The Stacks project

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15.39 Geometric regularity and formal smoothness

In this section we combine the results of the previous sections to prove the following characterization of geometrically regular local rings over fields. We then recycle some of our arguments to prove a characterization of formally smooth maps in the $\mathfrak m$-adic topology between Noetherian local rings.

Theorem 15.39.1. Let $k$ be a field. Let $(A, \mathfrak m, K)$ be a Noetherian local $k$-algebra. If the characteristic of $k$ is zero then the following are equivalent

  1. $A$ is a regular local ring, and

  2. $k \to A$ is formally smooth in the $\mathfrak m$-adic topology.

If the characteristic of $k$ is $p > 0$ then the following are equivalent

  1. $A$ is geometrically regular over $k$,

  2. $k \to A$ is formally smooth in the $\mathfrak m$-adic topology.

  3. for all $k \subset k' \subset k^{1/p}$ finite over $k$ the ring $A \otimes _ k k'$ is regular,

  4. $A$ is regular and the canonical map $H_1(L_{K/k}) \to \mathfrak m/\mathfrak m^2$ is injective, and

  5. $A$ is regular and the map $\Omega _{k/\mathbf{F}_ p} \otimes _ k K \to \Omega _{A/\mathbf{F}_ p} \otimes _ A K$ is injective.

Proof. If the characteristic of $k$ is zero, then the equivalence of (1) and (2) follows from Lemmas 15.37.2 and 15.37.5.

If the characteristic of $k$ is $p > 0$, then it follows from Proposition 15.34.1 that (1), (3), (4), and (5) are equivalent. Assume (2) holds. By Lemma 15.36.8 we see that $k' \to A' = A \otimes _ k k'$ is formally smooth for the $\mathfrak m' = \mathfrak mA$-adic topology. Hence if $k \subset k'$ is finite purely inseparable, then $A'$ is a regular local ring by Lemma 15.37.2. Thus we see that (1) holds.

Finally, we will prove that (5) implies (2). Choose a solid diagram

\[ \xymatrix{ A \ar[r]_{\bar\psi } \ar@{-->}[rd] & B/J \\ k \ar[u]^ i \ar[r]^\varphi & B \ar[u]_\pi } \]

as in Definition 15.36.1. As $J^2 = 0$ we see that $J$ has a canonical $B/J$ module structure and via $\bar\psi $ an $A$-module structure. As $\bar\psi $ is continuous for the $\mathfrak m$-adic topology we see that $\mathfrak m^ nJ = 0$ for some $n$. Hence we can filter $J$ by $B/J$-submodules $0 \subset J_1 \subset J_2 \subset \ldots \subset J_ n = J$ such that each quotient $J_{t + 1}/J_ t$ is annihilated by $\mathfrak m$. Considering the sequence of ring maps $B \to B/J_1 \to B/J_2 \to \ldots \to B/J$ we see that it suffices to prove the existence of the dotted arrow when $J$ is annihilated by $\mathfrak m$, i.e., when $J$ is a $K$-vector space.

Assume given a diagram as above such that $J$ is annihilated by $\mathfrak m$. By Lemma 15.37.5 we see that $\mathbf{F}_ p \to A$ is formally smooth in the $\mathfrak m$-adic topology. Hence we can find a ring map $\psi : A \to B$ such that $\pi \circ \psi = \bar\psi $. Then $\psi \circ i, \varphi : k \to B$ are two maps whose compositions with $\pi $ are equal. Hence $D = \psi \circ i - \varphi : k \to J$ is a derivation. By Algebra, Lemma 10.130.3 we can write $D = \xi \circ \text{d}$ for some $k$-linear map $\xi : \Omega _{k/\mathbf{F}_ p} \to J$. Using the $K$-vector space structure on $J$ we extend $\xi $ to a $K$-linear map $\xi ' : \Omega _{k/\mathbf{F}_ p} \otimes _ k K \to J$. Using (5) we can find a $K$-linear map $\xi '' : \Omega _{A/\mathbf{F}_ p} \otimes _ A K$ whose restriction to $\Omega _{k/\mathbf{F}_ p} \otimes _ k K$ is $\xi '$. Write

\[ D' : A \xrightarrow {\text{d}} \Omega _{A/\mathbf{F}_ p} \to \Omega _{A/\mathbf{F}_ p} \otimes _ A K \xrightarrow {\xi ''} J. \]

