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V is a variety of commutative semi group satisfying the identity $x^2 = x^3$. I need to prove that: $|F_V(\{x_1\dots,x_n\})|$ = $3^n -1$. Any hints on this ?

$F_V$ is V-free algebra.

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  • $\begingroup$ Judging from the result that you expect, I'm suspecting that you are after the variety of a commutative semigroup ring? Maybe one that satisfies $x_i^2 = x_i^3$ for each indeterminate $x_i$? Please rework this questions, at the moment it makes no sense. $\endgroup$ Apr 15, 2014 at 9:38
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    $\begingroup$ @Thomas I suspect you are confusing identities and relations. $\endgroup$
    – J.-E. Pin
    Apr 15, 2014 at 10:14
  • $\begingroup$ @J.-E.Pin You are right, I have no idea what the difference between an identity and a relation may be in the context here. $\endgroup$ Apr 15, 2014 at 13:50
  • $\begingroup$ Regarding the question on hold. $\endgroup$
    – Alvis
    Apr 15, 2014 at 15:25
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    $\begingroup$ It is phrased in terms of advanced mathematics, but its solution is accessible to students in high school. Have you tried determining the structure of the algebra even for small values of n? $\endgroup$ Apr 15, 2014 at 16:02

1 Answer 1

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Let $S_n = F_V(\{x_1, ..., x_n\})$. Since $S_n$ is commutative, its elements can be written in the form $x_1^{r_1} \cdots x_n^{r_n}$ where $r_1 + \dots + r_n > 0$. Morevover, since the semigroup $S_n$ satisfies the identity $x^2 = x^3$, you may assume that each $r_i$ is equal to $0$, $1$ or $2$. It follows that $|S_n| \leqslant 3^n -1$. To prove that this inequality is in fact an equality, it suffices to verify that the set of size $3^n -1$ $$ \{x_1^{r_1} \cdots x_n^{r_n} \mid 0 \leqslant r_i \leqslant 2 \text{ and } r_1 + \dots + r_n > 0\} $$ equipped with the product defined by $$ (x_1^{r_1} \cdots x_n^{r_n})(x_1^{s_1} \cdots x_n^{s_n}) = x_1^{\min\{2, r_1 + s_1\}} \cdots x_n^{\min\{2, r_n + s_n\}} $$ is a semigroup of $V$.

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