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## Re: [univalg] algebra with Pixley's term

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• Dear Ralph McKenzie thank you for the example. is it possible to obtain an example in the case that the algebra is minimal(i.e. has no subalgebra)
Message 1 of 5 , Feb 1, 2005
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Dear Ralph McKenzie thank you for the example. is it possible to obtain an
example in the case that the algebra <A;F> is minimal(i.e. <A;F> has no
subalgebra) and rigid(i.e. <A;F> has no proper isomorphism)?

Alomo Temgoua

>
>
> Let's try A = {0,1,2} and operations
>
> s(0)=1, s(1)=0, s(2)=2
>
> b(x,y,z) = Pixley(x,y,z) if {x,y,z} is a subset of {0,1};
> b(x,y,z)=2 otherwise
>
> Now for <A;s,b>, you have just the three congruences
> (0)(1)(2), (01)(2), (012) and operation b acts like Pixley's
> operation on equivalence classes of the non-trivial congruence
> (01(2).
>
> The variety generated by <A;F> is not arithmetical: for every term
> t(x_0,x_1,x_2), there is a variable x_i such that whenever 2
> is among {a_0,a_1,a_2} then t(a_0,a_1,a_2)=a_1.
>
> Nice question!
>
> Ralph McKenzie
>
>
>
> --On Monday, January 31, 2005 12:49 PM -0100 Temgoua Alomo Etienne
> R <rtemgoua@...> wrote:
>
>>
>> Can anybody know an example of finite algebra <A;F> satisfying
>> these properties:
>> (i) <A;F> has only three distinct congruences: {<a,a>, a in A},
>> \theta and {<a,b>; a,b are in A}.
>> (ii) <A;F> has a ternary term t(x,y,z) such that for all a,b in
>> A, <a,b> belongs to \theta implies
>> t^A(a,b,b)=t^A(a,b,a)=t^A(b,b,a)=a. (iii) The variety generates
>> by the algebra <A;F> is not arithmetical.
>>
>> Sincerely yours;
>> Temgoua Alomo
>> Departement of mathematics, University of yaounde I
>> Cameroon
>>
>>
>>
>>
>> Yahoo! Groups Links
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>>
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>>
>
>
>
> -----------------------------------------------------------------
> Mckenzie, Ralph N
> Vanderbilt University
> Email: ralph.n.mckenzie@...
>
>
>
> Yahoo! Groups Links
>
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• Can t we just modify my example, to the algebra with the three constants. Now there are no subalgebras or proper automorphisms. The
Message 2 of 5 , Feb 1, 2005
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Can't we just modify my example, to the algebra
<A, s, b, 0, 1, 2> with the three constants. Now there
are no subalgebras or proper automorphisms. The congruences
are the same. It is true (I think) that every term (or polynomial)
operation p(x,y,z,..) that actually depends on the variable x
will give p(2,..,..,...)=2, no matter what the values of
y,z, etc. Thus there can be no Pixley term.

Sincerely,
Ralph McKenzie

--On Tuesday, February 01, 2005 1:59 PM -0100 Temgoua Alomo Etienne
R <rtemgoua@...> wrote:

>
> Dear Ralph McKenzie thank you for the example. is it possible to
> obtain an example in the case that the algebra <A;F> is
> minimal(i.e. <A;F> has no subalgebra) and rigid(i.e. <A;F> has no
> proper isomorphism)?
>
>
> Alomo Temgoua
>
>>
>>
>> Let's try A = {0,1,2} and operations
>>
>> s(0)=1, s(1)=0, s(2)=2
>>
>> b(x,y,z) = Pixley(x,y,z) if {x,y,z} is a subset of {0,1};
>> b(x,y,z)=2 otherwise
>>
>> Now for <A;s,b>, you have just the three congruences
>> (0)(1)(2), (01)(2), (012) and operation b acts like Pixley's
>> operation on equivalence classes of the non-trivial congruence
>> (01(2).
>>
>> The variety generated by <A;F> is not arithmetical: for every
>> term t(x_0,x_1,x_2), there is a variable x_i such that whenever
>> 2 is among {a_0,a_1,a_2} then t(a_0,a_1,a_2)=a_1.
>>
>> Nice question!
>>
>> Ralph McKenzie
>>
>>
>>
>> --On Monday, January 31, 2005 12:49 PM -0100 Temgoua Alomo
>> Etienne R <rtemgoua@...> wrote:
>>
>>>
>>> Can anybody know an example of finite algebra <A;F> satisfying
>>> these properties:
>>> (i) <A;F> has only three distinct congruences: {<a,a>, a in A},
>>> \theta and {<a,b>; a,b are in A}.
>>> (ii) <A;F> has a ternary term t(x,y,z) such that for all a,b in
>>> A, <a,b> belongs to \theta implies
>>> t^A(a,b,b)=t^A(a,b,a)=t^A(b,b,a)=a. (iii) The variety generates
>>> by the algebra <A;F> is not arithmetical.
>>>
>>> Sincerely yours;
>>> Temgoua Alomo
>>> Departement of mathematics, University of yaounde I
>>> Cameroon
>>>
>>>
>>>
>>>
>>> Yahoo! Groups Links
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>
>>
>>
>> -----------------------------------------------------------------
>> Mckenzie, Ralph N
>> Vanderbilt University
>> Email: ralph.n.mckenzie@...
>>
>>
>>
>> Yahoo! Groups Links
>>
>>
>>
>>
>>
>>
>>
>
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> Yahoo! Groups Links
>
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>
>
>
>

-----------------------------------------------------------------
Mckenzie, Ralph N
Vanderbilt University
Email: ralph.n.mckenzie@...
• Dear Mckenzie your example solve one part of my problem. Here is my other preoccupation: is it possible to have an example of finite algebra satisfying
Message 3 of 5 , Feb 3, 2005
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Dear Mckenzie your example solve one part of my problem. Here is my other
preoccupation: is it possible to have an example of finite algebra <A;F>
satisfying these properties:
(i) <A;F> has only three distinct congruences: {<a,a>, a in A}, \theta and
{<a,b>; a,b are in A}; and \theta have only two equivalence classes C_1
and C_2.
(ii) <A;F> has a ternary term t(x,y,z) such that for all a,b in A, <a,b>
belongs to \theta implies t^A(a,b,b)=t^A(a,b,a)=t^A(b,b,a)=a.
(iii) The variety generates by the algebra <A;F> is not arithmetical.
(iv) If a subalgebra of A^2 is not contained in \theta then it is a
product of subalgebras of <A;F>.
(v) let a = (a_1,a_2,a_3), b = (b_1,b_2,b_3) and d = (d_1,d_2,d_3) in A^3
such that {a_1,a_2,a_3} is a subset of C_1 or C_2; {b_1,b_2,b_3} is a
subset of C_1 or C_2 and {d_1,d_2,d_3} is a subset of C_1 or C_2.
The subuniverse of the algebra A^3 generates by (a_1,b_1,d_1),
(a_2,b_2,d_2) and (a_3,b_3,d_3) is the set H = {(x_1,x_2,x_3) such that
for each i<j in the set {1,2,3},(x_i,x_j) belongs to the subuniverse
generates by (a_i,a_j), (b_i,b_j) and (d_i,d_j)}.

Sincerely yours;
Temgoua Alomo
Departement of mathematics, University of yaounde I
Cameroon
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