Finally, set $\psi ' = \psi - D' : A \to B$. The reader verifies that $\psi '$ is a ring map such that $\pi \circ \psi ' = \bar\psi $ and such that $\psi ' \circ i = \varphi $ as desired. $\square$

Example 15.39.2. Let $k$ be a field of characteristic $p > 0$. Suppose that $a \in k$ is an element which is not a $p$th power. A standard example of a geometrically regular local $k$-algebra whose residue field is purely inseparable over $k$ is the ring

\[ A = k[x, y]_{(x, y^ p - a)}/(y^ p - a - x) \]

Namely, $A$ is a localization of a smooth algebra over $k$ hence $k \to A$ is formally smooth, hence $k \to A$ is formally smooth for the $\mathfrak m$-adic topology. A closely related example is the following. Let $k = \mathbf{F}_ p(s)$ and $K = \mathbf{F}_ p(t)^{perf}$. We claim the ring map

\[ k \longrightarrow A = K[[x]],\quad s \longmapsto t + x \]

is formally smooth for the $(x)$-adic topology on $A$. Namely, $\Omega _{k/\mathbf{F}_ p}$ is $1$-dimensional with basis $\text{d}s$. It maps to the element $\text{d}x + \text{d}t = \text{d}x$ in $\Omega _{A/\mathbf{F}_ p}$. We leave it to the reader to show that $\Omega _{A/\mathbf{F}_ p}$ is free on $\text{d}x$ as an $A$-module. Hence we see that condition (5) of Theorem 15.39.1 holds and we conclude that $k \to A$ is formally smooth in the $(x)$-adic topology.

Lemma 15.39.3. Let $A \to B$ be a local homomorphism of Noetherian local rings. Assume $A \to B$ is formally smooth in the $\mathfrak m_ B$-adic topology. Then $A \to B$ is flat.

Proof. We may assume that $A$ and $B$ a Noetherian complete local rings by Lemma 15.36.4 and Algebra, Lemma 10.96.6 (this also uses Algebra, Lemma 10.38.9 and 10.96.3 to see that flatness of the map on completions implies flatness of $A \to B$). Choose a commutative diagram

\[ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } \]

as in Lemma 15.38.3 with $R \to S$ flat. Let $I \subset R$ be the kernel of $R \to A$. Because $B$ is formally smooth over $A$ we see that the $A$-algebra map

\[ S/IS \longrightarrow B \]

has a section, see Lemma 15.36.5. Hence $B$ is a direct summand of the flat $A$-module $S/IS$ (by base change of flatness, see Algebra, Lemma 10.38.7), whence flat. $\square$

Lemma 15.39.4. Let $A \to B$ be a local homomorphism of Noetherian local rings. Assume $A \to B$ is formally smooth in the $\mathfrak m_ B$-adic topology. Let $K$ be the residue field of $B$. Then the Jacobi-Zariski sequence for $A \to B \to K$ gives an exact sequence

\[ 0 \to H_1(\mathop{N\! L}\nolimits _{K/A}) \to \mathfrak m_ B/\mathfrak m_ B^2 \to \Omega _{B/A} \otimes _ B K \to \Omega _{K/A} \to 0 \]

