The Stacks project

Proof. Assume $I \not= 0$. Write $I = (f_1, \ldots , f_ r)$. As $A$ is a UFD (More on Algebra, Lemma 15.121.2) we can write $f_ i = fg_ i$ where $f$ is the gcd of $f_1, \ldots , f_ r$. Thus the gcd of $g_1, \ldots , g_ r$ is $1$ which means that there is no height $1$ prime ideal over $g_1, \ldots , g_ r$. Then either $(g_1, \ldots , g_ r) = A$ which implies $I = (f)$ or if not, then $\dim (A) = 2$ implies that $V(g_1, \ldots , g_ r) = \{ \mathfrak m\} $, i.e., $\mathfrak m = \sqrt{(g_1, \ldots , g_ r)}$.

Assume $A/I$ reduced, i.e., $I$ radical. If $f$ is a unit, then since $I$ is radical we see that $I = \mathfrak m$. If $f \in \mathfrak m$, then we see that $f^ n$ maps to zero in $A/I$. Hence $f \in I$ by reducedness and we conclude $I = (f)$.

Assume $A/I$ has depth $1$. Then $\mathfrak m$ is not an associated prime of $A/I$. Since the class of $f$ modulo $I$ is annihilated by $g_1, \ldots , g_ r$, this implies that the class of $f$ is zero in $A/I$. Thus $I = (f)$ as desired. $\square$

Proof. Assume (3). Then $A^\wedge $ is reduced because $ax^2 + bxy + cy^2$ is either irreducible or a product of two nonassociated prime elements. Hence $A \subset A^\wedge $ is reduced. It follows that (1) is true.

Assume (1). Then $A$ cannot be Artinian, since it would not be reduced because $\mathfrak m \not= (0)$. Hence $\dim (A) \geq 1$, hence $\text{depth}(A) \geq 1$ by Algebra, Lemma 10.157.3. On the other hand $\dim (A) = 2$ implies $A$ is regular which contradicts the existence of $q$ by Algebra, Lemma 10.106.1. Thus $\dim (A) \leq 1$ and we conclude $\text{depth}(A) = 1$ by Algebra, Lemma 10.72.3. It follows that (2) is true.

Assume (2). Since the depth of $A$ is the same as the depth of $A^\wedge $ (More on Algebra, Lemma 15.43.2) and since the other conditions are insensitive to completion, we may assume that $A$ is complete. Choose $\kappa \to A$ as in More on Algebra, Lemma 15.38.3. Since $\dim _\kappa (\mathfrak m/\mathfrak m^2) = 2$ we can choose $x_0, y_0 \in \mathfrak m$ which map to a basis. We obtain a continuous $\kappa $-algebra map

by the rules $x \mapsto x_0$ and $y \mapsto y_0$. Let $q$ be the class of $ax_0^2 + bx_0y_0 + cy_0^2$ in $\text{Sym}^2_\kappa (\mathfrak m/\mathfrak m^2)$. Write $Q(x, y) = ax^2 + bxy + cy^2$ viewed as a polynomial in two variables. Then we see that

for some $a_{ij}$ in $A$. We want to prove that we can increase the order of vanishing by changing our choice of $x_0$, $y_0$. Suppose that $x_1, y_1 \in \mathfrak m^2$. Then

Nondegeneracy of $Q$ means exactly that $2ax_0 + by_0$ and $bx_0 + 2cy_0$ are a $\kappa $-basis for $\mathfrak m/\mathfrak m^2$, see discussion preceding the lemma. Hence we can certainly choose $x_1, y_1 \in \mathfrak m^2$ such that $Q(x_0 + x_1, y_0 + y_1) \in \mathfrak m^4$. Continuing in this fashion by induction we can find $x_ i, y_ i \in \mathfrak m^{i + 1}$ such that

Since $A$ is complete we can set $x_\infty = \sum x_ i$ and $y_\infty = \sum y_ i$ and we can consider the map $\kappa [[x, y]] \longrightarrow A$ sending $x$ to $x_\infty $ and $y$ to $y_\infty $. This map induces a surjection $\kappa [[x, y]]/(Q) \longrightarrow A$ by Algebra, Lemma 10.96.1. By Lemma 53.19.2 the kernel of $k[[x, y]] \to A$ is principal. But the kernel cannot contain a proper divisor of $Q$ as such a divisor would have degree $1$ in $x, y$ and this would contradict $\dim (\mathfrak m/\mathfrak m^2) = 2$. Hence $Q$ generates the kernel as desired. $\square$

