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In the book The Structure of Finite Algebras, by Hobby and McKenzie, page 23, it says, in a paragraph before the exercises:

The first substantial result proved as an application of tame congruence theory was that $\mathbf{Sub\,A}_{4}$ (and many other tight lattices, in particular $\mathbf{M}_{n}$ if $n \geq 3$) cannot be isomorphic to the congruence lattice of any finite algebra with just one basic operation, such as a semigroup. (This result will not be proved in this book. The proof can be found in [22].)

[22] refers to a paper of McKenzie: Finite Forbidden Latices, in Universal Algebra and Lattice Theory, number 1004, Springer Lecture Notes.

Could this be a typo? and it's mean to be "$\mathbf{M}_n$ if $n > 3$"?
I'm asking because $\mathbf{M}_3$ is, up to isomorphism, the partition lattice on a $3$-element set, and I think I know (and it seems obvious to me) that every partition lattice is a congruence lattice of an algebra with the same underlying set.
For example the left-zero semigroup on that set (or the right-zero semigroup, or the null semigroup...). The reason is that in these semigroups all equivalences are congruences (are they not?).

I tried to find some errata, but didn't succeed...
Is there some subtlety that I'm missing here?

amrsa
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    Of course you are right. If every operation of an algebra is either constant or a projection, then every equivalence relation is a congruence relation. Must be a typo. – bof Sep 25 '16 at 11:33
  • Why don't you ask the authors? – bof Sep 25 '16 at 22:48

1 Answers1

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Even if you change $n\geq 3$ to $n\gt 3$, the statement is still incorrect. $M_n$ is the congruence lattice of a finite algebra with one basic operation if and only if $n$ is a prime power plus $1$. One can rule out other values of $n$ with the results of the Finite Forbidden Lattices paper. But when $n=q+1$, where $q$ is a prime power, one can represent $M_n$ as the congruence lattice of a 2-dimensional affine vector space over the $q$-element field. This algebra can be presented with one basic operation.

Keith Kearnes
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  • Of course! Now I noticed that even in the figure above that statement there is the lattice $\mathbf{M}_6$ as being isomorphic to $\mathbf{S}(5,2)$, a particular case of the result you are mentioning. I suppose the authors meant something like $\mathbf{M}_n$ for most values of $n$... Thanks – amrsa Sep 26 '16 at 09:31
  • Ok, actually there is no "of course", because the algebra being possible to present with only one basic operation is not immediate. – amrsa Sep 26 '16 at 11:33
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    Here is how to present a finite vec sp with one operation, $g$. Let $f(x_1,\dots,x_k) = c_1x_1+\dots + c_kx_k$ be a linear operation where every nonzero scalar appears exactly once as a coefficient, except that the scalar $1$ appears twice. Let $g(x_1,\dots,x_k,y)=f(x_1,\dots,x_k)-f(y,\dots,y)$. Claim: The operation $g$ generates all nonnullary vec sp operations. To see this, note that $g(x,\dots,x,x)=0(x)$ is the unary zero function. Now substitute $0(x)$ into $g$ in different ways to get $1x_1+1x_2$, $-1x_i$ and $r_jx_j$ for all $j$, so you get all the usual nonnullary operations. – Keith Kearnes Sep 26 '16 at 19:15