Let <mi mathvariant="bold">A be the algebra given by the following multiplication table <mta

Cristopher Knox 2022-06-29 Answered
Let A be the algebra given by the following multiplication table
0 1 2 3 0 0 0 0 0 1 0 0 0 1 2 0 0 1 2 3 0 1 2 3
I need to prove that the variety generated by A is exactly the variety of commutative semigroups satisfying x 3 x 4 .
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Answers (1)

Jamarcus Shields
Answered 2022-06-30 Author has 17 answers
Let V be the variety of all commutative semigroups satisfying x 3 x 4 . Let V ( A ) be the subvariety generated by A. If V ( A ) is a proper subvariety of V, then there is a law that holds in A that does not hold throughout V. Using the identities of V, we may reduce any such law to one of the form s t where s = x 1 a 1 x k a k , t = x 1 b 1 x k b k , and a i , b i { 0 , 1 , 2 , 3 } for all i. Here a power of the form x i 0 , with exponent 0, should be interpreted as the identity element of A, which is 3.
Since s t does not hold in V, there must be some index where the variables in these words have different exponents, say a j b j . Substitute the identity element 3 A for all variables except the jth, and substitute 2 for x j . You obtain from s t that 2 a j = 2 b j . But the possible powers of 2 are all distinct: 2 0 = 3 , 2 1 = 2 , 2 2 = 1 , 2 3 = 0. This makes it impossible to have a i , b i { 0 , 1 , 2 , 3 }, a j b j , and 2 a j = 2 b j .
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