Proof. Observe that $\mathfrak m_ B/\mathfrak m_ B^2 = H_1(\mathop{N\! L}\nolimits _{K/B})$ by Algebra, Lemma 10.132.6. By Algebra, Lemma 10.132.4 it remains to show injectivity of $H_1(\mathop{N\! L}\nolimits _{K/A}) \to \mathfrak m_ B/\mathfrak m_ B^2$. With $k$ the residue field of $A$, the Jacobi-Zariski sequence for $A \to k \to K$ gives $\Omega _{K/A} = \Omega _{K/k}$ and an exact sequence

\[ \mathfrak m_ A/\mathfrak m_ A^2 \otimes _ k K \to H_1(\mathop{N\! L}\nolimits _{K/A}) \to H_1(\mathop{N\! L}\nolimits _{K/k}) \to 0 \]

Set $\overline{B} = B \otimes _ A k$. Since $\overline{B}$ is regular the ideal $\mathfrak m_{\overline{B}}$ is generated by a regular sequence. Applying Lemmas 15.29.9 and 15.29.7 to $\mathfrak m_ A B \subset \mathfrak m_ B$ we find $\mathfrak m_ A B / (\mathfrak m_ AB \cap \mathfrak m_ B^2) = \mathfrak m_ A B / \mathfrak m_ A \mathfrak m_ B$ which is equal to $\mathfrak m_ A/\mathfrak m_ A^2 \otimes _ k K$ as $A \to B$ is flat by Lemma 15.39.3. Thus we obtain a short exact sequence

\[ 0 \to \mathfrak m_ A/\mathfrak m_ A^2 \otimes _ k K \to \mathfrak m_ B/\mathfrak m_ B^2 \to \mathfrak m_{\overline{B}}/\mathfrak m_{\overline{B}}^2 \to 0 \]

Functoriality of the Jacobi-Zariski sequences shows that we obtain a commutative diagram

\[ \xymatrix{ & \mathfrak m_ A/\mathfrak m_ A^2 \otimes _ k K \ar[d] \ar[r] & H_1(\mathop{N\! L}\nolimits _{K/A}) \ar[d] \ar[r] & H_1(\mathop{N\! L}\nolimits _{K/k}) \ar[d] \ar[r] & 0 \\ 0 \ar[r] & \mathfrak m_ A/\mathfrak m_ A^2 \otimes _ k K \ar[r] & \mathfrak m_ B/\mathfrak m_ B^2 \ar[r] & \mathfrak m_{\overline{B}}/\mathfrak m_{\overline{B}}^2 \ar[r] & 0 } \]

The left vertical arrow is injective by Theorem 15.39.1 as $k \to \overline{B}$ is formally smooth in the $\mathfrak m_{\overline{B}}$-adic topology by Lemma 15.36.8. This finishes the proof by the snake lemma. $\square$

Proposition 15.39.5. Let $A \to B$ be a local homomorphism of Noetherian local rings. Let $k$ be the residue field of $A$ and $\overline{B} = B \otimes _ A k$ the special fibre. The following are equivalent

  1. $A \to B$ is flat and $\overline{B}$ is geometrically regular over $k$,

  2. $A \to B$ is flat and $k \to \overline{B}$ is formally smooth in the $\mathfrak m_{\overline{B}}$-adic topology, and

  3. $A \to B$ is formally smooth in the $\mathfrak m_ B$-adic topology.

Proof. The equivalence of (1) and (2) follows from Theorem 15.39.1.

Assume (3). By Lemma 15.39.3 we see that $A \to B$ is flat. By Lemma 15.36.8 we see that $k \to \overline{B}$ is formally smooth in the $\mathfrak m_{\overline{B}}$-adic topology. Thus (2) holds.

