Correction to: Flat coset decompositions of skew lattices

1 Correction to: Semigroup Forum (2016) 92:361–376 https://doi.org/10.1007/s00233-015-9739-8

Abstract In this corrigendum, we correct the statement and the proof of Proposition 4.11, where the word “lower” is stated instead of “upper,” and vice versa.

Proposition 1

(Proposition 4.11) Let S be a skew lattice such that $S/{{\mathcal {D}}}$ is a distributive lattice.

• (i) Let S be upper symmetric. Then, S is upper cancellative, if and only if $M\vee x\vee M= M\vee x'\vee M \Leftrightarrow B \vee x \vee B= B \vee x'\vee B$ holds for all skew diamonds in S and all $x,x'\in A$.

• (ii) Let S be lower symmetric. Then, S is lower cancellative, if and only if $B\wedge x \wedge B=B\wedge x' \wedge B \Leftrightarrow J\wedge x \wedge J=J\wedge x' \wedge J$ holds for all skew diamonds in S and all $x,x'\in A$.

Proof

(i). A skew lattice is upper cancellative if and only if it is upper symmetric and simply cancellative. Assume first that S is not upper cancellative, i.e., it is not simply cancellative. By a result of [5], it follows that S contains a subalgebra ${S}'$ isomorphic to ${{\mathbf {N}}}{{\mathbf {C}}}_5$, given by the diagram below:

In ${{\mathbf {N}}}{{\mathbf {C}}}_5$ operations on $\{x_1, x_2\}$ are defined in one of the following ways: either $x_i\wedge x_j=x_j$ and $x_i\vee x_j=x_i$ which yields a right-handed structure; or $x_i\wedge x_j=x_i$ and $x_i\vee x_j=x_j$ yielding a left-handed structure. Let A, B, M and J denote the ${{\mathcal {D}}}$-classes of elements $x_1$, y, u and v in S, respectively. Note that $x_2\in A$. Since $x_1$ and $x_2$ are both contained in the image of u in A, they cannot lie in the same coset of M in A, i.e., we obtain $M\vee x_1\vee M\ne M\vee x_2\vee M$. On the other hand, the cosets $B\vee x_1\vee B$ and $B\vee x_2\vee B$ both contain v (because $y\vee x_1\vee y=v=y\vee x_2\vee y$) and hence coincide by Theorem 2.4. Therefore, the implication $B\vee x_1\vee B=B\vee x_2\vee B \Rightarrow M\vee x_1\vee M=M\vee x_2\vee M$ fails.

Assume next that S is upper cancellative and let be a skew diamond in S. Given any $x,x'\in A$, the direct implication $M\vee x\vee M=M\vee x'\vee M \Rightarrow B\vee x\vee B=B\vee x'\vee B$ holds by Remark 4.8. Assume that there exist $x_1,x_2\in A$ such that $B \vee x_1 \vee B= B \vee x_2\vee B$, but $M\vee x_1\vee M\ne M\vee x_2\vee M$. Take any $m\in M$ and consider $a_1=m \vee x_1\vee m$, $a_2=m \vee x_2\vee m$. Note that $a_1,a_2\in A$, $m<a_1$ and $m<a_2$. Take $b\in B$ such that $b>m$. Since $m<a_1$ and $m<b$, it follows that $a_1\wedge b=m=b\wedge a_1$, and likewise $a_2\wedge b=m=b\wedge a_2$. By upper symmetry, $a_1\vee b=b\vee a_1$ and $a_2\vee b=b\vee a_2$ follow. Denote $j_1=a_1\vee b$ and $j_2=a_2\vee b$. The assumption $B \vee x_1 \vee B= B \vee x_2\vee B$ together with $m<b$ implies $b\vee a_1 \vee b=b\vee m\vee x_1\vee m\vee b=b\vee x_1\vee b=b\vee x_2\vee b=b\vee m\vee x_2\vee m\vee b=b\vee a_2\vee b$. It follows that $j_1=j_2$, and the set $S'=\{m,a_1,a_2, b,j_1\}$ forms a subalgebra of S, given by the following diagram:

(Again, with the operations on $\{a_1,a_2\}$ defined either in a right-handed or in a left-handed fashion.) Subalgebra $S'$ is isomorphic to ${{\mathbf {N}}}{{\mathbf {C}}}_5$. By a result of [5], it follows that S is not simply cancellative and thus not upper cancellative, which is a contradiction.

The proof of (ii) is similar. $\square$