimal}} A_\mathfrak p$ by Algebra, Proposition 10.60.7. The existence of an $f$ is asserted in Algebra, Lemma 10.60.8.

Assume $A$ has depth $1$. (This is the maximum by Algebra, Lemma 10.72.3 and holds if $A$ is reduced by Algebra, Lemma 10.157.3.) Then $\mathfrak m$ is not an associated prime of $A$. Every minimal prime of $A$ is an associated prime (Algebra, Proposition 10.63.6). Hence the set of nonzerodivisors of $A$ is exactly the set of elements not contained in any of the minimal primes by Algebra, Lemma 10.63.9. Thus (4) is equivalent to (2). Part (5) is equivalent to (3) by Algebra, Lemma 10.25.4.

Then $A_\mathfrak p$ is a field for $\mathfrak p \subset A$ minimal, see Algebra, Lemma 10.25.1. Hence (3) is equivalent ot (6). $\square$

Proof. The total ring of fractions is essentially of finite type over $A$ hence $A \to A'$ is finite because $A$ is Nagata, see Algebra, Lemma 10.162.2. The ring $A'$ is normal for example by Algebra, Lemma 10.37.16 and 10.31.6. The ring $A'$ is Nagata for example by Algebra, Lemma 10.162.5. Choose $f \in \mathfrak m$ as in Lemma 33.39.1. As $A' \subset A_ f$ it is clear that $A_ f = A'_ f$. Hence the support of the finite $A$-module $A'/A$ is contained in $\{ \mathfrak m\} $. It follows that it has finite length by Algebra, Lemma 10.62.3. $\square$

Proof. If $A$ is a discrete valuation ring, then $A$ is normal and the ring $A'$ is equal to $A$. Conversely, if the $\delta $-invariant of $A$ is $0$, then $A$ is integrally closed in its total ring of fractions which implies that $A$ is normal (Algebra, Lemma 10.37.16) and this forces $A$ to be a discrete valuation ring by Algebra, Lemma 10.119.7. $\square$

Proof. Observe that $A^\wedge $ is reduced, see More on Algebra, Lemma 15.43.6. The rings $A^ h$ and $A^{sh}$ are reduced by More on Algebra, Lemma 15.45.4. The dimensions of $A$, $A^ h$, $A^{sh}$, and $A^\wedge $ are the same by More on Algebra, Lemmas 15.43.1 and 15.45.7.

Recall that a Noetherian local ring is Nagata if and only if the formal fibres of $A$ are geometrically reduced, see More on Algebra, Lemma 15.52.4. This property is inherited by $A^ h$ and $A^{sh}$, see the material in More on Algebra, Section 15.51 and especially Lemma 15.51.8. The completion is Nagata by Algebra, Lemma 10.162.8.

Now we come to the statement on integral closures. Before continuing let us pick $f \in \mathfrak m$ as in Lemma 33.39.1. Then the image of $f$ in $A^ h$, $A^{sh}$, and $A^\wedge $ clearly is an element satisfying properties (1) – (6) in that ring.

Since $A \to A'$ is finite we see that $A' \otimes _ A A^ h$ and $A' \otimes _ A A^{sh}$ is the product of henselian local rings finite over $A^ h$ and $A^{sh}$, see Algebra, Lemma 10.153.4. Each of these local rings is the henselization of $A'$ at a maximal ideal $\mathfrak m' \subset A'$ lying over $\mathfrak m$, see Algebra, Lemma 10.156.1 or 10.156.3. Hence these local rings are normal domains by More on Algebra, Lemma 15.45.6. It follows that $A' \otimes _ A A^ h$ and $A' \otimes _ A A^{sh}$ are normal rings. Since $A^ h \to A' \otimes _ A A^ h$ and $A^{sh} \to A' \otimes _ A A^{sh}$ are finite (hence integral) and since $A' \otimes _ A A^ h \subset (A^ h)_ f = Q(A^ h)$ and $A' \otimes _ A A^{sh} \subset (A^{sh})_ f = Q(A^{sh})$ we conclude that $A' \otimes _ A A^ h$ and $A' \otimes _ A A^{sh}$ are the desired integral closures.