Assume (2). Lemma 15.36.4 tells us formal smoothness is preserved under completion. The same is true for flatness by Algebra, Lemma 10.96.3. Hence we may replace $A$ and $B$ by their respective completions and assume that $A$ and $B$ are Noetherian complete local rings. In this case choose a diagram

\[ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } \]

as in Lemma 15.38.3. We will use all of the properties of this diagram without further mention. Fix a regular system of parameters $t_1, \ldots , t_ d$ of $R$ with $t_1 = p$ in case the characteristic of $k$ is $p > 0$. Set $\overline{S} = S \otimes _ R k$. Consider the short exact sequence

\[ 0 \to J \to S \to B \to 0 \]

Since $B$ is flat over $A$ we see that $J \otimes _ R k$ is the kernel of $\overline{S} \to \overline{B}$. As $\overline{B}$ and $\overline{S}$ are regular we see that $J \otimes _ R k$ is generated by elements $\overline{x}_1, \ldots , \overline{x}_ r$ which form part of a regular system of parameters of $\overline{S}$, see Algebra, Lemma 10.105.4. Lift these elements to $x_1, \ldots , x_ r \in J$. Then $t_1, \ldots , t_ d, x_1, \ldots , x_ r$ is part of a regular system of parameters for $S$. Hence $S/(x_1, \ldots , x_ r)$ is a power series ring over a field (if the characteristic of $k$ is zero) or a power series ring over a Cohen ring (if the characteristic of $k$ is $p > 0$), see Lemma 15.38.2. Moreover, it is still the case that $R \to S/(x_1, \ldots , x_ r)$ maps $t_1, \ldots , t_ d$ to a part of a regular system of parameters of $S/(x_1, \ldots , x_ r)$. In other words, we may replace $S$ by $S/(x_1, \ldots , x_ r)$ and assume we have a diagram

\[ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } \]

as in Lemma 15.38.3 with moreover $\overline{S} = \overline{B}$. In this case the map

\[ S \otimes _ R A \longrightarrow B \]

is an isomorphism as it is surjective and an isomorphism on special fibres, see Algebra, Lemma 10.98.1. Thus by Lemma 15.36.8 it suffices to show that $R \to S$ is formally smooth in the $\mathfrak m_ S$-adic topology. Of course, since $\overline{S} = \overline{B}$, we have that $\overline{S}$ is formally smooth over $k = R/\mathfrak m_ R$.

Choose elements $y_1, \ldots , y_ m \in S$ such that $t_1, \ldots , t_ d, y_1, \ldots , y_ m$ is a regular system of parameters for $S$. If the characteristic of $k$ is zero, choose a coefficient field $K \subset S$ and if the characteristic of $k$ is $p > 0$ choose a Cohen ring $\Lambda \subset S$ with residue field $K$. At this point the map $K[[t_1, \ldots , t_ d, y_1, \ldots , y_ m]] \to S$ (characteristic zero case) or $\Lambda [[t_2, \ldots , t_ d, y_1, \ldots , y_ m]] \to S$ (characteristic $p > 0$ case) is an isomorphism, see Lemma 15.38.2. From now on we think of $S$ as the above power series ring.

The rest of the proof is analogous to the argument in the proof of Theorem 15.39.1. Choose a solid diagram

\[ \xymatrix{ S \ar[r]_{\bar\psi } \ar@{-->}[rd] & N/J \\ R \ar[u]^ i \ar[r]^\varphi & N \ar[u]_\pi } \]

as in Definition 15.36.1. As $J^2 = 0$ we see that $J$ has a canonical $N/J$ module structure and via $\bar\psi $ a $S$-module structure. As $\bar\psi $ is continuous for the $\mathfrak m_ S$-adic topology we see that $\mathfrak m_ S^ nJ = 0$ for some $n$. Hence we can filter $J$ by $N/J$-submodules $0 \subset J_1 \subset J_2 \subset \ldots \subset J_ n = J$ such that each quotient $J_{t + 1}/J_ t$ is annihilated by $\mathfrak m_ S$. Considering the sequence of ring maps $N \to N/J_1 \to N/J_2 \to \ldots \to N/J$ we see that it suffices to prove the existence of the dotted arrow when $J$ is annihilated by $\mathfrak m_ S$, i.e., when $J$ is a $K$-vector space.