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

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

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

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

Proof. The “correct” proof of this lemma is to prove that this ideal is the $(n - r)$th Fitting ideal of a module of continuous differentials of $A/I$ over $k$. Here is a direct proof. If $g_1, \ldots g_ l$ is a second set of generators of $I$, then we can write $g_ s = \sum a_{sj}f_ j$ and we have the equality of matrices

The final term is zero in $A/I$. By the Cauchy-Binet formula we see that the ideal of minors for the $g_ s$ is contained in the ideal for the $f_ j$. By symmetry these ideals are the same. If $y_1, \ldots , y_ n \in \mathfrak m_ A$ is a second regular system of parameters, then the matrix $(\partial y_ j/\partial x_ i)$ is invertible and we can use the chain rule for differentiation. Some details omitted. $\square$

Proof. By Lemma 53.19.5 we may change our system of coordinates and the choice of generators during the proof.

If (1) holds, then we may change coordinates such that $x_1, \ldots , x_{n - 2}$ map to zero in $A/I$ and $A/I = k[[x_{n - 1}, x_ n]]/(a x_{n - 1}^2 + b x_{n - 1}x_ n + c x_ n^2)$ for some nondegenerate quadric $a x_{n - 1}^2 + b x_{n - 1}x_ n + c x_ n^2$. Then we can explicitly compute to show that both (2) and (3) are true.

Assume the $(n - 1) \times (n - 1)$ minors of the matrix $(\partial f_ j/\partial x_ i)$ generate $I + \mathfrak m_ A$. Suppose that for some $i$ and $j$ the partial derivative $\partial f_ j/\partial x_ i$ is a unit in $A$. Then we may use the system of parameters $f_ j, x_1, \ldots , x_{i - 1}, \hat x_ i, x_{i + 1}, \ldots , x_ n$ and the generators $f_ j, f_1, \ldots , f_{j - 1}, \hat f_ j, f_{j + 1}, \ldots , f_ m$ of $I$. Then we get a regular system of parameters $x_1, \ldots , x_ n$ and generators $x_1, f_2, \ldots , f_ m$ of $I$. Next, we look for an $i \geq 2$ and $j \geq 2$ such that $\partial f_ j/\partial x_ i$ is a unit in $A$. If such a pair exists, then we can make a replacement as above and assume that we have a regular system of parameters $x_1, \ldots , x_ n$ and generators $x_1, x_2, f_3, \ldots , f_ m$ of $I$. Continuing, in finitely many steps we reach the situation where we have a regular system of parameters $x_1, \ldots , x_ n$ and generators $x_1, \ldots , x_ t, f_{t + 1}, \ldots , f_ m$ of $I$ such that $\partial f_ j/\partial x_ i \in \mathfrak m_ A$ for all $i, j \geq t + 1$.

In this case the matrix of partial derivatives has the following block shape

Hence every $(n - 1) \times (n - 1)$-minor is in $\mathfrak m_ A^{n - 1 - t}$. Note that $I \not= \mathfrak m_ A$ otherwise the ideal of minors would contain $1$. It follows that $n - 1 - t \leq 1$ because there is an element of $\mathfrak m_ A \setminus \mathfrak m_ A^2 + I$ (otherwise $I = \mathfrak m_ A$ by Nakayama). Thus $t \geq n - 2$. We have seen that $t \not= n$ above and similarly if $t = n - 1$, then there is an invertible $(n - 1) \times (n - 1)$-minor which is disallowed as well. Hence $t = n - 2$. Then $A/I$ is a quotient of $k[[x_{n - 1}, x_ n]]$ and Lemma 53.19.2 implies in both cases (2) and (3) that $I$ is generated by $x_1, \ldots , x_{n - 2}, f$ for some $f = f(x_{n - 1}, x_ n)$. In this case the condition on the minors exactly says that the quadratic term in $f$ is nondegenerate, i.e., $A/I$ is as in Lemma 53.19.3. $\square$

Proof. First assume that $k$ is algebraically closed. In this case the equivalence of (1) and (3) is trivial. The equivalence of (1) and (3) with (2) holds because the only nondegenerate quadric in two variables is $xy$ up to change in coordinates. The equivalence of (1) and (6) is Lemma 53.16.1. After replacing $X$ by an affine neighbourhood of $x$, we may assume there is a closed immersion $X \to \mathbf{A}^ n_ k$ mapping $x$ to $0$. Let $f_1, \ldots , f_ m \in k[x_1, \ldots , x_ n]$ be generators for the ideal $I$ of $X$ in $\mathbf{A}^ n_ k$. Then $\Omega _{X/k}$ corresponds to the $R = k[x_1, \ldots , x_ n]/I$-module $\Omega _{R/k}$ which has a presentation