For the completion we argue in entirely the same manner. First, by Algebra, Lemma 10.97.8 we have

The local rings $A'_{\mathfrak m'}$ are normal and have dimension $1$ (by Algebra, Lemma 10.113.2 for example or the discussion in Algebra, Section 10.112). Thus $A'_{\mathfrak m'}$ is a discrete valuation ring, see Algebra, Lemma 10.119.7. Hence $(A'_{\mathfrak m'})^\wedge $ is a discrete valuation ring by More on Algebra, Lemma 15.43.5. It follows that $A' \otimes _ A A^\wedge $ is a normal ring and we can conclude in exactly the same manner as before. $\square$

Proof. Let us do this in case of the completion $B = A^\wedge $; the other cases are proved in exactly the same manner. Let $A'$, resp. $B'$ be the integral closure of $A$, resp. $B$ in its total ring of fractions. Then $B' = A' \otimes _ A B$ by Lemma 33.39.5. Hence $B'/B = A'/A \otimes _ A B$. The equality now follows from Algebra, Lemma 10.52.13 and the fact that $B \otimes _ A \kappa _ A = \kappa _ B$. $\square$

Proof. Set $A = \mathcal{O}_{X, x}$ and $B = \mathcal{O}_{Y, y}$. By Lemma 33.6.2 we see that $A$ is geometrically reduced. Hence $B$ is a localization of $A \otimes _ k K$. Let $A \to A'$ be as in Lemma 33.39.2. Then

is finite over $B$ and $B \to B'$ induces an isomorphism on total rings of fractions. Namely, pick $f \in \mathfrak m_ A$ satisfying (1) – (6) of Lemma 33.39.1; since $\dim (B) = 1$ we see that $f \in \mathfrak m_ B$ playes the same role for $B$ and we see that $B_ f = B'_ f$ because $A_ f = A'_ f$. Let $B''$ be the integral closure of $B$ in its total ring of fractions as in Lemma 33.39.2. Then $B' \subset B''$. Thus the $\delta $-invariant of $Y$ at $y$ is $\text{length}_ B(B''/B)$ and

by Algebra, Lemma 10.52.13 since $A \to B$ is flat (as a localization of $A \to A \otimes _ k K$). Since $\text{length}_ A(A'/A)$ is the $\delta $-invariant of $X$ at $x$ and since $\text{length}_ B(B/\mathfrak m_ A B) \geq 1$ the lemma is proved. $\square$

Proof. Set $A = \mathcal{O}_{X, x}$ and $B = \mathcal{O}_{Y, y}$. By Lemma 33.27.6 we see that $A$ is geometrically reduced. Hence $B$ is a localization of $A \otimes _ k K$. Let $A \to A'$ be as in Lemma 33.39.2. By Lemma 33.27.6 we see that $A' \otimes _ k K$ is normal. Hence

is normal, finite over $B$, and $B \to B'$ induces an isomorphism on total rings of fractions. Namely, pick $f \in \mathfrak m_ A$ satisfying (1) – (6) of Lemma 33.39.1; since $\dim (B) = 1$ we see that $f \in \mathfrak m_ B$ playes the same role for $B$ and we see that $B_ f = B'_ f$ because $A_ f = A'_ f$. It follows that $B \to B'$ is as in Lemma 33.39.2 for $B$. Thus we have to show that $\text{length}_ A(A'/A) = \text{length}_ B(B'/B) = \text{length}_ B((A'/A) \otimes _ A B)$. Since $A \to B$ is flat (as a localization of $A \to A \otimes _ k K$) and since $\mathfrak m_ B = \mathfrak m_ A B$ (because $B/\mathfrak m_ A B$ is zero dimensional by the remarks above and a localization of $K \otimes _ k \kappa (x)$ which is reduced as $\kappa (x)$ is separable over $k$) we conclude by Algebra, Lemma 10.52.13. $\square$