Assume given a diagram as above such that $J$ is annihilated by $\mathfrak m_ S$. As $\mathbf{Q} \to S$ (characteristic zero case) or $\mathbf{Z} \to S$ (characteristic $p > 0$ case) is formally smooth in the $\mathfrak m_ S$-adic topology (see Lemma 15.38.1), we can find a ring map $\psi : S \to N$ such that $\pi \circ \psi = \bar\psi $. Since $S$ is a power series ring in $t_1, \ldots , t_ d$ (characteristic zero) or $t_2, \ldots , t_ d$ (characteristic $p > 0$) over a subring, it follows from the universal property of power series rings that we can change our choice of $\psi $ so that $\psi (t_ i)$ equals $\varphi (t_ i)$ (automatic for $t_1 = p$ in the characteristic $p$ case). Then $\psi \circ i$ and $\varphi : R \to N$ are two maps whose compositions with $\pi $ are equal and which agree on $t_1, \ldots , t_ d$. Hence $D = \psi \circ i - \varphi : R \to J$ is a derivation which annihilates $t_1, \ldots , t_ d$. By Algebra, Lemma 10.130.3 we can write $D = \xi \circ \text{d}$ for some $R$-linear map $\xi : \Omega _{R/\mathbf{Z}} \to J$ which annihilates $\text{d}t_1, \ldots , \text{d}t_ d$ (by construction) and $\mathfrak m_ R \Omega _{R/\mathbf{Z}}$ (as $J$ is annihilated by $\mathfrak m_ R$). Hence $\xi $ factors as a composition

\[ \Omega _{R/\mathbf{Z}} \to \Omega _{k/\mathbf{Z}} \xrightarrow {\xi '} J \]

where $\xi '$ is $k$-linear. Using the $K$-vector space structure on $J$ we extend $\xi '$ to a $K$-linear map

\[ \xi '' : \Omega _{k/\mathbf{Z}} \otimes _ k K \longrightarrow J. \]

Using that $\overline{S}/k$ is formally smooth we see that

\[ \Omega _{k/\mathbf{Z}} \otimes _ k K \to \Omega _{\overline{S}/\mathbf{Z}} \otimes _ S K \]

is injective by Theorem 15.39.1 (this is true also in the characteristic zero case as it is even true that $\Omega _{k/\mathbf{Z}} \to \Omega _{K/\mathbf{Z}}$ is injective in characteristic zero, see Algebra, Proposition 10.152.9). Hence we can find a $K$-linear map $\xi ''' : \Omega _{\overline{S}/\mathbf{Z}} \otimes _ S K \to J$ whose restriction to $\Omega _{k/\mathbf{Z}} \otimes _ k K$ is $\xi ''$. Write

\[ D' : S \xrightarrow {\text{d}} \Omega _{S/\mathbf{Z}} \to \Omega _{\overline{S}/\mathbf{Z}} \to \Omega _{\overline{S}/\mathbf{Z}} \otimes _ S K \xrightarrow {\xi '''} J. \]

Finally, set $\psi ' = \psi - D' : S \to N$. The reader verifies that $\psi '$ is a ring map such that $\pi \circ \psi ' = \bar\psi $ and such that $\psi ' \circ i = \varphi $ as desired. $\square$

As an application of the result above we prove that deformations of formally smooth algebras are unobstructed.

Lemma 15.39.6. Let $A$ be a Noetherian complete local ring with residue field $k$. Let $B$ be a Noetherian complete local $k$-algebra. Assume $k \to B$ is formally smooth in the $\mathfrak m_ B$-adic topology. Then there exists a Noetherian complete local ring $C$ and a local homomorphism $A \to C$ which is formally smooth in the $\mathfrak m_ C$-adic topology such that $C \otimes _ A k \cong B$.