(See Algebra, Sections 10.131 and 10.134.) The first Fitting ideal of $\Omega _{R/k}$ is thus the ideal generated by the $(n - 1) \times (n - 1)$-minors of the matrix $(\partial f_ j/\partial x_ i)$. Hence (2), (4), (5) are equivalent by Lemma 53.19.6 applied to the completion of $k[x_1, \ldots , x_ n] \to R$ at the maximal ideal $(x_1, \ldots , x_ n)$.

Now assume $k$ is an arbitrary field. In cases (2), (4), (5), (6) the residue field $\kappa (x)$ is separable over $k$. Let us show this holds as well in cases (1) and (3). Namely, let $Z \subset X$ be the closed subscheme of $X$ defined by the first Fitting ideal of $\Omega _{X/k}$. The formation of $Z$ commutes with field extension (Divisors, Lemma 31.10.1). If (1) or (3) is true, then there exists a point $\overline{x}$ of $X_{\overline{k}}$ such that $\overline{x}$ is an isolated point of multiplicity $1$ of $Z_{\overline{k}}$ (as we have the equivalence of the conditions of the lemma over $\overline{k}$). In particular $Z_{\overline{x}}$ is geometrically reduced at $\overline{x}$ (because $\overline{k}$ is algebraically closed). Hence $Z$ is geometrically reduced at $x$ (Varieties, Lemma 33.6.6). In particular, $Z$ is reduced at $x$, hence $Z = \mathop{\mathrm{Spec}}(\kappa (x))$ in a neighbourhood of $x$ and $\kappa (x)$ is geometrically reduced over $k$. This means that $\kappa (x)/k$ is separable (Algebra, Lemma 10.44.1).

The argument of the previous paragraph shows that if (1) or (3) holds, then the first Fitting ideal of $\Omega _{X/k}$ generates $\mathfrak m_ x$. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is flat and since $\mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is reduced and has depth $1$, we see that (4) and (5) hold (use Algebra, Lemmas 10.164.2 and 10.163.2). Conversely, (4) implies (5) by Algebra, Lemma 10.157.3. If (5) holds, then $Z$ is geometrically reduced at $x$ (because $\kappa (x)/k$ separable and $Z$ is $x$ in a neighbourhood). Hence $Z_{\overline{k}}$ is reduced at any point $\overline{x}$ of $X_{\overline{k}}$ lying over $x$. In other words, the first fitting ideal of $\Omega _{X_{\overline{k}}/\overline{k}}$ generates $\mathfrak m_{\overline{x}}$ in $\mathcal{O}_{X_{\overline{k}, \overline{x}}}$. Moreover, since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is flat we see that $\text{depth}(\mathcal{O}_{X_{\overline{k}}, \overline{x}}) = 1$ (see reference above). Hence (5) holds for $\overline{x} \in X_{\overline{k}}$ and we conclude that (3) holds (because of the equivalence over algebraically closed fields). In this way we see that (1), (3), (4), (5) are equivalent.

The equivalence of (2) and (6) follows from Lemma 53.19.4.

Finally, we prove the equivalence of (2) $=$ (6) with (1) $=$ (3) $=$ (4) $=$ (5). First we note that the geometric number of branches of $X$ at $x$ and the geometric number of branches of $X_{\overline{k}}$ at $\overline{x}$ are equal by Varieties, Lemma 33.40.2. We conclude from the information available to us at this point that in all cases this number is equal to $2$. On the other hand, in case (1) it is clear that $X$ is geometrically reduced at $x$, and hence

by Varieties, Lemma 33.39.8. Since in case (1) the right hand side is $1$, this forces the $\delta $-invariant of $X$ at $x$ to be $1$ (because if it were zero, then $\mathcal{O}_{X, x}$ would be a discrete valuation ring by Varieties, Lemma 33.39.4 which is unibranch, a contradiction). Thus (5) holds. Conversely, if (2) $=$ (5) is true, then assumptions (a), (b), (c) of Varieties, Lemma 33.27.6 hold for $x \in X$ by Lemma 53.19.4. Thus Varieties, Lemma 33.39.9 applies and shows that we have equality in the above displayed inequality. We conclude that (5) holds for $\overline{x} \in X_{\overline{k}}$ and we are back in case (1) by the equivalence of the conditions over an algebraically closed field. $\square$