Proof. Choose a diagram

\[ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } \]

as in Lemma 15.38.3. Let $t_1, \ldots , t_ d$ be a regular system of parameters for $R$ with $t_1 = p$ in case the characteristic of $k$ is $p > 0$. As $B$ and $\overline{S} = S \otimes _ A k$ are regular we see that $\mathop{\mathrm{Ker}}(\overline{S} \to B)$ is generated by elements $\overline{x}_1, \ldots , \overline{x}_ r$ which form part of a regular system of parameters of $\overline{S}$, see Algebra, Lemma 10.105.4. Lift these elements to $x_1, \ldots , x_ r \in S$. Then $t_1, \ldots , t_ d, x_1, \ldots , x_ r$ is part of a regular system of parameters for $S$. Hence $S/(x_1, \ldots , x_ r)$ is a power series ring over a field (if the characteristic of $k$ is zero) or a power series ring over a Cohen ring (if the characteristic of $k$ is $p > 0$), see Lemma 15.38.2. Moreover, it is still the case that $R \to S/(x_1, \ldots , x_ r)$ maps $t_1, \ldots , t_ d$ to a part of a regular system of parameters of $S/(x_1, \ldots , x_ r)$. In other words, we may replace $S$ by $S/(x_1, \ldots , x_ r)$ and assume we have a diagram

\[ \xymatrix{ S \ar[r] & B \\ R \ar[u] \ar[r] & A \ar[u] } \]

as in Lemma 15.38.3 with moreover $\overline{S} = B$. In this case $R \to S$ is formally smooth in the $\mathfrak m_ S$-adic topology by Proposition 15.39.5. Hence the base change $C = S \otimes _ R A$ is formally smooth over $A$ in the $\mathfrak m_ C$-adic topology by Lemma 15.36.8. $\square$

Remark 15.39.7. The assertion of Lemma 15.39.6 is quite strong. Namely, suppose that we have a diagram

\[ \xymatrix{ & B \\ A \ar[r] & A' \ar[u] } \]

of local homomorphisms of Noetherian complete local rings where $A \to A'$ induces an isomorphism of residue fields $k = A/\mathfrak m_ A = A'/\mathfrak m_{A'}$ and with $B \otimes _{A'} k$ formally smooth over $k$. Then we can extend this to a commutative diagram

\[ \xymatrix{ C \ar[r] & B \\ A \ar[r] \ar[u] & A' \ar[u] } \]

of local homomorphisms of Noetherian complete local rings where $A \to C$ is formally smooth in the $\mathfrak m_ C$-adic topology and where $C \otimes _ A k \cong B \otimes _{A'} k$. Namely, pick $A \to C$ as in Lemma 15.39.6 lifting $B \otimes _{A'} k$ over $k$. By formal smoothness we can find the arrow $C \to B$, see Lemma 15.36.5. Denote $C \otimes _ A^\wedge A'$ the completion of $C \otimes _ A A'$ with respect to the ideal $C \otimes _ A \mathfrak m_{A'}$. Note that $C \otimes _ A^\wedge A'$ is a Noetherian complete local ring (see Algebra, Lemma 10.96.5) which is flat over $A'$ (see Algebra, Lemma 10.98.11). We have moreover

  1. $C \otimes _ A^\wedge A' \to B$ is surjective,

  2. if $A \to A'$ is surjective, then $C \to B$ is surjective,

  3. if $A \to A'$ is finite, then $C \to B$ is finite, and

  4. if $A' \to B$ is flat, then $C \otimes _ A^\wedge A' \cong B$.

Namely, by Nakayama's lemma for nilpotent ideals (see Algebra, Lemma 10.19.1) we see that $C \otimes _ A k \cong B \otimes _{A'} k$ implies that $C \otimes _ A A'/\mathfrak m_{A'}^ n \to B/\mathfrak m_{A'}^ nB$ is surjective for all $n$. This proves (1). Parts (2) and (3) follow from part (1). Part (4) follows from Algebra, Lemma 10.98.1.


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