Proof. This follows from the discussion in Remark 53.19.9 and Lemma 53.19.7. $\square$

Proof. If $x$ is a closed point of $X$, then $y$ is too (look at residue fields). But conversely, this need not be the case, i.e., it can happen that a closed point of $Y$ maps to a nonclosed point of $X$. However, in this case $y$ cannot be a node. Namely, then $X$ would be geometrically unibranch at $x$ (because $x$ would be a generic point of $X$ and $\mathcal{O}_{X, x}$ would be Artinian and any Artinian local ring is geometrically unibranch), hence $Y$ is geometrically unibranch at $y$ (Varieties, Lemma 33.40.3), which means that $y$ cannot be a node by Lemma 53.19.7. Thus we may and do assume that both $x$ and $y$ are closed points.

Choose algebraic closures $\overline{k}$, $\overline{K}$ and a map $\overline{k} \to \overline{K}$ extending the given map $k \to K$. Using the equivalence of (1) and (3) in Lemma 53.19.7 we reduce to the case where $k$ and $K$ are algebraically closed. In this case we can argue as in the proof of Lemma 53.19.7 that the geometric number of branches and $\delta $-invariants of $X$ at $x$ and $Y$ at $y$ are the same. Another argument can be given by choosing an isomorphism $k[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \to \mathcal{O}_{X, x}^\wedge $ of $k$-algebras as in Varieties, Lemma 33.21.1. By Varieties, Lemma 33.21.2 this gives an isomorphism $K[[x_1, \ldots , x_ n]]/(g_1, \ldots , g_ m) \to \mathcal{O}_{Y, y}^\wedge $ of $K$-algebras. By definition we have to show that

if and only if

We encourage the reader to prove this for themselves. Since $k$ and $K$ are algebraically closed fields, this is the same as asking these rings to be as in Lemma 53.19.3. Via Lemma 53.19.6 this translates into a statement about the $(n - 1) \times (n - 1)$-minors of the matrix $(\partial g_ j/\partial x_ i)$ which is clearly independent of the field used. We omit the details. $\square$

Proof. By Lemma 53.19.12 we may base change to the algebraic closure of $k$. Then the residue fields of $x$ and $y$ are $k$. Hence the map $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge $ is an isomorphism (for example by Étale Morphisms, Lemma 41.11.3 or More on Algebra, Lemma 15.43.9). Thus the lemma is clear. $\square$

Proof. Follows immediately from the characterization of nodes in Lemma 53.19.7. $\square$

Proof. We urge the reader to find their own proof of this lemma; what follows is just putting together earlier results and may hide what is really going on.

Assume (2). Since $Z \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite (Morphisms, Lemma 29.35.10) we see that the residue fields of points $x \in Z$ are finite over $k$ (as well as separable) by Morphisms, Lemma 29.20.5. Hence each $x \in Z$ is a closed point of $X$ by Morphisms, Lemma 29.20.2. The local ring $\mathcal{O}_{X, x}$ is Cohen-Macaulay by Algebra, Lemma 10.135.3. Since $\dim (\mathcal{O}_{X, x}) = 1$ by dimension theory (Varieties, Section 33.20), we conclude that $\text{depth}(\mathcal{O}_{X, x})) = 1$. Thus $x$ is a node by Lemma 53.19.7. If $x \in X$, $x \not\in Z$, then $X \to \mathop{\mathrm{Spec}}(k)$ is smooth at $x$ by Divisors, Lemma 31.10.3.

Assume (1). Under this assumption $X$ is geometrically reduced at every closed point (see Varieties, Lemma 33.6.6). Hence $X \to \mathop{\mathrm{Spec}}(k)$ is smooth on a dense open by Varieties, Lemma 33.25.7. Thus $Z$ is closed and consists of closed points. By Divisors, Lemma 31.10.3 the morphism $X \setminus Z \to \mathop{\mathrm{Spec}}(k)$ is smooth. Hence $X \setminus Z$ is a local complete intersection by Morphisms, Lemma 29.34.7 and the definition of a local complete intersection in Morphisms, Definition 29.30.1. By Lemma 53.19.7 for every point $x \in Z$ the local ring $\mathcal{O}_{Z, x}$ is equal to $\kappa (x)$ and $\kappa (x)$ is separable over $k$. Thus $Z \to \mathop{\mathrm{Spec}}(k)$ is unramified (Morphisms, Lemma 29.35.11). Finally, Lemma 53.19.7 via part (3) of Lemma 53.19.3, shows that $\mathcal{O}_{X, x}$ is a complete intersection in the sense of Divided Power Algebra, Definition 23.8.5. However, Divided Power Algebra, Lemma 23.8.8 and Morphisms, Lemma 29.30.9 show that this agrees with the notion used to define a local complete intersection scheme over a field and the proof is complete. $\square$

Proof. The Gorenstein assertion follows from Lemma 53.19.15 and Duality for Schemes, Lemma 48.24.5. Or you can use that it suffices to check after passing to the algebraic closure (Duality for Schemes, Lemma 48.25.1), then use that a Noetherian local ring is Gorenstein if and only if its completion is so (by Dualizing Complexes, Lemma 47.21.8), and then prove that the local rings $k[[t]]$ and $k[[x, y]]/(xy)$ are Gorenstein by hand.

To see that $X$ is geometrically reduced, it suffices to show that $X_{\overline{k}}$ is reduced (Varieties, Lemmas 33.6.3 and 33.6.4). But $X_{\overline{k}}$ is a nodal curve over an algebraically closed field. Thus the complete local rings of $X_{\overline{k}}$ are isomorphic to either $\overline{k}[[t]]$ or $\overline{k}[[x, y]]/(xy)$ which are reduced as desired. $\square$

Proof. Since $X$ and $Y$ are reduced and equidimensional of dimension $1$, we see that $Y$ is the scheme theoretic union of a subset of the irreducible components of $X$ (in a reduced ring $(0)$ is the intersection of the minimal primes). Let $y \in Y$ be a closed point. If $y$ is in the smooth locus of $X \to \mathop{\mathrm{Spec}}(k)$, then $y$ is on a unique irreducible component of $X$ and we see that $Y$ and $X$ agree in an open neighbourhood of $y$. Hence $Y \to \mathop{\mathrm{Spec}}(k)$ is smooth at $y$. If $y$ is a node of $X$ but still lies on a unique irreducible component of $X$, then $y$ is a node on $Y$ by the same argument. Suppose that $y$ lies on more than $1$ irreducible component of $X$. Since the number of geometric branches of $X$ at $y$ is $2$ by Lemma 53.19.7, there can be at most $2$ irreducible components passing through $y$ by Properties, Lemma 28.15.5. If $Y$ contains both of these, then again $Y = X$ in an open neighbourhood of $y$ and $y$ is a node of $Y$. Finally, assume $Y$ contains only one of the irreducible components. After replacing $X$ by an open neighbourhood of $x$ we may assume $Y$ is one of the two irreducble components and $Z$ is the other. By Properties, Lemma 28.15.5 again we see that $X$ has two branches at $y$, i.e., the local ring $\mathcal{O}_{X, y}$ has two branches and that these branches come from $\mathcal{O}_{Y, y}$ and $\mathcal{O}_{Z, y}$. Write $\mathcal{O}_{X, y}^\wedge \cong \kappa (y)[[u, v]]/(uv)$ as in Remark 53.19.9. The field $\kappa (y)$ is finite separable over $k$ by Lemma 53.19.7 for example. Thus, after possibly switching the roles of $u$ and $v$, the completion of the map $\mathcal{O}_{X, y} \to \mathcal{O}_{Y, Y}$ corresponds to $\kappa (y)[[u, v]]/(uv) \to \kappa (y)[[u]]$ and the completion of the map $\mathcal{O}_{X, y} \to \mathcal{O}_{Y, Y}$ corresponds to $\kappa (y)[[u, v]]/(uv) \to \kappa (y)[[v]]$. The scheme theoretic intersection of $Y \cap Z$ is cut out by the sum of their ideas which in the completion is $(u, v)$, i.e., the maximal ideal. Thus (2)(a) and (2)(b) are clear. Finally, (2)(c) holds: the completion of $\mathcal{O}_{Y, y}$ is regular, hence $\mathcal{O}_{Y, y}$ is regular (More on Algebra, Lemma 15.43.4) and $\kappa (y)/k$ is separable, hence smoothness in an open neighbourhood by Algebra, Lemma 10.140.5. $\square$